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

Transcription

1

2 Written as per the revised syllabus prescribed by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune. STD. XI Sci. Precise Chemistry Salient Features Concise coverage of syllabus in Question Answer Format. Covers answers to all Textual Questions and Intext 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 chapter for effective preparation. Printed at: India Printing Works, Mumbai 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. P.O. No TEID: 12593_JUP

3 Preface 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. : PRECISE CHEMISTRY is a compact yet complete guide designed to boost students confidence and prepare them to face the conspicuous Std. XI final exam. This book is specifically aimed at Maharashtra Board students. The content of the book is framed in accordance with Maharashtra State board syllabus and collates each and every important concept in question and answer format. 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 few additional questions to ensure complete exam preparation. Solved Examples provide step-wise solution to various numerical problems. This helps students to understand the application of different concepts and formulae. Notes cover additional bits of relevant information on each topic. Quick Review and Formulae sections facilitate instant revision. Exercise helps the students to gain insight on the various levels of theory and numercial-based questions. Multiple Choice Questions and Topic Test assess the students on their range of preparation and the amount of knowledge of each topic. The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we ve nearly missed something or want to applaud us for our triumphs, we d love to hear from you. Please write to us on : mail@targetpublications.org A book affects eternity; one can never tell where its influence stops. Yours faithfully, Publisher Edition: First Best of luck to all the aspirants! 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. No. Topic Name Page No. 1 Some Basic Concepts of Chemistry 1 2 States of Matter (Gases and 23 Liquids) 3 Structure of Atom 61 4 Periodic Table 96 5 Redox Reactions Chemical Equilibrium Surface Chemistry Nature of Chemical Bond 229 Contents No. Topic Name Page No. 9 Hydrogen s - Block Elements p - Block Elements (Group 13 and 14) Basic Principles and Techniques in 371 Organic Chemistry 13 Alkanes Alkenes Alkynes Aromatic Compounds Environmental Chemistry 527 Note: All the Textual questions are represented by * mark All the Intext questions are represented by # mark

4 Std. XI Sci.: Precise Chemistry Brief history of the development of periodic table Periodic Table Syllabus 4.1 Introduction 4.2 Brief history of the development of periodic table 4.1 Introduction *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. 4.2 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 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. 12 C contain 12 units of hydrogen. b. 40 Ca 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 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. 96

5 Chapter 04: Periodic Table eg. a. The average of atomic masses of lithium and potassium is equal to the atomic mass of sodium. Element Atomic weight Li 7 Na 23 K 39 b. Atomic mass of strontium is close to the average atomic masses of calcium and barium. Element Ca Atomic weight 40 Sr 88 Ba 137 c. The average of atomic masses of chlorine and iodine is close to the atomic mass of bromine. Element Cl Atomic weight 35.5 Br 80 I 127 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 ahomologous series increase in a regular fashion. iii. This is shown in the table given below: Element Atomic weight Type of atomic weight Nitrogen ( 14 N) 14 (14) a Phosphorus ( 31 P) 31 (14+17) a+b Arsenic ( 75 As) 75 ( ) a+b+c Antimony ( 119 Sb) 119 ( ) a+b+2c Bismuth ( 207 Bi) 207 ( ) a+b+4c 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 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 Element Na Mg Al Si P S Cl Atomic weight Element K Ca Atomic weight 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). 97

6 Std. XI Sci.: Precise Chemistry 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. 4.3 Modern periodic law and present form of periodic table Q.12. 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. Q.13. 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. 98

7 Chapter 04: Periodic Table s-block Li Be Na Mg K Ca Rb Sr Cs Ba Fr 88 Ra (223) (226) * Lanthanide Series Actinide * Series 1 H 1.1 d-block Sc Ti V Cr Mn Fe Y Zr Nb Mo Tc Ru La * Hf Ta W Re Os d block Ac * Rf Db Sg Bh Hs (227) * (261) (262) (263) (264) (265) 58 Ce Th Pr Pa Nd U Pm Np Modern Periodic Table Atomic number (Z) Symbol Atomic mass (A) 62 Sm Pu Co Rh Ir Mt (268) 63 Eu Am Pt Ds (271) N O P S As Se p block Sb Te Bi Po Uup Uuh (288) (292) Q.14. 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. Q.15. 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 Ni Pd 106 f-block 29 Cu Ag Au Rg (272) 64 Gd 65 Tb Cm Bk Zn Cd Hg Cn (285) 66 Dy Cf B Al Ga In Tl Uut (284) 67 Ho Es C Si Ge Sn Pb Uuq (289) 68 Er Fm 257 p-block 69 Tm Md F Cl Br I At (210) 117 Uus - 70 Yb No He Ne Ar Kr Xe Rn (222) 118 Uuo - 71 Lu Lr

