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Tamil Nadu 11th Physics Model Question Papers 2020-2021 English Tamil Medium

February 4, 2021

Subject Matter Experts at SamacheerKalvi.Guide have created Samacheer Kalvi Tamil Nadu State Board Syllabus New Paper Pattern 11th Physics Model Question Papers 2020-2021 with Answers Pdf Free Download in English Medium and Tamil Medium of TN 11th Standard Physics Public Exam Question Papers Answer Key, New Paper Pattern of HSC 11th Class Physics Previous Year Question Papers, Plus One +1 Physics Model Sample Papers are part of Tamil Nadu 11th Model Question Papers.

Let us look at these Government of Tamil Nadu State Board 11th Physics Model Question Papers Tamil Medium with Answers 2020-21 Pdf. Students can view or download the Class 11th Physics New Model Question Papers 2021 Tamil Nadu English Medium Pdf for their upcoming Tamil Nadu HSC Board Exams. Students can also read Tamilnadu Samcheer Kalvi 11th Physics Guide.

TN State Board 11th Physics Model Question Papers 2020 2021 English Tamil Medium

Tamil Nadu 11th Physics Model Question Papers English Medium 2020-2021

Tamil Nadu 11th Physics Model Question Papers Tamil Medium 2020-2021

Tamil Nadu 11th Physics Model Question Paper 1 Tamil Medium
Tamil Nadu 11th Physics Model Question Paper 2 Tamil Medium
Tamil Nadu 11th Physics Model Question Paper 3 Tamil Medium
Tamil Nadu 11th Physics Model Question Paper 4 Tamil Medium
Tamil Nadu 11th Physics Model Question Paper 5 Tamil Medium

11th Physics Model Question Paper Design 2020-2021 Tamil Nadu

Types of Questions	Marks	No. of Questions to be Answered	Total Marks
Part-I Objective Type	1	15	15
Part-II Very Short Answers (Totally 9 questions will be given. Answer any Six. Any one question should be answered compulsorily)	2	6	12
Part-Ill Short Answers (Totally 9 questions will be given. Answer any Six. Any one question should be answered compulsorily)	3	6	18
Part-IV Essay Type	5	5	25
Total			70
Practical Marks + Internal Assessment (20+10)			30
Total Marks			100

Tamil Nadu 11th Physics Model Question Paper Weightage of Marks

Purpose	Weightage
1. Knowledge	30%
2. Understanding	40%
3. Application	20%
4. Skill/Creativity	10%

It is necessary that students will understand the new pattern and style of Model Question Papers of 11th Standard Physics Tamilnadu State Board Syllabus according to the latest exam pattern. These Tamil Nadu Plus One 11th Physics Model Question Papers State Board Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN HSLC Board Exams and Score More marks.

We hope the given Samacheer Kalvi Tamil Nadu State Board Syllabus New Paper Pattern Class 11th Physics Model Question Papers 2020 2021 with Answers Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding the Tamil Nadu Government 11th Physics State Board Model Question Papers with Answers 2020 21, TN 11th Std Physics Public Exam Question Papers with Answer Key, New Paper Pattern of HSC Class 11th Physics Previous Year Question Papers, Plus One +1 Physics Model Sample Papers, drop a comment below and we will get back to you as soon as possible.

Samacheer Kalvi 11th Commerce Guide Book Answers Solutions

January 22, 2021

Subject Matter Experts at SamacheerKalvi.Guide have created Tamilnadu State Board Samacheer Kalvi 11th Commerce Answers Solutions Guide Pdf Free Download in English Medium and Tamil Medium are part of Samacheer Kalvi 11th Books Solutions.

Let us look at these TN Board Samacheer Kalvi 11th Std Commerce Guide Pdf of Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank and revise our understanding of the subject.

Students can also read Tamil Nadu 11th Commerce Model Question Papers 2020-2021 English & Tamil Medium.

Samacheer Kalvi 11th Commerce Book Solutions Answers Guide

Samacheer Kalvi 11th Commerce Book Back Answers

Tamilnadu State Board Samacheer Kalvi 11th Commerce Book Back Answers Solutions Guide.

Unit 1 Fundamentals of Business

Unit 2 Forms of Business Organisation

Unit 3 Service Business – I

Unit 4 Service Business – II

Chapter 16 Emerging Service Business in India

Unit 5 Service Business – II

Unit 6 Business Finance

Unit 7 Trade

Unit 8 International Business

Unit 9 The Indian Contract Act

Unit 10 Direct and Indirect Taxes

We have also created Samacheer Kalvi 11th Commerce Notes for students to help them prepare for the exam like scenario.

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Samacheer Kalvi 12th Computer Applications Guide Book Answers Solutions

January 22, 2021

Subject Matter Experts at SamacheerKalvi.Guide have created Tamilnadu State Board Samacheer Kalvi 12th Computer Applications Answers Solutions Guide Pdf Free Download in English Medium and Tamil Medium are part of Samacheer Kalvi 12th Books Solutions.

Let us look at these TN Board Samacheer Kalvi 12th Std Computer Applications Guide Pdf of Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank and revise our understanding of the subject.

Students can also read Tamil Nadu 12th Computer Applications Model Question Papers 2020-2021 English & Tamil Medium.

Samacheer Kalvi 12th Computer Applications Book Solutions Answers Guide

Samacheer Kalvi 12th Computer Applications Book Back Answers

Tamilnadu State Board Samacheer Kalvi 12th Computer Applications Book Back Answers Solutions Guide.

We hope these Tamilnadu State Board Class 12th Computer Applications Book Solutions Answers Guide Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding TN State Board New Syllabus Samacheer Kalvi 12th Standard Computer Applications Guide Pdf Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank, Formulas, drop a comment below and we will get back to you as soon as possible.

Samacheer Kalvi 12th Computer Science Guide Book Answers Solutions

January 22, 2021

Subject Matter Experts at SamacheerKalvi.Guide have created Tamilnadu State Board Samacheer Kalvi 12th Computer Science Book Answers Solutions Guide Pdf Free Download in English Medium and Tamil Medium are part of Samacheer Kalvi 12th Books Solutions.

Let us look at these TN State Board New Syllabus Samacheer Kalvi 12th Std Computer Science Guide Pdf of Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank and revise our understanding of the subject.

