01 Some basic concepts of chemistry - Target Publications...Dobereiner’s Triads) in which the...

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STD. XI Sci.

Perfect Chemistry - I

Written as per the revised syllabus prescribed by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune.

Printed at: Prabodhan Prakashan Pvt. Ltd., Navi Mumbai

P.O. No. 126617

© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical

including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

TEID: 12540_JUP

Salient Features • Exhaustive coverage of syllabus in Question Answer Format.

• Covers answers to all Textual Questions, Intext Questions and relevant NCERT Questions.

• Includes Solved and Practice Numericals.

• Quick Review for instant revision and summary of the chapter.

• Exercise, Multiple Choice Questions and Topic test at the end of each chapterfor effective preparation.

• Important inclusions: NCERT Corner and Apply Your Knowledge

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In the case of good books, the point is not how many of them you can get through, but rather how many can get through to you. “Std. XI Sci. : PERFECT CHEMISTRY - I” is a treasure house of knowledge that’d not only prepare you to face the conspicuous Std. XI final exam but also equip you up on parallel ground to face the prospective NEET and JEE exam. This book is specifically aimed at Maharashtra Board students. The content of the book is framed in accordance with Maharashtra State board syllabus splattered with additional snippets of information from the NCERT syllabus. This lethal combination of apt material from both the boards makes it the ultimate reference material for Std. XI. This book has been developed on certain key features as detailed below: Question and Answer format of the book provides students with appropriate answers for all textual and intext questions.

We’ve also included additional questions to ensure complete coverage of every concept. Solved Examples provide step-wise solution to various numerical problems. This helps students to understand the

application of different concepts and formulae. NCERT Corner, Do You Know, Enrich Your Knowledge and Notes cover additional bits of relevant information on each

topic. Apply Your Knowledge, Brain Teasers and Check Your Grasp cover brain-storming questions to stengthen the students’

conceptual understanding. Quick Review and Formulae sections facilitate instant revision. Exercise helps the students to gain insight on the various levels of theory and numerical-based questions. Multiple Choice Questions and Topic Test assess the students on their range of preparation and the amount of knowledge

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Please write to us on : [email protected] A book affects eternity; one can never tell where its influence stops.

Best of luck to all the aspirants! Yours faithfully, Publisher Edition: Second

Disclaimer This reference book is transformative work based on textual contents published by Bureau of Textbook. We the publishers are making this reference book which constitutes as fair use of textual contents which are transformed by adding and elaborating, with a view to simplify the same to enable the students to understand, memorize and reproduce the same in examinations. This work is purely inspired upon the course work as prescribed by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune. Every care has been taken in the publication of this reference book by the Authors while creating the contents. The Authors and the Publishers shall not be responsible for any loss or damages caused to any person on account of errors or omissions which might have crept in or disagreement of any third party on the point of view expressed in the reference book. © reserved with the Publisher for all the contents created by our Authors. No copyright is claimed in the textual contents which are presented as part of fair dealing with a view to provide best supplementary study material for the benefit of students.

Contents No. Topic Name Page No.

1 Some Basic Concepts of Chemistry 1 2 States of Matter (Gases and Liquids) 34 3 Structure of Atom 84 4 Periodic Table 135 5 Redox Reactions 166 6 Chemical Equilibrium 213 7 Surface Chemistry 276 8 Nature of Chemical Bond 315 Periodic Table 391 Logarithms 392

'Chapters 9 to 17 are a part of Std. XI: Perfect Chemistry - II'

Note: All the Textual questions are represented by * mark All the Intext questions are represented by # mark

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Chapter 04: Periodic Table

Syllabus

Scientists Contributions John Dobereiner (1780-1849) (German chemist)

i. ii.

Known for his periodic law of chemical elements. He arranged the elements in sets of three (known as Dobereiner’s Triads) in which the atomic mass of the middle element was equal to the average atomic masses of other two elements.

Dimitri Ivanovich Mendeleev (1834-1907) (Russian chemist)

i. ii. iii.

Created the first version of periodic table of elements. Investigated the composition of oil fields and helped to establish first oil refinery in Russia. The crater Mendeleev on the moon and the element number 101 mendelevium were named after him.

Julius Lothar Meyer (1830-1895) (German chemist)

Showed that the plot of atomic weights against atomic volumes, consists of series of maxima and minima. The most electropositive elements are at the peak of the curve in the order of their atomic weights.

*Q.1. What is the need of classification of elements? Ans: i. Upto seventeenth century, only 31 elements were known due to which it was very easy to study and

remember the properties of these elements. At present 118 elements are known. ii. A great variation is observed in the physical and chemical properties of the elements. This makes the

study of these elements difficult. iii. Hence, it became necessary to arrange the elements in a systematic way. *Q.2. What is the basic theme of organization of elements in the periodic table? (NCERT) Ans: i. The basic theme of organization of elements in the periodic table is to simplify and systematize the

study of the numerous properties of all the elements and their compounds. ii. This has been done by arranging the elements in such a way that similar elements are placed together

while dissimilar elements are separated from one another. iii. This has made the study simpler and easier to remember because the properties of the elements are

now studied in the form of groups or families having similar properties rather than studying the elements individually.

4.0 Prominent scientists

4.1 Introduction

4.0 Prominent scientists

4.1 Introduction

4.2 Brief history of the development of periodic table

4.3 Modern periodic law and present form of periodic table 4.4 Periodic trends in properties

of elements

Periodic Table04

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Std. XI Sci.: Perfect Chemistry ‐ I

Q.3. Write a note on Unitary theory. Why was it ruled out? Ans: Unitary theory: i. In 1815, William Prout first suggested this theory. ii. Statement: The values of the atomic weights (atomic masses) of all the elements were whole numbers or

varied only slightly from the whole numbers, if hydrogen was considered the bases of all atomic weights. iii. As per this theory all the elements contain hydrogen atoms. eg. a. 12C contain 12 units of hydrogen. b. 40Ca contain 40 units of hydrogen. Limitations: i. This theory couldn’t explain the elements with fractional atomic weight such as copper with

atomic weight 63.5 and chlorine with atomic weight 35.5. ii. At that time, the existence of two isotopes of copper with atomic weight 63 and 65 and the existence

of two isotopes of chlorine with atomic weights 35 and 37, were not known. Hence, this theory was ruled out.

Q.4. Write a note on Dobereiner’s Triads. Ans: Dobereiner’s triads: i. In 1817, Dobereiner proposed the ‘law of triads’. ii. Statement: The elements could be arranged in a group of three called triad in such a way that the

middle element had an atomic weight almost the average of the other two. iii. The three elements of a triad had similar properties. eg. a. The average of atomic masses of lithium and potassium

is equal to the atomic mass of sodium.

b. Atomic mass of strontium is close to the average atomic

masses of calcium and barium.

c. The average of atomic masses of chlorine and iodine is

close to the atomic mass of bromine. iv. Limitation: This relationship worked for a few elements only. Hence, it was dismissed as coincidence. Q.5. What is Cooke’s Homologous series? Ans: Cooke’s Homologous series: i. In 1854, on the basis of physical and chemical properties, J.P.Cooke classified the elements in several

homologous series. ii. Statement: The atomic weights of the elements present in a homologous series increase in a regular

fashion. iii. This is shown in the table given below:

Element Atomic weight Type of atomic weightNitrogen (14N) 14 (14) a Phosphorus (31P) 31 (14+17) a+b Arsenic (75As) 75 (14+17+44) a+b+c Antimony (119Sb) 119 (14+17+88) a+b+2c Bismuth (207Bi) 207 (14+17+176) a+b+4c

Element Atomic weight Li 7 Na 23 K 39

Element Atomic weightCa 40 Sr 88 Ba 137

Element Atomic weightCl 35.5 Br 80 I 127

4.2 Brief history of the development of periodic table

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Q.6. What is Newland’s law of octaves? Give its limitations. Ans: Newland’s law of octaves: i. John A.R. Newland, proposed the “Law of Octaves” in 1865. ii. Statement: When the elements are arranged in the increasing order of their atomic weights, the

properties of every eighth element has properties similar to those of the first one. iii. Some of the octaves formed by Newland are shown in the following table:

Element Li Be B C N O F Atomic weight 7 9 11 12 14 16 19 Element Na Mg Al Si P S Cl Atomic weight 23 24 27 29 31 32 35.5 Element K Ca Atomic weight 39 40

iv. The relationship was just like every eighth note that resembles the first octaves of music. Limitations: i. Newland’s law of octaves worked for elements upto calcium. It failed for elements with higher atomic weights. ii. With the discovery of inert gases, it was found that they did not obey the law of octaves. Q.7. Give an account of Lothar Meyer arrangement of elements. Ans: i. In 1869, Lothar Meyer showed that when properties of the elements such as atomic volume, density

melting point, boiling point, thermal conductivity etc. were plotted against the atomic weights, they varied in a periodic manner.

ii. On this basis, Lothar Meyer developed a table of elements which closely resembled the Mendeleev’s periodic table.

Q.8. State Mendeleev’s periodic law. Ans: Mendeleev’s periodic law: “The physical and chemical properties of elements are the periodic function of

their atomic weights (atomic masses)”. Q.9. How did Mendeleev arrange all the known elements in a periodic table? Ans: i. Mendeleev arranged elements in horizontal rows and vertical columns of a table in order of their increasing

atomic masses. ii. The chemical and physical properties of elements showed repetition after certain intervals. iii. He placed these elements below the first row of elements to form the second row of elements. Thus,

the elements with similar properties were placed in the same vertical column or group. iv. He arranged all the known 63 elements according to their properties, to form the first periodic table. v. At some places in the periodic table, he ignored the increasing order of atomic weights. He placed

those elements together which had similar properties. eg. Iodine was placed in VII group along with other halogens. Although the atomic weight of iodine

was lower than that of tellurium, the properties of iodine and other halogens were similar. vi. He left several gaps in the periodic table keeping in mind that some of the elements were still undiscovered. eg. He left the gaps below Al and Si and called these elements as Eka-aluminium and Eka-silicon.

These gaps were filled after the discovery of gallium (Ga) and germanium (Ge). Note: i. Mendeleev predicted the existence of the elements (Ga and Ge) and estimated their properties.

