Maharashtra-HSC-d & f Paper-2 Target

Written according to the New Text book (2012-2013) published by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune.

Edition: August 2013

STD. XII Sci.

Perfect Chemistry - II Prof. Santosh B. Yadav

Prof. Anil Thomas

(M. Sc., SET, NET) Department of Chemistry R. Jhunjunwala College, Ghatkopar

(M.Sc., Chemistry)

Salient Features: ¾ Exhaustive coverage of syllabus in Question Answer Format. ¾ Covers answers to all Textual Questions and numericals. ¾ Textual Questions are represented by * mark. ¾ Intext Questions are represented by # mark. ¾ Charts for quick reference. ¾ Simple and Lucid language ¾ Includes Board Question Paper of March 2013.

TargetPublications PVT. LTD. Mumbai, Maharashtra Tel: 022 – 6551 6551 Website : www.targetpublications.org | email : [email protected]

Std. XII Sci. Perfect Chemistry - II

©

Target Publications PVT. LTD.

Sixth Edition : August 2013

Price : ` 240/-

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. Organic chemistry as a branch of science represents one of many pillars that give Chemistry its stand in the world. Without organic chemistry, learning such a subject is like supporting a dome of steel with pillars of straw. Organic chemistry deals with the scientific study of structure, properties, composition, reactions and preparation of chemical compounds. They form the basis of all earthly life processes. A science that calls for such deep knowledge of concepts and an intrinsic study of all aspects must obviously, not be easy to conquer. Hence to ease this task we bring to you “Std. XII Sci. : PERFECT CHEMISTRY-II” a complete and thorough guide critically analysed and extensively drafted to boost the students confidence. Topic wise classified ‘question and answer format’ of this book helps the student to understand each and every concept thoroughly. The book provides answers to all textual questions marked with * and intext questions marked with #. Important definitions, statements and laws are specified with Italic representation. Charts are given for quick reference.

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And lastly, I would like to thank all those people who have helped me in preparing exclusive guide to all students. There is always room for improvement and hence we welcome all suggestions and regret any errors that may have occurred in the making of this book. A book affects eternity; one can never tell where its influence stops.

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PAPER PATTERN •

There will be one written paper of 70 Marks in Chemistry.

•

Duration of the paper will be 3 hours.

•

Chemistry paper will have two parts viz: Part I of 35 marks and Part II of 35 marks

•

Same Answer Sheet will be used for both the parts.

•

In the question paper, for each part there will be 4 Questions.

•

Students have freedom to decide the sequence of answers.

•

The paper pattern as per the marking scheme for Part I and Part II will be as follows: Question 1:

(7 Marks)

There will be 7 multiple choice Questions (MCQs), each carrying 1 mark. Total marks = 7 Question 2:

(12 Marks)

There will be 8 Questions out of which 6 Questions are to be answered, each carrying 2 marks. Total marks = 12 Question 3:

(9 Marks)

There will be 4 Questions out of which 3 Questions are to be answered, each carrying 3 marks. Total marks = 9 (There will be 3 Questions based on numericals from part I) Question 4:

(7 Marks)

There will be 2 Question out of which 1 Question has to be answered. It will carry 7 marks. Total Marks = 7 (There will be 2/3 marks Questions based on numericals from Part I)

Distribution of Marks According to Type of Questions Type of Questions

Marks

Marks with option

Percentage (%)

Objectives

14

14

20

Short Answers

42

56

60

Brief Answers

14

28

20

Total

70

98

100

Contents No.

Topic Name

Page No.

Marks Without Option

Marks With Option

8

d and f−Block Elements

1

05

06

9

Coordination Compounds

35

03

04

10

Halogen Derivatives of Alkanes and Arenes

87

04

06

11

Alcohols, Phenols and Ethers

155

04

06

12

Aldehydes, Ketones and Carboxylic Acids

233

05

07

13

Compounds Containing Nitrogen

312

04

06

14

Biomolecules

371

04

06

15

Polymers

413

03

04

16

Chemistry in Everyday Life

439

03

04

Board Question Paper March 2013

465

-

-

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

08

d and f block Elements

Syllabus 8.0

Prominent Scientists

8.1

General Introduction and Electronic configuration

8.2

Occurrence and characteristics of transition elements

8.3

General trends in properties of the first row transition metals

8.4

Preparation and properties of K2Cr2O7 and KMnO4

8.5

General Introduction and Electronic Configuration

8.6

Lanthanoids

8.7

Actinoids

1

1

Std. XII Sci.: Perfect Chemistry - II

TARGET Publications

d and f-Block Elements

08 8.0

Prominent Scientists Scientists Friedrich Wilhelm Ostwald (1853-1932) (German Chemist)

Contribution Discovered the law of dilution, Gave the first modern definition of catalyst. Devised a method in 1900 to manufacture nitric acid by oxidizing ammonia. iv. Got the Nobel prize in 1909 in recognition of his work on catalysis. Glenn Theodore Seaborg (1912-1999) i. Discovered and isolated ten transuranic elements (American Chemist) which includes Plutonium, Americium, Curium, Berkelium, Californium, Einsteinium, Fermium, Mendelevium, Nobelium and Seaborgium. ii. Shared the Nobel prize with McMillon in 1951 for significant contributions in the field of transuranic elements. d-Block Elements 8.1

i. ii. iii.

General Introduction Electronic configuration

*Q.1. What are d-block elements? Ans: The elements in which the last electron enters the d-orbital of the penultimate shell i.e.(n−1)d orbital where n is the outermost shell, are called d-block elements. Their general electronic configuration is (n − 1)d1 − 10ns1 − 2. *Q.2. Explain the position of d-block elements in the periodic table. Ans: Position of d-block elements: i. The d-block elements lie in between s- and p-block elements, i.e. these elements are located in the middle part of the periodic table. ii. The d-block elements are present in 4th period (Sc to Zn 10 elements), 5th period (Y to Cd 10 elements), 6th period (La, Hf to Hg 10 elements), 7th period (Ac, Rf to Uub 10 elements) iii. d-block elements are present in the groups 3 to 12. Each series starts with elements of Group 3 and ends with elements of Group 12. Position of d-block elements in the periodic table 1 4 5

2

3

s-block 19 20 21 Sc 37 38 39 Y

6

55

56

7

87

88

57 La 89 Ac

4

5

6

7

8

22 Ti 40 Zr

23 V 41 Nb

24 Cr 42 Mo

d-block 25 26 Mn Fe 43 44 Tc Ru

72 Hf 104 Rf

73 Ta 105 Db

74 W 106 Sg

75 Re 107 Bh

d and f Block Elements

76 Os 108 Hs

9

10

11

12

13

14

15

16

17

18

35

36

27 Co 45 Rh

28 Ni 46 Pd

29 Cu 47 Ag

30 Zn 48 Cd

31

32

p-block 33 34

49

50

51

52

53

54

77 Ir 109 Mt

78 Pt 110 Ds

79 Au 111 Rg

80 Hg 112 Uub

81

82

83

84

85

86

113 114 115 116 117

118 1


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*Q.3. Explain the meaning of transition series. Ans: i. d-block elements are also known as transition elements. There are four series of transition elements in the long form of periodic table. These are known as transition series. ii. Four transition series are 3d, 4d, 5d and 6d series wherein orbitals of (n − 1)th main shell gets filled respectively. iii. The 3d series includes all the elements from Sc (Z = 21) to Zn (Z = 30) belonging to the 4th period. iv. The 4d series includes all the elements from Y (Z = 39) to Cd (Z = 48) belonging to the 5th period. v. The 5d series begins with element La (Z = 57) and includes all the elements from Hf (Z = 72) to Hg (Z = 80) belonging to the 6th period. vi. The 6d series begins with element Ac (Z = 89) and includes all the elements from Rf (Z = 104) to Uub (Z = 112) belonging to the 7th period. Q.4. Why are d-block elements called transition elements? Ans: i. Transition elements are defined as those elements which have partly or incompletely filled (n−1)d orbitals in their elementary state or in any of their common oxidation states. ii. The 3d, 4d, 5d and 6d subshells of the d-block elements are incomplete and the last electron enters the (n–1)d-orbital. iii. d-block elements are called transition elements as they shows transition in the properties from the most electropositive s-block elements to the less electropositive p-block elements. Q.5. Give the general electronic configuration of four series of d-block elements. Ans: The general electronic configuration of d-block elements is (n−1)d1–10ns1–2. The general electronic configuration of the four series of d-block elements are as mentioned below: i. 3d series: [Ar] 3d1−10 4s1−2 ii. 4d series: [Kr] 4d1−10 5s0−2 iii. 5d series: [Xe] 5d1−10 6s2 iv. 6d series: [Rn] 6d1−10 7s2 Q.6. Give the electronic configuration of all the elements belonging to the 3d series of d-block elements. Ans: 3d series or first row transition series (Sc to Zn) belonging to the fourth period has the general electronic configuration [Ar] 3d1−10 4s1−2. 3d series or First Row Transition Series (Sc to Zn) Elements Symbols Atomic Expected Electronic Observed Electronic Configuration number configuration Scandium Sc 21 [Ar] 3d1 4s2 [Ar] 3d1 4s2 2 2 Titanium Ti 22 [Ar] 3d 4s [Ar] 3d2 4s2 Vanadium V 23 [Ar] 3d3 4s2 [Ar] 3d3 4s2 4 2 Chromium Cr 24 [Ar] 3d 4s [Ar] 3d5 4s1 5 2 Manganese Mn 25 [Ar] 3d 4s [Ar] 3d5 4s2 Iron Fe 26 [Ar] 3d6 4s2 [Ar] 3d6 4s2 7 2 Cobalt Co 27 [Ar] 3d 4s [Ar] 3d7 4s2 8 2 Nickel Ni 28 [Ar] 3d 4s [Ar] 3d8 4s2 Copper Cu 29 [Ar] 3d9 4s2 [Ar] 3d10 4s1 10 2 Zinc Zn 30 [Ar] 3d 4s [Ar] 3d10 4s2 #Q.7. In which period of the periodic table, will an element, be found whose differentiating electron is a 4d electron? Ans: Fifth period of the periodic table consist of elements in which the differentiating electron is a 4d electron. Q.8. Silver atom has completely filled d-orbitals (4d10) in its ground state. How can you say that it is a transition element? (NCERT) Ans: The outer electronic configuration of Ag (Z = 47) is 4d105s1. In addition to +1, it shows an oxidation state of +2 (Eg. AgO and AgF2 exist). In +2 oxidation state, the configuration is d9, i.e. the d subshell is incompletely filled hence it is a transition element. Ag2+: [Kr]36 4d9. Therefore, Ag is a transition element. 2


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Q.9. In what way is the electronic configuration of transition elements different from that of the nontransition elements? (NCERT) Ans: Transition elements contain incompletely filled d subshell, i.e. their electronic configuration is (n−1) d1−10 ns1−2 whereas non-transition elements have no d subshell or their d subshell is completely filled and they have the outer electronic configuration of ns1−2 or ns2 np1−6. Q.10. Write down the electronic configuration of: (NCERT) i. Cr3+ ii. Cu+ iii. Co2+ iv. Mn2+ 3+ 2 2 6 2 6 3 + 2 2 6 Ans: i. Cr = 1s 2s 2p 3s 3p 3d ii. Cu = 1s 2s 2p 3s2 3p6 3d10 2+ 2 2 6 2 6 7 iii. Co = 1s 2s 2p 3s 3p 3d iv. Mn2+ = 1s2 2s2 2p6 3s2 3p6 3d5 *Q.11. Why does copper show abnormal electronic configuration? Ans: Copper has atomic number 29. Its expected and actual electronic configurations are: 2 2 6 2 6 9 2 29Cu (Expected): 1s 2s 2p 3s 3p 3d 4s 2 2 6 2 6 10 (Actual): 1s 2s 2p 3s 3p 3d 4s1 Explanation: i. The energy difference between the 3d and 4s orbitals is very low. ii. The d-orbital being degenerate, acquires more stability when it is half-filled (3d5) or completely filled (3d10). iii. Due to the interelectronic repulsion forces one 4s electron from the 4s orbital is transferred to the 3d orbital in Cu so that Cu has completely filled 3d10 orbital, thus acquiring more stability. *Q.12. Why chromium has electronic configuration 3d54s1 and not 3d44s2? Ans: Chromium has atomic number 24. Its expected and actual electronic configurations are: 2 2 6 2 6 4 2 24Cr (Expected): 1s 2s 2p 3s 3p 3d 4s 2 2 6 2 6 5 (Actual): 1s 2s 2p 3s 3p 3d 4s1 Explanation: i. The energy difference between the 3d and 4s orbitals is very low. ii. The d-orbital being degenerate, acquires more stability when it is half-filled (3d5) or completely filled (3d10). iii. Due to inter electronic repulsion forces, one 4s electron gets transferred to 3d orbital in Cr in order to acquire the half-filled orbital configuration i.e. 1s22s22p63s23p63d54s1 in which the 3d and 4s subshells have extra stability, thus resulting in the greater stability of the Cr atom. 8.2

