Chemestry for JEE

7/27/2019 Chemestry for JEE

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QUESTION BANK

CHEMISTRY

Gujarat Secondary and Higher

Secondary Education Board,

Gandhinagar

Price: `

70.00Published by :Secretary

Gujarat Secondary and Higher Secondary Education Board,

Gandhinagar

I

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.


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Contribution

1 Dr. Hasmukh Adhiya (IAS) Principal Secretary , Education Department Gandhinagar

2 Shr i 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 Teachers1. 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 Navchetan High School, Ahmedabad

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

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


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PREFACE

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 requestedthat 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 bythe 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.

M.I. Joshi R.R. Varsani (IAS)

Date: 02/ 01/ 2013 Secretary Chairman

III


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INDEX

IV

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


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Important Points[A] Important formulae :

1. .mass

No of molesMolar mass

2. .22.4

Volume at STPNo of moles of gas

3. 23.

.6.022 10

No of ParticlesNo of moles of Particles

4. . ( )o of moles of solute Molarity Vol L

5.M.W.of salt

Eq. wt. of a salt =Total + ve charge of metal ion

6. . .Atomic Weight

Eq wt of elementValency

7. . .

m a n bAvg at mass

m n

where, a + b are atomic masses

and m + n are precentage.

8. % of element in compound =( ) 100

. . n at massof element

M W of compound

where, n= No. of atoms of that element

9.1000

. . ( )

wMolarity

M W V ml

10.1000

. . ( )

wNormality

E W V ml

11.

1000

( )

wMolality

MW Wo gWo = Weight of solvent

12. ( )

nMole fraction X

n N

13.100

%W

W WW Wo

14.

6( ) 10

( )

weight vol of soluteppm

weight vol of solution

UNIT : 1 SOME BASIC CONCEPTS IN CHEMISTRY


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15. Molecalar weight = 2 V.D.

16.2

.. 1.008

Wt of metalEq wt of metal

wt of H displaced

17.2

. 11200.( )

Wt of metalEq wt of metalVol of H displaced at STP mL

18.. 35.5

.Wt of metal

Eq wt of metalWt of Chlorine combined

19.2

. 11200.

( )


Vol of Cl combined at STP mL

20.. 8

.

Wt of metal

Eq wt of metalWt of oxygen combined

21.2

. 5600.

( )


Vol of O displaced at STP mL

22.% 10W W density

MolorityMolecular weight

23.1 1 2 2

( )M V M V Molarity equation

24.1 1 2 2

( )V N V Normality equation

25.

Molecular weightn

Empirical formula Weight

26.0 09 ( ) 32

5F C

27. 0 273.15K C

28. 3 31 1 , 1 1 L dm mL cm

[B] Important Facts :

1. Antoine Lavoisier - Law of conservation of mass

2. Joseph proust - Law of definite proportions

3. John Dalton - Law of Multiple proportions

4. Richter - Law of combining weights.

5. Gay Lussac - Law of combining Volumes.

6. 1 amu = 1.6605 x 10-24 gram

7.12

231.9926 10 Mass of C atom gram

8. 23( ) 6.022 10A

Avogadro number N

9. AZT = Azido thymidine,drug used for aids victims.

10. 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.


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[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. 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.

2. In the reaction

3( ) ( )( ) ( ) 2( )2 6 6 3aq aqs aq gAl HCl Al Cl H

(AIEEE 2007)

(a) 6L ( )1 aqHC is consumed for every 3L, ( )2 gH produced.

(b) 33.6L( )2 g

H is produced regardless of temperature and pressure for every mole of Al that reacts.

(c) 67.2L( )2 g

H at STP, is produced for every mole Al that reacts.

(d) 11.2L( )2 g

H at STP, is produced for every mole ( )1 aqHC consumed.

3. Consider a titration of potassium dichromate solution with acidified Mohrs salt solution using diphenyl

amine as indicator. The number of moles of Mohrs salt required per mole of dichromate is (IIT JEE

2007)

(a) 3 (b) 4 (c) 5 (d) 6

4. 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)

5. 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

6. 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)6 8 2

C H N (b)6 10

C H N

(c) 5 6 3C H N (d) 4 18 3C H N


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7. Number of atoms in 560 g of Fe (atomic mass = 56) is (AIEEE 2002)

(a) twice that of 70 g N. (b) half that of 20 g H

(c) both (a) and (b) (d) None of these

8. In the standardization of2 2 3

a S O using2 2 7

K Cr O by iodometry, the equivalent weight of2 2 7

K Cr O is

(IIT JEE 2001)

(a)2

Molar mass(b)

6

Molar mass

(c)3

Molar mass(d) same as molar mass

9. Mixture X=0.02 mole of [Co(NH3)

5SO

4]

Br and 0.02 mole of

[Co(NH

3)

5Br]SO

4was prepared in 2L of

Solution

31 L of mixture X excess AgNO Y

21 L of mixture X excess BaCl Z

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

10. How many moles of electron weight one kilogram ? (IIT JEE 2002)

(a) 236.023 10 (b)311 10

9.108

(c)546.023 10

9.108 (d)

81 109.108 6.023

11. An Oxide of metal contains 60% of the metal. What will be the equivalent weight of the metal ?

(a) 12 (b) 40 (c) 24 (d) 48

12. A container is filled with 2L of water. What will be the volume of water in m3?

(a) 32 10 (b) 31 10 (c) 32 10 (d) 31 10

13. 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

14. The drug which is used for treating AIDS victims is

(a) Azidothymidine (b) Cis- platin

(c) Taxol (d) All of these

15. Chose the incorrect statement .

(a) The constituents of a compound cannot be separated into simpler substances by physical meth-

ods.