When these elements were later discovered, the prediction of Mendeleev proved to be remarkably correct.

ii. This made him and his periodic table famous. *Q.10. Which important property did Mendeleev used to classify the elements in his periodic table? (NCERT) Ans: i. Mendeleev studied several physical and chemical properties of the elements and several compounds. ii. In his periodic table, he classified the elements on the basis of their atomic masses. *Q.11. How many periods and groups were present in Mendeleev’s Periodic table? Ans: Mendeleev’s periodic table had 12 periods and 9 groups (0 to VIII, including inert elements in group 0). Note: Mendeleev’s original periodic table had only 8 groups since inert elements were not discovered then.

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Q.12. Why was atomic number considered more fundamental than atomic mass? Ans: i. Henry Moseley in 1913, studied the relationship between X-ray spectra and the atomic number of the

elements. ii. He observed direct proportionality between the frequencies of X-rays emitted from the elements and

the atomic number (Z).

iii. A plot of ν (where is the frequency of X-rays emitted) versus atomic number (Z) gave a straight line.

Hence, Moseley concluded that atomic number and not atomic mass is the fundamental property of the atoms. He suggested that atomic number instead of atomic mass should be the basis of classification of elements. Hence, Mendeleev’s periodic law was modified.

Q.13. State and explain Modern periodic law. Ans: Statement:“The physical and chemical properties of the elements are periodic functions of their

atomic numbers”. Explanation: i. As per the law, the physical and chemical properties of the elements are dependant upon

atomic number and this dependence shows periodicity. ii. Thus, when the elements are arranged in the order of increasing atomic numbers, the elements with

similar properties should recur after regular intervals. Note: i. Based on the modern periodic law, several new periodic tables were proposed. ii. The long form (the extended form) of the periodic table is the most useful. iii. The modern periodic table contains 18 groups and 7 periods. Lanthanide and Actinide series are

placed separately at the bottom of the periodic table. Q.14. Give the advantages of the long form of periodic table. Ans: The important advantages of the long form of the periodic table are given below: i. The long form of the periodic table or extended form of the periodic table is based on the fact that the

physical and chemical properties of the elements are the periodic functions of their atomic numbers. ii. Since, this classification is based on the atomic number and not on the atomic mass, the position of

placing isotopes at one place is fully justified. iii. The position of elements in the periodic table is governed by the electronic configurations, which

determine their properties. iv. It is easy to remember and reproduce. v. The systematic grouping of elements into four blocks; s, p, d and f has made the study of the elements

more simple. vi. The position of some elements which were misfit on the basis of atomic mass is now justified on the

basis of atomic number. eg. Argon proceeds potassium because argon has atomic number 18 and potassium has 19. vii. The lanthanoids and actinoids which have properties different from other groups are placed separately

at the bottom of the periodic table. Q.15. State the characteristics of periods in the long form of the periodic table. Ans: Periods: i. The horizontal rows in the periodic table are called periods. ii. The long form of a periodic table has seven periods, numbered as 1, 2, 3, 4, 5, 6 and 7. iii. The first period contains two elements and second and third periods contain eight elements each.

These periods are called as short periods. iv. The fourth, fifth and sixth periods are called long periods. They contain 18, 18 and 32 elements respectively. v. The fourteen lanthanoids are placed in a series at the bottom of the periodic table. They belong to the

sixth period. vi. The seventh period is incomplete. It contains 23 elements which include 14 members of the

actinoid series placed at the bottom of the periodic table.

4.3 Modern periodic law and present form of periodic table

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Q.16. State the characteristics of groups in the long form of the periodic table. Ans: Groups: i. The vertical columns in periodic table are called groups or families. ii. The elements with similar physical and chemical properties are present in a group. iii. The long form of periodic table consists of 18 groups numbered as IA, IIA, IIIB to VIIB, VIII, IB,

IIB, IIIA to VIIA and 0 group. With recommendations of IUPAC, these groups are numbered from 1 to 18.

iv. Elements of groups 1, 2 and 13 to 17, are called normal or representative elements. Elements of group 18 are noble gases or inert gas elements.

v. Elements from group 3 to 12 are transition elements. vi. Lanthanoids and actinoids are placed at the bottom of the periodic table. These are two series of

fourteen elements each. Note: i. Lanthanoids are placed in the third group and sixth period. ii. Actinoids are placed in the third group and seventh period. iii. Lanthanoids and actinoids are known as inner transition elements or rare earth elements.

1

d block

p block

2

3 4 5 6 7 8 9 10 11 12

13 14 15

16

17

18

87 Fr

(223)

88 Ra

(226)

89 Ac

(227)

104 Rf

(261)

105 Db

(262)

106 Sg

(263)

107 Bh

(264)

108 Hs

(265)

110 Ds

(271)

111 Rg

(272)

112 Cn

(285)

113 Uut

(284)

114 Uuq (289)

115 Uup (288)

109 Mt

(268)

116 Uuh(292)

117 Uus

-

118 Uuo

-

84 Po 210

85 At

(210)

86 Rn

(222)

83 Bi

209

82 Pb 207

81 Tl

204

80 Hg 201

79 Au 197

78 Pt

195

77 Ir

192

76 Os 190

75 Re 186

74 W

183.9

73 Ta

181.0

72 Hf

178.5

57 La

138.9

56 Ba

137.3

55 Cs

132.9

37 Rb 85.5

38 Sr

87.6

39 Y

88.9

40 Zr

91.2

41 Nb 92.9

42 Mo 95.9

43 Tc 98

44 Ru 101

45 Rh 103

46 Pd 106

47 Ag 108

48 Cd 112

49 In

115

50 Sn 119

51 Sb 122

52 Te

127.6

53 I

126.9

54 Xe

131.3

19 K

39.1

20 Ca

40.1

21 Sc

45.0

22 Ti

47.8

23 V

50.9

24 Cr

52.0

25 Mn 54.9

26 Fe

55.8

27 Co

58.9

28 Ni

58.7

29 Cu 63.5

30 Zn

65.4

31 Ga 69.7

32 Ge

72.6

33 As

74.9

34 Se

79.0

35 Br

79.9

36 Kr

83.8

11 Na

23.0

12 Mg 24.3

3 Li 6.9

4 Be 9.0

13 Al

27.0

14 Si

28.1

15 P

31.0

16 S

32.1

17 Cl

35.5

18 Ar

39.9

5 B

10.8

6 C

12.0

7 N

14.0

8 O

16.0

9 F

18.9

10 Ne

20.2

2 He 4.0

s-block

p-block

1 H 1.1

Atomic number (Z)

*

* *

d-block

Symbol

Atomic mass (A)

Modern Periodic Table

* Lanthanide Series

Actinide Series

* *

f-block

58 Ce

140.1

59 Pr

140.9

60 Nd

144.2

61 Pm 145

62 Sm

150.4

63 Eu

152.0

64 Gd

157.3

65 Tb

159.0

66 Dy

163.0

67 Ho

165.0

68 Er

167.3

69 Tm

169.0

70 Yb

173.0

71 Lu

175.090 Th

232.0

91 Pa

231.0

92 U

238.0

93 Np

237.0

94 Pu 244

95 Am243

96 Cm247

97 Bk 247

98 Cf 251

99 Es 252

100 Fm 257

101 Md 258

102No 259

103Lr 262

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Q.17. State the relationship between long form of the periodic table and electronic configuration. Ans: i. The long form of the periodic table is based upon the atomic numbers. The long form of

periodic table and electronic configuration of the elements are closely related. ii. Each period of the periodic table corresponds to a particular shell. iii. A period begins with filling of a particular shell and ends when the shell is completely filled. iv. The number of period corresponds to the principal quantum number of the valence shell. v. The total number of elements present in a particular period is equal to the number of electrons that

can be accommodated in the valence shell. Note: A group constitutes a series of elements with same outermost electronic configuration. Q.18. Explain how does the filling of electrons takes place in i. First period ii. Second period iii. Third period Ans: i. Filling of electrons in first period: a. The first period corresponds to the filling of the first shell, i.e., n = 1 shell. b. The first shell contains only one orbital (1s orbital). Hence, it can accomodate a maximum of

two electrons only. c. Therefore, the first period has two elements hydrogen (1s1) and helium (1s2). d. In He, the first shell (i.e., K shell) is completely filled. Hence, it is not possible to accommodate

any more element in this period. ii. Filling of electrons in second period: a. The second period corresponds to the filling of the second shell i.e. n = 2. This shell contains

four orbitals (2s,2px,2py,2pz). Hence, it can accomodate a maximum of eight electrons. b. Therefore, the second period contains eight elements. c. In Li and Be, 2s orbital is filled whereas in the other elements (B, C, N, O, F and Ne) 2p orbital

is filled. d. In neon, the valence shell is completely filled. iii. Filling of electrons in third period: a. The third period corresponds to the filling of the third shell i.e. n = 3. This shell contains nine

orbitals (one 3s, three 3p and five 3d orbitals). b. The energy of 3d orbitals is higher than that of 4s orbital. Hence, only one 3s and three 3p orbitals

can be filled before 4s shell begins to be filled. c. Due to filling of 3s, 3px, 3py and 3pz orbitals, the third period contains a maximum of eight elements. d. In Na and Mg, 3s orbital is filled whereas in the other elements (Al, Si, P, S, Cl, Ar), 3p orbital is filled. e. Thus, first element Na has the configuration [Ne] 3s1 and the last element Ar has the

configuration [Ne] 3s2 3p6. Q.19. There cannot be more than 18 elements in the fourth period. Explain. Ans: i. The fourth period corresponds to the filling of fourth shell, n = 4. Therefore, 4s orbitals are filled first. ii. The energy of 3d orbital is lower than that of 4p orbitals. Hence, electrons enter into 3d shell till it is

completely filled. iii. Electrons then enter into 4p orbitals. iv. 4d and 4f orbitals have higher energy. Hence, they cannot be filled before filling of 5s orbitals. v. Thus, the elements in fourth period are: 2 elements (with 4s orbitals), 10 elements (with 3d orbitals)

and 6 elements (with 4p orbitals). vi. Hence, fourth period cannot have more than 18 elements and is completed at Kr with fully filled

4p orbitals. Q.20. Justify - “There are only 18 elements in the fifth period.” Ans: i. The fifth period corresponds to the filling of fifth shell i.e., n = 5. First 5s orbital is filled in

Rb (5s1) and Sr (5s2) and then 4d orbitals are filled Y(4d1,5s2)…Cd(4d10,5s2). ii. When the 4d orbitals are completely filled, the electrons enter into 5p orbitals from In(4d10 5s2 5p1) to

Xe(4d10 5s2 5p6). iii. The 4f, 5d and 5f orbitals are of higher energy. Hence, they cannot be filled before filling of 6s orbital. iv. Thus, the fifth period gets completed at Xe and contains only eighteen elements.