Occurrence and characteristics of transition elements

Q.13. Explain the occurrence of d-block elements. Ans: i. d-block elements occur in free state as well as in combined states. ii Among the d-block elements, soft elements occur as sulphide minerals and more electropositive hard metals occur as oxides. iii. Occurrence of some d-block elements: a. Titanium occurs in the combined state. It is found in small quantities of coal, clay, rocks, sand etc. Eg. Rutile (TiO2), Ilmenite (FeTiO3). b. Vanadium occurs in the ores vanadinite and carnotite. c. Chromium occurs in the ores as Chromite (FeO, Cr2O3), Chrom ochre (Cr2O3), Crocoisite (PbCrO4). d. Manganese does not occur in the free state but mainly occurs in pyrolusite ore (MnO2) and other minerals. e. Iron occurs in free as well as combined states. f. Nickel is found in cobalt ores. g. Copper is found in rocks, soil, sea, mineral waters and in ores like cuprite (Cu2O), malachite [CuCO3, Cu(OH)2], azurite [2CuCO3, Cu(OH)2] etc. h. Zinc is found in the ores of silver, copper, lead and platinum. 3 d and f Block Elements


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Q.14. State the general characteristics of transition elements or d-block elements. Ans: Characteristics of transition elements: i. Most of the transition elements are metals and thereby they show metallic properties such as ductility, malleability, electrical conductivity, high tensile strength and metallic lustre. ii. Except mercury which is liquid at room temperature, other transition elements have typical metallic structures. iii. Their compounds generally contain unpaired electrons, hence they are paramagnetic in nature and form coloured compounds. iv. They exhibit variable oxidation states. v. They have tendency to form large number of complexes. vi. They have higher densities as compared to the metals of IA and IIA group (s-block). vii. They are heavy metals with higher melting and boiling point as well as higher heats of vaporisation. viii. Transition elements are less reactive than s-block elements due to their higher ionisation energy. ix. Most of the transition metals such as Mn, Ni, Co, Cr, V, Pt, etc. and their compounds are used as catalysts. x. They are good conductors of heat and electricity. xi. They form alloys with different metals. xii. They form interstitial compounds with elements such as hydrogen, boron, carbon, nitrogen etc. xiii. They form organometallic compounds. Q.15. Why do transition metals possess high density and high melting and boiling points? Ans: i. The densities of d-block elements are relatively higher as compared to the s-block elements due to the decrease in the size of the atoms and the consequent increase in the nuclear charge, which results in the compact structure of the elements. ii. The density of the atoms increases with the decrease in the size of the atom. Therefore the density of the elements goes on increasing from left to right across a period. iii. Transition elements form strong metallic bond in which both (n–1) and ns electrons take part. Due to the notable covalent character of the strong metallic bond, considerable amount of energy is required to break the metallic bond in order to melt the metal. Hence these metals possess high melting and boiling points. Note: Atomic properties of 3d series: Element Sc − 21

Density (g cm−3) 2.99

Radii/pm M+ M2+ 164 −

Ti − 22

4.50

147

V − 23

5.96

Cr − 24

M 73

Ionization enthalpy kJmol−1 631

67

656

135

− 79

64

650

7.20

129

82

65

653

Mn − 25

7.21

137

82

65

717

Fe − 26

7.86

126

77

65

762

Co − 27

8.90

125

74

61

758

Ni − 28

8.90

125

70

60

736

Cu − 29

8.92

128

73

−

745

Zn − 30

7.14

137

75

−

906

4

3+


TARGET Publications


Q.16. All d-block elements are not transition elements. Explain. Ans: i. The d-block elements are those in which the last electron enters the d orbital. ii. The transition elements are those elements which have incompletely filled (partly filled) d-subshells in their elementary state or in any one of their oxidation states. iii. Hence, only those d-block elements which have completely filled d orbitals, (n−1)d10 are not transition elements. iv. Eg. Zn, Cd and Hg atoms have completely filled d-orbitals (3d10) in their ground state as well as in their oxidised states, hence they are d-block elements but not transition elements. 8.3

General trends in properties of the first row transition metals

Q.17. Explain the metallic characters of the d-block elements. Ans: i. Except mercury which is a liquid, all the transition elements are metallic in nature. ii. They are hard, lustrous, malleable and ductile with high melting and boiling points, and having good thermal and electrical conductivities. iii. The metallic character of transition elements may be attributed to their low ionization enthalpies and the presence of several vacant orbitals in their outermost shell, which favours the formation of metallic bonds in them. iv. In addition to the electrons from outermost energy level, the unpaired d-electrons also contribute for the formation of metallic bond. v. So, greater the number of unpaired d-electrons, stronger is the bonding due to the overlapping of the d-orbitals containing unpaired electrons. Q.18. Why are Cr, Mo and W hard metals while Zn, Cd and Hg are not very hard metals? Ans: i. The hardness of the metals indicates the presence of covalent bonds in them. ii. The d-orbitals containing unpaired electrons may overlap to form covalent bonds. iii. Greater the number of unpaired d-electrons present in a transition metal atom, more is the number of covalent bonds formed by it and thus greater is the strength and hardness of the metal. iv. Cr, Mo and W have maximum number of unpaired d-electrons and are, therefore, hard metals whereas Zn, Cd and Hg are not very hard metals due to the absence of unpaired electrons. Q.19. Explain the trends in melting and boiling points of first row transition metals. Ans: i. The high melting and boiling points of transition metals are due to their close-packed structures, in which the transition metals are held together by strong metallic bonds which have significant covalent character. ii. Considerable amount of energy is required to break the metallic bonds in order to melt the metal, therefore, these metals have very high melting points and boiling points. iii. The strength of metallic bonds depends upon the number of unpaired electrons. Greater the number of valence electrons, stronger is the metallic bonding, and consequently, melting points are higher. iv. Therefore, as we move along a particular series, the metallic strength increases upto the middle with increasing availability of unpaired electrons upto d5 configuration (Eg. Sc has 1, Ti has 2, V has 3, Cr has 5 unpaired electrons) and then decreases with decreasing availability of unpaired d-electrons (Eg. Fe has 4, Co has 3 unpaired electrons and so on). v. Thus the melting and boiling points increase from Sc to Cr. Chromium has the maximum melting point in the first transition series. After Cr, the melting and boiling points decrease from Cr to Cu. vi. Zn, Cd, Hg have no unpaired electrons. Thus these metals have low melting and boiling points. Mercury (Hg) is liquid at room temperature and has melting point of 234 K. The unexpectedly lower melting point of Mn and Tc are probably due to their complicated lattice structure. These metals have low values for enthalpies of atomization. Q.20. Define Ionisation enthalpy. Ans: It is the amount of energy required to remove the outermost electron completely from a gaseous atom in its ground state. 5 d and f Block Elements


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First ionisational(kJ mol-1) enthalpy

Q.21. Explain the trends observed in the ionisation enthalpies of the d-block elements. Ans: i. The ionization enthalpies of transition metals are quite high and lie between those of s-block and p-block Ir 900 Au elements. Pt Os W ii. The first ionization enthalpy increases irregularly with Fe Pd 800 Ta Hf Re increase in atomic number across a given transition Cu Co Mn Ni Ag series. The added electron enters into (n−1) d-subshell Ru Mo 700 Zr Nb Rh Tc and shields the valence electrons from the nucleus. Ti V Cr Thus, the effect of increased nuclear charge is opposed 600 Transition Elements by the screening effect of (n−1) d-electrons. Therefore, The first ionization enthalpies the increase in ionization enthalpy with the increase in of Transition elements atomic number is rather small and not very regular. iii.

The first ionization enthalpies of elements of third transition series are higher than those of the elements of first and second transition series, due to the poor shielding effect of the electrons present in the filled 4-f orbitals.

Note: The energy required to remove first electron from gaseous atom in its ground state is called first ionisation enthalpy, to remove second electron, it is called second ionisation enthalpy and for removal of third electron it is called third ionisation enthalpy. Q.22. Explain thermodynamic stability of transition metal compound on the basis of ionization enthalpy. Ans: The smaller the sum of ionization enthalpies needed for transition metals to attain a particular oxidation state, greater is the stability of the compounds of that metal in that particular oxidation state. Eg. Nickel (II) compounds, are thermodynamically more stable than platinum (II) compounds, whereas platinum (IV) compounds are more stable than nickel (IV) compounds. Explanation: i. Sum of first and second ionization enthalpies (IE1 + IE2) for nickel is lesser as compared to that of platinum (Pt). This implies that ionization of Ni to Ni2+ ion requires lesser energy (enthalpy) than the energy required by Pt to form Pt2+ ion. Therefore Ni(II) compounds are thermodynamically more stable than Pt (II) compounds. ii. On the other hand, the sum of first four ionization enthalpies (IE1 + IE2 + IE3 + IE4) for Pt(9.36 × 103 kJmol−1) is lesser than that required by Ni to form Ni4+. Therefore, smaller amount of energy is required to ionize Pt to Pt4+ ion. iii. Thus Pt(IV) compounds are thermodynamically more stable than Ni(IV) compounds. *Q.23. Explain the oxidation states of first row elements of transition series. Ans: i. The oxidation states of transition elements are related to the electronic configuration. Transition elements have last two orbitals incompletely filled i.e. They contain ns and (n–1)d orbitals. ii. Thus these elements possess +2 and +3 common oxidation states. iii. iv. v.

vi. 6

Since (n−1)d-electrons and ns-electrons have nearly equal energies, both are available for chemical bonding. Hence, the transition metals form compounds showing more than one (variable) oxidation states. Since the transition metals of the first series have two electrons in their outermost shell with the exception of Cr and Cu (4s1), the lowest oxidation states of these elements are +1 or +2, which is due to their 4s electrons. As the 3d electrons take part in the chemical bonding one after another, in addition to the +2 oxidation state, there are number of other oxidation states as illustrated below. d and f Block Elements


TARGET Publications

Oxidation states of first transition (3d) series elements: Sc +2 +3

vii.

Ti +2 +3 +4

V +2 +3 +4 +5

Cr +1 +2 +3 +4 +5 +6

Mn +2 +3 +4 +5 +6 +7

Fe +2 +3 +4 +5 +6

Co +2 +3 +4 +5

Ni +2 +3 +4

Cu +1 +2

Zn +2

As the number of unpaired electrons in the 3d orbitals increases, the number of oxidation state also increases.

Q.24. Compare the stability of +2 oxidation state for the elements of the first transition series. (NCERT) Ans: i. In the beginning of 3d transition series, Sc2+ is virtually not known or in other words it is not stable in comparison to Sc3+. Ti2+, V2+, Cr2+ are known but less stable in comparison to their most common oxidation state of +3. ii. In the middle of the 3d transition series, Mn2+, Fe2+, Co2+ are known and quite common. Mn2+ and Mn7+ are most stable in Mn. Fe2+ is less stable in comparison to Fe3+ due to fact that Fe3+ tends to lose one electron to acquire d5 structure, which has extra stability. Co2+ is less stable as compared to Co3+. Ni2+ is most common and most stable among its +2, +3, +4 states. Cu2+ is more stable and is most common species as compared to Cu1+. iii. At the end of the 3d transition series, Zn forms only Zn2+ which is highly stable as it has 3d10 configuration. Q.25. Write the different oxidation states of manganese. Why +2 oxidation state of manganese is more stable? [March 2013] Ans: Oxidation states of Mn: +2, +3, +4, +5, +6 and +7. Electronic configuration of Mn: 1s2 2s2 2p6 3s2 3d5 4s2 Due to the presence of half filled ‘d’ orbital, the +2 oxidation state of Manganese is more stable Q.26. To what extent do the electronic configurations decide the stability of oxidation states in the first series of transition elements? Illustrate with example. (NCERT) Ans: In a transition series, the oxidation states which lead to noble gas or exactly half-filled or completely filled d-orbitals are more stable. Eg. In the first transition series, electronic configuration of Mn (Z = 25) is [Ar] 3d54s2. It shows oxidation states +2 to +7 but Mn (II) is most stable because it has the half-filled configuration [Ar] 3d5. Similarly, Sc3+ and Zn2+ are more stable as illustrated below: Sc = [Ar] 3d14s2, Sc3+ = [Ar] i.e. noble gas configuration. Zn = [Ar] 3d104s2, Zn2+ = [Ar] 3d10 i.e. completely filled configuration. Q.27. What may be the stable oxidation state of the transition element with the following d-electron configurations in the ground state of their atoms: 3d3, 3d5, 3d8 and 3d4? (NCERT) Ans: The stable oxidation state of transition elements with the d-electron configuration 3d3, 3d5, 3d8 and 3d4 in ground state of atoms is as follows: Sr. No. 1. 2. 3. 4.

d-electron configuration 3d3 3d4 3d5 3d8


Symbol of Element V (3d34s2) Cr (3d54s1) Mn (3d54s2) Ni (3d84s2)