(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.


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16. 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.

17. What is the symbol of S.I. unit for the amount of substance ?

(a) NA (b) n (c) mole (d) mol18. What is the symbol of a multiple 109 ?

(a) G (b) E (c) n (d) Z

19. Find the correct relation.

(a) 9 ( ) 325

o oF C (b)5

( 32)9

o oC F

(c) Both & (a) and (b) (d) Neither (a) nor (b)

20. 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

21. What is the correct scientific notation for 0.00016 ?

(a) 41.6 10 (b) 516 10

(c) 30.16 10 (d)cannot be determined.

22. How many significant digits are there in 0.25 ?

(a) 1 (b) 2 (c) 3 (d) cannot be determined.

23. Which of the following number contains there significant digits ?

(a) 0.200 (b) 0.030 (c) 0.0052 (d) 0.00224. What is the number of neutrons in Zn2+ ion

(Atomic mass namber = 70) (IITJEE 1979)

(a) 34 (b) 36 (c) 38 (d) 40

25. 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

26. 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 g27. Find the total number of electrons in one molecule of carbon dioxide.

(a) 22 (b) 44 (c)66 (d) 88

28. 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

29. Identify the incorrect unit conversion factor.

(a)

31

1

cm

mL(b)

1

10

cm

mm(c)

60

1min

s(d) None of these


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30. 90 g KClO3on 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) Avogadros law.

31. 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

32. Two oxides of a metal M contain 27.6% and 30.0% of oxygen respectively. If the formula of the first

oxide is M3

O4 ,

find that of the second.

(a)2 3

M O (b)2

M O (c)2

MO (d)3 2

M O

33. 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%

34. 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%

35. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass.

(a) FeO (b) Fe2O

3(c) Fe

3O

4(d) Fe

3O

2

36. 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

37. 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

38. In a reaction : 2 2 3( ) ( ) ( )3 2g g gH NH 2,2000g N reacts with 21000 H

which reactant will left unreacted ? How much ?

(a)2, 2428g (b)

2, 428.6H g

(c)2, 571.4g (d)

2, 571.4H g

39. Calculate the number of sulphate ions in 100mL of 0.001M ammonium sulphate solution.

(a) 196.022 10 (b) 196.022 10

(c) 206.022 10 (d) 206.022 10

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


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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 (b) TTTT (c) TTFT (d)FTFT

42. The normality of 0.3M phosphorous acid is (IITJEE 1999)

(a) 0.1 (b) 0.9 (c) 0.3 (d) 0.6

43. 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

44. The pair of the compounds in which both the metals are in the highest possible oxidation state is

(a) 3 3

6 6,Fe CN Co CN

(b) 2 2 4,CrO Cl MnO

(c)3 2

,TiO MnO (d) 3

36,Co CN MnO

45. 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)2 2

10.62% , 3.58%Na O K O (b)2 2

3.58% ,10.62%Na O K O

(c)2 2

10.62% , 35.8%a O K O (d)2 2

35.8% ,10.62%Na O K O

46. 5.5 g of a mixture of FeSO4.7H

2O and Fe

2(SO

4)

39H

2O requires 5.4 mL of 0.1N KMnO

4solution for

complete oxidation. Calculate the number of mole of Fe2(SO

4)

39H

2O in the mixture.

(a) 0.0095 (b) 0.15 (c) 0.0952 (d) 1.52

47. 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

48. The density of a 3M Na2|S

2O

3solution is 1.25 g per mL, What is the molalities of Na+ and S

2O

3

2-ions ?

(a) 3.865, 7.732 (b) 7.732, 3.865 (c) 1.933, 7.732 (c) 7.732, 1.933


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49. 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

50. 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

51. The unit J Pa -1 is equivalent to

(a) m3 (b) cm3 (c) dm3 (d) none of these

52. The density of Al metal is 2.7 gcm-3. 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

53. 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%

54. 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

55. What weight of CuO will be required to provide 200Kg copper

(a) 200Kg (b) 79.5Kg (c) 250Kg (d) 100Kg

56. 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 (P2O

3)is half than that of its dimer phosphorous

hexoxide (P4O

6) for each of the elements present in them.

(a) T, F (b)F,T (c) T, T (d) F, F

57. 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%

58. If the atomic mass of carbon were set at 50 amu, what would be the value of Avog adros number ?

(a) 5.01 x 1024

(b) 6.022 x 1023

(c) 1.66 x 1024

(d) none of these59. For which of the following compounds molecular weigh cannot be determined from atomic weights ?

(a) 4 6 Fe Fe CN (b) 2TiO

(c)1.12

TiO (d) none of these

60. 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


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61. Which of the following chemical equation is incorrectly balanced ?

(a)2 3 3 6 2

12 2 3 Sb S HCl H SbCl H S

(b)3 3 4

3 4 3 IBr NH NI NH Br

(c)2 2 2

2 2 2 2 4 KrF H O Kr O HF

(d)3 2 3 3

3 3 PCl H O H PO HCl

62. Match the following

Column -I Column -II

(i) Cl2O

3(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)

63. How many g of NaOH can be obtained by reaction of 1 Kg of Na2CO

3with Ca(OH)

2?

(a) 106 g (b) 850 g (c) 755g (d) 943 g

64. 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

65. Calculate the amount of NaOH required to neutralize 100 mL 0.1M H2SO

4.