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Q.21. Explain how does the filling of electrons takes place in i. sixth period ii. seventh period. Ans: i. Filling of electrons in sixth period: a. The sixth period corresponds to the filling of sixth shell i.e., n = 6. First 6s orbital is filled in

Cs (6s1) and Ba (6s2). Then 5d orbital is filled in La with configuration [Xe] 5d16s2. b. The energy of 4f orbital is lowered when the electron enters into 5d orbitals. Hence, 4f orbitals

are filled in elements Ce (4f15d16s2) ….… Lu (4f145d16s2). c. These fourteen elements (i.e., from Ce to Lu) are lanthanoids placed in a separate series

(called as lanthanoid series) at the bottom of the periodic table. d. When 4f orbitals are completely filled, 5d orbital starts filling till Hg (4f14 5d10 6s2). e. Then 6p subshell is filled in elements Tl(4f14 5d10 6s2 6p1) ……… Rn (4f14 5d10 6s2 6p6). f. Sixth period is completed. It contains a total of thirty two elements. ii. Filling of electrons in seventh period: a. The seventh period corresponds to the filling of seventh shell, i.e., n = 7. The seventh shell is

filled similar to sixth shell. b. The 7s orbital is filled in elements Fr (7s1) and Ra (7s2). Then 6d subshell is filled in

Ac (6d1 7s2). c. Then 5f orbitals are filled from Th (5f1 6d1 7s2) to Lr (5f14 6d1 7s2). These fourteen elements are

actinoids and are placed in a separate series (called actinoid series) at the bottom of the periodic table.

d. Now 6d and 7p orbitals are filled in the remaining elements of this period. e. The seventh period can accomodate a maximum of 32 elements. It is yet incomplete. Q.22. Which period is incomplete period? Ans: The seventh period can accommodate a maximum of 32 elements. However, only 29 elements are present.

Hence, it is incomplete period. Q.23. What are lanthanoid and actinoid series? Ans: i. Lanthanoid series: The elements after Lanthanum i.e., from Cerium (4f1 5d1 6s2) to

Lutetium (4f14 5d1 6s2) in total fourteen elements are called lanthanoids and are kept in separate series called Lanthanoid series at the bottom of the periodic table.

ii. Actinoid series: The fourteen elements after Actinium i.e., from Thorium (5f1 6d1 7s2) to Lawrencium (5f14 6d1 7s2) are kept in separate series called actinoid series at the bottom of the periodic table.

Q.24. State the drawbacks of long form of the periodic table. Ans: i. Hydrogen is not properly placed. It resembles with alkali metals and halogens. However, it has been

placed with alkali metals. ii. Lanthanoids and actinoids are not included in the main body of the periodic table.

Helium belongs to the s-block but it is placed in the p-block along with other group 18 elements. This is justified because it has a completely filled valence shell (1s2) and as a result, exhibits

properties characteristic of other noble gases. Hydrogen has only one s-electron and hence can be placed in group 1 (alkali metals). It can also

gain an electron to achieve a noble gas configuration and hence it can behave similar to a group 17element. Since, it is a special case, hydrogen is placed separately at the top of the periodic table.

NCERT Corner

Position of hydrogen and helium:

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*Q.25. Give an account of history of the development of periodic table. Ans: Classification of elements into groups and development of periodic law and periodic table are the

consequences of numerous attempts by scientists through their observations and experiments. Various prominent attempts made in this field are summarized below: i. Unitary theory: William Prout first suggested this theory in 1815. As per this theory, all the elements

contain hydrogen atoms. ii. Dobereiner’s traids: In 1817, Dobereiner arranged elements with similar properties in a group of

three elements (called triads) in such a way that the middle element had an atomic weight almost the average of the other two.

iii. Cooke’s homologous series: In 1854, on the basis of physical and chemical properties, J. P. Cooke classified the elements in several homologous series.

iv. Newland’s octaves: In 1865, John A. R. Newland, arranged the elements in the increasing order of atomic weights so that the properties of every eighth element were similar to those of the first one.

v. Lothar Meyer’s arrangement of elements: In 1869, Lothar Meyer arranged elements according to physical properties (atomic volume, density, melting point, boiling point, thermal conductivity, etc., and showed periodic variations in these properties.

vi. Mendeleev’s periodic table: In 1869, Dimitri Mendeleev arranged the elements on the basis of their atomic masses.

vii. Modern periodic table: In modern periodic table (also known as long form or extended form of the periodic table), the elements are arranged on the basis of their atomic numbers. This periodic table is most useful.

Q.26. Explain why the elements with Z > 100 should be named according to the IUPAC rules? Ans: i. The elements with atomic numbers greater than 100 were discovered during artificial transmutations

of elements. ii. Being highly unstable, they were obtained in minute quantities. iii. This resulted in several complications. Some elements were discovered by more than one scientist.

Due to this different scientists assigned different names to same newly discovered element. eg. Element having atomic number 104 was discovered by both American as well as Soviet

scientists. The American scientist named it as Rutherfordium, whereas Soviet scientist named it Kurchatovium.

iv. To avoid complications, the IUPAC, has recommended that a newly discovered element be named according to the rules till discovery is confirmed and approved by it.

Rules for IUPAC nomenclature of elements with Z > 100: i. The name of element is to be derived from atomic number. The following numerical roots are to be used. ii. The roots are written together in the order of digits present in the atomic number. They are terminated

by the suffix ‘ium’. If ‘enn’ occurs before ‘nil’, the second ‘n’ of ‘enn’ is dropped. Similarly the final ‘i’ of ‘bi’ and ‘tri’ is dropped when they occur before ‘ium’.

iii. The symbol of element is derived by composing the initial letters of the numerical roots (abbreviations) which constitute the name.

IUPAC Nomenclature system

Digits present in atomic number 0 1 2 3 4 5 6 7 8 9

Numerical Root nil un bi tri quad pent hex sept oct enn

Abbreviation n U b t q p h s o e

IUPAC nomenclature of elements with Z > 100

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eg. The systematic IUPAC name of the element with Z = 106 is derived as follows: The atomic number 106 consists of digit 1, 0 and 6. The numerical roots of these digits, are written to

get unnilhex. The remaining suffix ‘ium’ is added to this, to get unnilhexium. This is the systematic IUPAC name of the element having Z = 106. As per rule (iii), its symbol would be ‘Unh’. iv. The systematic IUPAC official names of some elements having Z > 100 are given below:

Atomic Number Name Symbol IUPAC official Name IUPAC symbol 101 Unnilunium Unu Mendelevium Md 102 Unnilbium Unb Nobelium No 103 Unniltrium Unt Lawrencium Lr 104 Unnilquadium Unq Rutherfordium Rf 105 Unnilpentium Unp Dubnium Db 106 Unnilhexium Unh Seaborgium Sg 107 Unnilseptium Uns Bohrium Bh 108 Unniloctium Uno Hassnium Hs 109 Unnilennium Une Meitnerium Mt 110 Ununnilium Uun Darmstadium Ds 111 Unununium Uuu Rontgenium Rg 112 Ununbium Uub Copernicium Cn 113 Ununtrium Uut - - 114 Ununquadium Uuq - - 115 Ununpentium Uup - - 116 Ununhexium Uuh - - 117 Ununseptium Uus - - 118 Ununoctium Uuo - -

- Elements yet to be discovered #Q.27. What would be the IUPAC name and symbol of the element with atomic number 119? Ans: i. The element has atomic number 119. From the IUPAC system, roots of 1, 1, 9 are un, un and enn

respectively. ii. The numerical roots of these digits are written together to get, ‘ununenn’. iii. The suffix ‘ium’ is added to this, to get the name of the element ‘ununennium’. iv. The symbol of the element is ‘Uue’. Q.28. Given an account of four blocks of the periodic table. Ans: The elements in the periodic table are classified into s-block, p-block, d-block and f-block. This is on the

basis of the type of orbital in the valence shell in which the last electron enters. i. s-block elements: a. The last electron enters s-orbital of the valence shell. b. They have electronic configuration ns1 and ns2. They belong to group-1 and group-2 respectively. c. They are placed on the extreme left of the periodic table. ii. p-block elements: a. The last electron enters p-orbital of the valence shell. b. They have electronic configuration ns2np1 to ns2np6. They belong to groups 13 to 18 (except He). c. They are placed on extreme right of periodic table. iii. d-block elements: a. The last electron enters d-orbital of penultimate shell, i.e., (n 1) d-orbital. b. They have electronic configuration (n 1) d1ns1 2 to (n 1) d10ns12. They belong to groups 3 to 12. c. They are placed in the middle portion of the periodic table. d. There are four d series elements (3d, 4d, 5d and 6d series).

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iv. f-block elements: a. The last electron enters into f-orbital of pre-penultimate shell, i.e., (n 2) f-orbital. b. They have electronic configuration (n 2) f1 (n 1) d01 ns2 to (n 2) f14 (n 1) d01 ns2. c. They are placed at the bottom of the periodic table. d. This block consists of series of Lanthanides and Actinides. Note: Q.29. How can a period, group and block of the element be determined? Ans: The group, period and the block of the element can be determined on the basis of its electronic configuration. i. Period: The principal quantum number of the valence shell corresponds to the period of the element. eg. The principal quantum number of the valence shell (3s1) of Na (1s2 2s2 2p6 3s1) is 3.