Stable oxidation States +2, +3, +5 +2, +3, +6 +2 , +7 +2, +4 7


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Q.28. How is the variability of oxidation states of the transition elements different from that of the nontransition elements? Illustrate with examples. (NCERT) Ans: i. The variability in oxidation states of transition metals is due to the incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity. Eg. Fe2+ and Fe3+, Cu+ and Cu2+ etc. In case of non-transition elements, the oxidation states differ by units of two. Eg. Pb2+ and Pb4+, Sn2+ and Sn4+ etc. ii. In transition elements, the higher oxidation states are more stable for heavier elements in a group. Eg. In group 6, Mo (VI) and W (VI) are more stable than Cr (VI). In p-block elements, the lower oxidation states are more stable for heavier members due to inert pair effect, Eg. In group 16, Pb (II) is more stable than Pb (IV). Q.29. Why does scandium show only +2 and +3 oxidation states? Ans: Scandium (Sc) has electronic configuration, Sc: 1s22s22p63s23p63d14s2 Due to the loss of two electrons from the 4s-orbital, Sc acquires +2 oxidation state. Sc2+ : 1s22s22p6 3s23p63d1 By the loss of one more electron from the 3d-orbital, it acquires +3 oxidation state. Sc3+ : 1s22s22p63s23p6 Since Sc3+ acquires extra stability of inert element [Ar], it does not form higher oxidation state. #Q.30.Why is manganese more stable in the +2 state than in the +3 state and the reverse is true for iron? OR 2+ Why Mn compounds are more stable than Fe2+ towards oxidation to their +3 state? (NCERT) Ans: Electronic configuration of Mn2+ is [Ar] 3d5 which is half filled and hence it is stable. So third ionisation enthalpy is very high i.e third electron cannot be easily removed. Whereas in case of Fe2+, the electronic configuration is 3d6. Therefore, Fe2+ can easily lose one electron to acquire 3d5 stable electronic configuration. Q.31. Why Zn does not exists in variable oxidation states? Ans: The oxidation state of an element depends on its electronic configuration. Zn: 1s22s22p63s23p63d104s2 When Zn loses two electrons from 4s-orbital, it gets a very stable electronic configuration wherein all the electrons are paired in d-orbital, giving a +2 oxidation state. Zn2+: 1s22s22p63s23p63d10 Since all the orbitals are completely filled with paired electrons, it has a very stable electronic configuration. Therefore, Zn shows only one oxidation state of +2. #Q.32.Which element of 3d series of the transition metals exhibits the largest number of oxidation states and Why? Ans: i. Among transition metals, Manganese exhibits the largest number of oxidation states ranging from +2 to +7. ii. The electronic configuration of Mn is: Mn: 1s22s22p63s23p63d54s2 iii. Different oxidation states shown by Mn are: +2

+3

+4

+5

+6

+7

Mn Cl 2 , Mn 2 O3 , Mn O 2 , Mn F5 , K 2 Mn O 4 , K Mn O 4 .

iv. v. vi. 8

Mn2+ state is very stable due to extra stability of half filled 3d5 subshell. Mn7+ is a good oxidising agent since it can be easily reduced to Mn4+ and Mn2+. As the number of unpaired electrons in 3d subshell increases, the number of oxidation state increases. d and f Block Elements


TARGET Publications

#Q.33. Explain why iron, cobalt and nickel do not show the expected highest oxidation states of +8, +9 and +10 respectively. Ans: i. The oxidation states of an element depends on its electronic configuration and the number of unpaired electrons. ii. The electronic configuration of Fe, Co and Ni are as follows: Fe: [Ar] 3d64s2 Co: [Ar] 3d74s2 Ni: [Ar] 3d84s2 iii. iv.

v.

vi.

Available electrons for excitation 6+2=8 7+2=9 8 + 2 = 10

Hence the expected oxidation states of Fe, Co and Ni shall be +8, +9 and +10 respectively. Fe by the loss of two electrons from 4s subshell forms Fe2+ and further loss of one more electron from 3d-orbital forms Fe3+, which is very stable due to half filled orbital, 3d5. Hence, there is no further loss of electrons from the 3d subshell and thus Fe does not show +8 oxidation. Co forms Co2+ due to the loss of two electrons from 4s subshell and further loss of one more electron from 3d-orbital forms Co3+. In cobalt +2 and +3 oxidation states are more stable and it does not lose electrons further. Hence Co does not show +9 oxidation state. Ni forms Ni2+ by the loss of two electrons from 4s-orbital and by the loss of one more electron from 3dorbital, it forms Ni3+. Now there are 7 electrons in 3d-orbital which give extra stability. Hence, there is no further loss of electrons and therefore Ni does not show +10 oxidation state.

Q.34.Which metal in the first transition series exhibits +1 oxidation state most frequently and why? (NCERT) Ans: Copper has electronic configuration [Ar] 3d104s1. It can easily lose one (4s1) electron to give stable 3d10 configuration. Q.35. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? (NCERT) Ans: Oxygen and fluorine have small size and high electronegativity so they can easily oxidise the metal to its highest oxidation state. Q.36. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number. : Oxidation state of Mn = +7 (Group No = 7) Ans: MnO4− CrO42− : Oxidation state of Cr = +6 (Group No = 6) Q.37. Comment on the statement that the “elements of the first transition series possess many properties different from those of heavier transition elements”. (NCERT) Ans: The given statement is true because: i. Atomic radii of the heavier transition elements (4d and 5d series) are larger than the corresponding elements of the first transition series though those of 4d and 5d series are very close to each other. ii. For first transition series, +2 and +3 oxidation states are more common whereas for heavier transition elements, higher oxidation states are more common. iii. Melting and boiling points of heavier transition elements are greater than those of the first transition series due to stronger intermetallic bonding (M−M bonding). iv. Ionisation enthalpies of 5d series are higher than the corresponding elements of 3d and 4d series. v. The elements of the first transition series form low spin or high spin complexes depending upon the strength of the ligand field. However, the heavier transition elements form low spin complexes irrespective of the strength of the ligand field. *Q.38. Explain the trend of atomic radii in transition elements. Ans: i. The atomic radii for transition metals are smaller than their corresponding s-block elements. Transition metals

s-block elements 4th period At. Radius (pm) d and f Block Elements

K

Ca

Sc

Cu

203

174

144

117 9


TARGET Publications

The atomic radii of the elements of a given series decrease with the increase in atomic number but this decrease becomes small after midway. Element Atomic radii (pm) ii.

iii. iv.

v. vi.

vii.

Sc 144

Ti 132

V 122

Cr 118

Mn 117

Fe 117

Co 116

Ni 115

Cu 117

Zn 125

In the first transition series, the atomic radius gradually decreases from scandium to chromium but from chromium to copper, it is nearly the same. Similar behaviour has been observed in the second and third transition series. The decrease in atomic radii in each series, in the beginning, is due to an increase in nuclear charge across the period, which tends to pull the ns electrons inward, i.e. it tends to reduce the size. The addition of extra electrons to (n−1) d-orbitals also provides the screening effect. As the number of d-electrons increases, the screening effect increases. Thus, there are two operating effects namely screening effect and nuclear charge effect which oppose each other. In the midway onwards of the series, both these effects become nearly equal and thus, there is no change in atomic radii inspite of the fact that atomic number increases gradually. The values of atomic radii at the end of each series are slightly higher which is due to electronelectron repulsions among (n−1) d-electrons. These repulsions become predominant at the end of each series and thus resulting in the increase in size. In a vertical row, the atomic radii is expected to increase from top to bottom. Therefore, the atomic radii of transition metals of second series have larger values than those of the first transition series. However, the transition metals of third series except the first member, lanthanum, have nearly the same radii as metals of second transition series above them which is due to the Lanthanoid contraction.

*Q.39. Explain the trend of the ionic radii in transition elements? Ans: i. In the table given below ionic radii of the oxidation states of +2 and +3 are given. In general ionic radii decreases with increase in the oxidation state. For the same oxidation state, the ionic radii generally decrease with increase in the nuclear charge in a given transition series. This trend is more pronounced for the divalent ions of the elements belonging to the first transition series. The gradual decrease in ionic radii in going from Ti2+ ion to Cu2+ ion is due to an increase in the effective nuclear charge.

ii.

Ni Cu Zn 72 72 74 − − − The ionic radii decreases with the increase in the oxidation state. This is due to the increase in the effective nuclear charge. Ionic radii of the transition elements are smaller than those of the representative elements of the same period. M2+ M3+

Sc − 81

Ti 90 76

V 88 74

Cr 84 69

Q.40. Explain the splitting of d-orbitals in transition elements. Ans: i. In a free metal ion, all the orbitals at the same energy level are called degenerate orbitals. ii. However, when they are in the form of compounds, the five d-orbitals split into two groups. iii. One group consist of three orbitals namely dxy, dyz, dxz and they are of lower energy level. iv. The second group consists of two orbitals d x − y and d z and are of slightly higher energy 2

v. vi. 10

2

2

level. The splitting of d-orbitals is caused by the groups linked to the metal ions. The difference in energy (∆E) between the two sets of d-orbitals is small.

Mn 80 66

Fe 76 64

Co 76 63

d x2 -y2 d 2 z

Energy

Element Ionic radii in pm

∆E d-orbitals dxy dyzdxz Splitting of d-orbitals



TARGET Publications

*Q.41. Explain, most of the transition metal compounds are remarkably coloured. OR Explain, most of the transition elements form coloured compounds. Ans: i The colour of a substance depends upon the absorption of light of a particular wavelength of visible lights. ii. The colour of the compounds of the transition metals may be attributed to the presence of incomplete (n−1) d subshell. In these compounds, the energies of the five d-orbitals in the same sub-shell do not remain same. iii. The amount of energy required to excite some of the electrons to the higher energy states within the same d-subshell, is quite small. iv. In case of transition metal ions, the electrons can be easily promoted from one energy level to other within the same d-subshell. This transition is called the d-d transition. v. The energy corresponding to such small transitions is available within the visible range. vi. Therefore, the transition metal ions absorb certain radiations from visible region and appear coloured. Q.42. What are the factors responsible for colour of a transition metal ions? Ans: Factors responsible for colour of a transition metal ions: i. The presence of incompletely filled d-orbitals in metal ions. ii. The presence of unpaired electrons in d-orbitals. iii. d-d transitions of electrons due to absorption of radiation in the visible region. iv. Nature of groups (anions or ligands) linked to the metal ion in the compound or a complex. v. Type of hybridisation in metal ion in the complex. vi. Geometry of the complex formed by the metal ion. Note: Colour of 3d transition metal ions: Ion Sc3+ Ti3+ Ti4+ V3+ Cr3+ Mn2+ Mn3+

Outer electronic configuration 3d0 3d1 3d0 3d2 3d3 3d5 3d4

Number of Colour unpaired electrons 0 Colourless 1 Purple 0 Colourless 2 Green 3 Violet 5 Light Pink 4 Violet

Ion Fe2+ Fe3+ Co2+ Ni2+ Cu2+ Cu1+ Zn2+

Outer electronic configuration 3d6 3d5 3d3 3d8 3d9 3d10 3d10

Number of unpaired electrons 4 5 3 2 1 0 0

Colour

Pale green Yellow Pink Green Blue Colourless Colourless

Q.43. Why some of the transition metal ions are colourless? Ans: i. Transition metal ions exhibit colour due to the presence of unpaired electrons in (n−1)d-orbitals which undergo d-d transition. ii. The metal ions which does not have unpaired electrons (n−1)d0 or which have completely filled d-orbital (n−1)d10 do not absorb radiations in visible region, since d-d transitions are not possible. Hence, they are colourless ions. Eg. Cu+ (3d10), Ag+ (4d10), Zn2+ (3d10), Cd2+ (4d10), Hg2+ (5d10), etc. #Q.44. Why salts of Sc3+, Ti4+, V5+ are colourless? Ans: i. Electronic configurations of Sc3+, Ti4+, V5+ are: Sc3+: [Ar] 3d0 Ti4+: [Ar] 3d0 V5+: [Ar] 3d0 ii. Since the ions Sc3+, Ti4+ and V5+ have completely empty d orbitals i.e. no unpaired electron, their salts are colourless. d and f Block Elements