Students can also read Tamil Nadu 12th Computer Science Model Question Papers 2020-2021 English & Tamil Medium.

Samacheer Kalvi 12th Computer Science Book Solutions Answers Guide

Samacheer Kalvi 12th Computer Science Book Back Answers

Tamilnadu State Board Samacheer Kalvi 12th Computer Science Book Back Answers Solutions Guide.

Unit 1 Problem Solving Techniques

Unit 2 Core Python

Unit 3 Modularity and OOPS

Unit 4 Database Concepts and MySql

Unit 5 Integrating Python with MySql and C++

We hope these Tamilnadu State Board Samacheer Kalvi Class 12th Computer Science Book Solutions Answers Guide Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding TN State Board New Syllabus Samacheer Kalvi 12th Standard Computer Science Guide Pdf Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank, Formulas, drop a comment below and we will get back to you as soon as possible.

Tamil Nadu 12th Computer Science Model Question Papers 2020-2021 English Tamil Medium

January 21, 2021

Subject Matter Experts at SamacheerKalvi.Guide have created Samacheer Kalvi Tamil Nadu State Board Syllabus New Paper Pattern 12th Computer Science Model Question Papers 2020-2021 with Answers Pdf Free Download in English Medium and Tamil Medium of TN 12th Standard Computer Science Public Exam Question Papers Answer Key, New Paper Pattern of HSC 12th Class Computer Science Previous Year Question Papers, Plus Two +2 Computer Science Model Sample Papers are part of Tamil Nadu 12th Model Question Papers.

Let us look at these Government of Tamil Nadu State Board 12th Computer Science Model Question Papers Tamil Medium with Answers 2020-21 Pdf. Students can view or download the Class 12th Computer Science New Model Question Papers 2021 Tamil Nadu English Medium Pdf for their upcoming Tamil Nadu HSC Board Exams. Students can also read Tamilnadu Samcheer Kalvi 12th Computer Science Guide.

TN State Board 12th Computer Science Model Question Papers 2020 2021 English Tamil Medium

Tamil Nadu 12th Computer Science Model Question Papers English Medium 2020-2021

Tamil Nadu 12th Computer Science Model Question Papers Tamil Medium 2020-2021

Tamil Nadu 12th Computer Science Model Question Paper 1 Tamil Medium
Tamil Nadu 12th Computer Science Model Question Paper 2 Tamil Medium
Tamil Nadu 12th Computer Science Model Question Paper 3 Tamil Medium
Tamil Nadu 12th Computer Science Model Question Paper 4 Tamil Medium
Tamil Nadu 12th Computer Science Model Question Paper 5 Tamil Medium

12th Computer Science Model Question Paper Design 2020-2021 Tamil Nadu

Types of Questions	Marks	No. of Questions to be Answered	Total Marks
Part-I Objective Type	1	15	15
Part-II Very Short Answers (Totally 9 questions will be given. Answer any Six. Any one question should be answered compulsorily)	2	6	12
Part-III Short Answers (Totally 9 questions will be given. Answer any Six. Any one question should be answered compulsorily)	3	6	18
Part-IV Essay Type	5	5	25
Total			70
Practical Marks + Internal Assessment (20+10)			30
Total Marks			100

Tamil Nadu 12th Computer Science Model Question Paper Weightage of Marks

Purpose	Weightage
1. Knowledge	30%
2. Understanding	40%
3. Application	20%
4. Skill/Creativity	10%

It is necessary that students will understand the new pattern and style of Model Question Papers of 12th Standard Computer Science Tamilnadu State Board Syllabus according to the latest exam pattern. These Tamil Nadu Plus Two 12th Computer Science Model Question Papers State Board Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN HSLC Board Exams and Score More marks.

We hope the given Samacheer Kalvi Tamil Nadu State Board Syllabus New Paper Pattern Class 12th Computer Science Model Question Papers 2020 2021 with Answers Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding the Tamil Nadu Government 12th Computer Science State Board Model Question Papers with Answers 2020 21, TN 12th Std Computer Science Public Exam Question Papers with Answer Key, New Paper Pattern of HSC Class 12th Computer Science Previous Year Question Papers, Plus Two +2 Computer Science Model Sample Papers, drop a comment below and we will get back to you as soon as possible.

Samacheer Kalvi 12th Economics Guide Book Answers Solutions

January 18, 2021

Subject Matter Experts at SamacheerKalvi.Guide have created Tamilnadu State Board Samacheer Kalvi 12th Economics Answers Solutions Guide Pdf Free Download in English Medium and Tamil Medium are part of Samacheer Kalvi 12th Books Solutions.

Let us look at these TN Board Samacheer Kalvi 12th Std Economics Guide Pdf of Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank, Formulas and revise our understanding of the subject.

Students can also read Tamil Nadu 12th Economics Model Question Papers 2020-2021 English & Tamil Medium.

Samacheer Kalvi 12th Economics Book Solutions Answers Guide

Samacheer Kalvi 11th Economics Book Back Answers

Tamilnadu State Board Samacheer Kalvi 12th Economics Book Back Answers Solutions Guide.

We have also created Samacheer Kalvi 12th Economics Notes for students to help them prepare for the exam like scenario.

We hope these Tamilnadu State Board Class 12th Economics Book Solutions Answers Guide Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding TN State Board New Syllabus Samacheer Kalvi 12th Standard Economics Guide Pdf Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank, Formulas, drop a comment below and we will get back to you as soon as possible.

Samacheer Kalvi 11th History Guide Book Answers Solutions

January 13, 2021

Subject Matter Experts at SamacheerKalvi.Guide have created Tamilnadu State Board Samacheer Kalvi 11th History Answers Solutions Guide Pdf Free Download in English Medium and Tamil Medium are part of Samacheer Kalvi 11th Books Solutions.

Let us look at these TN Board Samacheer Kalvi 11th Std History Guide Pdf of Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank and revise our understanding of the subject.

Students can also read Tamil Nadu 11th History Model Question Papers 2020-2021 English & Tamil Medium.

Samacheer Kalvi 11th History Book Solutions Answers Guide

Tamilnadu State Board Samacheer Kalvi 11th History Book Back Answers Solutions Guide.

Samacheer Kalvi 11th History Book Back Answers

We hope these Tamilnadu State Board Class 11th History Book Solutions Answers Guide Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding TN State Board New Syllabus Samacheer Kalvi 11th Standard History Guide Pdf Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank, Formulas, drop a comment below and we will get back to you as soon as possible.