This corresponds to third period. ii. Block: The subshell in which the differentiating electron enters, corresponds to the block of the

elements (with exception being He). eg. The subshell 3d (in which the differentiating electron enters) for Sc(1s2 2s2 2p6 3s2 3p6 3d1 4s2)

corresponds to 3d block. iii. Group: The group of the element is determined on the basis of number of electrons present in the

outermost or penultimate (next to outermost, i.e. (n1)) shell: a. For s-block elements, number of the group = number of valence electrons. b. For p-block elements, number of the group = 2 + 10 + number of electrons in p subshell. c. For d-block elements, number of the group = 2 + number of (n1) d electrons. Q.30. Predict the block, periods and groups to which the following elements belong. i. Mg (Z = 12) ii. V (Z = 23) iii. Sb (Z = 51) iv. Rn (Z = 86) v. Na (Z = 11) vi. Cl (Z = 17) Ans: i. Mg (Z = 12): Atomic number of Mg is 12. Electronic configuration is 1s2 2s2 2p6 3s2. Block: The subshell 3s (in which the differentiating electron enters) corresponds to s block. Period: The principal quantum number of the valence shell (3s) is 3. This corresponds to third period. Group: For s block element, number of the group = number of valence electrons = 2. Hence it

belongs to group 2. ii. V (Z = 23): Atomic number of V is 23. Electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d3 4s2. Block: The subshell 3d (in which the differentiating electrons enters) corresponds to d block. Period: The principal quantum number of the valence shell (4s2) is 4. This corresponds to fourth period. Group : For d block elements, group = 2 + number of (n 1) d electrons = 2 + 3 = 5. Hence it

belongs to group 5.

s-block elements

d-block elements

p-blockelements

f-block elements

1

2

3 4 5 6 7 8 9 10 11 12

13 14 15 16 17

18

Position of s, p, d, f blocks in long form of periodic table

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iii. Sb (Z = 51): The atomic number of Sb is 51. Electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p3. Block: The subshell 5p (in which the differentiating electron enters) corresponds to p block. Period: The principal quantum number of the valence shell (5s25p3) is 5. This corresponds to

fifth period. Group: For p block elements, group = 2 + 10 + number of electrons in p subshell = 2 + 10 + 3 = 15.

Hence it belongs to group 15. iv. Rn (Z = 86): The atomic number of Rn is 86. Electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6p6.

Block: The subshell 6p (in which the differentiating electrons enters) corresponds to p block. Period: The principal quantum number of the valence shell (6s26p6) is 6. This corresponds to sixth period. Group: For p block elements, group = 2 + 10 + number of electrons in p subshell = 2 + 10 + 6 = 18.

Hence it belongs to group 18. v. Na (Z = 11): Atomic number of Na is 11. Electronic configuration is 1s2 2s2 2p6 3s1. Block: The subshell 3s (in which the differentiating electron enters) corresponds to s-block. Period: The principal quantum number of the valence shell (3s) is 3. This corresponds to third period. Group: For s-block element, number of the group = number of valence electrons = 1. Hence it

belongs to group 1. vi. Cl (Z = 17): Atomic number of Cl is 17. Electronic configuration is 1s2 2s2 2p6 3s2 3p5. Block: The subshell 3p (in which the differentiating electron enters) corresponds to p block. Period: The principal quantum number of valence shell (3s23p5) is 3. This corresponds to third period. Group: For p block elements, group = 2 + 10 + number of electrons in p subshell = 2 + 10 + 5 = 17.

Hence it belongs to group 17. Q.31. What are periodic properties? Ans: i. The elements in long form of periodic table are arranged in such a way that on moving across a period

or down the group, several properties of elements vary in regular fashion. These properties are called periodic properties.

eg. boiling point, melting point, heat of fusion and vapourisation, energy of atomization, etc. are several physical properties of elements show periodic variations.

ii. Atomic and ionic radii, ionization enthalpy, electron gain enthalpy, electronegativity, valency and oxidation states are some of the other important periodic properties.

Q.32. Define following terms: i. Periodicity *ii. Atomic radius Ans: i. Periodicity: The periodic recurrence of elements having similar properties after regular intervals is

called periodicity. ii. Atomic radius: Atomic radius (atomic size) of an atom may be regarded as the distance from the

centre of the nucleus of an atom to the outermost shell (valence shell) of electrons. Units: It is expressed in the terms of Angstrom or picometre where 1Å = 1010 m and 1 pm = 1012 m Q.33. Define covalent bond length. Ans: Internuclear distance in a diatomic molecule of an element is known as its covalent bond length. Q.34. How is atomic size of a non-metallic element estimated? Ans: i. The atomic size of a non-metallic element is estimated from the distance between the two atoms

bound together by a single covalent bond. From this the covalent radius of the element is estimated. ii. The internuclear distance in a diatomic molecule of an element is its covalent bond length. Half the

covalent bond length gives the covalent radius.

covalent

Internuclear distancebetween two bondedatomsr =

2

eg. Bond distance in the chlorine molecule (Cl2) is 198 pm. The half of this distance (99 pm) is the atomic radius of chlorine.

4.4 Periodic trends in properties of elements

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Q.35. Define metallic radius. Ans: Metallic radius is defined as ‘half the internuclear distance separating the metal ions in the metallic

crystal’. OR One half of the distance between the centres of nucleus of the two adjacent atoms of a metallic crystal is

called as a metallic radius. eg. The distance between two adjacent copper atoms in solid copper is 256 pm. The half of this distance

(128 pm) is the metallic radius of copper. Q.36. Write a note on van der Waals radii. Ans: i. In elements where the atoms are not chemically bound to each other, the only attractive forces are

van der Waals forces. ii. The shortest distance to which the atoms of the element can approach before their electron clouds

start repelling each other is called van der Waals radii. iii. van der Waals radii are usually larger than the covalent radii because orbital overlap during

hybridization in covalent molecules brings the atoms much closer. iv. Noble gases which are monoatomic in nature have very large values of atomic radii

(non-bonded radii). Ideally these atomic radii should not be compared with covalent radii but with the van der Waals radii of other elements.

Q.37. Write a note on atomic radius. Ans: Atomic radius (atomic size) of an atom may be regarded as the distance from the centre of the nucleus of

an atom to the outermost shell (valence shell) of electrons. Units: It is expressed in the terms of Angstrom or picometre where 1Å = 1010 m and 1 pm = 1012 m respectively. i. The size of atom is small (1.2 1010 m). Atomic radius is a very important property because many

physical and chemical properties are related to it. ii. The term atomic radius refers to both covalent radius or metallic radius depending on whether the

element is a non-metal or a metal. iii. Measurement of atomic radii can be done by X-ray or other spectroscopic methods. *Q.38. Explain the factors affecting atomic radius. Ans: Atomic size depends upon the following factors: i. Number of shells: Atomic radius is directly proportional to the number of electronic shells present in

an atom of the element. It increases with increase in the number of electronic shells. Atomic size (or radius) Number of shells present in an atom.

ii. Nuclear charge: a. Atomic radius is inversely proportional to the nuclear charge. It decreases with increase in the

nuclear charge. b. When the nuclear charge is more, the nucleus attracts the electrons towards it and atomic size decreases.

Atomic size (or radius) 1

nuclear charge

iii. Screening effect or shielding effect: a. For a given quantum shell, in an atom having many electrons, the electrons in the inner shells

tend to prevent the attractive influence of the nucleus from reaching the outermost electron. Thus, they act as a screen or shield between the nuclear attraction and outermost or valence electrons and the effect is known as screening effect or shielding effect.

b. The shielding ability of inner electrons decreases in the order of s > p > d > f. c. The atomic radius is directly proportional to shielding effect. As the screening or

shielding effect increases, the atomic radius also increases. atomic radius shielding effect *Q.39. Explain the trends or variation in atomic radius. Ans: i. Trends or variation in atomic radius along a period: a. As we move across a period from left to right in the periodic table, the atomic radius of an

element decreases with the increase in atomic number.

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b. This is because, as the atomic number increases, nuclear charge increases gradually but addition of electrons takes place in the same shell.

c. Hence, the attraction between nucleus and valence electrons increases and atomic size decreases. d. Therefore, in a period from left to right in the periodic table, the atomic size is largest for

alkali metals, decreases gradually and it becomes smallest for the halogen elements. ii. Trends or variation in atomic radius down a group: a. As we move down the group from top to bottom in the periodic table, the atomic radius

increases with the increase in atomic number. b. This is because, as the atomic number increases, nuclear charge increases but simultaneously

the number of shells in the atoms also increases. c. As a result, the effective nuclear charge (attraction between nucleus and valence electrons)

decreases due to the increase in the size of the electron cloud and hence the atomic size increases in a group from top to bottom.

Note: 1. Atomic radii (pm) across the periods:

2. Atomic radii down the groups:

Atom (group 1) Atomic radius (pm) Atom (group 17) Atomic radius (pm) Li 152 F 72 Na 186 Cl 99 K 231 Br 114 Rb 244 I 133 Cs 262 At 140

3. 4.

Atom (Period 2)

Li Be B C N O F Atom

(Period 3)Na Mg Al Si P S Cl

Atomic radius

152 111 88 77 74 66 64 Atomic

radius 186 160 143 117 110 104 99

50

100

150

200

250

300

Ato

mic

rad

ius

(pm

)

Atomic number (Z)

Li(152) Na(186)

K(231) Rb(244)Cs(262)

I(133)Br(114)C1(99)

F(72)

Variation of atomic radius withatomic number of alkali metals and halogens

The variation of atomic radius alkali metals and halogens in their family is shown in the following figure:

Variation of atomic radius with atomic numberacross the second period is shown in the followingfigure:

Li

Be

B

C N

O F

2 6 8 104

Ato

mic

rad

ius

(pm

)

60

80

100

120

140

160

Atomic number (Z)Variation of atomic radius with atomic

number across the second period

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*Q.40. Define ionic radius. Ans: Ionic radius is defined as the distance of valence shell of electrons from the centre of the nucleus in an ion. Units: It is expressed in the terms of Angstrom or picometre where 1Å = 1010 m and 1 pm = 1012 m

respectively. Q.41. Explain the trends or variation in ionic radius. Ans: i. Trends or variation in ionic radius along a period: a. As we move across a period from left to right in the periodic table, the ionic radius of an ion

decreases with the increase in atomic number. b. This is because, as the atomic number increases, nuclear charge increases gradually but

addition of electrons takes place in the same shell. c. Hence, the attraction between nucleus and valence electrons increases and ionic size decreases. ii. Trends or variation in ionic radius down a group: a. As we move down the group from top to bottom in the periodic table, the ionic radius increases

with the increase in atomic number. b. This is because, as the atomic number increases, nuclear charge increases but simultaneously

the number of shells in the ions also increases. c. However, the effective nuclear charge (attraction between nucleus and valence electrons)

decreases due to the increase in the size of the electron cloud and hence the ionic size increases in a group from top to bottom.