11


TARGET Publications

*Q.45. Explain why the compounds of copper (II) are coloured but those of zinc are colourless. Ans: i. The electronic configuration of Cu: [Ar] 3d104s1 and Cu2+: [Ar] 3d9 ii. In copper compounds, Cu2+ ions have incompletely filled 3d-orbital (3d9). iii. Due to the presence of one unpaired electron in 3d-orbital, Cu2+ ions absorb red light from visible spectrum and emit blue radiation due to d-d transition. Therefore copper compounds are coloured. iv. In case of zinc, the configuration is Zn: [Ar] 3d104s2 and Zn2+: [Ar] 3d10 v. Since 3d subshell is completely filled and there are no unpaired electrons, d-d transition is not possible and hence Zn2+ ions do not absorb radiation in visible region. Therefore the compounds of Zn are colourless. Q.46. Explain why the solution of Ti3+ salt is purple in colour? Ans: i. In [Ti(H2O)6]3+, Ti3+ has 3d1 configuration. ii. This one electron will occupy one of the orbital of lower energy and the complex (ion) will absorb suitable wavelength of white light and promote the electron from lower energy level to higher energy level. iii. Since the complex absorbs light around 500 nm region, yellow and green lights are absorbed to excite the electron and the transmitted light is of the complementary colour i.e. red blue (purple). iv. Hence solution containing hydrated Ti3+ ion is purple in colour. Q.47. Explain why copper sulphate is blue in colour when dissolved in water but turns yellow when treated with concentrated HCl. Ans: i. The electronic configuration of Cu: [Ar] 3d104s1 and Cu2+: [Ar] 3d9. ii. In the aqueous solution of copper sulphate, Cu2+ forms blue hydrated complex [Cu(H2O)6]+2 because it has one unpaired electron in 3d subshell. The complex has octahedral geometry. iii. When CuSO4 solution is treated with concentrated HCl solution, it forms a new complex [CuCl4]2−. − [Cu(H2O)6]2+ + 4Cl ⎯→ [CuCl4]2− + 6H2O iv. The complex [CuCl4]2− has a tetrahedral geometry with one unpaired 3d1 electron. Due to a change in geometry and hybridisation of the complex, the colour of the solution changes from blue to yellow. *Q.48. Explain why transition elements show tendency to form large number of complexes. Ans: i. All the transition metals have a tendency to form complexes. ii. The tendency arises due to the following reasons: a. They have small ionic radii b. They have high effective ionic charge. c. Hence they have high ratio of ionic charge to ionic radius. d. Transition metals and ions have vacant d-orbitals which can accomodate the lone pairs of electrons from the ligands to form coordination compounds. e. Transition metals show variable oxidation states. f. After accepting the electrons from the ligands, metal ions acquire a stable electronic configuration of the nearest inert element and form stable complexes. g. Eg. [Cu(NH3)4]2+, [Co(NH3)6]3+, [Ni(CN)4]2−, etc. iii. The stability of these complexes depends upon the nature of the metal ion, ligands and their bonding. *Q.49. Explain, why Pt(IV) complexes are generally octahedral while Pt(II) complexes are square planar. Ans: i. In case of Pt(II) complexes the central metal atom has d8 configuration and it shows coordination number 4. According to valence bond theory, central ion undergoes dsp2 hybridisation which results in formation of square planar complexes. ii. In the Pt (IV) complexes the central metal atom has d6 configuration and it shows coordination number 6. According to valence bond theory, central ion undergoes d2sp3 or sp3d2 hybridisation which results in the formation of octahedral complexes. 12



TARGET Publications

Q.50. Explain catalytic activity of transition metals. Ans: i. Many transition metals and complexes are used as catalysts which influence the rate of chemical reaction. The rate of a chemical reaction increases by the decrease in activation energy of the reactants. This decrease is caused by the catalyst which probably alter the path of the reaction. ii. Catalysts at solid surface involve the formation of bonds between reactant molecules and atoms on the surface of the catalyst (first row transition metals utilise 3d and 4s-electrons for bonding). This results in the formation of reaction intermediates which provides path of lower activation energy and therefore, increase the rate of the reaction. A + B + C ⎯→ [A−B−C] ⎯→ A−B + C Reactants

iii. iv. v.

Catalyst

Intermediate

Product

Catalyst

These reaction intermediates readily decompose yielding the products and regenerating the original substance. Many transition metals are used as catalysts for reactions. The commonly used transition metals as catalysts are Fe, Co, Pt, Cr, Mn, etc. Eg. a. MnO2 acts as a catalyst for decomposition of KClO3 to O2. b. Nickel acts as a catalyst in hydrogenation of oils to fats. c. Vanadium pentoxide is used as catalyst in the manufacture of H2SO4 by contact process. d. Fe(III) catalyzes the reaction between iodide and persulphate ions. 2I− + S2 O82− ⎯→ I2 + 2 SO 24− e. Titanium tetrachloride (in Ziegler-Natta catalyst) is used in the manufacture of high density polythene. Zeigler − Natta catalyst nCH2 = CH2 ⎯⎯⎯⎯⎯⎯⎯ → [−CH2 − CH2 −]n f. In the manufacture of ammonia, Fe with Mo is used as a catalyst. g. Co-Th alloy is used in Fischer Tropsch process in the synthesis of gasoline.

Q.51. Explain catalytic property of 3d series elements. Ans: The catalytic property of 3d series elements can be explained as follows: i. Because of variable oxidation states, transition metals easily absorb and re-emit wide range of energies to provide necessary activation energy. 3+

ii.

Fe Mechanism of 2I− + S2 O82− ⎯⎯⎯ → I2 + 2 SO 24− is explained as a. 2Fe3+ + 2I− ⎯→ 2Fe2+ + I2

b. iii.

2Fe2+ + S2 O82− ⎯→ 2Fe3+ + 2 SO 24−

Because of the presence of free valencies on the surface, they can absorb the reacting molecules, thereby increasing the concentration of the reactants on the surface and hence resulting in increase of the rate of reaction.

Q.52. Transition metals and many of their compounds show paramagnetic behaviour. Explain. Ans: i. Paramagnetism is a property due to the presence of unpaired electrons. ii. In case of transition metals, as they contain unpaired electrons in the (n−1)d orbitals, most of the transition metal ions and their compounds are paramagnetic in nature. iii. As the number of unpaired electrons increases from one to five, paramagnetic character also increases. Note: i. Paramagnetism: The substances which are attracted by the applied magnetic field are called paramagnetic substances. This property of substances is known as paramagnetism. (They have unpaired electrons) Eg. Magnesium, Lithium ii. Diamagnetism: The substances which are repelled by the applied magnetic field are called diamagnetic substances. This property of substances is known as diamagnetism. (They have paired electrons) Eg. Copper, Silver, Gold d and f Block Elements

13

Std. XII Sci.: Perfect Chemistry - II iii.

TARGET Publications

Ferromagnetism: The substances which exhibits permanent magnetism even when the magnetic field is removed are called ferromagnetic substances. This property of substances is known as ferromagnetism. Eg. Iron, Nickel, Cobalt

Q.53. Explain magnetic moment for transition elements and give formula for its calculation. Ans: i. The magnetic moment of an electron is partly due to its orbital motion and partly due to its spin motion. ii. In transition metal ions, the orbital magnetic moment is suppressed by the electrostatic field of other atoms, ions or molecules surrounding the metal ion. Thus the effective magnetic moment arises mainly from the spin of electrons. iii. The expression for the effective magnetic moment is, µ = n(n + 2) B.M where n is number of unpaired electrons. B.M stand for Bohr magneton (unit of magnetic moment) iv. Magnetic moment of a substance varies only with the number of unpaired electrons present. Q.54. The d-electron configuration of Co2+ and Cu2+ is d7 and d9 respectively. Which one of these ions will be more paramagnetic? Ans: Co2+, (electronic configuration: 3d7) contains three unpaired electrons while Cu2+, (electronic configuration: 3d9) contains only one unpaired electron. More the unpaired electrons present, more paramagnetic is the substance. So, Co2+ is more paramagnetic. Q.55. Why does Mn(II) ion show maximum paramagnetic character amongst bivalent ions of first transition series? Ans: The electronic configuration of Mn(II) ion (atomic number of Mn = 25) is [Ar] 3d5 4s0 It has five unpaired electrons in its d-orbitals which is a maximum value among the first transition metal ions. As the paramagnetic character is due to the presence of unpaired electron, Mn(II) ions shows maximum paramagnetic character. #Q.56. Why are zinc metal and Zn++ ions diamagnetic? Ans: The atomic number of zinc is 30 and the electronic configuration of Zn is [Ar] 3d104s2 and Zn++ is [Ar] 3d10. In zinc metal and Zn++ all electrons are paired and hence they are diamagnetic. Q.57. Calculate the magnetic moment of Fe3+ ion (atomic number of Fe = 26). Ans: Electronic configuration of Fe3+ ion is [Ar] 3d5 4s0 So d-orbital has following distribution of electrons,

It has 5 unpaired electrons, ∴ Magnetic moment, µ = n(n + 2) = 5(5 + 2) = 5.9 B.M Q.58. Calculate the magnetic moment of divalent ion in aqueous solution if its atomic number is 24. Ans: With atomic number 24, the divalent ion in aqueous solution will have d4 configuration (Four unpaired electrons). The magnetic moment µ is µ = 4(4 + 2) = 4.89 B.M #Q.59. Calculate the ‘Spin only’ magnetic moment of M2+(aq) ion (Z = 26). Ans: Atomic number = 26 Electronic configuration of M atom = [Ar] 3d6 4s2. Electronic configuration of M2+ = [Ar] 3d6 So d-orbital has following distribution of electrons.

There are 4 unpaired electrons. 14



TARGET Publications

∴ Spin only magnetic moment (µ) =

n(n + 2) B.M.

=

4(4 + 2) B.M.

= 4.89 B.M. 2+

Q.60.Calculate magnetic moment of Fe Ans: Refer Q.59

(aq)

ion (Z = 26)

[March 2013]

Q.61. Calculate the ‘Spin only’ magnetic moment of M2+(aq) ion (Z = 27). (NCERT) Ans: Atomic number = 27 Electronic configuration of M atom = [Ar] 3d7 4s2. Electronic configuration of M2+ = [Ar] 3d7 So d-orbital has following distribution of electrons.

There are 3 unpaired electrons. ∴ Spin only magnetic moment (µ) =

n(n + 2) B.M.

3(3 + 2) B.M. = 3.87 B.M.

=

Q.62. What are interstitial compounds? Why interstitial compounds are well known for transition metals? (NCERT) Interstitial compounds are those which are formed when small atoms like H, C, N, B etc. are trapped Ans: i. inside the crystal lattice of metals. They are generally non-stoichiometric and neither typically ionic, nor covalent. ii. Most of the transition metals form interstitial compounds with small non-metals such as H, C, N, B etc. iii. These small atoms enter into the void sites between the packed atoms of crystalline transition metals and form chemical bonds with transition metals. Eg. steel and cast iron become hard by forming interstitial compound with carbon. iv. The presence of vacant (n−1) d orbitals in transition elements and their ability to make bonds with trapped small atoms is the main cause of interstitial compound formation. Q.63. State the characteristics of interstitial compounds. Ans: Characteristics of interstitial compounds: i. Their chemical properties are same as that of parent metal. ii. These interstitial compounds are hard and show electrical and thermal conductivity and lustre. iii. They have high melting points, higher than those of pure metals as the metal-nonmetal bonds are stronger than metal-metal bonds in pure metals. iv. Their densities are lesser than the parent metal. v. Hydrides of transition metals are used as powerful reducing agents. vi. The metallic carbides are chemically inert and extremely hard as diamond. Q.64. Explain alloy formation in case of transition elements. Ans: i. Alloys are formed by metals whose atomic radii differ by not more than 15% so that the atoms of one metal can easily take up the positions in crystal lattice of the other. ii. The transition metals have similar atomic radii and other characteristics, hence they form alloys very readily. Alloys are generally harder, have high melting points and more resistant to corrosion than the individual metals. iii. The metals chromium, vanadium, molybdenum, tungsten and manganese are used in the formation of alloy steels and stainless steels. Ferrous alloys are the most common alloys. iv. Some alloys of transition metals with non-transition metals are also very common. Eg. Brass (Cu + Zn) and Bronze (Cu + Sn) d and f Block Elements

15


8.4

TARGET Publications

Preparation and properties of K2Cr2O7 and KMnO4

Q.65. Indicate the steps involved in the preparation of K2Cr2O7 from chromite ore (FeO.Cr2O3). OR How is potassium dichromate prepared from chrome iron? Ans: Preparation of K2Cr2O7 from chrome iron (FeO.Cr2O3): Step I: Concentration of ore: Powdered chromite ore is concentrated by washing with current of water in a hydraulic classifier. Lighter gangue is carried away by water while the heavier chromite ore settles at the bottom thus resulting in its separation. Step II: Conversion of chromite ore into sodium chromate (Roasting): Concentrated ore is mixed with soda ash (sodium carbonate) and lime stone (calcium carbonate) and roasted with excess of air in reverberatory furnace. The mass turns yellow because of the formation of sodium chromate. Lime stone is added to keep the mass porous so that the oxidation by air becomes easy. 4FeO.Cr2O3 + O2 ⎯→ 2Fe2O3 + 4Cr2O3 [2Cr2O3 + 4Na2CO3 + 3O2 ⎯→ 4Na2CrO4 + 4CO2] × 2 ______________________________________________ 4FeO.Cr2O3 + 8Na2CO3 + 7O2 ⎯→ 2Fe2O3 + 8Na2CrO4 + 8CO2

Roasted mass is extracted with water. Sodium chromate goes into solution and insoluble substances are separated by filtration. Step III: Conversion of sodium chromate into sodium dichromate: The filtrate containing sodium chromate solution is treated with concentrated sulphuric acid during which sodium chromate is converted into sodium dichromate. 2Na2CrO4 + H2SO4 ⎯→ Na2Cr2O7 + Na2SO4 + H2O Sodium

(conc.)

chromate

Sodium dichromate

Sodium sulphate being less soluble crystallizes out as decahydrate, Na2SO4.10H2O and is removed. The crystals are separated by filtration and concentrated solution of sodium dichromate is obtained. Step IV: Conversion of sodium dichromate into potassium dichromate: Hot concentrated solution of sodium dichromate is treated with calculated amount of potassium chloride. Double decomposition takes place forming potassium dichromate and sodium chloride. Na2Cr2O7 + 2KCl ⎯→ K2Cr2O7 + 2NaCl Sodium

Potassium

dichromate

dichromate

The solution is concentrated and the hot solution is cooled. The less soluble potassium dichromate separates as orange red crystals which is separated by filtration and purified by recrystallization, while sodium chloride remains in the solution. Q.66. State properties of K2Cr2O7. Ans: Properties of K2Cr2O7 : i. Physical properties: It is orange colored crystalline solid, soluble in water (readily soluble in hot water) and melts at 669 K. ii. Action of alkali: When potassium hydroxide solution is added to orange red coloured solution of potassium dichromate, a yellow coloured solution of potassium chromate is obtained. H2O K2Cr2O7 + 2KOH ⎯→ 2K2CrO4 +

16

Potassium

Potassium

dichromate

chromate



TARGET Publications

iii.