8 Std. XI Sci.: Precise Chemistry 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. Q.16. 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.17. 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 (1s 1 ) and helium (1s 2 ). 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,2p x,2p y,2p z ). 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, 3p x, 3p y and 3p z orbitals, the third period contains a maximum of eight elements. d. InNa 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] 3s 1 and the last element Ar has the configuration [Ne] 3s 2 3p 6. Q.18. 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. 100

9 Chapter 04: Periodic Table Q.19. 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 (5s 1 ) and Sr (5s 2 ) and then 4d orbitals are filled Y(4d 1,5s 2 ) Cd(4d 10,5s 2 ). ii. When the 4d orbitals are completely filled, the electrons enter into 5p orbitals from In(4d 10 5s 2 5p 1 ) to Xe(4d 10 5s 2 5p 6 ). 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. Q.20. 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 (6s 1 ) and Ba (6s 2 ). Then 5d orbital is filled in La with configuration [Xe] 5d 1 6s 2. b. The energy of 4f orbital is lowered when the electron enters into 5d orbitals. Hence, 4f orbitals are filled in elements Ce (4f 1 5d 1 6s 2 ). Lu (4f 14 5d 1 6s 2 ). 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 (4f 14 5d 10 6s 2 ). e. Then 6p subshell is filled in elements Tl(4f 14 5d 10 6s 2 6p 1 ) Rn (4f 14 5d 10 6s 2 6p 6 ). 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 (7s 1 ) and Ra (7s 2 ). Then 6d subshell is filled in Ac (6d 1 7s 2 ). c. Then 5f orbitals are filled from Th (5f 1 6d 1 7s 2 ) to Lr (5f 14 6d 1 7s 2 ). 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.21. 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.22. What are lanthanoid and actinoid series? Ans: i. Lanthanoid series: The elements after Lanthanum i.e., from Cerium (4f 1 5d 1 6s 2 ) to Lutetium (4f 14 5d 1 6s 2 ) 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 (5f 1 6d 1 7s 2 ) to Lawrencium (5f 14 6d 1 7s 2 ) are kept in separate series called actinoid series at the bottom of the periodic table. Q.23. 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. *Q.24. 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 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. 101

10 Std. XI Sci.: Precise Chemistry 102 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.25. 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. 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. #Q.26. 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.27. 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 ns 1 and ns 2. 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 ns 2 np 1 to ns 2 np 6. 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) d 1 ns 1 2 to (n 1) d 10 ns 1 2. 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). 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) f 1 (n 1) d 0 1 ns 2 to (n 2) f 14 (n 1) d 0 1 ns 2. c. They are placed at the bottom of the periodic table. d. This block consists of series of Lanthanides and Actinides. Q.28. 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 (3s 1 ) of Na (1s 2 2s 2 2p 6 3s 1 ) is 3. This corresponds to third period.

11 Chapter 04: Periodic Table ii. iii. 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(1s 2 2s 2 2p 6 3s 2 3p 6 3d 1 4s 2 ) corresponds to 3d block. 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. (n 1)) shell: a. For s-block elements, number of the group = number of valence electrons. b. For p-block elements, number of the group = number of electrons in p subshell. c. For d-block elements, number of the group = 2 + number of (n 1) d electrons. Q.29. 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) Ans: i. Mg (Z = 12): Atomic number of Mg is 12. Electronic configuration is 1s 2 2s 2 2p 6 3s 2. 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 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2. Block: The subshell 3d (in which the differentiating electrons enters) corresponds to d block. Period: The principal quantum number of the valence shell (4s 2 ) is 4. This corresponds to fourth period. Group : For d block elements, group = 2 + number of (n 1) d electrons = = 5. Hence it belongs to group 5. iii. Sb (Z = 51): The atomic number of Sb is 51. Electronic configuration is 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 3. Block: The subshell 5p (in which the differentiating electron enters) corresponds to p block. Period: The principal quantum number of the valence shell (5s 2 5p 3 ) is 5. This corresponds to fifth period. Group: For p block elements, group = number of electrons in p subshell = = 15. Hence it belongs to group 15. iv. Rn (Z = 86): The atomic number of Rn is 86. Electronic configuration is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 4d 10 4f 14 5s 2 5p 6 5d 10 6s 2 6p 6. Block: The subshell 6p (in which the differentiating electrons enters) corresponds to p block. Period: The principal quantum number of the valence shell (6s 2 6p 6 ) is 6. This corresponds to sixth period. Group: For p block elements, group = number of electrons in p subshell = = 18. Hence it belongs to group Periodic trends in properties of elements Q.30. 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.31. Define the 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 = m and 1 pm = m Q.32. Explain the term covalent bond length. Ans: i. Internuclear distance in a diatomic molecule of an element is known as its covalent bond length. 103