As we move across a period from left to right in the periodic table, the atomic radius of an elementdecreases with the increase in atomic number. i. Trends across the period in s and p-block elements: a. In ‘s’ and ‘p’ block elements, along a period with increase in atomic number, the electron cloud

density of the inner shell remains unchanged because the electrons are added to the outermostshell.

b. Therefore, screening effect of the inner electrons remains the same and the nuclear chargeincreases.

c. Hence, the effective nuclear charge on the outermost electron also increases. d. Therefore, the atomic radius decreases significantly as the atomic number increases along a

period. ii. Trends across the period in d-block elements (transition elements): a. In the case of d-block elements, along a period, as the atomic number increases, the electrons are

added to the penultimate shell [(n 1)d]. b. It leads to the steady increase in the electron cloud density of the inner shells and screening effect

of inner electrons increases. c. Thus, though the nuclear charge progressively increases with increase in atomic number, the

effective nuclear charge on the outer electron decreases due to screening effect. d. As a result, atomic radius decreases to a lesser extent than that among the ‘s’ and ‘p’ block

elements. iii. Trends across the period in f-block elements (inner-transition elements): a. In the case of f-block elements, along a period, with increase in atomic number, the electrons are

added to the pre-penultimate shell [(n 2)f]. b. It leads to the steady increase in the electron cloud density in pre-penultimate shell and thereby

increases the screening effect by the larger number of electrons. c. Thus, the increase in the effective nuclear charge on the outer electrons decreases more than that

in d-block elements. d. As a result, the decrease in the atomic radius amongst f-block elements is extremely less.

Enrich Your Knowledge

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*Q.42. Radius of cation is smaller and that of anion is larger than that of the corresponding atom. How is this behaviour accounted? Illustrate with example.

Ans: i. The gaseous atom loses one or more electrons to form the corresponding cation. M Mn+ + ne ; Here, n represents the number of electrons lost. ii. Due to this, the number of electrons remaining decreases but the nuclear charge remains same. Thus

the same nuclear charge now acts on lesser number of electrons. This increases the effective nuclear charge per electron. Hence outermost electrons experience greater pull towards the nucleus. This decreases the size of the cation.

eg. Sodium atom and sodium cation (Na+), have radii equal to 186 pm and 95 pm respectively. iii. The gaseous atom loses one or more electrons to form the corresponding anion. A + ne An ; Here, n represents the number of electrons gained. iv. Due to this the number of electrons increases but the nuclear charge remains same. Thus the same

nuclear charge now acts on more number of electrons. This decreases the effective nuclear charge per electron. Hence outermost electrons experience lesser pull towards the nucleus. Also the inter electronic repulsion increases and the size of the anion increases.

eg. Fluorine atom and fluoride ion (F) have radii 64 pm and 136 pm respectively. *Q.43. Explain why the size of isoelectronic species decreases with the increase in atomic number. Ans: i. The isoelectronic species (atoms or ions) are those which have the same number of electrons. ii. For isoelectronic species, the size decreases with an increase in atomic number. This is because with

increase in atomic number, nuclear charge increases. iii. However, this increased nuclear charge acts on the same number of electrons in species and results in

decrease of size. eg. a. Consider the isoelectronic species, Na+, Mg2+, Al3+ and Si4+.

Ions Na+ Mg2+ Al3+ Si4+

Number of electrons 10 10 10 10 Nuclear charge (At.Number) 11 12 13 14 Ionic radii in pm 95 72 53.5 40

The nuclear charge increases from Na+ to Si4+ but the number of electrons remains the same in

each ion. Hence, the size decreases from Na+ to Si4+. b. Consider the species; O2, F, Na+ and Mg2+. All of these have the same number of electrons

i.e., (10). Their radii would be different because of their different nuclear charges.

Ions O2 F Na+ Mg2+

Number of electrons 10 10 10 10

Nuclear charge (Atomic number) 8 9 11 12 The cations with greater positive charge will have smaller radius because of the greater

attraction of electrons to the nucleus. Anions with greater negative charge will have the larger radius. In this case, the net repulsion of the electrons will outweigh the nuclear charge and the ion will expand in size.

Therefore, the size decreases in the order: O2 > F > Na+ > Mg2+ Q.44. Name a species that will be isoelectronic with each of the following atoms or ions. i. F ii. Ar iii. Mg2+ iv. Rb+ (NCERT) Ans: i. F possesses 10 electrons, O2 is its isoelectronic species. ii. Ar possesses 18 electrons, Cl is its isoelectronic species. iii. Mg2+ possesses 10 electrons, Na+ is its isoelectronic species. iv. Rb+ possesses 36 electrons, Kr is its isoelectronic species.

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Q.45. Which of the following species will have largest and the smallest size? Mg, Mg2+, Al, Al3+ Ans: i. Atomic radii decreases across the period. Hence, the atomic radius of Mg is larger than that of Al. ii. Parent atoms have larger radius than their corresponding cations. Hence, the radius of Mg is larger

than that of Mg2+ and the radius of Al is larger than that of Al3+. iii. Among iso-electronic species, the one with larger positive nuclear charge will have smaller radius.

Mg2+ and Al3+ are isoelectronic with Al3+ having larger positive nuclear charge. Hence, the ionic radius of Al3+ is smaller than Mg2+.

Hence, the decreasing order of radii is Mg > Al > Mg2+ > Al3+. iv. Therefore, species with the largest size is Mg and with smallest size is Al3+. Q.46. Consider the species: N3, O2, F, Na+, Mg2+ and Al3+ i. What is common in them? ii. Arrange them in the order of increasing ionic radii. (NCERT) Ans: i. All these ions have same number of electrons (10). Therefore, these are isoelectronic species. ii. Since, the number of electrons are same, the ionic size increases with decrease in nuclear charge

(i.e., positive charge). Therefore, the ions can be arranged in increasing order of ionic radii as, Al3+ < Mg2+ < Na+ < F < O2 < N3

*Q.47. Define ionization enthalpy. Ans: Ionization enthalpy is defined as the amount of energy required to remove most loosely bound electron

from an isolated gaseous atom of an element to form positive gaseous ion in its ground state. *Q.48. Give reasons “Second ionization enthalpy is greater than first ionization enthalpy”. Ans: i. The energy required to remove first electron from a gaseous atom of an element is called

first ionization enthalpy. X(g) X+

(g) + e ; 1H ii. The energy required to remove an electron (i.e., second electron) from singly positively charged

gaseous cation of an element is called second ionization enthalpy. X+

(g) X2+(g) + e ; 2H

iii. Energy is always required to remove electrons from an atom. Hence, ionization enthalpies have positive value.

iv. It is more difficult to remove an electron from a positively charged ion than from a neutral atom. Hence, the second ionization enthalpy (IE2) is higher than the first ionization enthalpy (IE1).

v. In general, if IE1, IE2 and IE3 are first, second and third ionization enthalpies respectively then,

IE3 > IE2 > IE1. *Q.49. Explain the factors affecting ionization enthalpy. Ans: Ionization enthalpy depends on the following factors: i. Size (radius) of atom: Larger the size of an atom, lesser is the attraction between the nucleus and

outermost electron. Hence it is easier to remove electron from it. Therefore, as atomic size increases, ionization enthalpy decreases.

Ionization enthalpy 1

atomic radius

ii. Nuclear charge: Greater the charge on the nucleus of an atom, greater will be the attraction between the nucleus and outermost electron and it will be more difficult to remove electron from an atom. Hence as the nuclear charge increases, ionization enthalpy increases.

iii. The shielding or screening effect of inner electrons: a. The nuclear attraction on the outermost electrons decreases with the increase in shielding or

screening effect. b. Due to screening effect of inner electrons, the effective nuclear charge decreases, therefore

ionization enthalpy decreases.

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c. The shielding effect of inner electrons for the given quantum number decreases in the order of s > p > d > f, hence the ionization enthalpy also decreases in the same order. d. Therefore more energy will be required for the removal of s electron than p, d and f electrons

for the same principal shell or for the removal of s electron, ionization enthalpy will be more and it will be least for f-electrons.

e. Shielding is effective when the orbitals in the inner shells are completely filled. iv. Nature of electronic configuration: a. The atoms of the elements having extra stability due to half-filled or completely filled orbitals

have higher ionisation enthalpy values. b. Therefore inert elements which have complete octet and extra stability have high ionisation enthalpies. *Q.50. What is screening effect? How does it govern the ionization enthalpy of an atom? Ans: Screening Effect: The inner shell electrons in an atom screen or shield the outermost valence electrons

from the nuclear attraction, and this effect is called screening effect or shielding effect. Effect on ionization enthalpy: Refer Q.49.iii. *Q.51. Explain the trends or variation in ionization enthalpy. Ans: i. Trends or variation in ionization enthalpy along a period: a. As we move across a period from left to right in the periodic table, the first ionization enthalpy

of an element increases (with few exceptions) with the increase in atomic number. b. This is because, as the atomic number increases, nuclear charge increases gradually but

addition of electrons takes place in the same shell. c. The added electrons poorly shield each other from the nucleus. Hence, the attraction between

nucleus and valence electrons increases and atomic size decreases. d. With the decrease in atomic size it becomes more difficult to remove the electron from valence

shell and hence, the first ionization enthalpy increases across a period. e. Therefore, the first ionization enthalpy for alkali metals has the lowest value while that for the

inert gas elements has the highest value. f. There are some irregularities in general trend of ionization energy. eg. Ionization energy of boron is less than that of beryllium. Similarly, ionization energy of

nitrogen is greater than that of oxygen. g. In the case of transition and inner transition elements, ionization energy increases gradually

across the period. ii. Trends or variation in ionization enthalpy down a group: a. As we move down the group from top to bottom in the periodic table, the first ionization

enthalpy decreases with the increase in atomic number. b. This is because, as the atomic number increases, nuclear charge increases but simultaneously

the number of shells in the atoms also increases. c. However, the effective nuclear charge (attraction between nucleus and valence electrons)

decreases due to the increase in the size of the electron cloud and therefore, the first ionization enthalpy decreases down the group.