Samacheer Kalvi 12th History Guide Book Answers Solutions

January 13, 2021

Subject Matter Experts at SamacheerKalvi.Guide have created Tamilnadu State Board Samacheer Kalvi 12th History Answers Solutions Guide Pdf Free Download in English Medium and Tamil Medium are part of Samacheer Kalvi 12th Books Solutions.

Let us look at these TN Board Samacheer Kalvi 12th Std History Guide Pdf of Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank and revise our understanding of the subject.

Students can also read Tamil Nadu 12th History Model Question Papers 2020-2021 English & Tamil Medium.

Samacheer Kalvi 12th History Book Solutions Answers Guide

Tamilnadu State Board Samacheer Kalvi 12th History Book Back Answers Solutions Guide.

Samacheer Kalvi 12th History Book Back Answers

We hope these Tamilnadu State Board Class 12th History Book Solutions Answers Guide Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding TN State Board New Syllabus Samacheer Kalvi 12th Standard History Guide Pdf Text Book Back Questions and Answers, Notes, Chapter Wise Important Questions, Model Question Papers with Answers, Study Material, Question Bank, Formulas, drop a comment below and we will get back to you as soon as possible.

Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry

January 4, 2021

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 5 Coordination Chemistry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 5 Coordination Chemistry

12th Chemistry Guide Coordination Chemistry Text Book Questions and Answers

Part – I Text Book Evaluation

I. Choose the correct answer

1. The sum of primary valence and secondary valance of the metal M in the complex [M(en))₂ (Ox)] Cl is L
a) 3
b) 6
c) -3
d) 9
Answer:
d) 9

2. An excess of silver nitrate is added to 100ml of aO.OlM solution of pentaaquachlori- dochromium (III) chloride. The number of moles of AgCl precipitated would be
a) 0.02
b) 0.002
c) 0.01
d) 0.2
Answer:
b) 0.002

3. A complex has a molecular formula MSO₄Cl.6H₂O. The aqueous solution of it gives white precipitate with Barium chloride solution and no precipitate is obtained when it is treated with silver nitrate solution. If the secondary valence of the metal is six, which one of the following correctly represents the complex?
a) [M(H₂O)₄Cl] SO₄.2H₂O
b) [M(H₂O)₆] SO₄
c) [M(H₂O)₅Cl]SO₄.H₂O
d) [M (H₂O)₃Cl] SO₄.3H₂O
Answer:
c) [M(H₂O)₅Cl]SO₄.H₂O

4. Oxidation state of Iron and the charge on the ligand NO in [Fe (H₂O)₅ NO] SO₄ are
a) +2 and 0 respectively
b) +3 and 0 respectively
c) +3 and -1 respectively
d) +1 and +1 respectively
Answer:
d) +1 and +1 respectively

5. As per IUPAC guidelines, the name of the complex [Co(en)2 (ONO)Cl] Cl is
a) chlorobisethylenediaminenitritocobalt(III) chloride
b) chi or id obis (e thane-1, 2-diamine) nitrito K-Ocobaltate(III) chloride
c) chloridobis (ethane-1, 2-diammine) nitrito K -Ocobalt(II) chloride
d) chloridobis (ethane-1, 2-diamine) nitrito K -Ocobalt(III) chloride
Answer:
d) chloridobis (ethane-1,2-diamine) nitro K -Ocobalt(III) chloride

6. IUPAC name of the complex K₃[Al(C₂O₄)₃] is
a) potassiumtrioxalatoaluminium(III)
b) potassiumtrioxalatoaluminate(II)
c) potassiumtrisoxalatoaluminate(III)
d) potassiumtrioxalatoaluminate(III)
Answer:
d) potassiumtrioxalatoaluminate(III)

7. A magnetic moment of 1.73BM will be shown by one among the following (NEET)
a) TiCl₄
b) [CoCl₆]^4-
c) [CU(NH₃)₄]²⁺
d) [Ni(CN)₄]^2-
Answer:
c) [CU(NH₃)₄]²⁺

8. Crystal field stabilization energy for high spin d⁵ octahedral complex is
a) -0.6∆₀
b) 0
c) 2(P – ∆₀)
d) 2(P + ∆₀)
Answer:
b) 0

9. In which of the following coordination entities the magnitude of ∆₀ will be maximum?
a) [Co(CN)₆]^3-
b) [Co(C₂O₄)₃]^3-
c) [Co(H₂O)₆]³⁺
d) [Co(NH₃)₆]³⁺
Answer:
a) [Co(CN)₆]^3-

10. Which one of the following will give a pair of enantiomorphs?
a) [Cr(NH₃)₆][Co(CN))₆]
b) [Co(en)₂Cl₂]Cl
c) [Pt(NH₃)₄][FtCl₄]
d) [CO(NH₃)₄Cl₂]NO₂
Answer:
b) [Co(en)₂Cl₂]Cl

11. Which type of isomerism is exhibited by [Pt(NH₃)₂Cl₂]?
a) Coordination isomerism
b) Linkage isomerism
c) Optical isomerism
d) Geometrical isomerism
Answer:
d) Geometrical isomerism

12. How many geometrical isomers are possible for [Pt(Py)(NH₃)(Br)(Cl)]?
a) 3
b) 4
c) 0
d) 15
Answer:
a) 3

13. Which one of the following pairs represents linkage isomers?
a) [Cu(NH₃)₄] [PtCl₄] and [Pt (NH₃)₄] [CuCl₄]
b) [Co(NH₃)₅ (NO₃)] SO₄ and [CO(NH₃)₅ (ONO)]
c) [Co(NH₃)₄ (NCS)₂] Cl and [Co(NH₃)₄ (SCN)₂]Cl
d) both (b) and (c)
Ans:
c) [Co(NH₃)₄ (NCS)₂] Cl and [Co(NH₃)₄ (SCN)₂]Cl

14. Which kind of isomerism is possible for a complex [Co(NH₃)₄Br₂]Cl? (PTA -3)
a) geometrical and ionization
b) geometrical and optical
c) optical and ionization
d) geometrical only
Answer:
a) geometrical and ionization