12 Std. XI Sci.: Precise Chemistry 104 ii. Half the covalent bond length gives the covalent radius. Internuclear distancebetween two bondedatoms r covalent = 2 eg. Bond distance in the chlorine molecule (Cl 2 ) is 198 pm. The half of this distance (99 pm) is the atomic radius of chlorine. Q.33. 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.34. 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.35. 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. 1 Atomic size (or radius) 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.36. 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. 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.

13 Chapter 04: Periodic Table 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. Q.37. Explain briefly the trends in atomic radius along a period in d-block and f-block. Ans: i. In a period from left to right in the periodic table, the atomic radius of an element decreases with the increase in atomic number. ii. In the case of d-block elements, though the nuclear charge progressively increases with increase in atomic number, the effective nuclear charge on the outer electron decreases due to screening effect. As a result, atomic radius decreases to a lesser extent than that among the s and p block elements. iii. In the case of f-block elements, the effective nuclear charge on the outer electrons decreases more than that in d-block elements. As aresult, the decrease in the atomic radius amongst f-block elements is extremely less. *Q.38. 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 = m and 1 pm = m respectively. Q.39. 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. *Q.40. 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 M n+ + 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 A n ; 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. 105

14 Std. XI Sci.: Precise Chemistry *Q.41. 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. Consider the isoelectronic species, Na +, Mg 2+, Al 3+ and Si 4+. Ions Na + Mg 2+ Al 3+ Si 4+ Number of electrons Nuclear charge (At.Number) Ionic radii in pm The nuclear charge increases from Na + to Si 4+ but the number of electrons remains the same in each ion. Hence, the size decreases from Na + to Si 4+. Q.42. Name a species that will be isoelectronic with each of the following atoms or ions. i. F ii. Ar iii. Mg 2+ iv. Rb + (NCERT) Ans: i. F possesses 10 electrons, O 2 is its isoelectronic species. ii. Ar possesses 18 electrons, Cl is its isoelectronic species. iii. Mg 2+ possesses 10 electrons, Na + is its isoelectronic species. iv. Rb + possesses 36 electrons, Kr is its isoelectronic species. Q.43. Which of the following species will have largest and the smallest size? Mg, Mg 2+, Al, Al 3+ 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 Mg 2+ and the radius of Al is larger than that of Al 3+. iii. Among iso-electronic species, the one with larger positive nuclear charge will have smaller radius. Mg 2+ and Al 3+ are isoelectronic with Al 3+ having larger positive nuclear charge. Hence, the ionic radius of Al 3+ is smaller than Mg 2+. Hence, the decreasing order of radii is Mg > Al > Mg 2+ > Al 3+. iv. Therefore, species with the largest size is Mg and with smallest size is Al 3+. Q.44. Consider the species: N 3, O 2, F, Na +, Mg 2+ and Al 3+ 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, Al 3+ < Mg 2+ < Na + < F < O 2 < N 3 *Q.45. 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.46. 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 ; 1 H 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) X 2+ (g) + e ; 2 H iii. Energy is always required to remove electrons froman atom. Hence, ionization enthalpies havepositive 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 (IE 2 ) is higher than the first ionization enthalpy (IE 1 ). v. In general, if IE 1, IE 2 and IE 3 are first, second and third ionization enthalpies respectively then, IE 3 > IE 2 > IE 1.