(a) 40g (b) 0.4 g (c) 80 g (d) 0.8 g

66. 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

67. 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

68. The number of molecules in 100 mL of each of O2,NH

3and CO

2at STP are

(a) 2 2 3CO O NH (b) 3 2 2NH O CO

(c) NH3

= CO2

< O2

(d) 3 2 2NH O CO

69. Which of the following represents the formula of a compound which contains 26% nitrogen and

74% oxygen ?

(a) N2O (b) NO (c) NO

2(d) N

2O

5

70. NKg-1 is the unit of

(a) momentum (b) velocity (c) Pressure (d) accelaration


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71. 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.

72. A balanced chemical equation is in accordance with

(a) Avogadros law.

(b) Law of constant proportions

(c) Law of conservation of mass

(d) Law of gaseous volumes.

73. 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 (b) a (c) 2b (d) (2

a)

74. If the components of air are N2,78%, O

2, 21% ; Ar, 0.9% and CO

2, 0.1% by volume, what will be the

molecular weight of air ?

(a) 28.9 (b) 32.4 (c) 16.4 (d) 14.5

75. Calculate the molarity of a solution obtained by mixing 50mL of 0.5M H2SO

4and 75 mL of 0.25M

H2SO

4.

(a) 0.375M (b) 0.35M (c) 0.045M (d) 0.45M

76. Which of the following has the highest normality ?(a) 1M H

2SO

4(b) 1M H

3PO

3(c) 1M H

3PO

4(d) 1M HNO

3

77. In an experiment, 4 gm of M2O

xoxide 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

78. 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

79. 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

80. 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


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11

81. 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

82. 100 mL of PH3on heating forms P and H

2. 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

83. 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

84. 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

85. Calculate the number of moles in 1m3 gas at STP.

(a) 4.46 (b) 44.6 (c) 446 (d) 4460

86. An ore contains 1.24% of the mineral argentite Ag2S by mass. How many grams of this ore would haveto 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

87. 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

88. Which of the following is not a homogeneous mixture ?

(a) smoke (b) air (c) Brass (d) Aqueous solution of sugar

89. 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 Cu90. 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

91. 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

92. Which of the following pairs of substances illustrates the law of multiple proportions ?(a) CO and CO2

(b) NaCl and NaBr

(c) H2O and D

2O (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.


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93. A : Normality of 0.1M H2SO

4is 0.2N.

R : H2SO

4is a dibasic acid.

94. A : 1 Gram molecule of sulphar also represents 1 gram atom of sulphur.

R : Atomicity of sulphur is one.

95. A : In the equation 3 4 NH HCl NH Cl, Gay-Lussacs law is not applicable to NH4Cl.R : NH

4Cl 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 of12C.

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 of12C.

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 O2and 1 L of O

3contains the same number of moles under identical conditions.

R : Under identical conditions, 1 L of O2and 1 L of O

3contain 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.

ANSWER KEY

1 c 2 d 3 d 4 a 5 b 6 a 7 c 8 b 9 a 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 89 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


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13

SOLUTIONS/HINTS

3. (d)

The redox reaction between potassiumdichromate and Mohrs salt is :

2 2 3 32 7 26 14 6 2 7 Fe Cr O H Fe Cr H O

4. (a) Number of particles Number of moles

No. of moles of carbon24

212

5. (b)

3 22 3 2 6 BC H B HC

2 mol 3 mol 2 mol

= 21.6 g

3 0.0821 273

67.21

nRT

V LP

6. (a)

mass of carbon =9

108 7213.5

72

No.of carbon atoms 612

similarly, no of H and N atoms are 8 and 2 respectively.8. (b)

During the reaction,

2

2 7Cr O

changes to Cr

3+

. Hence the change in oxidation number of Cr is 6.

6

Molar mass

Equivalent weight

9. (a)

In 2L solution, there are 0.02 mol Br-ions and 0.02 mole 2

4so

1 L of mixture X contains 0.01 mol Brand 0.01 mol2

4

SO ions.

Hence, Y= 0.01 mol Ag BrZ= 0.01 mol BaSO

4

10 (d)

Mass of an electron 319.108 10 Kg

311

. s in 19.108 10

No of electron Kg

31 23 1

8

1

9.108 10 6.023 10

10

9.108 6.023

mol

mol


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14

11. (a)

mass of metal = 60 g

mass of oxygen = 40 g

mass of oxygen = mass of metal40 g = 60g

8 g = (?)

=8 60

1240

25. (a)

(i)2 4 4 2

Zn H SO ZnSO H

(ii)2 2 2

2Zn NaOH Na ZnO H

26. (c)

( ) ( ) ( )2 3 2 2s s gAg CO Ag O CO

1 mol 1mol

0.01 mol 0.01 mol

There fore mass of residue (Ag2O) 20.01 molarmass of Ag O

0.01 232 2.32g

28. (c)

The ratio by weight =1

4

Ratio of moles

128 7

324 4 32 3228

32. (a)

Let x be the atomic mass of metal M In the oixde M

3O

4, the mass of M = 72.4 and that of O = 27.6

3 4

72.4 27.6

16

M O

x

M O

72.4 27.6: 3: 4

16x

56x

For second oxide, the mass of M = 70 and that of O =30

70 30

56 16

M O

1.25 1.875M O

1 1.5 2 3ORM O M O


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33. (a)

Let the % of isotope with atomic mass 10.01 be x

% of isotope with atomic mass 11.01 = 100-x

Avg at mass =10.01 (100 )11.01

10.81( )100

x xGiven

37. (a)

69% w/w means 100 g nitric acid soution contain 69 g of nitric acid by mass.

369

1.09563

moles of HNO

Vol. of 100 g nitric acid solution100

0.070921.41

L

1.09515.44

0.07092 moles per litre

38. (d) 2 2 33 2H NH

28 g 6 g

2000 g (?)