15. Which one of the following complexes is not expected to exhibit isomerism?
a) [Ni(NH₃)₄ (H₂O)₂]²⁺
b) [Pt(NH₃)₂Cl₂]
c) [Co(NH₃)₅SO₄]Cl
d) [FeCl₆ ]^3-
Answer:
d) [FeCl₆]^3-

16. A complex in which the oxidation number of the metal is zero is
a) K4 [Fe(CN)₆]
b) [Fe(CN)₃ (NH₃)₃]
c) [Fe(Co)₅]
d) both (b) and (c)
Answer:
c) Fe(CO)s]

17. Formula of tris(ethane-l,2-diamine) iron (II) phosphate
a) [Fe(CH₃ – CH(NH₂)₂)₃](PO₄)₃
b) [Fe(H₂N – CH₂ – CH₂ – NH₂)₃] (PO₄)
c) [Fe(H₂N-CH₂-CH₂-NH₂)₃](PO₄)₂
d) [Fe(H₂N-CH₂-CH₂-NH₂)₃]₃(PO₄)₂
Answer:
d) [Fe(H₂N-CH₂-CH₂-NH₂)₃]₃(PO₄)₂

18. Which of the following is paramagneticin nature? (PTA -5)
a) [Zn(NH₃)₄]²⁺
b) [CO (NH₃)₆]³⁺
c) [Ni(H₂O)₆]²⁺
d) [Ni (CN)₄]^2-
Answer:
c) [Ni(H₂O)₆]²⁺

19. Fac-mer isomerism is shown by
a) [Co (en)₃]³⁺
b) [Co(NH₃)₄(Cl)₂]⁺
c) [Co (NH₃)₃ (Cl)₃]
d) [Co (NH₃)₅ Cl]SO₄
Answer:
c) [Co (NH₃)₃ (Cl)₃]

20. Choose the correct statement.
a) Square planar complexes are more stable than octahedral complexes
b) The spin only magnetic moment of [Cu (Cl)₄]₄^2- is 1.732 BM and it has square planar structure.
c) Crystal field splitting energy (∆_o) of [FeF₆]^4- is higher than the (Ao) of [Fe (CN)6]4
d) Crystal field stabilization energy of [V(H₂O)f₆]²⁺ is higher than the crystal field stabilization of [Ti(H₂O)₆]²⁺
Answer:
d) Crystal field stabilization energy of [V(H₂O)f₆]²⁺ is higher than the crystal field stabilization of [Ti(H₂O)₆]²⁺

II. Answer the following questions

Question 1.
Write the IUPAC names for the following complexes. (PTA -3)

Na₂ [Ni (EDTA)]
[Ag(CN)₂]^–
[Co(en)₃]₂ (SO₄)₃
[Co (ONO) (NH₃)₅]²⁺
[Pt(NH₃)₂ Cl(NO₂)]

Answer:

Sodium 2, 2¹, 2¹¹, 2¹¹¹-(ethane-1, 2-diyldinitrilo) tetra acetato nickelate (II)
Dicyanido k-c argentate (I) ion
Tris (ethane -1, 2 – diamine) cobalt (III) sulphate
Pentaammine nitrito _ kO cobalt (III) ion
Diammine chlorido nitrito – kN Platinum (II)

Question 2.
Write the formula for the following coordination compounds.

Potassiumhexacyanidoferrate(II)
pentacarbonyl iron(O)
pentaamminenitrito -k -Ncobalt(III)ion
hexaamminecobalt(III) sulphate
sodium tetra fluoridodi hydroxidochromate (III)

Answer:

K₄[Fe(CN)₆]
[Fe(CO)₅]
[CO(NH₃)₅ (NO₂)]²⁺
[CO[NH₃)₆]₂ (SO₄)₃
Na₃[CrF₄(OH)₂]

Question 3.
Arrange the following in order of increasing molar conductivity
i) Mg[Cr(NH₃)(Cl)₅]
ii) [Cr(NH₃)₅Cl]₃[CoF₆]₂
iii) [Cr(NH₃)₃Cl₃]
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 1

iii) [Cr(NH₃)₃Cl₃] – Neutral complex – No ions Molar conductivity increases as the number of ions increase
∴ Molar conductivity is in the increasing order. [Cr(NH₃)₃Cl₃] < Mg (Cr(NH₃)Cl₅] < [Cr(NH₃)₅Cl]₃[CoF₆]₂

Question 4.
Give an example of a coordination compound used in medicine and two examples of biologically important coordination compounds.
Answer:
Medical uses of coordination compounds:

Ca-EDTA chelate is used in the treatment of lead and radioactive poisoning. That is for removing lead and radioactive metal ions from the body.
Cis-platin is used as an antitumor drug in cancer treatment.

Biological importance of coordination compounds:

1. A red blood corpuscles (RBC) is composed of heme group, which is Fe²⁺ Porphyrin complex.it plays an important role in carrying oxygen from lungs to tissues and carbon dioxide from tissues to lungs.

2. Chlorophyll, a green pigment present in green plants and algae, is a coordination complex containing Mg²⁺ as a central metal ion surrounded by a modified Porphyrin ligand called corrin ring. It plays an important role in photosynthesis, by which plants convert CO, and water into carbohydrates and oxygen.

3. Vitamin B₁₂ (cyanocobalamin) is the only vitamin consist of a metal ion. it is a coordination complex in which the central metal ion is CO⁺ surrounded by Porphyrin ligand.

4. Many enzymes are known to be metal complexes, they regulate biological processes. For example, Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in digestion contains a zinc ion coordinated to the protein.

Question 5.
Based on VB theory explain why [Cr(NH₃)₆]³⁺ is paramagnetic, while [Ni(CN)₄]^2- is diamagnetic.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 3

Question 6.
Draw all possible geometrical isomers of the complex [Co(en)₂Cl₂]⁺ and identify the optically active isomer.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 6

Question 7.
[Ti (H₂O)₆]³⁺ is coloured, while [Sc (H₂O)₆]³⁺ is colourless – Explain. (PAT -4; MARCH 2020)
Answer:

In [Ti (H₂O)₆]³⁺ the – outer electronic configuration of metal ion Ti³⁺ is 3d¹.
This single electron present in lower energy t_2g orbitals in the octahedral aqua ligand field absorbs light and excited to higher energy eg level.
This is known as d-d transition.
For this excitation absorption maximum is at 20000 cm^-1 corresponding to the CFSE ∆₀ 239.7 KJ mol^-1
The transmitted colour associated with this absorption in purple and hence the complex appears purple in colour. [Ti(H20)6]³⁺
But in [Sc(H₂O)₆]³⁺, the outer electronic configuration of metal ion Sc³⁺ is 3d°
Since there is no electron in d orbital, d-d transition is not possible.
Hence [Sc(H₂O)₆]³⁺ is colourless.