Oxidizing properties: K2Cr2O7 is a powerful oxidising agent in acidic medium where dichromate is reduced to chromium sulphate. Its oxidizing action can be represented as Cr2 O72 − + 14 H+ + 6e−⎯→ 2Cr3+ + 7H2O a. Action on ferrous sulphate: In acidic medium potassium dichromate oxidizes ferrous sulphate to ferric sulphate, potassium dichromate is reduced to chromic sulphate. K2Cr2O7 + 4H2SO4 ⎯→ K2SO4 + Cr2(SO4)3 + 4H2O + 3(O) [2FeSO4 + (O) + H2SO4 ⎯→ Fe2(SO4)3 + H2O] × 3

K2Cr2O7 + 6FeSO4 + 7H2SO4 ⎯→ K2SO4 + Cr2(SO4)3 + 3Fe2(SO4)3 + 7H2O b.

c.

d.

e.

Action on potassium iodide: Acidified potassium dichromate oxidizes potassium iodide to iodine. K2Cr2O7 + 6KI + 7H2SO4 ⎯→ 4K2SO4 + Cr2(SO4)3 + 7H2O + 3I2 liberated Iodine turns the solution brown. Action on hydrogen sulphide: It oxidizes H2S to a pale yellow precipitate of sulphur. Potassium dichromate is reduced to chromic sulphate and the orange colour of the solution turns green. K2Cr2O7 + 4H2SO4 + 3H2S ⎯→ K2SO4 + Cr2(SO4)3 + 7H2O + 3S Action on sulphur dioxide: It oxidizes sulphur dioxide to sulphuric acid. Potassium dichromate is reduced to chromic sulphate and the orange colour of the solution turns green. K2Cr2O7 + 3SO2 + H2SO4 ⎯→ K2SO4 + Cr2(SO4)3 + H2O Action on alcohols: Primary and secondary alcohols are oxidized to aldehyde and ketones respectively. Aldehydes are further oxidized to carboxylic acids. CH3CH2OH + (O) ⎯→ CH3CHO + H2O Ethyl alcohol

Acetaldehyde

CH3CHO + (O) ⎯→ CH3COOH Acetaldehyde

Acetic acid

CH3−CHOH − CH3 + (O) ⎯→ CH3− CO − CH3 + H2O Isopropyl alcohol

iv.

Acetone

Formation of chromyl chloride (CrO2Cl2): When a mixture of potassium dichromate crystals and sodium chloride (or KCl) is heated with concentrated sulphuric acid, red vapours of chromyl chloride is evolved. On condensing, it gives oily red liquid. K2Cr2O7 + 6H2SO4 + 4NaCl ⎯→ 2KHSO4 + 4NaHSO4 + 2CrO2Cl2 + 3H2O

Q.67. Describe the oxidizing action of potassium dichromate and write the ionic equations for its reaction with: (NCERT) a. iodide b. iron (II) solution c. H2S. Ans: Potassium dichromate, K2Cr2O7 is a strong oxidizing agent and is used as a primary standard in volumetric analysis involving oxidation of iodides, ferrous ion and S2- ions etc. a. It oxidizes potassium iodide to iodine. K2Cr2O7 + 6KI + 7H2SO4 ⎯→ 4K2SO4 + Cr(SO4)3 + 7H2O + 3I2. b. It oxidizes iron (II) salt to iron (III) salt K2Cr2O7 + 6FeSO4 +7H2SO4 ⎯→ K2SO4 + Cr2(SO4)3 + 3Fe(SO4)3 + 7H2O c. It oxidizes H2S to S K2Cr2O7 + 4H2SO4 + 3H2S ⎯→ K2SO4 + Cr(SO4)3 + 7H2O + 3S d and f Block Elements

17


TARGET Publications

Q.68. Explain structure of chromate ion and dichromate ion. Ans: Structure of chromate and dichromate ions 2− O O O O

Cr O

O O Chromate ion

O

Cr

126°

Cr

O

2−

O O

Dichromate ion

The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedral sharing one corner with the Cr−O−Cr bond angle of 126°. Q.69. State the uses of potassium dichromate. Ans: Uses of K2Cr2O7: K2Cr2O7 is used i. in volumetric analysis, as a primary standard for the estimation of Fe2+ (ferrous ions) and I− (iodides) in redox titrations. ii. in the manufacture of chromium compounds such as lead chromate and chrome alum. iii. in calico printing and dyeing. iv in photography and in hardening gelatine film. v. as an oxidising agent. vi. in the detection of chloride ion by chromyl chloride test. vii. in the manufacture of pigments and in the manufacture of inks. Q.70. Indicate the steps involved in the preparation of KMnO4 from pyrolusite ore (MnO2) (NCERT) OR How is potassium permanganate prepared from pyrolusite ore? Ans: Preparation of KMnO4 from pyrolusite ore (MnO2): Step I: Conversion of MnO2 into potassium manganate. The finely powdered pyrolusite mineral is fused with potassium hydroxide or potassium carbonate in the presence of air or oxidising agent such as potassium nitrate or potassium chlorate giving green coloured potassium manganate. 2MnO2 + 4KOH + O2 ⎯→ 2K2MnO4 + 2H2O Potassium manganate

2MnO2 + 2K2CO3 + O2 ⎯→ 2K2MnO4 + 2CO2 MnO2 + 2KOH + KNO3 ⎯→ K2MnO4 + KNO2 + H2O 3MnO2 + 6KOH + KClO3 ⎯→ 3K2MnO4 + KCl + 3H2O The fused mass obtained containing K2MnO4 is treated with water and then converted into KMnO4 either by oxidation or by electrolysis. Step II: Oxidation of potassium manganate to potassium permanganate. There are two methods for oxidation of potassium manganate, a. Chemical oxidation: Oxidation of K2MnO4 to KMnO4, disproportionation is carried out by H2SO4, by passing CO2 or Cl2 or O3 through the solution. 3K2MnO4 + 2H2SO4 ⎯→ 2K2SO4 + 2KMnO4 + 2H2O + MnO2 3K2MnO4 + 4CO2 + 2H2O ⎯→ 2KMnO4 + MnO2 + 4KHCO3 2K2MnO4 + Cl2 ⎯→ 2KMnO4 + 2KCl 2K2MnO4 + H2O + O3 ⎯→ 2KMnO4 + 2KOH + O2 The carbon dioxide process is uneconomical as one third of the original manganate is reconverted to manganese dioxide. However, this process has the advantage that the potassium carbonate formed as a byproduct can be used for the oxidative fusion of manganese dioxide. In the chlorine process, potassium chloride obtained as a byproduct is lost. 18



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b.

Electrolytic oxidation: For manufacturing potassium permanganate commercially, the method of electrolytic oxidation is preferred. The alkaline manganate solution obtained in step (I) is electrolysed between iron electrodes separated by diaphragm. The reactions taking place are as follows: ZZX 2K+ + MnO 24 − K2MnO4 YZZ

ZZX H+ + OH− H2O YZZ −

MnO 24 ⎯→ MnO −4 + e− At anode: At cathode: 2H+ + 2e− ⎯→ H2 Thus, manganate ions are oxidized to permanganate at the anode and hydrogen gas is liberated at the cathode. 2K2MnO4 + H2O + O ⎯→ K2MnO4 + 2KOH After the oxidation is complete, the solution is filtered and evaporated under controlled conditions to obtain the deep purple black crystals of potassium permanganate. Note: Laboratory Preparation: In the laboratory, potassium permanganate is prepared by oxidation of manganese (II) ion salt by peroxodisulphate. − − 5S2 O82 + 8H2O ⎯→ 2MnO −4 + 10SO 24 + 16H+ 2Mn2+ + Peroxodisulphate ion

Permanganate ion

Q.71. State the properties of KMnO4. Ans: Properties of KMnO4: i. Physical properties: It is a deep purple crystalline solid, moderately soluble in water at room temperature and it is more soluble in hot water. ii. KMnO4 when heated to 473 K, readily decomposes giving oxygen. heat 2KMnO4 ⎯⎯⎯ → K2MnO4 + MnO2 + O2 At red heat, potassium permanganate decomposes into potassium manganate (K2MnO3) and oxygen. Re d heat 2K2MnO4 ⎯⎯⎯⎯ → 2K2MnO3 + O2 iii. With well cooled conc H2SO4, KMnO4 gives Mn2O7, which on warming decomposes to MnO2. 2KMnO4 + 2H2SO4 ⎯→ Mn2O7 + 2KHSO4 + H2O 2Mn2O7 ⎯→ 4MnO2 + 3O2 With warm conc. H2SO4, O2 gas is evolved. 4KMnO4+ 6H2SO4 ⎯→ 2K2SO4 + 4MnSO4 + 6H2O + 5O2 iv. Solid KMnO4 when heated in current of H2, gives KOH, MnO and water vapours. ∆ → 2KOH + 2MnO + 4H2O 2KMnO4 + 5H2 ⎯⎯ v. KMnO4 is a powerful oxidising agent. a. In neutral medium: 2KMnO4 + H2O ⎯→ 2KOH + 2MnO2 + 3[O] MnO4− + 2H2O + 3e− ⎯→ MnO2 + 4OH− Manganous salt is oxidised to MnO2; 2KMnO4 + 3MnSO4 + 2H2O ⎯→ 5MnO2 + K2SO4 + 2H2SO4 b. In alkaline medium: 2KMnO4 + 2KOH ⎯→ 2K2MnO4 + H2O + [O] or MnO4− + e− ⎯→ MnO42− In the presence of a reducing agent, K2MnO4 is further reduced to manganese dioxide. K2MnO4 + H2O ⎯→ MnO2 + 2KOH + O or MnO42− + 2H2O + 2e− ⎯→ MnO2 + 4OH− Hence the complete equation can be written as: MnO4− + 2H2O + 3e− ⎯→ MnO2 + 4OH− 19 d and f Block Elements

Std. XII Sci.: Perfect Chemistry - II c.

TARGET Publications

In acidic medium: In the presence of dil. H2SO4, it acts as oxidizing agent because of the reaction 2KMnO4 + 3H2SO4 ⎯→ K2SO4 + 2MnSO4 + 3H2O + 5[O] or MnO4− + 8H+ + 5e− ⎯→ Mn2+ + 4H2O

Q.72. How does acidified permanganate solution react with iii. oxalic acid? (NCERT) i. iron (II) ions ii. SO2 Ans: i. Acidified permanganate solution oxidizes iron (II) salt to iron (III) salts. 2MnO42− + 16H+ + 10Fe2+ ⎯→ 2Mn2+ + 8H2O + 10Fe3+ ii. It oxidizes sulphur dioxide to sulphuric acid. 2MnO4− + 5SO2 + 2H2O ⎯→ 5SO42− + 2Mn2+ + 4H+ iii. It oxidizes oxalic acid to CO2 and H2O. 2MnO4− + 16H+ + 5C2O42− ⎯→ 2Mn2+ + 8H2O + 10CO2 Q.73. Explain the structures of manganate and permanganate ion. Ans: Structures of manganate and permanganate ion: The manganate and permanganate ions have tetrahedral structures. The manganate ion is green in colour and is paramagnetic with one unpaired electron while the permanganate ion is purple in colour and is diamagnetic. The π-bonding takes place by overlap of p-orbitals of oxygen with d-orbitals of manganese. O− O