2000 6428.6

28g

But we are given 1000 g H2There fore 1000 - 4286 =

571.4 g H2will left.

39. 3(b)

No of moles of (NH4)2SO4 = molarity

vol(L)= 0.001 0.1 = 0.0001

2 23 19

4. 0.0001 6.022 10 6.022 10No of SO ions

40. (a)

2

( )

( )ETOH

ETOH

ETOH

H O

nX

n n

( )

( )

0.0455.55

ETOH

ETOH

n

n

( )2.31

ETOHn

42 (d)

phosphorous acid 3 3( )H PO is a dibasic acid. Its structure is as follows : Normality = basicity x Molarity

= 2 x 0.3 = 0.6

43. (a)

Equivalents of 2 2 4 2H C O . 2H O in 10ml = Equivalents of NaOH

6.3 1,0000

63 250 0.1

40

V

mL

O

P

OHOHH


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44. (b)

The oxidation states of various metals are :

(a) 3, 3Fe Co

(b) 6, 7Cr Mn

(c) 6, 4Ti Mn

(d) 3, 6Co Mn

45. (b)

Suppose amount of NaCl in the mixture = x g The amount of KCl in the mixture = (0.118 - x) g

NaCl + AgNO3 AgCl + NaNO

3

58.5 143.5

x143.5

.........................( )58.5

xi

Similarly AgCl obtained from KCl =143.5 (0.118 )

.........................( )74.5

xii

But (i) + (ii) = 0.2451 g (Given)

Amount of NaCl = 0.0338 gAmount of KCl = 0.0842 g

Now, 2NaCl = Na2O

117 62

0.03380.0338 62

0.0179117

g

2

0.0179 10 0% 3.58%

0.5of Na O .....

46. (a)

Weight of 4 25.4 0.1 278

7 0.1501000

FeSO H O g

Moles of 2. 4 3 25.35

( ) . 9 0.0095562

Fe SO H O

47. (a)

AtomicWeightEquivalent weight Valency

48. (b)

1000

,1000 s

Mm

d MM

M Molarity of solution

d density of solution

sM Molar mass of solute

1000 33.865

1000 1.25 3 158


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17

51. (a)

3

2Pr

J Work Nmm

Pa essure Nm

52. (c)

m

dv

340 14.8

2.7

mV cm

d

Water level in cylinder = 50+ 14.8 = 64.8 mL65. (d)

2 4 2 40.1 1000 solution contains 0.1MH SO mL mol H SO

2 4100 0.01mL solution contains mol H SO

2 4mass of 0.01 98 0.98H SO g

2 4 2 4 22 2

2(40) 98

(?) 0.98

80 0.980.8

98

NaOH H SO Na SO H O

g g

g

g

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

3 833.25

5 35.5

EE

E

67. (b)

4 8 2. .M F C H O

88Molar mass

88

442Vapour density

68. (b)

Equal volumes under identical conditions contain equal no. of molecules

69. (d)

26 74 1.85 4.625 2 5

14 16

O N O N O

70. (d)

F ma

1f Na NKg m Kg


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18

73. (a)

No of moles of X20

a

. ( )20

20

aNo of atoms of X N b given

ba

N2

.40

2.

40

2 20

40

aNo of moles of Y

aNo of atoms of Y N

bN b

N

74. (a)

78 28 21 32 0.9 40 0.1 44. .

78 21 0.9 0.1Mol wt of air

75. (b)

No.of moles of 0.05L H2SO

4= 0.5 x 0.05 = 0.025

No.of moles of 0.075L H2SO

4= 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

0.043750.35

0.125Molarity M

77. (c)

1 Mol M2O

x= (2 56 + 16x) gm

, (2 56 16 ) 112

112 44

112 16

ow x gm of oxide gm metal

gm of oxide gm metalx

112 4

2.8 ( )112 16

But given

xx = 3

79. (c) Molecular formula of ammonium dichromate is

4 2 2 7( )NH Cr O

80. (b)

2

11200.

.

wt of metalEq wt of metal

vol of H in ml displaced at STP


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82. (a)

( ) ( )3 ( ) 22 2 3

g gsPH P H

2 3mL mL

100 (?)ml100 3

1502

ml

50Increase mL

85. (b)

31 1000m L

, 22.4 1At STP L mol

10001000 44.622.4

L

86. (c)

2

248 216

(?) 1.0

1.148

Ag S Ag

g g

g

2

100 1.24

(?) 1.148

92.58 92.6

ore Ag S

g g

g

g

87. (a)

3 9. 0.3327

No of moles of Al

3 23 23. 0.33 6.022 10 2 10No of Al ions

19 23Electric charge 3 1.602 10 10 9.6 Coulomb


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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

attractive forces. This force is universal. This force of attraction is exerted upto 4.50

A distance in sub-

stance. 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 forcesare 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 non-

polar 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 Boyles law. At constant temperature for a fixed amount gas, pressure (P) varies

inversely with its volume (V). Mathematically the Boyles law is written as PV = K or P 1V1 = P2V2.

The equation d/P = K devised from Boyles 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

UNIT : 2 STATES OF MATTER

Important Points


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21

law. Mathematically it is written asV

= KT

or1 2

1 2

V V=

T T. The relation between pressure and absoulte

temperature (T) is obtained on the basis of experimental orbservations by scientist Gay Lussac and is

known as Gay Lussacs law. Mathematically it is written asP

= K

T

or1 2

1 2

P P=

T T

. The relation between

volume of a gas and number of molecules was given by Avogadro, which is known as Avogadros 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 Avogadros number respectively. Combining Boyles law and Charles

law, the relation obtaingedPV

= KT

or1 1 2 2

1 2

P V P V=

T Tis known as combined gas equation. The ideal

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

2

2

anP + (V nb) = nRT

V

- and this equation is also known as van der Waals equation. The gas can beliquefied 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 Maxwells distribution curve.