Question 8.
Give an example for complex of the type [Ma₂b₂C₂] where a, b, c are monodentate ligands and give the possible isomers.
Answer:

Example for the complex of the type [Ma₂b₂c₂] is [Cr(NH₃)₂ (H₂O)₂Br₂]⁺
There are totally 6 isomers possible Geometrical isomers – 5

Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 7

Question 9.
Give one test to differentiate [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅ SO₄]Cl
Answer:

[CO(NH₃)₅Cl]SO₄ → [CO(NH₃)₅Cl]⁺² + SO₄^2-
[CO(NH₃)₅SO₄] Cl → [CO(NH₃)₅Cl] SO₄

1. Aqueous solution of (a) gives sulphate ion. When the addition of BaCL, solution (a) gives a white precipitate of BaS04. But (b) does not give any precipitate.

2. Aqueous solution of (b) gives chloride ion. When the addition of AgNO₃ solution (b) gives curdy white precipitate of AgCl. But (a) does not give any precipitate.

Question 10.
In an octahedral crystal field, draw the figure to show the splitting of d orbitals.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 9 a

Question 11.
What is linkage isomerism? Explain with an example.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 10
Linkage isomerism arises when an ambidentate ligand is bonded to the central metal atom/ion through either of its two different donor atoms.
(eg) [Co(NH₃)₅NO₂]²⁺
[Co(NH₃)₅NO₂]²⁺
Pentaammine nitrito – kN Cobalt (III) ion – Nattached [CO(NH₃)₅ONO]²⁺
Pentaammine nitrito – kO Cobalt (III) ion – O attached

Question 12.
Classify the following ligand based on the number of donor atoms,
a) NH₃
b) en
c) ox^2-
d) triaminotriethylamine
e) pyridine
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 11

Question 13.
Give the difference between double salts and coordination compounds.
Answer:

Double Salts	Coordination Compounds
1. Lose their identity	Do not lose llieii identity
2. Dissociate into their constituent simple ions in solutions	Never dissociate to give simple ions.
3. (eg) Mohr’s salt FeSO₄(NH₄)₂ SO₄.6H₂O	K₃[Fe(SCN)₆]
4. Answer the tests for simple ions Fe2+, NH⁴⁺,SO₄^2- -ions.	Does not answer for simple ions Fe³⁺, SCN^–

Question 14.
Write the postulates of Werner’s theory?
Answer:
In a complex, every metal atom possesses two types of valency.
(i) Primary Valency
(ii) Secondary Valency

Primary Valency	Secondary Valency
1. Refers to the Oxidation State of metal ion	Refers to the Coordination number of metal ion
2. Always satisfied by negative ions	Satisfied bv negative ions, neutral molecules or positive ions.
3. These ions are generally written outside the bracket are called counter ions.	These ions written inside the bracket are called ligands.
4. The outer sphere in which these ions present are called ionisation sphere.	The inner sphere in which these ions present are called coordination sphere
5. The groups present in this sphere are loosely bound to the central metal ion and can be separated into ions.	The groups present in this sphere are firmly attached to the central metal atom and can not be separated into ions.
6. This Valency is ionisable	This valency is non-ionisable.
7. This valency is non-directional	This valency is directional and determines the geometry of the complex.

15. Why tetrahedral complexes do not exhibit geometrical isomerism.
Answer:
Cis-trans isomerism is not possible in tetrahedral complexes because all the four ligands are adjacent to one another [Ma₂b₂]
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 12

Question 16.
Explain optical isomerism in coordination compounds with an example.
Answer:

Coordination compounds which possess chirality exhibit optical isomerism similar to organic compounds.
The pair of two optically active isomers which are mirror images of each other are called enantiomers.
Their solutions rotate the plane of the plane polarised light.
If the rotation is in the clockwise direction the isomer is called dextro rotatory ‘d’ form.
If the rotation is in the anti-clockwise direction the isomer is called laevo rotatory T form.
The octahedral complexes exhibit optical isomerism.
Examples: [CoCl₂(en)₂]⁺

Two Cis isomers are optically active. One trans isomer is optically inactive
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 13

Question 17.
What are hydrate isomers? Explain with an example. (PTA- 6; March 2020)
Answer:

The exchange of free solvent molecules such as water, ammonia, alcohol etc., in the crystal lattice with a ligand in the coordination entity will give Solvate isomers.
If the solvent molecule is water, then these isomers are called hydrate isomers.
Example: CrCl₃.6H₂O has three hydrate isomers

Complex	Colour	Number of Chloride ions in solutions
[Cr(H₂O)₆]Cl₃	Violet	3
[Cr(H₂O)₅Cl]Cl₂.H₂O	Pole green	2
[Cr(H₂O)₄Cl₂]Cl.2H₂O	Dark green	1

Question 18.
What is crystal field splitting energy?
Answer:
The orbitals lying along the axes dx²-y² and dz² orbitals will experience strong repulsion and raise in energy to a greater extent than the orbitals with lobes directed between the axes (d_xy, d_yz and d_zx). Thus the degenerated orbitals now split into two sets and the process is called crystal field splitting.