Mn O

Mn −

O O MnO42− Tetrahedral Manganate (green) ion

O

O−

O MnO4− Tetrahedral permanganate (purple) ion

Q.74. What is meant by disproportionation? Give two examples of disproportionation reaction in aqueous solution. (NCERT) Ans: Sometimes transition metal species in same oxidation state undergo a chemical change in such a way that some species get oxidized and some other get reduced. Eg. Cr (V) and Mn (VI) species undergo disproportionation reaction in acidic medium as follows: 3CrO43− + 8H+ ⎯→ 2CrO42− + Cr3+ + 4H2O (Cr in +5 o.s.) (Cr in +6 o.s.) (Cr in +3 o.s.) 2MnO4− + MnO2 + 2H2O 3MnO42− + 4H+ ⎯→ (Mn in +6 o.s.) (Mn in +7 o.s.) (Mn in + 4 o.s.) Cr in +5 oxidation state undergo disproportionation into its +6 and +3 states. Similarly Mn in +6 oxidation state undergoes disproportionation into +7 and +4 oxidation states. Q.75. State the uses of potassium permanganate. Ans: Uses of KMnO4: KMnO4 is used in volumetric analysis for the estimation of ferrous salts, oxalates, iodides and hydrogen peroxide. i. ii. as a strong oxidising agent in the laboratory as well as an effective oxidising agent in organic synthesis. Alkaline potassium permanganate is used for testing unsaturation in organic compounds and is known as Baeyer’s reagent. iii. as a disinfectant and germicide. A very dilute solution of permanganate is used for washing wounds and gargling for mouth sore. It is also used for purifying water of stinking wells. iv. for bleaching of wool, cotton, silk and other textile fibres because of its strong oxidizing power and also for decolourisation of oils. *Q.76. Write the preparation, properties and uses of potassium permanganate. Ans: Refer Q.70, 71 and 75 20



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f-Block elements 8.5

General Introduction and Electronic Configuration

*Q.77. What are f-block elements? Ans: i. The elements in which the last electron enters into (n−2) f-orbital of the atoms are called f-block elements. ii. In these elements, the last electron enters in to the prepenultimate (n−2) shell (called antepenultimate). iii. The general electronic configuration is: (n–2)f 1−14(n–1)d0−1 ns2. *Q.78. Explain the meaning of inner transition series. OR What are inner-transition elements? The last electron in the f-block elements enters into (n−2) f-orbitals, i.e. inner to the penultimate Ans: i. energy level and they form a transition series within the transition series (d-block elements). Hence the f-block elements are known as inner transition series. ii. There are two series of inner transition elements: a. Lanthanoids (58-71) and b. Actinoids (90-103). Q.79. What are rare earth elements? Ans: Rare earth elements are a collection of seventeen chemical elements in the periodic table, namely scandium, yttrium and the fifteen lanthanoids (4f-series). They are originally isolated as oxides and are referred as ‘earths’ and ‘rare’ because of their isolation from relatively rare minerals. Q.80. Write the general electronic configuration of 4f and 5f series elements. [March 2013] Ans: General electronic configuration 4f series −[Xe]4f1 − 145d0 − 16s2 5f series −[Rn]5f1 − 146d0 − 17s2 8.6

Lanthanoids

Q.81. Define lanthanoids. Ans: The series involving the filling of 4f-orbitals following lanthanum La (Z = 57) is called lanthanoid series. The elements present in this series are called lanthanoids. *Q.82. Explain the position of lanthanoids in the periodic table. Ans: Position of lanthanoids in the periodic table: i. The 14 elements (from 58 to 71) of lanthanoid series have been placed along with lanthanum (at no. 57) in the IIIB group (3rd column) and sixth period of the periodic table. ii. In the periodic table, as we move from one element to other, either from left to right or from top to bottom, the properties exhibit a gradual change. But these fifteen are so similar to one another, that they cannot be placed one after the other or one below the other. iii. As the fourteen elements i.e. Ce(58) to Lu(71) are closely similar to La(57), the best place for them is along with Lanthanum (La) i.e. IIIB group (3rd column) and sixth period in the periodic table. iv. In case, these elements are given different position in order of their increasing atomic numbers, the symmetry of the periodic table would be disrupted. v. Due to this reason, the lanthanoids are placed at the bottom of the periodic table with a reference to the IIIB group in the sixth period. d and f Block Elements

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IA IIA 1 2 Transition Elements

s-block elements

1 2 PERIODS

3 4 5

IIIB

IVB

VB VIB VIIB

3

4

5

6

8

9

10

IB

IIB

11

12

IVA

VA

VIIIA VIA VIIA 18

13

14

15

16

17

p-block elements d-block elements

La *

6

7

VIIIB

IIIA

Ac * *

7

* Lanthanoids ** Actinoids

58 Ce 90 Th

59 Pr 91 Pa

60 Nd 92 U

61 Pm 93 Np

INNER TRANSITION ELEMENTS f-block elements 62 68 69 63 64 65 66 67 Sm Eu Gd Tb Dy Ho Er Tm 94 95 96 97 98 99 100 101 Pu Am Cm Bk Cf Es Fm Md

70 Yb 102 No

71 Lu 103 Lr

Q.83. Briefly explain why electronic configurations of lanthanoids are not known with certainty. Ans: In the lanthanoids, 4f and 5d subshells are very close in energy. The outermost 6s-orbital remains filled with two electrons (6s2). The electrons can easily jump from 4f to 5d or vice-versa. Further, irregularities in electronic configurations are also related to the stabilities of f0, f7 and f14 occupancy of f-orbitals. Hence, their electronic configurations are not known with certainty. Q.84. State the general electronic configuration of 4f-series (lanthanoids). Ans: Electronic configuration of 4f-series: The elements of 4f-series begin from La to Lu. The electronic configuration of these elements can be expressed in terms of its nearest inert gas Xe (54). Electronic configuration of Xe (54) = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 Therefore, general electronic configuration of 4f-series is [Xe] 4f1-14 5d0-1 6s2. Atomic Expected electronic Observed electronic Element Symbol Number configuration configuration Lanthanum La 57 [Xe] 4f0 5d1 6s2 [Xe] 4f0 5d1 6s2 Cerium Ce 58 [Xe] 4f1 5d1 6s2 [Xe] 4f2 5d0 6s2 2 1 2 Praseodymium Pr 59 [Xe] 4f 5d 6s [Xe] 4f3 5d0 6s2 Neodymium Nd 60 [Xe] 4f3 5d1 6s2 [Xe] 4f4 5d0 6s2 4 1 2 Promethium Pm 61 [Xe] 4f 5d 6s [Xe] 4f5 5d0 6s2 5 1 2 Samarium Sm 62 [Xe] 4f 5d 6s [Xe] 4f6 5d0 6s2 6 1 2 Europium Eu 63 [Xe] 4f 5d 6s [Xe] 4f7 5d0 6s2 Gadolinium Gd 64 [Xe] 4f7 5d1 6s2 [Xe] 4f7 5d1 6s2 8 1 2 Terbium Tb 65 [Xe] 4f 5d 6s [Xe] 4f9 5d0 6s2 9 1 2 Dysprosium Dy 66 [Xe] 4f 5d 6s [Xe] 4f10 5d0 6s2 Holmium Ho 67 [Xe] 4f10 5d1 6s2 [Xe] 4f11 5d0 6s2 11 1 2 Erbium Er 68 [Xe] 4f 5d 6s [Xe] 4f12 5d0 6s2 12 1 2 Thulium Tm 69 [Xe] 4f 5d 6s [Xe] 4f13 5d0 6s2 13 1 2 Ytterbium Yb 70 [Xe] 4f 5d 6s [Xe] 4f14 5d0 6s2 Lutetium Lu 71 [Xe] 4f14 5d1 6s2 [Xe] 4f14 5d1 6s2 Mnemonic: Lazy College Professors Never Produce Superior Excellent Graduates To Dramatically Help Executives Trim Yearly Losses. 22


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Q.85. Write down the electronic configuration of i. Pm3+ ii. Ce4+ iii. Lu2+ (NCERT) 3+ 2 2 6 2 6 10 2 6 10 4 2 Ans: i. Pm = 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p6.

ii.

Ce4+ = 1s22s22p63s23p63d104s24p64d105s25p6.

iii.

Lu2+ = 1s22s22p63s23p63d104s24p64d104f145s25p65d1.

Q.86. Why lanthanum, gadolinium and lutetium show different electronic configurations? Ans: i. It is observed that lanthanum, gadolinium and lutetium show different electronic configuration because the 5d and 4f-orbitals are nearly of the same energy and the distinction between the two is difficult. ii. Due to this, some extra stability is achieved when the 4f is half-filled and completely filled so the next electron goes in 5d-orbital instead of 4f-orbital. Q.87. State the features of electronic configuration for lanthanoids. Ans: Features of electronic configuration of lanthanoids: i. Electronic configuration of Ba (Z = 56) is [Xe] 6s2. ii. Lanthanum has electronic configuration [Xe] 4f0 5d1 6s2. It does not have any 4f electrons. iii. The next incoming electron after lanthanum does not enter into 5d-subshell but enters into 4fsubshell. iv. As atomic number increases successively by one unit from La to Lu, the fourteen electrons are filled in the 4f-subshell progressively. v. Elements La, Gd and Lu possess single electron in 5d-subshell. In case of other lanthanoids 5d-orbital is empty. vi. In idealised electronic configuration filling up of 4f-orbitals is regular. In observed electronic configuration filling of 4f-orbitals is not regular. Observed configuration involves moving of the single electron from 5d to 4f sub-shell. vii. f0, f7, f14 electronic configurations achieve extra stability due to empty, half filled and completely filled f-orbitals respectively. Eg. La(4f0), Gd (4f7) and Lu (4f14). viii. The 4f electrons in pre-penultimate i.e. (n−2) shell are totally shielded by higher outer orbitals. Q.88. Explain the trends in atomic and ionic sizes of lanthanoids Ans: Trends in atomic and ionic sizes of lanthanoids: i. In the lanthanoid series, with increasing atomic number, the atomic and ionic radii decrease from one element to another but the decrease is very small. ii. For example, on moving from Ce to Lu, the atomic radius decreases from 183 pm to 173 pm and the decrease is only 10 pm. iii. Similarly, the ionic radius decreases from 103 pm to 85 pm on moving from Ce3+ to Lu3+ ions and the decrease is only 18 pm. iv. Thus, for an increase of atomic number 14, the decrease in atomic radii or ionic radii are very small i.e., only 10 pm and 18 pm respectively in comparison to elements of other groups and periods,. v. The steady decrease in atomic and ionic sizes of lanthanoid elements with increasing atomic number is called lanthanoid contraction. *Q.89. Explain, the trends in atomic and ionic sizes of lanthanoids. Ans: Refer Q.88. *Q.90. Explain the oxidation states of lanthanoids. Ans: i. All lanthanoids exhibit a common stable oxidation state of +3. ii. In addition, some lanthanoids show +2 and +4 oxidation states also. iii. These are shown by those elements which by doing so attain the stable f0, f7 or f14 configurations i.e. empty, half-filled and completely filled 4f sub-shells. Eg. Ce and Tb exhibit +4 oxidation states, Eu and Yb exhibit +2 oxidation states, La, Gd and Lu exhibit only +3 oxidation states. d and f Block Elements

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Oxidation states of Lanthanoids are as shown in following table: Element

Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thullium Ytterbium Lutetium

Outer electronic configuration

4f0 5d1 6s2 4f2 5d0 6s2 4f3 5d0 6s2 4f4 5d0 6s2 4f5 5d0 6s2 4f6 5d0 6s2 4f7 5d0 6s2 4f7 5d1 6s2 4f9 5d0 6s2 4f10 5d0 6s2 4f11 5d0 6s2 4f12 5d0 6s2 4f13 5d0 6s2 4f14 5d0 6s2 4f14 5d1 6s2

Oxidation states M2+ M3+ M4+ 3+ 3+ 4+ 3+ 4+ 2+ 3+ 4+ 3+ 2+ 3+ 2+ 3+ 3+ 3+ 4+ 3+ 4+ 3+ 3+ 2+ 3+ 2+ 3+ 3+ -

Q.91. Explain why Eu and Yb show oxidation state +2. Ans: i. In the +2 oxidation state, Eu donates two electrons from its 6s-orbitals making 4f-orbitals half-filled i.e. 4f7 which is a more stable state than partially filled. ii. Similarly in the +2 oxidation state, Yb donates two electrons from its 6s-orbitals making 4f-orbitals completely filled i.e. 4f14 which is more stable than partially filled state. Therefore Eu and Yb show +2 oxidation states. Q.92. La, Gd and Lu show only +3 oxidation state. Explain. The electronic configuration of 57La, 64Gd and 71Lu in +3 oxidation state are as follows: Ans: i. La+3 : [Xe] 4f0 Gd+3 : [Xe] 4f7 Lu+3 : [Xe] 4f14. ii. f-orbital acquires extra-stability when it is half filled or completely filled. Hence, La, Gd and Lu exhibit only +3 oxidation state. Q.93. Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configuration of these elements. (NCERT) Ans: + 4 = Ce, Pr, Nd, Tb, Dy

+ 2 = Nd, Sm, Eu, Tm,Yb + 2 oxidation state is exhibited when the lanthanoid has the configuration 5d0 6s2 so that 2 electrons are easily lost. + 4 oxidation state is exhibited when the configuration left is close to 4f0 (Eg. 4f0, 4f1, 4f2) or close to 4f7 (Eg. 4f7 or 4f8) *Q.94. Why do lanthanoids form coloured compounds? Ans: Unpaired electrons are present in the outermost 4f subshell of lanthanoid ions. Therefore, they can undergo f-f transitions and hence form coloured compounds. 24



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*Q.95. Explain why Gd3+ is colourless. Gd3+ has electronic configuration [Xe] 4f7. Ans: i. ii. Lanthanoids show colour due to partially filled f orbitals which allow f-f transitions. iii. Gd3+ ion has exactly half filled electronic electronic configuration. So, electrons for f-f transition are absent due to which it is colourless. Q.96. Explain the chemical reactivity of lanthanoids. Ans: i. Carbides: When lanthanoids are heated around 2500° C with carbon, they form carbides having formulae Ln3C, LnC2 and Ln2C3 D

2500 C Ln + C ⎯⎯⎯⎯ → Lanthanoid carbide

ii.