The total pressure of the mixture of two or more than two gases is obtained by the Daltons law. Totalpressure (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 A% by volume of gas A total pressure

p =100

. The Grahams law of gaseous diffusion is

1r

d and using formula the ratio of rate of diffuson of NH3 and HCl gas was obtained practically as

1.46 + .01. The application of Grahams law of gaseous diffusion are as given in the text. The Avogadros

hypothesis is useful to calculate the number of molecules, atoms and total number of atoms in given amount

of gas.

The liquid state has its physical properties like fixed volume, fluidity, non- compressibility, diffusion,

evaporation, vapour pressure, surface tension and viseosity.


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1. What is value of Boyles temperature of ethane gas when a= 5.489 dm6 atm mol-2 and b = 0.0638 dm3

atm mol-1

(a) 1048K (b) 104.8K (c) 209.6K (d)290.6K

2. The value of universal gas constant R depends upon the

(a) Temperature of the gas (b) Volume of the gas

(c) Number moles of the gas (d) none of these

3. The Boyles temperature for the ideal gases is given by

(a)a

R(b)

a

bR(c)

2a

bR(d) none of these

4. The ratio of van der Waals constants a and bhas the dimensions of

(a) atm mole-1 (b) L mole-1 (c) atm . L mole-1 (d) atm mole-2

5. 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 Boyles temperature

6. 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

7. To which of the following gaseous mixture is Daltons law not applicable?

(a)He + Ne + SO2

(b)NH3

+ HCl + HBr (c) 2 2 2+ +O N CO (d) 2 2 2N H O+ +

8. The degree of dissociation of cl2at 1500K is 0.45 according to the reaction Cl

2 2Cl assumig that

both Cl2and 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 (b) 0.496 g. l-1 (c) 0.696 g. l-1 (d) 0.396 g. l-1

9. A gas is kept at 1 atm pressure. To compress it to 1/4thof its initial volume, the pressure to be applied is

(a) 1 atm (b) 2.0 atm (c) 4.0 atm (d)1

4atm

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 ?

(a) 200 C (b) 300 C (c) 400K (d) 300K

11. A mixture contains N2O

4and NO

2in the ratio 2 : 1 by volume. The vapour density of the mixture is

(a) 45.4 (b) 49.8 (C) 32.6 (d) 38.3

M.C.Q.


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12. At extremely low pressure, the vander waals equation for one mole of a gas may be written as

(a) PV =RT + pb (b)a

PV RTv

= - (c) PV = RT (d) ( )( )a

p v b RTv

+ - =

13. The compressibility of a gas is less than unity at STP Therefore

(a) Vm > 22.4 L (b) Vm < 22.4 Litre (c) Vm = 22.4 L (d) Vm = 44.8 Litre

14. The correct order for magnitude of vanderwaals constant b should be

(a) 2 6 2C H CO CO He< < < (b) 2 6 2CO C H He CO< < 1.0

(b) CO2gas is more compressive than the ideal gas as Z > 1.0

(c) CH4gas 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, CH

4and CO

2Q-75 gases which graph is related for fH

2

and CH4compared to an ideal gases

1.0

PVZ

RT=

D

B

A

C

(a) D and A (b) A and B (c) A and D (d) B and C

76. which state of matter whose the intermolecular attractive force do not exist ?

(a) solid (b) liquid (c) gas (d) none

77. Which word of the following does not used for states of matter ?

(a) Bose - Einstin (b) Boyle - Einstein (c) plasma (d) solid, gas, liquid

78. 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 ?

(a) 2.5 atm (b) 1.25 atm (c) 5.0 atm (d)14.2 2.5

7.1 V

79. When the unit of R = 8.314 x 107 erg. mole-1k-1 . 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

80. unit of R in CGS system is

(a) 7 1 18.314 10 erg K mole- - (b) 1 18.314 JK mole- -

(c) 1 10.082litre atm mole K- - (d) 1 11.987 cal mole K- -

81. What is the value of gas constant R cal.mol-1K-1 ?

(a) 0.082 (b) 1.987 (c) 8.314 (d) 8.314 x 107


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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 ofPV

nRTknown as compressibility factor

(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 (b) PV = iRT

(c)

2

2

( )( )am

P V nb CRTv

= - = (d)2

2

( )( )am

P V nb nRTv

= - =

84. Proportion of O2,U

2and N

2gases are 3 :2: 5, and the Total pressure of this gas mixture container is 50

bar, What are the partial pressure of cl2and N

2gases respectively?

(a) 10 bar, 25 bar (b) 10 bar, 15 bar (c) 15 bar, 25 bar (d) 25 bar, 15 bar

85. Which factor is the deciding factor of physical state of matter ?

(a) inter molecular forces (b) molecular interaction

(c) effect of thermal energy on the motion of partcles (d) Given all.

86. Whichphysical state is accuired by water in between temperature above than 273 K and below 373 K?

(a) plasma (b) liquid (c) solid (d) Gas

87. Which physical state of water is more compressable applying pressure at constant temperature ?

(a) Ice (b) water (c) Vapour (d) Plasma

88 Which sabstance can be easily poored form one container into the other at room temperature ?

(a) Kerosene (b) Ice (c) salt (d) all

89. 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


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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. What is negative electrical charge on F atom in HF the permanent dipole molecule ?