Question 19.
What is crystal field stabilization energy? (CFSE)? (PTA -1)
Answer:
Crystal field stabilisation energy (CFSE) is defined as the energy difference of electronic configurations in the ligand field (E_LF) and the isotropic field/barycentre (E_iso).
CFSE (∆Eo) = {FLF}-{Eiso} = {[nt2g(-0.4)+neg(0.6)] ∆o + npP} – {n’p P}
Where
nt_2g = No. f electrons in t_2g orbitals
n_eg = No. of electrons in e_g orbitals
n_p = No. of electron in the ligand field
n¹_p = No. of electron pairs in the isotropic field (barycentre)
P = pairing energy

Question 20.
A solution of [Ni(H₂O)₆]²⁺ is green, whereas a solution of [Ni(CN)₄]^2- is colorless – Explain.
Answer:

In [Ni(H₂O)₆]²⁺, the Ni²⁺ ion has two unpaired electrons that do not pair up in presence of weak ligand H₂O.
Hence d-d-the transition is possible.
Red colour is absorbed and complementary colour green is emitted. So it is green in colour.
In [Ni(CN)₄]^2-, the Ni²⁺ ion has no unpaired electrons because of the strong ligand CN^– pairs up the electrons.
Hence no d-d transition is possible.
So [Ni(CN)₄]^2- is colourless

Question 21.
Discuss briefly the nature of bonding in metal carbonyls. (PTA – 2)
Answer:

In metal carbonyls, the bond between metal atom and the carbonyl ligand consists of two components.
The first component is an electron pair donation from the carbon atom of carbonyl ligand into a vacant d-orbital of central metal atom.
This electron pair donation forms M←CO sigma bond.
This sigma bond formation increases the electron density in metal d-orbitals and makes the metal electron-rich.
In order to compensate for this increased electron density, a filled metal d-orbital interacts with the empty π* orbital on the carbonyl ligand and transfers the added electron density back to the ligand.
This second component is called n-back bonding.
Thus in metal carbonyls, electron density moves from ligand to metal through sigma bonding and from metal to ligand through pi bonding.
This synergic effect accounts for a strong M←CO bond in metal carbonyls.
This is shown diagrammatically as follows.

Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 14

Question 22.
What is the coordination entity formed when an excess of liquid ammonia is added to an aqueous solution of copper sulphate?
Answer:
When excess liquid ammonia is added to an aqueous solution of copper sulphate tetraammine copper (II) sulphate is formed.
CuSO₄ + 4NH₃ → [Cu(NH₃)₄]SO₄. The Coordination entity is [Cu(NH₃)₄]²⁺.

Question 23.
On the basis of VB theory explain the nature of bonding in [Co(C₂O₄)₃]^3-
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 15

Question 24.
What are the limitations of VB theory?
Answer:
Limitations of VB – Theory:
1. It does not explain the colour of the complex

2. It considers only the spin only magnetic moments and does not consider the other components of magnetic moments.

3. It does not provide a quantitative explanation as to why certain complexes are inner orbital complexes and the others are outer orbital complexes for the same metal. For example, [Fe(CN)₆]^4- is diamagnetic (low spin) whereas [Fe(CN)₆]^4- is paramagnetic (high spin).

Question 25.
Write the oxidation state, coordination number, nature of ligand, magnetic property and electronic configuration in octahedral crystal field for the complex K₄[Mn(CN)₆].
Answer:

(i) Oxidation State	+2 (Mn²⁺⁾
(ii) Coordination Number	6
(iii) Nature of ligand	Strong field (N^–)
(iv) Magnetic property	Paramagnetic
(v) Electronic configuration	t⁵_2g,e⁰_g

III. Evaluate yourself

Question 1.
When a coordination compound CrCl₃.4H₂O is mixed with silver nitrate solution, one mole of silver chloride is precipitated per mole of the compound. There are no free solvent molecules in that compound. Assign the secondary valence to the metal and write the structural formula of the compound.
Answer:
Since 1 mole of silver chloride is precipitated only one Cl^– ion is present outside the coordination sphere.
∴ The formula is [Cr(H₂O)₄Cl₂]Cl
Secondary valence of the metal is 6

Question 2.
In the complex, [Pt(NO₂)(H₂O)(NH₃)₂]Br, identify the following
i. Central metal atom/ion
ii. Ligand(s) and their types
iii. Coordination entity
iv. Oxidation number of the central metal ion
v. Coordination number
Answer:
[Pt(NO₂)(H₂O)(NH₃)₂]Br
Central metal atom/ ion : Pt (II)
Ligands and their types :
NO₂^– nitrito-k N – Anionic ligand
H₂O aqua – Neutral ligand
NH₃ ammine – Neutral ligand
Coordination entity: [Pt(NO₂)(H₂O)(NH₃)₂]⁺
Oxidation number of the central metal ion x + (-1) + 0 + 2 (0) = +1; x = +2
Coordination number = 4

Question 3.
Write the IUPAC name for the following compounds.
(i) K₂[Fe(CN)₃(Cl)₂(NH₃)]
(ii) [Cr(CN)₂(H₂O)₄][Co(OX)₂(en)]
(iii) [Cu(NH₃)₂Cl₂]
(iv) [Cr(NH₃)₃(NC)₂(H₂O)]⁺
(v) [Fe(CN)₆]^4-
Answer:
(i) K₂[Fe(CN)₃(Cl)₂(NH₃)]
Potassium amminedichloridotricyanido- kC ferrate (III)
(ii) [Cr(CN)₂(H₂O)₄][Co(OX)₂(en)]
Tetra aqua dicyanido kc chromium (III) (ethane -1,2-diamine) dioxalato cobaltate (III)
(iii) [Cu(NH₃)₂Cl₂]
Diammine dichlorido copper (II)
(iv) [Cr(NH₃)₃(NC)₂(H₂O)]⁺
Triammine aqua dicyanido -kN chromium (III) ion
(v) [Fe(CN)₆]^4- Hexacyanido-kc ferrate (II) ion

Question 4.
Give the structure for the following compounds.
Answer:
(i) diamminesilver(I) dicyanidoargentate(I)
(ii) Pentaammine nitrito-KNcobalt (III) ion
(iii) hexafluorido cobaltate (III) ion
(iv) dichloridobis (ethylenediamine)
Cobalt (III) sulphate
(v) Tetracarbonylnickel (0)
Answer:
(i) diamminesilver(I) dicyanidoargentate(I) : [Ag(NH₃)₂] [Ag(CN)₂]
(ii) Pentaammine nitrito-KNcobalt (III) ion : (CO(NH₃)₅ (NO)₂]²⁺
(iii) hexafluorido cobaltate (III) ion : [CoF₆]^3-
(iv) dichloridobis (ethylenediamine) Cobalt (III) sulphate : [Co(en)₂Cl₂]₂SO₄
(v) Tetracarbonylnickel (0): [Ni(CO)₄]

Question 5.
A solution of [Co(NH₃)₄I₂]Cl, when treated with AgNO3, gives a white precipitate. What should be the formula of isomer of the dissolved complex that gives yellow precipitate with AgNO₃. What are the above isomers called?
Answer:

Since yellow precipitate is obtained with AgN03 iodide ion is present outside the coordination sphere.
Hence the formula is [Co(NH₃)₄I Cl]I
This complex is an ionisation isomer of the complex [Co(NH₃)₄I₂]Cl

Question 6.
Three compounds A, B and C have empirical formula CrCl₃.6H₂O. They are kept in a container with a dehydrating agent and they lost water and attaining constant weight as shown below.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 16

Question 7.
Indicate the possible type of isomerism for the following complexes and draw their isomers,
i) [Co(en)₃][Cr(CN)₆]
ii) [Co(NH₃)₅(NO₂)]²⁺
iii) [Pt(NH₃)₃(NO₂)]Cl
Answer:
i) [Co(en)₃][Cr(CN)₆] and [Cr(en)₃][Co(CN)₆] are coordination isomers
ii) [CO(NH₃)₅(NO₂)]²⁺ and [Co(NH₃)₅ONO]²⁺ are linkage isomers
iii) [Pt(NH₃)₃(NO₂)]Cl and [Pt(NH3)3Cl] N02 are ionisation isomers
iv) [Pt(NH₃)₃(NO₂)] and [Pt(NH₃)₃ONO)]Cl are linkage isomers.

Evaluate yourself: 8
Question 1.
Draw all possible stereoisomers of a complex Ca[Co(NH₃)Cl(Ox)₂]
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 17

Evaluate yourself: 9
Question 1.
The spin-only magnetic moment of the Tetrachloridomanganate(II)ion is 5.9 BM. On the basis of VBT, predict the type of hybridisation and geometry of the compound.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 18

Question 2.
Predict the number of unpaired electrons in [C0Cl₄]^2- ion on the basis of VBT. [C0Cl₄]^2-
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 19

Question 3.
A metal complex having composition Co(en)₂Cl₂Br has been isolated in two forms A and B. (B) reacted with silver nitrate to give a white precipitate readily soluble in ammonium hydroxide. Whereas A gives a pale yellow precipitate. Write the formula of A and B. state the hybridization of Co in each and calculate their spin only magnetic moment. [Co(en)₂Cl₂Br.
Answer:
Since B gives white precipitate with AgNO₃, it contains free Cl^– ion outside the Coordination sphere. Hence B is [Co(en)₂Cl Br]Cl

Since A gives pale yellow precipitate with AgNO₃ it contains free Br^– ion outside the coordination sphere. Hence A is [Co(en)₂Cl₂]Br
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 20

Evaluate yourself: 10
Question 1.
The mean pairing energy and octahedral field splitting energy of [Mn(CN)₆]^3- are 28,800 cm^-1 and 38500 cm^-1 respectively. Whether this complex is stable in low spin or high spin?
Answer:
High spin Mn³⁺ complex
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 21

Question 2.
Draw energy level diagram and indicate the number of electrons in each level for the complex [Cu(H₂O)₆]²⁺. Whether the complex is paramagnetic or diamagnetic?
Answer:
[Cu(H₂O)₆]²⁺

Since one unpaired electron is present, the complex is paramagnetic and C.N is 6 the geometry is Octahedral.

Question 3.
For the [CoF₆]^3- ion the mean pairing energy is found to be 21000 cm^-1. The magnitude of ∆₀ is 13000cm^-1. Calculate the crystal field stabilization energy for this complex ion corresponding to low spin and high spin states. [CoF₆]^3-
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 23

Since the high spin complex possesses a large nature CFSE value. It is more favoured.

12th Chemistry Guide Coordination Chemistry Additional Questions and Answers

Part II – Additional Questions

I. Choose the best answer.

1. A theory of co-ordination compounds was proposed by …………….
a) Rutherford
b) J.J.Thomson
c) Alfred Werner
d) Neils Bohr
Answer:
c) Alfred Werner

2. The primary valence of a metal ion refers to …………….
a) atomic number
b) oxidation state
c) atomic mass
d) Co-ordination number
Answer:
b) oxidation state

3. The secondary valence of a metal ion refers to …………….
a) atomic number
b) oxidation state
c) atomic mass
d) Co-ordination number
Answer:
d) Co-ordination number

4. Primary valence of a metal ion is always satisfied by …………….
a) Positive ions
b) negative ions
c) neutral molecules
d) all of the above
Answer:
b) negative ions

5. Among the following complexes, which one shows Zero Crystal field stability: energy (CFSE) is (PTA- 6)
a) [Mn(H₂O)₃]³⁺
b) [Fe(H₂O)₆]3³⁺
c) [Co(H₂O)₆]²⁺
d) [CO(H₂O)₆]³⁺
Answer:
b) [Fe(H₂O)₆]3³⁺

6. The sphere in which the metal ion and ligands are firmly attached is called
a) outer sphere
b) ionization sphere
c) Coordination sphere
d) none of the above
Answer:
c) Coordination sphere

7. The sphere in which the metal ion and other ions are loosely bound is called
a) Inner sphere
b) Ionization sphere
c) Coordination sphere
d) none of the above
Answer:
b) ionization sphere

8. The groups which satisfy the secondary valence are called
a) Central metal ion
b) ligands
c) Complexion
d) Complex
Answer:
b) ligands

9. In a complex a central metal ion acts as a
a) Lewis acid
b) Lewis base
c) Bronsted acid
d) Bronsted base
Answer:
a) Lewis acid

10. In a complex a ligand acts as a
a) Lewis acid
b) Lewis base
c) Bronsted acid
d) Bronsted base
Answer:
b) Lewis base

11. In a complex primary valence of a metal ion is
a) ionizable
b) non-ionisable
c) directional
d) Coordination number
Answer:
a) ionizable

12. In a complex secondary valence of a metal ion is
a) ionizable
b) non-ionisable
c) non-directional
d) Oxidation state
Answer:
b) non – ionisable

13. In a complex primary valence of a metal ion is
a) non-ionizable
b) directional
c) non-directional
d) Co-ordination number
Answer:
c) non-directional

14. According to spectrochemical series which of the following ligand produces strongest field and cause maximum splitting? (PTA -1)
a) F^–
b) CO
c) H₂O
d) C^–
Answer:
b) CO