Note: Ionization enthalpies for the second period elements are shown in the following table:

Element Li Be B C N O F Ne First ionization Enthalpy (kJ mol1)

520 899 801 1086 1402 1324 1681 2080

The ionization enthalpies down the group of alkali metals are shown in the following table:

Elements Li Na K Rb Cs First ionization Enthalpy (kJ mol1)

520 496 419 403 374

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Q.52. Give reasons “Across a period in periodic table, alkali metals have lowest value of first ionization enthalpy while that of noble gases is highest.”

Ans: i. Alkali metals have only one electron in their valence shell which can be easily lost resulting in the stable noble gas configuration.

ii. However, in case of inert gas elements or noble gases, orbitals are completely filled (ns2np6) and the amount of energy required to remove the electrons from these orbitals is comparatively more.

Hence, across a period in periodic table, alkali metals have lowest value of first ionization enthalpy while that of noble gases is highest.

Note: The trends in ionization enthalpies are shown in the following graphs. Q.53. Why is the first ionization enthalpy of boron less than that of beryllium? Ans: i. Boron has electronic configuration 1s2 2s2 2p1. Beryllium has electronic configuration 1s2 2s2. ii. Ionization of boron requires removal of an electron from 2p subshell and that of beryllium requires

removal of an electron from 2s subshell. iii. For the same principal quantum shell, the removal of an electron from p subshell requires lower

energy than the removal of an electron from s subshell. Hence, the first ionization enthalpy of boron is less than that of beryllium.

Q.54. Why is the first ionization enthalpy of oxygen less than that of nitrogen? Ans: i. Nitrogen has electronic configuration 1s2 2s2 2p3. Oxygen has electronic configuration 1s2 2s2 2p4. ii. Ionization of nitrogen requires removal of an electron from 2p3 subshell which has extra stability as it

is half filled. iii. Ionization of oxygen requires removal of an electron from 2p4 subshell which is more than half filled. iv. The removal of electron from a subshell which is more than half filled requires less energy than the

removal of electron from half filled subshell. Hence, the first ionization enthalpy of oxygen is less than that of nitrogen.

H

(kJ

mol

1)

550

500

450

400

3500 10 20 30 40 50 60

Atomic number (Z)

Rb(403)

Na(496)

Li(520)

K(419)

Cs(374)

First ionization enthalpies 1H of alkalimetals as a function of Z.

H

(kJ

mol

1)

0 2 4 6 8 10 500

1000

1500

2000

2500

Ne(2080)

F(1681)

N(1402)O(1324)

C(1086)

B(801)Be(899)

Li(520)

First ionization enthalpies (1H) of elements of the second period as a function of Z

Atomic number (Z)Variation of first ionization enthalpies (1H) with atomic number (for the elements with Z = 1 to 60)

Atomic number (Z)

H

(kJ

mol

1)

2500

2000

1500

1000

500

0 0 10 20 30 40 50 60

Li Na K Rb Cs

He Ne

Ar Kr Xe

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Q.55. Why is the first ionization potential of transition elements nearly same? Ans: i. In transition elements, the added electron successively enters the inner shell i.e., penultimate d-orbital. ii. The outermost orbital (ns) of each element remains intact. In the ionization process, electrons are

removed from the outermost orbital in every case. iii. Hence, the first ionization potential of all transition elements is nearly same due to screening effect. #Q.56. Why do noble gases have high values of ionization enthalpy? Ans: i. Noble gases have completely filled outermost shell. ii. They have completely filled orbitals with complete octet (ns2 np6) and thus they acquire extra stability

(except He with two electrons). iii. Hence, more energy is required to remove the outermost electron from stable electronic configuration. Therefore, noble gases have high values of ionization enthalpy. *Q.57. Define electron gain enthalpy. Ans: When an electron is added to a neutral gaseous atom (X) to convert it into negative ion, the enthalpy

change accompanying the process is defined as the electron gain enthalpy (egH). X(g) + e X

(g) ; egH. Q.58. Justify “Halogens have negative electron gain enthalpy”. Ans: i. Electron gain enthalpy gives a measure of ease with which an electron adds to an atom to form an anion. ii. Halogens have outer electronic configuration as ns2 np5. They require just one electron to complete

their octet and attain stable configuration. Hence, halogens (group 17 elements) have very high negative electron gain enthalpy. *Q.59. Explain the trends or variation in electron gain enthalpy. Ans: i. Trends or variation in electron gain enthalpy along a period: a. As we move across a period from left to right in periodic table, the electron gain enthalpies

become negative with increase in the atomic number across the period. b. This is because, as the atomic number increases, nuclear charge increases gradually but

addition of electrons takes place in the same shell. c. Hence, the attraction between nucleus and valence electrons increases and atomic size decreases and electron gain enthalpy will be more negative. ii. Trends or variation in electron gain enthalpy down a group: a. As we move down the group from top to bottom in the periodic table, the electron gain enthalpy

becomes less negative. b. This is because, as the atomic number increases, nuclear charge increases but simultaneously

the number of shells in the atoms also increases. c. However, the effective nuclear charge (attraction between nucleus and valence electrons)

decreases due to the increase in the size of the electron cloud. Hence, electron gain enthalpy will be less negative.

Note: Electron gain enthalpies (in kJ mol1) of some elements are given in the following table:

Gr.1 egH Gr. 16 egH Gr. 17 egH Gr. 18 egH H 73 He + 48

Li 60 O 141 F 328 Ne + 116

Na 53 S 200 Cl 349 Ar + 96

K 48 Se 195 Br 325 Kr + 96

Rb 47 Te 190 I 295 Xe + 77

Cs 46 Po 174 At 270 Rn + 68

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*Q.60. Give reason “ Fluorine has less negative electron gain enthalpy than chlorine”. Ans: i. Electron gain enthalpy becomes less negative down the group. ii. But there is an exception in case of fluorine and chlorine in group 17. iii. This is due to the smaller size of fluorine. Adding an electron to the 2p orbital in fluorine leads to

greater repulsion than adding an electron to the larger 3p orbital of chlorine. iv. Hence, in chlorine the negative electron gain enthalpy is greater than fluorine. v. The values of negative electron gain enthalpy for fluorine F = 328 while for chlorine Cl = 349. Note: Chlorine is the element with most negative electron gain enthalpy. Q.61. Would you expect the second electron gain enthalpy of O as positive, more negative or less negative

than the first? Justify your answer. (NCERT) Ans: i. The second electron gain enthalpy of O is expected to be positive. ii. An electron will be added to oxygen atom to form negatively charged ion. This corresponds to first

electron gain enthalpy.

iii. For second electron gain enthalpy, an electron should be added to negatively charged O anion. iv. Therefore, energy has to be supplied to force the second electron into the anion and hence second

electron gain enthalpy would be positive. *Q.62. Define electronegativity. Ans: The qualitative measure of the ability of an atom in a chemical compound to attract shared pair of electrons

towards itself is called electronegativity. Q.63. Write a note on importance of electronegativity. Ans: i. Concept of electronegativity is very important as it gives an idea about the metallic and non-metallic

properties of elements. ii. Many properties (like bond dissociation energy, bond polarity and ionic character of covalent bonds)

the energy and the distribution of charge in bonds can be predicted and explained by the help of concept of electronegativity.

Note: i. Different numerical scales of electronegativity of elements are Pauling scale, Mulliken-Jaffe scale, Allred-Rochow scale, etc.

ii. Pauling scale is most widely used. Linus Pauling, an American scientist, in 1922 assigned arbitrarily a value of 4.0 to fluorine. This element has the greatest ability to attract electrons.

i. Electron gain enthalpy may have negative or positive value. a. When addition of an electron to form an anion is exothermic, energy is released and the electron

gain enthalpy has negative value. eg. Halogens have very high negative electron gain enthalpy. b. When addition of an electron to form an anion is endothermic, energy is required and the electron

gain enthalpy has positive value. eg. Noble gases have large positive electron gain enthalpies. ii. The variation in electron gain enthalpies of elements is less systematic. iii. Elements with negative electron gain enthalpy have more tendency to gain an electron whereas elements

with positive electron gain enthalpy have less tendency to gain an electron.

Do You Know?

i. The negative of the electron gain enthalpy is defined as the electron affinity (EA). ii. If energy is released when an electron is added to an atom, then EA is positive. iii. If energy needs to be supplied to add an electron to an atom, then EA is negative.

NCERT Corner

Electron affinity:

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*Q.64. Explain the factors affecting electronegativity.

Ans: Electronegativity depends upon the following factors:

i. Atomic radius (size): As the atomic radius increases, the electronegativity decreases.

ii. Nuclear charge: Electronegativity is directly proportional to nuclear charge. As the nuclear charge increases the electronegativity increases.

iii. Screening effect : As the screening effect or shielding effect increases, the electronegativity decreases. *Q.65. Explain the trends or variation in electronegativity.