[A] higher than 1.6 x 10-19C [A] less than 1.6 x 10-10C

[C] less than 1.6 x 10-19C [D] higher than 1.6 x 10-10C

95. State Figure showing dipole - Induced dipole forces in the following ?

(a)A Bd d+

+(b) H Cl

d d++ H Cl

d+


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35

(c)+ + +

Nonpolar

Atom-I

Nonpolar

Atom-II

(d) H

d

O

H

H O

H

d+d +d

96. 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

97. 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 S8has metallic properties while P

4is 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


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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 (IIT 1990)

(a) NTP (b) O0C, 2 atm (c) 2730C, 1 atm (d) 2730C, 2 atm

103. Pressure of a mixture of 4 g of O2and 2g of H

2combined 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 NH3and 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 NH3bottle (d) through out the length of the tube

106. At 1000C 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

(a) 2 (b)1

2(c) 4 (d)

1

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.


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110. Density ratio of O2and H

2gas 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)

(a)

1

2A

A B

B

D Dr

r=

(b)

1

2B

A B

A

D Dr

r=

(c)

1

2A

A B

B

D Dr

r=

(d)

1

2B

A B

A

D Dr

r=

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

a) Molecular interaction between atoms andPV

InRT

>

b) Molecular interaction between atoms and

d) Finite size of the atoms and V

y(d) V

x= 2V

y

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 Z and 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,Z containers 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 H2gas and N

2gas 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.


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(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

( )

( )

2

2

2

1a

P v b RT for moles andV

anP v nb nRT for n moles

V

+ - =

+ - =

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 PnRT

here quantity z is

called compressbility factor.

for H2and He gases

PVz

nRT= is always > 1. They show he deviation but for N

2,O

2CH

4andCo

2gases

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

2(3 )

27

8R

27

c c

c

ab

PV bz

aT

Rb

3 = = =

8

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

2

8, 3

27 27c c c

a aT P and V b

Rb b

= = =

alter natively, constants

28

, 3 ,3 3

c ccc

c

P VVb a Pc V R

T= = =

The units of a : atm L2mole-2

b : l mole-1

question (1)


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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 compared to total volume in the derivation of vander waals equation

(d) vander waals proposed the gas equation for 1 note gas is

( )2

aP v nb RT

n

+ - =

(A) a,b (B) b,c (C) c,d

(D)a,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 graphPV

nRTvs P

(A) a,b (B) b,c (C) c,d (D) a,d

125. When a graph Z plotted against P for CH4

and CO2gases then the graph obtained as ...

(a)

P

Z = 1.0

CO2

CH4

(b)

P

CO2

CH4

Z = 1

(c)

P

CO2

CH4

Z = 1.0

(d)

P

CO2

CH4

Z = 1.0


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126. What is indicated by the given graph for the behaviour of H2and CO

2gases correctly ?

P

H2

CO2

Z = 1.0

(a) H2has value z > 1 show positive deviation from ideal behaviour

(b) Value ofPV

znRt

= for H2gas is greater than zero but less then 1

(c) value of =PV

znRT

for CO2gas is greater than zero but less than 1 shows its negative deviation of its

ideal behaviour

(d) value of Z for H2is greater than 1 and for co

2it is less then one hence at high pressure co

2gas 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) 28

, , 327 27

a ab

Rb b(b)

2

3 8 3, ,

8 27

a a b

Rb b v(c)

2

2

8, , 3

27 27

a anb

Rb b(d)

2

3 8 3, ,

8 27

a a b

Rb b 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) 1

PVz

RT=

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 SO2with bigger size of its volume d) P( V - nb) = nRT

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


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129. column I column II

1) If force of attraition among the a) ( )2a

P v b RTv

+ - =

gaseous molecules are negligible

2) If the volume of the gas molecules b)a

PV RTv

= -

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 T, the

charles law ?

(a)

T(K)

V

T

(b)

T(K)

V

(c)

T(K)

V

(d)V

T(C)

(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) H2he gases are more compressible than ideal gas for all values of pressure

(c) compressiblity factor

=

PVz

nRTis less than 1 for all real gases at low pressure except H

2and He

(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 4( ) 4 ( )qM M g in a closed rigid container having volume 22.4L at

273K If the initialmoles of M4taken befor dissoliciation is 1 then.


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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) 4 1PV

z of MRT

= > (b) 4 1PV

z of MRT

=

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

1 a 2 d 3 b 4 c 5 b 6 a 7 b

8 d 9 c 10 c 11 d 12 b 13 b 14 d

15 a 16 a 17 b 18 b 19 c 20 d 21 b22 a 23 a 24 b 25 d 26 b 27 d 28 a

29 b 30 c 31 b 32 c 33 b 34 b 35 c

36 c 37 d 38 b 39 a 40 a 41 a 42 b

43 c 44 c 45 c 46 c 47 a 48 b 49 c

50 d 51 c 52 b 53 c 54 b 55 c 56 c

57 d 58 c 59 a 60 a 61 a 62 d 63 a

64 d 65 b 66 c 67 b 68 b 69 a 70 b

71 a 72 d 73 a 74 d 75 c 76 c 77 b

78 a 79 a 80 a 81 b 82 d 83 d 84 c

85 d 86 b 87 c 88 a 89 a 90 b 91 a

92 d 93 d 94 c 95 a 96 b 97 a 98 a

99 b 100 c 101 a 102 b 103 a 104 b 105 b

106 b 107 c 108 c 109 a 110 c 111 d 112 d

113 a 114 c 115 c 116 a 117 d 118 c 119 a

120 d 121 b 122 c 123 a 124 c 125 d 126 c

127 a 128 c 129 a 130 c 131 a 132 b 133 b

134 b

Answer Key


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Composition of atom

Electron (1eo

)(1) It was discovered byJ.J. Thomson (1897) and is negatively charged particle.