Hydrides: The metals combine with hydrogen when gently heated in the hydrogen gas. ∆

→ 2LnH3 4Ln + 3H2 ⎯⎯ iii.

Oxides: When burnt in oxygen, they forms oxides, with formula Ln2O3. ∆

→ 2Ln2O3 2Ln + 3O2 ⎯⎯ The oxide Ln2O3 reacts with water to form insoluble hydroxides. Ln2O3 + 3H2O ⎯→ 2Ln(OH)3 and with CO2 they form carbonates. Ln2O3 + 3CO2 ⎯→ Ln2(CO3)3 iv.

Reaction with nitrogen: On heating with nitrogen they form nitrides. ∆

→ 2LnN 2Ln + N2 ⎯⎯ v.

Reaction with mineral acids: Lanthanoids when treated with mineral acids liberate H2 gas as they have reduction potential of −2.0 to −2.4V 2Ln + 6HCl ⎯→ 2LnCl3 + 3H2

vi.

Reaction with water: When treated with water they form consequently hydroxides and liberates H2 gas. The hydroxides are ionic and basic in nature. Ln + 3H2O ⎯→ Ln(OH)3 + 3H2

vii.

Reaction with sulphur: Heating lanthanoids with sulphur forms corresponding sulphides 2Ln + 3S ⎯→ Ln2S3

Q.97. Write a short note on lanthanoid contraction. Ans: Lanthanoid contraction: i. The atomic and ionic radii of lanthanoids show gradual decrease with increase in atomic number. It is known as Lanthanoid contraction. ii. In Lanthanoids, after Lanthanum (La), the electrons are added to prepenultimate shell i.e. 4f-orbital. iii. There are 14 Lanthanoids from Ce to Lu. iv. For each electron, one proton is also added to the nucleus of the atom of the element. Hence from Ce to Lu as atomic number increases, nuclear charge increases, therefore nuclear attraction increases. v. As atomic number increases, atomic volume or radius decreases as observed with all the elements along the period. vi. But in case of Lanthanoids this decrease in atomic volume or radius is comparatively very small. This is explained in terms of Lanthanoid contraction. d and f Block Elements

25


TARGET Publications

Ionic radius (in pm)

Q.98. Explain the causes of the lanthanoid contraction. Ans: i. As the atomic number of the members of lanthanoids 110 series increases, the positive charge on nucleus La 3++3 increases by +1 unit and one more electron enters in Ce 3+ 105 Pr 3+ the same 4f subshell. Nd 3+ Pm 3+ ii. There is inadequate shielding of one electron by 100 Sm 3+ Eu 3+ another electron in the same 4f subshell. Gd 3+ 95 Tb 3+ iii. The extent of shielding for electrons is less in 4f Dy 3+ Ho 3+ subshell as compared to electrons in d subshell, 4f 90 Er 3+ Tm 3+ valence shell have diffused shapes. Due to which Yb 85 Lu 3+ with increase of nuclear charge the valence shell is pulled slightly towards nucleus. iv. Because of this pull, the size of M3+ ions goes on 57 59 61 63 65 67 69 71 decreasing with increase in atomic number. However Atomic number ⎯→ decrease in size is very small. v. In the complete f series only 10 pm decrease in atomic radii and 18 pm decrease in ionic radii is observed. vi. The graph indicates lathanoid contraction of various lanthanoids. vii. Though the contraction in size from one element to another is very small, the net contraction over the fourteen elements from Ce to Lu is appreciable. viii. Atomic radii show some irregularities but ionic radii decrease steadily from La to Lu. Q.99. Explain the effects of lanthanoid contraction. Ans: i. Decrease in Basicity: a. Because of lanthanoid contraction size of lanthanoid ions decreases regularly with increase in atomic number. As a result of decrease in size, the covalent character between lanthanoids ion and OH− ions increases from La3+ to Lu3+. b. The ionic character of M−OH bond decreases and covalent character of M−OH bond gradually increases.Therefore, the basic strength of the hydroxides decreases with increase in atomic number. Thus, Lu(OH)3 is least basic while La(OH)3 is most basic. c. Basicity and Ionic character of hydroxides decreases in the order of La(OH)3 > Ce(OH)3 >….. Lu(OH)3. ii. Ionic radii of post lanthanoids : The elements which follow the lanthanoids in the third transition series are known as postlanthanoids. There is a regular increase in size from Sc to Y to La. Similarly there will be increase in size in other groups as: Ti ⎯→ Zr ⎯→ Hf V ⎯→ Nb ⎯→ Ta But after the lanthanoids the increase in radii from second to third transition series almost vanishes. Pairs of elements such as Zr–Hf (group 4), Nb–Ta (group 5), Mo–W (group 6) and Tc-Re (group 7) possess almost same size. These pair of elements are called ‘chemical twins’. The properties of these elements are also similar. So due to lanthanoid contraction elements of second and third series resemble each other. iii. Similarity among lanthanoids: Lanthanoids show very small change in radii so their chemical properties are quite similar. Thus it is very difficult to separate the elements in pure state. *Q.100. What are chemical twins elements? Give their examples. Ans: A pair of elements having similar properties due to similar atomic radii and almost same size are called chemical twin elements. This effect arises due to lanthanoid contraction. Eg: Zr and Hf; Nb and Ta; Mo and W; Tc and Re are a pair of chemical twins elements. 26



TARGET Publications

Q.101. Write a short note on uses of lanthanoids. Ans: Uses of lanthanoids: i. Lanthanoids do not find any use in the pure state. The most important use of lanthanoids is in the production of alloy steels to improve the strength and workability of steel. ii. Their oxides (Eg. La2O3) are used in glass industry, for polishing glass and for making coloured glasses for goggles as they give protection against UV light and as phosphor for television screens and similar fluorescing surfaces. Mixed oxides of lanthanoids are used as catalysts in petroleum cracking. iii. Because of their paramagnetic and ferromagnetic properties, their compounds are used in making magnetic and electronic devices. iv. Ceric sulphate is a well known oxidizing agent in volumetric analysis. Q.102. What is Misch metal? Give its one use. Ans: Misch metal is an alloy of a lanthanoid metal and iron and traces of S, Ca, C or Al. It is used in making bullets and lighter flint. 8.7

Actinoids

*Q.103. What are actinoids? Ans: The series of elements from Thorium (Z = 90) to Lawrencium (Z = 103) in which 5f orbitals are progressively filled are called actinoids. The elements are called actinoid because many physical and chemical properties are similar to actinium. Note: Actinium (Z = 89) has electronic configuration [Rn] 5f° 6d1 7s2. There is no electron in 5f orbital. *Q.104. What are trans-uranic elements? Write their names. [March 2013 old course] Ans: The man made elements having atomic number higher than Uranium-92 are collectively called as transuranic elements. Elements from Rf(Z = 104) to Uub (Z = 112) have been identified. Elements upto atomic number 118 are also synthesized as shown in the table given below: Sr. No 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

Name

Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium Rutherfordium Dubnium

Symbol

Np Pu Am Cm Bk Cf Es Fm Md No Lr Rf Db

Atomic Number 93 94 95 96 97 98 99 100 101 102 103 104 105

Sr. No 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.

Name

Seaborgium Bohrium Hassium Meitnerium Darmstadtium Rontgenium/Unununium Copernicium/Ununbium Ununtrium Ununquadium Ununpentium Ununhexium Ununseptium Ununoctium

Symbol

Sg Bh Hs Mt Ds Rg/Uuu Cn/Uub Uut Uuq Uup Uuh Uus Uuo

Atomic number 106 107 108 109 110 111 112 113 114 115 116 117 118

*Q.105. Explain the position of actinoids in the periodic table. Ans: Position of actinoids in the periodic table. i. Actinoids belongs to the IIIB group of periodic table in the seventh period. ii. In the periodic table, as we move from one element to another, either from left to right or from top to bottom, the properties exhibit a gradual change. But these fifteen elements are so similar to one another, that they cannot be placed one after the other or one below the other. iii. As the fourteen elements i.e. Th(90) to Lr(103) are closely similar to Ac(89), the best place for them is along with actinium(89) i.e. IIIB group (3rd column) and seventh period in the periodic table. 27 d and f Block Elements

Std. XII Sci.: Perfect Chemistry - II iv.

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In case these elements are give different positions in order of their increasing atomic numbers, the symmetry of the periodic table would be disrupted. Due to this reason, the actinoids are placed at the bottom of the periodic table with a reference to the IIIB group in the seventh period i.e the position of actinium. For position refer figure shown in Q. 82

Q.106. Explain the electronic configuration of actinoids. Ans: Electronic configuration of actinoids: The elements of 4f-series begin from Ac to Lr. The electronic configuration of these elements can be expressed in terms of Rn (86). Electronic configuration of Rn (86) = 1s22s22p63s23p63d104s24p64d105s25p65d105f146s26p6.

Therefore, General electronic configuration of actinoids : [Rn] 5f1−14 6d0 − 1 7s2 Element

Symbol

Atomic Number

Actinium Thorium Protactinium Uranium Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium

Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

89 90 91 92 93 94 95 96 97 98 99 100 101 102 103

Electronic configuration Expected [Rn] 5f0 6d1 7s2 [Rn] 5f1 6d1 7s2 [Rn] 5f2 6d1 7s2 [Rn] 5f3 6d1 7s2 [Rn] 5f4 6d1 7s2 [Rn] 5f5 6d1 7s2 [Rn] 5f7 6d0 7s2 [Rn] 5f7 6d1 7s2 [Rn] 5f8 6d1 7s2 [Rn] 5f9 6d1 7s2 [Rn] 5f10 6d1 7s2 [Rn] 5f11 6d1 7s2 [Rn] 5f12 6d1 7s2 [Rn] 5f14 6d0 7s2 [Rn] 5f14 6d1 7s2

Electronic configuration Observed [Rn] 5f0 6d1 7s2 [Rn] 5f0 6d2 7s2 [Rn] 5f2 6d1 7s2 [Rn] 5f3 6d1 7s2 [Rn] 5f4 6d1 7s2 [Rn] 5f6 6d0 7s2 [Rn] 5f7 6d0 7s2 [Rn] 5f7 6d1 7s2 [Rn] 5f9 6d0 7s2 [Rn] 5f10 6d0 7s2 [Rn] 5f11 6d0 7s2 [Rn] 5f12 6d0 7s2 [Rn] 5f13 6d0 7s2 [Rn] 5f14 6d0 7s2 [Rn] 5f15 6d1 7s2

Q.107. Write down the electronic configuration of Th4+. (NCERT) Ans: Electronic configuration of Th4+ = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 5f1 6s2 6p6 6d1 7s2. Q.108. The electronic configurations of actinoid elements are not known with certainty. Explain. Ans: In actinoids, 5f and 6d subshells are close in energy. The outermost 7s orbital remains filled with 2 electrons (7s2). The electron can easily jump from 5f to 6d or vice versa. Further, irregularities in electronic configurations are also related to the stabilities of f0, f7 and f14 occupancy of the 5f-orbitals. Hence, they show a large number of oxidation states (Moreover, they are radioactive with short half-lives. Hence, their properties cannot be studied easily). *Q.109. Explain the oxidation states of actinoids OR Write a short note on oxidation states of actinoids. Ans: i. The dominant oxidation state of actinoid elements is +3. Besides +3, actinoids also exhibit +4 oxidation state. ii. Some actinoids show still higher oxidation states. iii. The maximum oxidation state first increases upto middle of the series and then decreases. 28



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Eg.