Ans: i. Trends or variation in electronegativity along a period:

a. As we move across a period from left to right in the periodic table, the electronegativity increases with decrease in atomic radius and increase in atomic number of an element.

b. This is because, as the atomic number increases, nuclear charge increases gradually but addition of electrons takes place in the same shell.

c. Hence, the attraction between nucleus and valence electrons increases and atomic size decreases.

d. Hence due to the increase in effective nuclear charge, the tendency to attract shared electron pair in a covalent bond i.e., electronegativity increases from left to right across a period.

eg. Li < Be < B < C < N < O < F.

ii. Trends or variation in electronegativity down a group:

a. As we move down the group from top to bottom in the periodic table, the electronegativity decreases with the increase in the atomic radius and increase in the atomic number of an element.

b. This is because, as the atomic number increases, nuclear charge increases but simultaneously the number of shells in the atoms also increases.

c. However, the effective nuclear charge (attraction between nucleus and valence electrons) decreases due to the increase in the size of the electron cloud and hence the atomic size increases in a group from top to bottom.

d. Thus the tendency to attract shared electron pair of a chemical bond decreases, decreasing the electronegativity down the group.

eg. F > Cl > Br > I > At. Note: Electronegativity across 2nd and 3rd period:

Atom (Period 2) Li

1.0

Be

1.5

B

2.0

C

2.5

N

3.0

O

3.5

F

4

Atom (Period 3) Na

0.9

Mg

1.2

Al

1.5

Si

1.8

P

2.1

S

2.5

Cl

3 Electronegativity in alkali metals (group 1) and halogen family (group 17):

Atom (Group 1) Electronegativity value Atom (Group 17) Electronegativity value

Li 1.0 F 4.0

Na 0.9 Cl 3.0

K 0.8 Br 2.8

Rb 0.8 I 2.5

Cs 0.7 At 2.2

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*Q.66. Answer the following : Which member of each of the following pairs of atoms/ions- i. has larger radius and why? a. Li : Na b. Br : Kr c. Na : Mg d. Be : B ii. has higher first ionization enthalpy and why? a. Na : K b. Cl : Ar c. Zr : Ti d. Be : B iii. has higher electronegativity? a. H : Na b. F : Cl c. C : N d. N : P

Ans: i. a. Li : Na Sodium (Na) has larger radius than Lithium (Li) as atomic radius increases down the group and lithium is placed above sodium in group 1.

b. Br : Kr Krypton (Kr) has larger atomic radius than Bromine (Br) as the inert gas in a particular period has the highest atomic radius due to screening effect and electronic repulsion of completely filled orbitals.

c. Na : Mg Sodium (Na) has larger atomic radius than Magnesium (Mg) as atomic radius decreases across the period from left to right till the halogens.

d. Be : B Beryllium (Be) has large atomic radius than Boron (B) as Beryllium is towards the left of boron in period II and atomic radius decreases across a period from left to right till halogens.

ii. a. Na : K Sodium (Na) has higher first I.E values than Potassium (K) as I.E. value decreases down the group.

b. Cl : Ar Argon (Ar) has higher I.E values than Chlorine (Cl) as removal of electron from completely filled shell in Ar will require more energy than in Cl.

c. Zr : Ti Titanium (Ti) has higher first I.E values than Zirconium (Zr) as I.E value decreases down the group.

d. Be : B Boron(B) has higher I.E. value than Beryllium (Be) as I.E value increases across the period from left to right.

iii. a. H : Na Electronegativity decreases down the group. Hence electronegativity of Hydrogen (H) is higher than that of Sodium (Na).

b. F : Cl Electronegativity decreases down the group. Hence electronegativity of Fluorine (F) is higher than that of Chlorine (Cl).

c. C : N Nitrogen (N) has higher electronegativity than Carbon (C) as electronegativity increases along a period.

d. N : P Nitrogen (N) has higher electronegativity than Phosphorus (P) as electronegativity decreases down the group.

*Q.67. What is the basic difference between the terms electron gain enthalpy and electronegativity? Ans: i. Electron gain enthalpy refers to the tendency of an atom in its gaseous isolated state to accept an

additional electron to form a negative ion (anion). ii. Electronegativity refers to the tendency of an atom in a compound to attract the shared pair of

electrons towards itself. Thus, electron gain enthalpy is the property of gaseous isolated atoms whereas electronegativity is the property of atoms in molecules.

*Q.68. Define valence of an element. Ans: Valence of an element is defined as, the number of hydrogen atoms or number of any other univalent atoms

which can combine with an atom of the given element. OR Valence of an element is defined as, the number of hydrogen atoms or chlorine atoms or the number of

oxygen atoms that combine with an atom of the given element.

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*Q.69. Explain the trends or variation in valence. Ans: i. Trends along a period: In a period from left to right, the valence with respect to hydrogen and

chlorine increases from 1 to 4 and then decreases from 4 to 0. ii. Trends down a group: a. All the elements present in a group possess the same number of valence electrons. b. Therefore, on moving down the group, there is no variation in the valence of elements. c. All the elements present in the group show same valence. eg. All the elements in group-1 have valence equal to one while those present in group-2 show

valence equal to two. Q.70. What do you mean by oxidation state of an element?

Ans: i. Oxidation state of an element in a particular compound is defined as the charge acquired by its atom on the basis of electronegative consideration from other atoms in the molecule.

ii. It is the apparent charge present on an atom of an element in a compound.

iii. It may have positive, negative or zero value.

iv. An element may have one or more oxidation states in various compounds.

eg. Transition elements, lanthanoids and actinoids.

Note: i. In compound OF2: Fluorine has higher electronegativity than oxygen. Fluorine has 1 oxidation state whereas oxygen has +2 oxidation state.

ii. In Na2O: Oxygen has higher electronegativity than sodium. Oxygen has 2 oxidation state and sodium has +1 oxidation state.

i. The electrons of the valence shell determine the chemical behaviour of an element. These electrons participate in bond formation and decide its combining capacity. This combining capacity is known as valence.

ii. The valence of an element is equal to the number of electrons in valence shell or equal to eight minus number of electrons in valence shell.

iii. The valence of elements of second period is given below:

Valence of representative elements of second period Element Li Be B C N O F Ne

Electronic configuration

1s2 2s1 1s2 2s2 1s2 2s2 2p1 1s2 2s2 2p2 1s2 2s2 2p3 1s2 2s2 2p4 1s2 2s2 2p5 1s2 2s2 2p6

Number of electrons in Valence shell

1 2 3 4 5 6 7 8

Valence 1 2 3 4 8 5 = 3 8 6 = 2 8 7 = 1 8 8 = 0 iv. Variation of valence of elements with respect to H and Cl in second and third period is shown below: II Period Element Li Be B C N O F Hydride LiH BeH2 BH3 CH4 NH3 H2O HF Valence with respect to H 1 2 3 4 3 2 1 Chloride LiCl BeCl2 BCl3 CCl4 NCl3 Cl2O CIF Valence with respect to Cl 1 2 3 4 3 2 1 III Period Element Na Mg Al Si P S Cl Hydride NaH MgH2 AlH3 SiH4 PH3 H2S HCl Valence with respect to H 1 2 3 4 3 2 1

Do You Know?

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Q.71. The first ionization enthalpies of 5 elements of second period are given below:

Element 1st IE values (kJ mol–1) I 520 II 1681 III 1086 IV 2080 V 899

Based on the above data, answer the following questions: i. Identify the element having highest atomic number. ii. If element I is lithium, how will you explain its low value of first ionization enthalpy? iii. Explain why ionization enthalpies are always positive. Ans: i. Element IV. The first ionization enthalpy increases with increase in atomic number along a period.

Hence, the element IV having highest IE will have highest atomic number among the given elements. ii. Refer.Q.52.i. iii. Refer.Q.48.iii.

Apply Your Knowledge

A diagonal relationship is said to exist between certain pairs of diagonally adjacent elements in thesecond and third period of the periodic table. These pairs such as Lithium (Li) and Magnesium (Mg),Beryllium (Be) and Aluminium (Al), etc. exhibit similar properties. Such relationship occursbecause moving across the period and down the group has opposite effects.

NCERT Corner

The first member of each groups 1, 2 and 13-17 exhibits different chemical behaviour as comparedto that of the subsequent members in the group. This is due to their small size, large charge toradius ratio and high electronegativity. Other factors include the following:

i. The maximum covalency of the first member of each group is 4, while the other members of thegroup can expand their valence shell to accommodate more than four electron pairs.

ii. The first member of p-block elements displays greater ability to form pπ–pπ multiple bondscompared to subsequent members of the same group.

i. The chemical reactivity is highest at the two extremes of a period and is lowest in the centre. Themaximum chemical reactivity at the extreme left is exhibited by the loss of an electron leading to theformation of a cation while the maximum chemical reactivity at the extreme right is exhibited by the gainof an electron forming an anion. This can be related to the metallic and non-metallic character of elements.

ii. Metallic character decreases from left to right across the period, while it increases down the group. iii. Non-metallic character increases from left to right across the period while it decreases down the group.

i. Elements on the left side of periodic table form basic oxides. ii. Elements on the right side of periodic table form acidic oxides. iii. Elements in the centre of the periodic table form amphoteric or neutral oxides.

i. Reducing property of the elements decreases while oxidizing property increases across the periodfrom left to right.

ii. Reducing property increases while oxidizing property decreases down the group.

Diagonal relationship:

Anomalous properties of first member of a group:

Periodic trends and chemical reactivity:

Nature of oxides:

Reducing and oxidizing property:

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Q.72. From the elements K, Ar, Cl, Sr, P and S, choose one that fits each of the below given descriptions: i. An element having one valence electron. ii. An element having properties similar to that of O. iii. A noble gas. iv. An alkaline earth metal. v. An element having electronic configuration 1s2 2s2 2p6 3s2 3p3. Ans: i. Potassium (K)

ii. Sulphur (S) iii. Argon (Ar) iv. Strontium (Sr) v. Phosphorus (P) Classification of modern periodic table:

Block Last electron enters Contains elements of

‘s’ s-orbital

(Maximum electrons = 2)

Group 1 (alkali metals)

Group 2 (alkaline earth metals)

[Normal or representative elements]

‘p’ p-orbital


Group 13 to group 17 elements [Normal or representative elements] and

Group 18 elements [Noble gases or inert gas elements]

‘d’ d-orbital


Group 3 to group 12 elements

[Transition elements]

‘f’ f-orbital


Lanthanoids and actinoids series

[Inner transition elements] Factors affecting atomic radius:

Quick Review

Shielding effect or

Screening effect The inner shell electrons prevent(shield) the outermost electronsfrom the attractive influence ofnucleus. This is called shieldingeffect. Atomic radius increaseswith increase in shielding effect. Atomic radius Shielding effect

Factors affecting atomic radius

Nuclear charge

With increase in effective nuclear

charge, Nucleus attracts electrons

strongly and so atomic radius

decreases.