(2) Electron is a component particle of cathode rays.

(3) Cathode rays were discovered by William Crookes &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 XCu, rays are produced.

(v) The nature of these rays does not depend upon the nature of gas or the cathode material usedin discharge tube.

(vi) The e/m (charge to mass ratio) for cathode rays was found to be the same as that for an e8

1076.1( coloumb pergm). 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 Thomsonsexperiment and observed the formation of anode rays. These rays also termed aspositive orcanal

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 byJames Chadwick (1932) according to the following nuclear

reaction, 11264294 nCHeBe o or 114742115 nNHeB o

(2) Neutron is an unstable particle. It decays as follows,

oantinutrin

00

electon

01

Proton

11

neutron

10 eHn

UNIT : 3 STRUCTURE OF ATOM

Important Points


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Name of constant Unit Electron(e

) Proton(p+) Neutron(n)

Mass (m)

Amu

Kg

Relative

0.000546

9.109 1031

1/1837

1.00728

1.673 1027

1

1.00899

1.675 1027

1

Charge(e)

Coulomb (C)

Esu

Relative

1.602 1019

4.8 1010

1

+1.602 1019

+4.8 1010

+1

Zero

Zero

ZeroSpecific charge (e/m) C/g 1.76 10

89.58 10

4Zero

Density Gram / cc 172.17 10

14

1.114 10 14

1.5 10

The atomic mass unit (amu) is 1/12 of the mass of an individual atom of 126 C , i.e. kg2710660.1 .

Other non fundamental particles

Particle Symbol Nature Charge

esu

1010

Mass

(amu)

Discovered by

Positron

0

,1 ,e e

+ + 4.8029 0.0005486 Anderson (1932)

Neutrino 0 0 max 2 22 1( )

n n1 22 2

n n Rl

Rn 21

min l 21

22

2

2

min

max

nn

n

l

l

(1) Lymen

series

Ultraviole

t region

11 n

....4,3,22n

2and1 21 nn

R34max l

21 and1 nn

R1min l

34

(2) Balmer

series

Visible

region

21 n

....5,4,32n

3and2 21 nn

R536

maxl

21 and2 nn

R4

minl

5

9

(3) Paschen

series

Infra red

region

n1 = 3

....6,5,42n

4and321

nn

R7144

maxl

21

and3 nn

R9

minl 7

16

(4) Brackett

series

Infra red

region

41

n

....7,6,52n

5and421

nn

R9

2516max

l

1 24 andn n

min 16Rl 9

25

(5) Pfund

series

Infra red

region

51

n

....8,7,62n

6and521

nn

R113625

max

l

21

and5 nn

R

25min l 11

36

(6) Humphrey

series

Far

infrared

region

61 n

....8,72n

7and6 21 nn

R134936

max

l

21 and6 nn

R36

min l 13

49

(iv) If an electron from nth excited state comes to various energy states, the maximum spectral lines

obtained will be .2)1( nn

n principal quantum number..

As n 6 than total number of spectral lines .152

30

2

)16(6

BohrSommerfeilds model

It is an extension of Bohrs model. The electrons in an atom revolve around the nuclei in elliptical orbit. Thecircular path is a special case of ellipse. Association of elliptical orbits with circular orbit explains the fineline spectrum of atoms.

Dual nature of electron

(1) In 1924, the French physicist,Louis de Broglie suggested that if light has both particle and wavelike nature, the similar duality must be true for matter. Thus an electron, behaves both as a materialparticle 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 ,mvh

where h Plancks constant.


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(3) This was experimentally verified byDavisson and Germerby observing diffraction effects with anelectron beam. Let the electron is accelerated with a potential ofV than the Kinetic energy is

21

2mv eV ; 2 2 2m v eVm 2mv eVm P ; 2

h

eVm

2 .

h

k E m

(4) If Bohrs theory is associated with de-Broglies equation then wave length of an electron can bedetermined in bohrs orbit and relate it with circumference and multiply with a whole number

22

rr n or

n

From de-Broglie equation,

h

mv .

Thus2h r

mv n

or

2

nhmvr

(5) The de-Broglie equation is applicable to all material objects but it has significance only in case ofmicroscopic particles.

Heisenbergs uncertainty principle

This principle states It is impossible to specify at any given moment both the position and momentum(velocity) of an electron.

Mathematically it is represented as , .4

hx p

Where x uncertainty is position of the particle, p uncertainty in the momentum of the particle

Now since p m v

So equation becomes, .4

hx m v

or 4

hx v

m

In terms of uncertainty in energy, E and uncertainty in time ,t this principle is written as,

.4

hE t

Schrdinger wave equation

(1) Schrodinger wave equation is given byErwin Schrdingerin 1926 and based on dual nature ofelectron.

The Schrodinger wave equation is,2 2 2 2

2 2 2 2

8( ) 0

mE V

x y z h

Where yx, andz are the 3 space co-ordinates, m mass of electron, h Plancks constant, ETotal energy, Vpotential energy of electron, amplitude of wave also called as wave function,

2

2

is mathematical operation to be performed on YY

(2) The Schrodinger wave equation can also be written as,2

2

2

8( ) 0

mE V

h

Where laplacian operator..


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(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. )......,,( timeszyx

(ii) For a single particle, the square of the wave function 2 at any point is proportional to theprobability of finding the particle at that point.

(iii) If 2 is maximum than probability of finding eis maximum around nucleus and the placewhere probability of finding e is maximum is called electron density, electron cloud or anatomic orbital. It is different from the Bohrs orbit.