Ac +2 +3 +4

Oxidation state increases from +4 for the Th to +5, +6 and +7 for Pa, U and Np but decreases in the succeeding elements. Oxidation states of actinium and actinoids Th +2 +3 +4 +5

Pa +3

U +3

Np +3

Pu +3

+4 +5

+4 +5 +6

+4 +5 +6 +7

+4 +5 +6 +7

Am +2, +3 +4 +5 +6

Cm +3

Bk +3

+4

+4

Cf +3

Es +3

Fm +3

Md +3

No Lr +3 +3

Q.110. The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements. (NCERT) Ans: Lanthanoids show limited number of oxidation state, viz., +2, +3 and +4 (out of which +3 is most common). This is because of large energy gap between 4f and 5d-subshells. The dominant oxidation state of actinoids is also +3 but they show a number of other oxidation states also. Eg. uranium (Z = 92) and plutonium (Z = 94), show +3, +4, +5 and +6, neptunium (Z = 93) shows +3, +4, +5 and +7 etc. This is due to small energy difference between 5f, 6d and 7s-subshells of the actinoids. Q.111. Which is the last element in the series of the actinoids? Write the electronic configuration of this elements, comment on the possible oxidation state of this element. (NCERT) Ans: Last element : Lawrencium (Z = 103) Electronic Configuration : [Rn] 5f146d17s2 Possible oxidation state : +3 Q.112.Compare the chemistry of actinoids with that of the lanthanoids with special reference to : (NCERT) i. electronic configuration ii. oxidation state iii. atomic size and ionic sizes iv. chemical reactivity Ans: Characteristics Lanthanoids Actinoids x y i. Electronic It may be represented by [Xe]4f 5d It may be represented by [Rn]5fx 6dy configuration 6s2, where x varies from 0 to 14 and 7s2, where x varies from 0 to 14 and y = 0 or 1. y = 0 or 1. ii. Oxidation state Show +3 oxidation state only, except Show higher oxidation states such in few cases where it is +2 or +4. as +4, +5, +6, +7 also in addition to They never show more than +4 +3 oxidation state. oxidation state. iii. Atomic and ionic sizes The ionic radii of M3+ ions in There is a greater and gradual lanthanoids series show a regular decrease in the size of atoms or M3+ decrease in size of ions with increase ions across the series. This greater in atomic number. This decrease is decrease is known as actinoid contraction. known as lanthanoid contraction. iv. Chemical reactivity These are less reactive metals and These are highly reactive metals form oxides, sulphides, nitrides, especially in finely divided state. hydroxides and halides etc. These They form a mixture of oxide and also form H2 with acids. They show hydride by action of boiling water. a lesser tendency for complex They combine with non-metals even formation. at moderate temperature. They show a greater tendency for complex formation. 29 d and f Block Elements


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*Q.113. Differentiate between lanthanoids and actinoids. Ans: No. Lanthanoids i. Differentiating electron enters in 4f orbitals ii. Belong to sixth period and form a part of the third transition series. They constitute first inner transition series. iii. Except promethium, all elements occur in nature.

iv. v. vi. vii. viii. ix. x. xi.

Actinoids Differentiating electron enters in 5f orbitals Belong to seventh period and form a part of the fourth transition series. They constitute second inner transition series. Except Uranium and Thorium which occur in nature, all other elements are synthesized in laboratory. Binding energy of 4f orbitals are higher Binding energy of 5f orbital are lower. Only promethium is radioactive. All the members of the series are radioactive Besides +3 oxidation state lanthanoids show Besides +3 oxidation state, actinoids show higher +2 and +4 oxidation states in few cases. oxidation states such as +4, +5, +7 also. Lanthanoids show less tendency to form Actinoids show greater tendency to form complexes. complexes. Some of the ions of lanthanoids are fairly Most of the ions of actinoids are deeply coloured coloured. U3+ (red) , U4+(green) Lanthanoids cannot form oxocations. Actinoids form oxocations such as UO 2+ , 2

PuO2+, UO+. Lanthanoids hydroxides are less basic in Actinoids hydroxides are more basic in nature. nature. Contraction is relatively less. Contraction is greater in this series due to poor shielding of 5f electrons.

#Q.114. The extent of actinoid contraction is greater than lanthanoid contraction, Explain why? OR Actinoid contraction is greater from element to element than lanthanoid contraction. Why?(NCERT) Ans: Actinoid contraction is greater than lanthanoid contraction: i. The size of the atoms or ions of actinoids decrease regularly along the series with the increase in atomic number from actinium to lawrencium. ii. This steady decrease in the ionic radii with the increase in atomic number is called actinoid contraction. iii. The actinoid contraction is due to the imperfect shielding of 5f-electron. iv. Despite of the imperfect shielding of 5f-orbitals, the effective nuclear charge increases which results in contraction of the size. v. It may be noted that in actinoid contraction, there are bigger jumps in ionic size between the consecutive members as compared to lanthanoids. vi. This is due to lesser shielding of 5f-electrons which results in greater increase in the effective nuclear charge and therefore, larger attraction. Q.115. Write a short note on uses of actinoids. Ans: Uses of actinoids: The three most important actinoids which find uses as such or in the form of their compounds are thorium, uranium and plutonium. i. Thorium: It is used in atomic reactors and in the treatment of cancer. Its salts are used in making incandescent gas mantles. ii. Uranium: It is used as a nuclear fuel. Its salts are used in glass industry (for imparting green colour), textile industry, ceramic industry and in medicines. iii. Plutonium: It is used as a fuel for atomic reactors as well as for making atomic bombs. 30



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Points to Remember

¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾

¾ ¾ ¾ ¾

Transition elements are classified as those whose atoms or ions have partly filled d-orbitals. General electronic configuration of transition elements is (n-1) d1–10 ns1–2. Transition metals possess high density and high melting and boiling points Within a group, the atomic sizes of elements first increase from 3d to 4d and then decrease from 4d to 5d. This is due to lanthanoid contraction in 5d elements. The ionization enthalpies of transition elements increase across a period, but this increase is not regular. This is again due to imperfect shielding power of (n−1) d electrons. The atomic radii for transition metals are smaller than their corresponding s-block elements. An important feature of transition elements is the colour of their compounds. The colour is linked with the electronic configurations of ions and d-d transitions. Ions with d0 and d10 configurations are colourless while d5 ions show light colours. Transitions elements exhibit a variety of oxidation states. The highest oxidation states, for example, +6 for Cr in CrO42– and +7 for MnO4−, are attained by highly electronegative elements oxygen and fluorine. Transition metals and their compounds find applications as catalysts in various industrial chemical processes. Many transition metals and their compounds show magnetic behaviour. The number of unpaired electrons present can be obtained from the measurement of magnetic moments (µ) as µ = n(n + 2) B.M, where n is the number of unpaired electrons. Two series of f-block elements (4f and 5f) are embedded in transition elements. These are called inner transition elements. Elements in 4f series are known as lanthanoids and those of 5f series as actinoids. In the lanthanoid series, with increasing atomic number, the atomic and ionic radii decreases from one element to another but the decrease is very small. This is called Actinoid contraction. Actinoid contraction is greater from element to element than lanthanoid contraction. Elements of f-series also behave in a manner similar to transition elements in many respects. Multiple choice Questions

1.

2.

3.

The general outer electronic configuration of transition elements is (A) (n−1) d1−10 ns1 (B) (n−1) d10 ns2 (C) (n−1) d1−10 ns1−2 (D) (n−l) d5 ns1 Which of the following is a lanthanoid? (A) Ta (B) Th (C) Lu (D) Rh Which of the following belongs to the actinoid series? (A) U (B) Yb (C) Lu (D) Tb

4.

Lanthanoid contraction implies (A) decrease in density (B) decrease in mass (C) decrease in ionic radii (D) decrease in radioactivity

5.

Which one of the following does not show different oxidation states? (A) Iron (B) Copper (C) Zinc (D) Manganese


6.

Paramagnetism is the property of (A) filled electronic sub-shells (B) unpaired electrons (C) non-transition elements (D) m.p and b.p of elements.

7.

In KMnO4 oxidation number of Mn is (A) +2 (B) +4 (C) +6 (D) +7

8.

Which of the diamagnetic? (A) Cu2+ (C) Cd2+

following (B) (D)

would

be

Ni2+ Ti3+

9.

Which among the following pairs is not a [March 2013 old course] chemical twin? (A) Mo – W (B) Nb - Mo (C) Nb – Ta (D) Zr - Hf

10.

Which of the following has the maximum number of unpaired electrons? (A) Fe2+ (B) Cr3+ 3+ (C) Fe (D) Co2+ 31


11.

12.

Which one of the following characteristic of the metals is associated with their catalytic activity? (A) Paramagnetic behaviour (B) Colour of hydrated ions (C) Variable oxidation states (D) High enthalpy of atomisation Which of the following ions has the highest magnetic moment? (B) Sc3+ (A) Ti3+ 2+ (D) Zn2+ (C) Mn

13.

Which lanthanoid has the smallest atomic radius? (A) Gadolinium (B) Scandium (C) Lutetium (D) Cerium

14.

The lanthanoid contraction is responsible for the fact that (A) Zr and ,Y have about same radius (B) Zr and Nb have similar oxidation state (C) Zr and Hf have about same radius (D) Zr and Zn have same oxidation state

15.

16.

17.

The atomic number of an element is 22. The highest oxidation state exhibited by it in its compounds is (A) 1 (B) 2 (C) 3 (D) 4 Reduction in atomic size with increase in atomic number is a characteristic of elements of (A) radioactive series (B) p-Block (C) d-Block (D) f-Block The element having general electronic configuration (n−1)d1−10ns1−2 is (A) Noble gas (B) Non-metal (C) Metalloid (D) Transition metal

18.

Variable valency is shown by (A) Typical elements (B) Normal elements (C) Transition elements (D) None of these

19.

Zinc is a member of IIB or 12th group of periodic table. The other members of this group are (A) Boron and aluminium (B) Cadmium and mercury (C) Silver and gold (D) Tin and lead

32

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20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

Which one of the following ions is colourless? (B) Co2+ (A) Cu+ 2+ (C) Ni (D) Fe3+ The d-block elements include (A) metals and non-metals (B) only non-metals (C) only metals (D) metals, non-metals and metalloids The most abundant transition metal is (A) Zn (B) Fe (C) Hg (D) Au Which of the following statement is NOT TRUE? (A) Colourless compounds of transition elements are paramagnetic (B) Coloured compounds of transition elements are paramagnetic (C) Colourless compounds of transition elements are diamagnetic (D) Transition elements form the complex compounds The maximum oxidation state shown by actinoids is (A) +6 (B) +7 (C) +5 (D) +4 The element with the electronic configuration [Xe]54 4f14 5d1 6s2 is a (A) representative element (B) transition element (C) lanthanoids (D) actinoid With increase in atomic number the ionic radii of actinoids, (A) contract slightly (B) increase gradually (C) show no change (D) change irregularly The general electronic configuration of lanthanoids is (A) [Xe] 4f1−14 5d0−1 6s2 (B) [Xe] 4f0−14 5d1−2 6s1 (C) [Xe] 4f0−14 5d0−1 6s1−2 (D) None of these The Lanthanoid ions are coloured due to (A) d-d transition (B) d-f transition (C) f-f transition (D) All of these Actinoids (A) are all synthetic elements (B) include element 104 (C) have only short lived isotopes (D) have variable valency d and f Block Elements


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30.

31.

32.

33.

34.

The catalytic activity of the transition metals and their compounds is ascribed to (A) their chemical reactivity (B) their unfilled d−orbitals (C) their ability to adopt multiple oxidation states and their complexing ability (D) less metallic character

37.

Contraction in atomic and ionic radii is given by (A) lanthanoids but not actinoids (B) actinoids but not lanthanoids (C) both lanthanoids and actinoids (D) neither lanthanoids nor actinoids.

38.

Which ion has the highest ionic radii? (A) Cr3+ (B) Mn3+ (C) Fe3+ (D) Co3+

Chemical formula of pyrolusite is (B) MnO3 (A) Mn2O3 (D) Mn2O7 (C) MnO2

39.

Which one of the following ions is coloured? (B) Ti4+ (A) Sc3+ 2+ (C) Zn (D) V2+

40.

In which of the following pair highest oxidation states of transition metals are found? [March 2013] (A) nitriles and chlorides (B) fluorides and chlorides (C) fluorides and oxides (D) nitriles and oxides

Actinoid and Lanthanoid are placed respectively in (A) III B group and 6th and 7th period of periodic table (B) III A group and 7th and 6th period of periodic table (C) III B group and 7th and 6th period of periodic table (D) Both III A and III B group of the periodic table _____ is paramagnetic in nature. (A) La3+ (B) Lu3+ 3+ (C) Gd (D) Ce4+ Which of the following factors may be regarded as the main cause of lanthanoid contraction? (A) Poor shielding of one of 4f electron by another in the subshell (B) Effective shielding of one of 4f electrons by another in the subshell (C) Poorer shielding of 5d electrons by 4f electrons (D) Greater shielding of 5d electron by 4f electrons

35.

Cerium can show the oxidation state of +4 because, (A) it resembles alkali metals (B) it has very low value of I.E. (C) of its tendency to attain noble gas configuration of xenon (D) of its tendency to attain 4f7 configuration

36.

Which is the strongest base among the following? (A) La(OH)3 (B) Lu(OH)3 (C) Ce(OH)3 (D) Yb(OH)3


Answers to Multiple Choice Question

1.

(C)

2.

(C)

3.

(A)

4.

(C)

5.

(C)

6.

(B)

7.

(D)

8.

(C)

9.

(B)

10. (C)

11. (C)

12. (C)

13. (C)

14. (C)

15. (D)

16. (D)

17. (D)

18. (C)

19. (B)

20. (A)

21. (C)

22. (B)

23. (A)

24. (B)

25. (C)

26. (A)

27. (A)

28. (C)

29. (D)

30. (C)

31. (A)

32. (C)

33. (C)

34. (A)

35. (C)

36. (A)

37. (C)

38. (C)

39. (D)

40. (C)

33

Maharashtra-HSC-d & f Paper-2 Target

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