15 Chapter 04: Periodic Table *Q.47. 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. 1 Ionization enthalpy 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. 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.48. 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.47.iii. *Q.49. 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. 107

16 Std. XI Sci.: Precise Chemistry 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. Q.50. Give reason 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 (ns 2 np 6 ) 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. Q.51. Why is the first ionization enthalpy of boron less than that of beryllium? Ans: i. Boron has electronic configuration 1s 2 2s 2 2p 1. Beryllium has electronic configuration 1s 2 2s 2. 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.52. Why is the first ionization enthalpy of oxygen less than that of nitrogen? Ans: i. Nitrogen has electronic configuration 1s 2 2s 2 2p 3. Oxygen has electronic configuration 1s 2 2s 2 2p 4. ii. Ionization of nitrogen requires removal of an electron from 2p 3 subshell which has extra stability as it is half filled. iii. Ionization of oxygen requires removal of an electron from 2p 4 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. Q.53. 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.54. 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 (ns 2 np 6 ) 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.55. 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 ( eg H). X (g) + e X (g) ; eg H. Q.56. Justify Halogens have negative electron gain enthalpy. 108 Ans: i. ii. Electron gain enthalpy gives a measure of ease with which an electron adds to an atom to form an anion. Halogens have outer electronic configuration as ns 2 np 5. 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.

17 Chapter 04: Periodic Table *Q.57. 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 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. *Q.58. 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.59. 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.60. 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.61. 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. 109

18 Std. XI Sci.: Precise Chemistry *Q.62. Explain the factors affecting electronegativity. Ans: Electronegativity depends upon the following factors: i. Atomic radius (size): As the atomic radius increases, the electronegativity decreases. 110 ii. iii. Nuclear charge: Electronegativity is directly proportional to nuclear charge. As the nuclear charge increases the electronegativity increases. Screening effect : As the screening effect or shielding effect increases, the electronegativity decreases. *Q.63. 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. *Q.64. Answer the following : Which member of each of the following pairs of atoms/ionsi. 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.

19 Chapter 04: Periodic Table 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.65. 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.66. 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. *Q.67. 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. Note: 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. Q.68. 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 OF 2 : Fluorine has higher electronegativity than oxygen. Fluorine has 1 oxidation state whereas oxygen has +2 oxidation state. ii. In Na 2 O: Oxygen has higher electronegativity than sodium. Oxygen has 2 oxidation state and sodium has +1 oxidation state. 111

20 Std. XI Sci.: Precise Chemistry Quick Review Classification of modern periodic table: Block Last electron enters Contains elements of s p d f s-orbital (Maximum electrons = 2) p-orbital (Maximum electrons = 6) d-orbital (Maximum electrons = 10) f-orbital (Maximum electrons = 14) Factors affecting atomic radius: 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 Factors affecting ionization enthalpy (IE): Size (radius) of atom Group 1 (alkali metals) Group 2 (alkaline earth metals) [Normal or representative elements] Group 13 to group 17 elements [Normal or representative elements] and Group 18 elements [Noble gases or inert gas elements] Group 3 to group 12 elements [Transition elements] Lanthanoids and actinoids series [Inner transition elements] Shielding effect or Screening effect IE decreases with increase in shielding effect or screening effect. 1 IE Shielding effect IE increases with decrease in size of atom. 1 IE Sizeof atom Factors affecting atomic radius Nuclear charge With increase in effective nuclear charge, Nucleus attracts electrons strongly and so atomic radius decreases. 1 Atomic radius nuclear charge Factors affecting ionization enthalpy (IE) Nuclear charge IE increases with increase in effective nuclear charge. IE Nuclear charge Shielding effect or Screening effect The inner shell electrons prevent (shield) the outermost electrons from the attractive influence of nucleus. This is called shielding effect. Atomic radius increases with increase in shielding effect. Atomic radius Shielding effect 112