(iv) The solution of this equation provides a set of number calledquantum numberswhich describespecific or definite energy state of the electron in atom and information about the shapes andorientations 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 ofquantisation, and the conclusions of de-Broglies principle and Heisenbergs principle in his model.

In this model, the behaviour of the electron in an atom is described by the mathematical equation

known as Schrodinger Wave Equation. )2 2 2

2 2 2 2

8 m(E V)

x y z h

+

(Here x, y and z are three space coordinates, m mass of electron, h Plancks constant, E Totalenergy, V Potential energy and 1/J Wave function of electron wave)

The above expression can also be expressed as 22

8 m(E V)

h

The permitted solutions of Schrodinger equation are known as wave functions which correspondto a definite energy state called orbital. Thus, the discrete Bohr orbits are replaced by orbitals 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,potentialenergy E, velocity v, has wave like characteristics associated with it, with amplitude given bywave function.

Probability Distribution

In wave mechanics a moving electron is represented by wave function, j. It has on physical significance andrefers 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. Anorbital is considered as a diffused electron cloud having more density close to the nucleus. The probabilityof finding an electron in a given volume is understood best in the form of radial probability distributioncurves. The probability curves for some orbitals are given in the figure. The distance of maximum radialprobability is radius of an atom. There are two humps for 2p-orbital which means that the 2s electronpenetrates a little closer to nucleus. The point at which radial probability becomes zero is known as nodalpoint.


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The radial probability plots for some orbitals are shown in the given figure.

1sDISTANCE

RAD

AL

ROVAB

2sDISTANCE

RADIALPROVABILITY

2p

DISTANCE

RADIALPROVABILITY

ORBITAL WAVE FUNCTIONS AND SHAPES OF ORBITALS

According to wave mechanics, atomic orbitals are described bywave functions known as orbital wavefunctions. 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 wavefunction depends upon the direction given by the angles with respect to different co-ordinate axis. It isfound that the wave function for s-orbital is independent of angles and, therefore, s-orbitals do not haveangular dependence. Thus, all s-orbitals are spherically symmetrical. However, all other types of orbitals

(p, dorf) 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 fromnucleus), we obtain graphs as shown below. These graphs are for ls, 2sand 2p-orbitals of hydrogen atom.This type of dependence is known as radial dependence. These plots show radial dependence on onlyone side of the nucleus. These plots do not have any direct physical significance, but are useful in molecularstructure because atomic wave functions are needed to build molecular wave functions. It is clear from thegraph, that in ls radial wave function,j, is positive everywhere, but for others orbitals i.e., for 2s or 3s-orbitals it is positive in some regions and negative in others. It may be noted that +ve and - ve signs haveonly 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 unitvolume 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 earliergraphs as R2 is positive throughout.


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(In case R is not real, IRI2 can be taken in place of R2. In such cases IRI2R. .)

1s

R2

r

R2

2s

r

2p

r

R2

Graph between R2 (radialprobabUity density) and r

R dv2

r1s

R dv2

r

2s

r2p

R dv2

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,R2dVgives the probability of finding the given electron in a volume dV. The productR2dVis also known as radialprobability distribution function. The graph ofR2dVversus r(distance from nucleus) is known as radialprobability distribution function graphs.

If we observe the radial probability distributive graph of1S, we find it is quite different from the radialprobability density graph. The radial probability density is maximum close to the nucleus, but the radialprobability is least. This is due to the fact that volume of the spherical shell (dV) near the nucleus is verysmall resulting in a small value of radial probabilityR2dV. At the nucleus (i.e., r 0), dVis zero, henceR2dVas also zero, although R2 is very large at this point. As the distance from nucleus (r) increases, thevolume of the shell dV (4r2dr) increases while R2 decreases. As a result, the radial probability keeps onincreasing gradually and reaches a maximum value and then decreases gradually. The maximum in thecurve indicates the most probable value and the corresponding distance from the nucleus is called distanceof maximum probability(r

0). 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 thenucleus for hydrogen atom in ground state. However, according to wave mechanical model the electron ismost likely to be found at this distance and there is probability of finding the electron at distances bothshorter and longer than this.

The radial probability distribution curve for 2s-orbital (n 2, l 0) shows two maxima, a smaller onenearer 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 anelectron in the 1s orbital.

The radial probability distribution curves of three 2p-orbitals (n 2, l 1) are identical. It shows only onemaximum. 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 2s-orbital. In other words, 2s-orbital penetrates closer to the nucleus, than a 2p-orbital. Thus, an electron in2s-orbital has a lower energy than an electron in a 2p-orbital.


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Some Note worth points

1. The radius of maximum probability of 1s electron is 0.529.

2. The number of regions of maximum probability for 1s, 2p, 3d and 4f is one. For 2s, 3p, 4d and 5fthese 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 theenergy of an orbital.

5. In these curves, the first orbital of cash type (1s, 2p, 3d, 4f) has one region of maximum probabilityand no node. Whereas the first orbital of each type (2s, 3p, 4d, 5f) has two regions of maximumprobability 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 isdesignated 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 .2 2n(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) It tells about the shape of subshells.

(iv) It also expresses the energies of subshells fdps (increasing energy).

(v) The value of )1( nl always.

(vii) It represent the orbital angular momentum. Which is equal to ( 1)2

hl l

(viii) The maximum number of electrons in subshell 2(2 1)l

(3) Magnetic quantum number (m)

(i) It was proposed byZeeman and denoted by m.

(ii) It gives the number of permitted orientation of subshells.

(iii) The value ofm varies from to +

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