Atomic radius 1

nuclear charge

Number of orbits or shells

Atomic size increases with increase

in number of orbits and hence

atomic radius also increases. Atomic radius Number of shells

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Factors affecting ionization enthalpy (IE):

Factors affecting electronegativity: Periodic trends:

Electron Gain Enthalpy

Atomic Radius

Electronegativity

Ionization Enthalpy

Ioni

zati

on E

ntha

lpy

Ele

ctro

n G

ain

Ent

halp

y

Ele

ctro

nega

tivi

ty

Ato

mic

Rad

ius

The periodic trends of elements in the periodic table

Shielding effect or

Screening effect

IE decreases with increase in shielding effect or screening effect.

IE 1

Shielding effect

Size (radius) of atom IE increases with decrease in size of atom.

IE 1

Sizeof atom

Nuclear charge IE increases with increase in effective nuclear charge. IE Nuclear charge

Factors affecting ionization enthalpy (IE)

Factors affecting electronegativity

Atomic size (radius) Electronegativity decreases with

increase in atomic radius.

Electronegativity 1

Atomicradius

Shielding effect or

Screening effect

Electronegativity decreases with increase in shielding or screening effect.

Electronegativity 1

Shielding effect

Nuclear charge Electronegativity increases with

increase in effective nuclear charge.

Electronegativity Nuclear charge

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1. State Newland’s law of octaves. Ans: Refer Q.6.ii. 2. Define oxidation state. Ans: Refer Q.70.i. 3. Which groups of the modern periodic table

contains transition elements?

Ans: Groups 3 to 12. 4. Explain why van der Waals radius is usually

larger than covalent radius. Ans: Refer Q.36.iii. 5. Give the drawbacks of Newland’s law of

octaves. Ans: Refer Q.6. (Limitations).

1. Explain how screening effect affects the

atomic radius.

Ans: Refer Q.38.iii. 2. What are ‘groups’ and ‘periods’ in the modern

periodic table?

Ans: Refer Q.16.i. and Q.15.i. 3. Describe any two factors that affect the atomic

radius. Ans: Refer Q.38. 1. Explain the terms: i. First ionization enthalpy ii. Second ionization enthalpy How do their magnitude vary? Ans: Refer Q.48. 2. i. Give reason: Krypton has larger atomic

radius than bromine. ii. Explain how nuclear charge affects the

ionization enthalpy of an atom. Ans: i. Refer Q.66.i.b. ii. Refer Q.49.ii.

3. i. Name the element that Mendeleev called ‘eka-silicon’.

ii. The size of an anion is larger than that of the corresponding parent atom. Justify the statement.

Ans: i. Germanium

ii. Refer Q.42.iii. and iv. 1. i. Arrange the following species in

decreasing order of their radii.

Mg, Mg2+, Al, Al3+

ii. Explain the effect of following factors on the ionization enthalpy.

a. Size of atom

b. Nuclear charge

c. Nature of electronic configuration

Ans: i. Refer Q.45.i.to iii.

ii. a. Refer Q.49.i.

b. Refer Q.49.ii.

c. Refer Q.49.iv. 2. i. ‘Electronegativity decreases down the

group from top to bottom in the periodic table’. Justify the statement.

ii. Determine the block to which an element with Z = 12 belongs.

iii. Mention any four advantages of modern periodic table.

Ans: i. Refer Q.65.ii.b. to d.

ii. Refer Q.30.i.

iii. Refer Q.14. 3. i. Define isoelectronic species.

ii. Explain why cations are smaller than their parent atoms.

iii. State Dobereiner’s law of triads. Give an example.

Ans: i. Refer Q.43.i.

ii. Refer Q.42.i. and ii.

iii. Refer Q.4.ii. and any one example.

Exercise

One Mark Questions

Two Marks Questions

Three Marks Questions

Five Marks Questions

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1. At present, how many elements are known?

(A) 118 (B) 110

(C) 114 (D) 120 2. In Unitary theory, the values of the atomic

weights of all the elements were _______.

(A) prime numbers

(B) even numbers

(C) odd numbers

(D) whole numbers 3. Which of the following represents

Dobereiner’s triad?

(A) Li, Ca, I (B) Li, Na, K

(C) Cl, I, Ba (D) Na, Ca, Ba 4. As per the Newland’s law of octaves, the

properties of every eighth element were similar to those of _______.

(A) first (B) second

(C) third (D) seven 5. Lothar Meyer plotted physical properties like

atomic volume, density etc against _______.

(A) atomic number

(B) atomic weights

(C) molecular weight

(D) molecular number 6. Mendeleev predicted the existence of _______.

(A) aluminium

(B) silicon

(C) tellurium

(D) gallium and germanium 7. According to Mendeleev’s periodic law, the

physical and chemical properties of elements are the periodic function of their _______.

(A) atomic weights

(B) atomic numbers

(C) molecular formulas

(D) molecular weights

8. Eka-aluminium and Eka-silicon are known as _______.

(A) gallium and germanium

(B) aluminium and silicon

(C) iron and sulphur

(D) proton and silicon 9. Which Scientist proved that atomic number is

more fundamental property of an element than its atomic mass?

(A) Mendeleev (B) Moseley

(C) Newland (D) Cooke 10. According to periodic law of elements, the

variation in properties of elements is related to their _______.

(A) densities

(B) atomic masses

(C) atomic sizes

(D) atomic numbers 11. The long form of the periodic table consists of

how many periods?

(A) 5 (B) 8

(C) 10 (D) 7 12. The fourth, fifth and sixth periods are long

periods and contain _______.

(A) 18, 18 and 36

(B) 18, 28 and 32

(C) 18, 15 and 31

(D) 18, 18 and 32 13. Elements from group 3 to 12 are called

_______.

(A) transition elements

(B) inert gas elements

(C) normal elements

(D) inner transition elements 14. The name ‘rare earths’ is used for_______.

(A) lanthanides only

(B) actinides only

(C) both lanthanides and actinides

(D) alkaline earth metals

Multiple Choice Questions

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15. Atomic number of V is 23 and its electronic configuration is _______.

(A) 1s2 2s2 2p6 3p6 3d3 4s2

(B) 1s2 2s2 2d3 3p6 2p6 4s2

(C) 2s2 1s2 2p6 3s2 3d3 4s2

(D) 1s2 2s2 2p6 3s2 3p6 3d3 4s2 16. Which pair of atomic numbers represents

s-block elements?

(A) 7, 15 (B) 6, 12

(C) 9, 17 (D) 3, 12 17. The elements with atomic numbers 9, 17, 35,

53, 85 are all _______.

(A) noble gases (B) halogens

(C) lanthanoids (D) actinoids 18. Aluminium belongs to _______ elements.

(A) s-block (B) p-block

(C) d-block (D) f-block 19. In P3, S2 and Cl ions, the increasing order of

size is_______.

(A) Cl, S2, P3

(B) P3, S2, Cl

(C) S2, Cl, P3

(D) S2, P3, Cl

20. The metallic and non-metallic properties of

elements can be judged by their _______.

(A) electron gain enthalpy

(B) ionization enthalpy

(C) electronegativity

(D) valence 21. Which of the following species will have the

largest size Mg, Mg+2, Fe, Fe3+ ?

(A) Mg

(B) Mg+2

(C) Fe

(D) Fe3+

22. The CORRECT order of increasing radii of

the elements Na, Si, Al and P is _______.

(A) Si, Al, P, Na

(B) Al, Si, P, Na

(C) P, Si, Al, Na

(D) Al, P, Si, Na

23. Which element has the most negative electron gain enthalpy?

(A) Sulphur (B) Fluorine (C) Chlorine (D) Hydrogen 24. Which of the properties remain unchanged on

descending a group in the periodic table? (A) Atomic size (B) Density (C) Valence electrons (D) Metallic character 25. Which one of the following is CORRECT

order of the size?

(A) I > I >I+

(B) I > I+ > I

(C) I+ > I > I

(D) I > I > I+ 26. The maximum valence of an element with

atomic number 7 is _______. (A) 2 (B) 5 (C) 4 (D) 3 27. The CORRECT order of radii is _______. (A) N < Be < B

(B) F < O2 < N3 (C) Na < Li < K (D) Fe3+ < Fe2+ < Fe4+

1. (A) 2. (D) 3. (B) 4. (A) 5. (B) 6. (D) 7. (A) 8. (A) 9. (B) 10. (D) 11. (D) 12. (D) 13. (A) 14. (C) 15. (D) 16. (D) 17. (B) 18. (B) 19. (A) 20. (C) 21. (C) 22. (C) 23. (C) 24. (C) 25. (D) 26. (B) 27. (B)

Answers to Multiple Choice Questions

SAMPLE C

ONTENT

165


Choose the correct alternative: 1. Identify the element with most negative electron gain enthalpy. (A) Li (B) Na (C) K (D) Rb 2. In the modern periodic table, the period number indicates the value of _______. (A) atomic number (B) atomic mass number (C) principal quantum number (D) oxidation number 3. Which of the following is isoelectronic with Mg2+? (A) O– (B) Na (C) F– (D) Al+ Answer the following: 4. Give the symbol of the element with Z = 119. 5. State modern periodic law. 6. Explain the trend in electron gain enthalpy along a period. 7. Why is it necessary to arrange elements in a systematic way? 8. Describe how ionic radius vary along a period. OR Give a short account of d-block elements. 9. Write a note on factors affecting electronegativity. 10. i. With the help of an example, explain how can the period of an element be determined from its

electronic configuration. ii. Define van der Waals radii. 11. Define valence of an element. Explain its trends along a period and down a group. OR State Newland’s law of octaves. What are lanthanoid and actinoid series? 12. i. An element ‘X’ has atomic number 16. Predict the block, period and group to which it belongs in the

modern periodic table. ii. Give reason: First ionization enthalpy of B is less than that of Be. OR i. Explain why the second period contains only eight elements. ii. Define periodicity. iii. Why do noble gases have high values of ionization enthalpy?

TOPIC TEST Total : 25 Marks

Section A (1 × 5 = 5 Marks)

Section B (2 × 3 = 6 Marks)

Section C (3 × 3 = 9 Marks)

Section D (5 × 1= 5 Marks)

SAMPLE C

ONTENT

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