21 Chapter 04: Periodic Table Factors affecting electronegativity: Periodic trends: Exercise Atomic size (radius) Electronegativity decreases with increase in atomic radius. 1 Electronegativity Atomicradius One Mark Questions Electronegativity 1. State Newland s law of octaves. Ans: Refer Q.6.ii. 2. Define oxidation state. Ans: Refer Q.68.i. Electron Gain Enthalpy Ionization Enthalpy Atomic Radius Electronegativity 3. Which groups of the modern periodic table contains transition elements? Ans: Groups 3 to Explain why van der Waals radius is usually larger than covalent radius. Ans: Refer Q.34.iii. 5. Give the drawbacks of Newland s law of octaves. Ans: Refer Q.6. (Limitations). Atomic Radius Factors affecting electronegativity Nuclear charge Electronegativity increases with increase in effective nuclear charge. Electronegativity Nuclear charge The periodic trends of elements in the periodic table Ionization Enthalpy Electron Gain Enthalpy Two Marks Questions 1. Explain how screening effect affects the atomic radius. Ans: Refer Q.35.iii. 2. What are groups and periods in the modern periodic table? Ans: Refer Q.15.i. and Q.14.i. 3. Describe any two factors that affect the atomic radius. Ans: Refer Q.35. Three Marks Questions Shielding effect or Screening effect Electronegativity decreases with increase in shielding or screening effect. 1 Electronegativity Shielding effect 1. Explain the terms: i. First ionization enthalpy ii. Second ionization enthalpy How do their magnitude vary? Ans: Refer Q

22 Std. XI Sci.: Precise Chemistry 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.64.i.b. ii. Refer Q.47.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.40.iii. and iv. Five Marks Questions 1. i. Arrange the following species in decreasing order of their radii. Explain. Mg, Mg 2+, Al, Al 3+ 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.43.i.to iii. ii. a. Refer Q.47.i. b. Refer Q.47.ii. c. Refer Q.47.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.63.ii.b. to d. ii. Refer Q.29.i. iii. Refer Q 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.41.i. ii. Refer Q.40.i. and ii. iii. Refer Q.4.ii. and any one example. 114 Multiple Choice Questions 1. Which of the following represents Dobereiner s triad? (A) Li, Ca, I (B) Li, Na, K (C) Cl, I, Ba (D) Na, Ca, Ba 2. 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 3. 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 4. Eka-aluminium and Eka-silicon are known as respectively. (A) gallium and germanium (B) aluminium and silicon (C) iron and sulphur (D) proton and silicon 5. 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 6. The long form of the periodic table consists of how many periods? (A) 5 (B) 8 (C) 10 (D) 7

23 Chapter 04: Periodic Table 7. 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 Elements from group 3 to 12 are called. (A) transition elements (B) inert gas elements (C) normal elements (D) inner transition elements 9. The name rare earths is used for. (A) lanthanides only (B) actinides only (C) both lanthanides and actinides (D) alkaline earth metals 10. Atomic number of V is 23 and its electronic configuration is. (A) 1s 2 2s 2 2p 6 3p 6 3d 3 4s 2 (B) 1s 2 2s 2 2d 3 3p 6 2p 6 4s 2 (C) 2s 2 1s 2 2p 6 3s 2 3d 3 4s 2 (D) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s Which pair of atomic numbers represents s-block elements? (A) 7, 15 (B) 6, 12 (C) 9, 17 (D) 3, In P 3, S 2 and Cl ions, the increasing order of size is. (A) Cl, S 2, P 3 (B) P 3, S 2, Cl (C) S 2, Cl, P 3 (D) S 2, P 3, Cl 13. Which of the following species will have the largest size Mg, Mg +2, Fe, Fe 3+? (A) Mg (B) Mg +2 (C) Fe (D) Fe 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 15. Which element has the most negative electron gain enthalpy? (A) Sulphur (B) Fluorine (C) Chlorine (D) Hydrogen 16. 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 17. 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 The maximum valence of an element with atomic number 7 is. (A) 2 (B) 5 (C) 4 (D) The CORRECT order of radii is. (A) N < Be < B (B) F < O 2 < N 3 (C) Na < Li < K (D) Fe 3+ < Fe 2+ < Fe 4+ Answers to Multiple Choice Questions 1. (B) 2. (A) 3. (A) 4. (A) 5. (D) 6. (D) 7. (D) 8. (A) 9. (C) 10. (D) 11. (D) 12. (A) 13. (C) 14. (C) 15. (C) 16. (C) 17. (D) 18. (B) 19. (B) 115

24 Std. XI Sci.: Precise Chemistry TOPIC TEST Total : 25 Marks Section A (1 5 = 5 Marks) 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 Mg 2+? (A) O (B) Na (C) F (D) Al + Answer the following: 4. Give the symbol of the element with Z = State modern periodic law. Section B (2 3 = 6 Marks) 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. Section C (3 3 = 9 Marks) 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? Section D (5 1= 5 Marks) 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? 116

25