15. In the complex K₄[Fe(CN)₆] the central metal ion is
a) K⁺
b) Fe²⁺
c) Fe³⁺
d) CN^–
Answer:
b) Fe²⁺

16. In the complex K₄[Fe(CN)₆] the ligand is
a) K⁺
b) Fe²⁺
c) Fe³⁺
d) CN^–
Answer:
d) CN^–

17. In the complex K₄[Fe(CN)₆] the primary valence of the central metal ion is
a) +2
b) +3
c) +4
d) +6
Answer:
a) +2

18. In the complex K₄[Fe(CN)₆] the secondary valence of the central metal ion is
a) 2
b) 3
c) 4
d) 6
Answer:
d) 6

19. The geometry of the complex K₄[Fe(CN)₆] is
a) Square planar
b) tetrahedral
c) Octahedral
d) trigonal bipyramidal
Answer:
c) Octahedral

20. K [PtCl₃(C₂H₄)] was called as
a) Wilkinson Catalyst
b) Zeiglar Natta catalyst
c) Magnus’s green salt
d) Zeise’s salt
Answer:
d) Zeise’s salt

21. Which among the following is not a neutral ligand?
a) aqua
b) ammine
c) Oxalato
d) Pyridine
Answer:
c) Oxalato

22. Which is a monodentate ligand?
a) Carbonato
b) Oxalato
c) Cyanido
d) en
Ans :
c) Cyanido

23. [Co (NH₃)₅ NO₂]²⁺ ion exhibits
a) Ionisation isomerism
b) Linkage isomerism
c) Coordination isomerism
d) solvate isomerism
Answer:
b) Linkage isomerism

24. [Pt (NH₃)₄] [Pd(Cl)₄] complex exhibits
a) Ionisation isomerism
b) Linkage isomerism
c) Coordination isomerism
d) solvate isomerism
Answer:
c) Coordination isomerism

25. [Cr(NH₃)₄ClBr] NO₂ and [Cr(NH₃)₄ClNO₂] Br are
a) Linkage isomers
b) Ionisation isomers
c) Coordination isomers
d) solvate isomers
Answer:
b) Ionisation isomers

26. Which among the following gives a curdy white precipitate with silver nitrate solution?
a) [Co (NH₃)₄ Cl Br] Br
b) [Co (NH₃)₄ Br₂] Cl
c) [Cr (NH₃)₄ Cl Br] NO₂
d) [Cr (NH₃)₄ ClNO₂] Br
Answer:
b) [Co (NH₃)₄ Br₂] Cl

27. Which among the following gives three moles of silver chloride with silver nitrate solution?
a) [Cr(H₂O)₄ Cl₂]Cl. 2H₂O
b) [Cr(H₂O)₅ Cl]Cl₂. 2H₂O
c) [Cr(H₂O)₆]Cl₃
d) [Cr(H₂O)₃ Cl₃]. 3H₂O
Answer:
c) [Cr(H₂O)₆]Cl₃

28. Which among the following gives a white precipitate with barium chloride solution?
a) [Co (NH₃)₅ Cl] Cl₂
b) [Cu(NH₃)₄] SO₄
c) [Ag (NH₃)₂] Br
d) [Co (NO₂)₃ (NH₃)₃]
Answer:
b) [Cu (NH₃)₄] SO₄

29. According to VBT, in a Coordination complex the bonding between central metal ion and ligands is
a) ionic
b) covalent
c) metallic
d) vander waals force
Answer:
b) covalent

30. In which of the following complexes the central metal ion undergoes SP3 hybridisation
a) [Ni(CN)₄]^2-
b) [Pt(NH₃)₄]²⁺
c) [NiCl₄]^2-
d) [Cu (NH₃)₄]²⁺
Ans :
c) [[NiCl₄]^2-

31. If the (n-1) d orbitals are involved in hybridisation, the complex is called as
a) Outer orbital complex
b) Spin paired complex
c) High spin complex
d) Spin free complex
Answer:
b) Spin paired complex

32. If the n d orbitals are involved in hybridisation, the complex is called as
a) Inner orbital complex
b) low spin complex
c) high spin complex
d) spin paired complex
Answer:
c) high spin complex

33. Which statement is incorrect?
a) [Ni(CO)₄] – Tetrahedral, Paramagnetic
b) [Ni(CN)₄]^2- – Square planar, diamagnetic
c) [Ni(CO)₄] – Tetrahedral, diamagnetic
d) [Ni(Cl)₄]^2- – Tetrahedral, Paramagnetic
Answer:
a) [Ni(CO)₄] – Tetrahedral, Paramagnetic

34. In [Ni(CO)₄], the hybridisation of central metal ion is
a) dsp²
b) sp³
c) d²sp³
d) dsp³
Answer:
b) Sp³

35. Co-ordination number of Ni in [Ni(C₂O₄)₃]^4- is (PTA – 4)
a) 3
b) 6
c) 4
d) 2
Answer:
b) 6

36. According to CFT, the bonding between central metal ion and ligands is
a) ionic
b) covalent
c) metallic
d) Vanderwaal’s force
Answer:
a) ionic

37. Which is a strong field ligand?
a) I^–
b) CN^–
c) Cl^–
d) S^2-
Answer:
b) CN^–

38. Which is a weak field ligand?
a) NO^–₂
b) NH³
c) CO
d) SCN^–
Answer:
d) SCN^–

39. The observed colour of a coordination compound can be explained using
a) Valence bond theory
b) Werner’s theory
c) Crystal field theory
d) Molecular orbital theory
Answer:
c) Crystal Field theory

40. Which among the following is a mononuclear carbonyl?
a) [C0₂(C0)₈]
b) [Fe₃(CO)₁₂]
c) [Fe(CO)₅]
d) [Fe₂(CO)₉]
Answer:
c) [Fe (CO)₅]

41. If the instability constant value of [Cu(NH₃)₄]²⁺ is 1.0 × 10^-12, its stability constant value is:
a) 1.0 × 10^-12
b) 1.0 × 10¹²
c) 12
d) -12
Answer:
b) 1.0 × 10¹²

42.

Complexion	Instability constant (∝)
[Fe (SCN)]²⁺	1.0 × 10^-3
[Cu(NH₃)₄]²⁺	1.0 × 10^-12
[Ag (CN)₂]^–	1.8 × 10^-19
[Co (NH₃)₆]³⁺	6.2 × 10^-36

From the above table which of the following complex is most stable?
a) [Cu (NH₃)₄]²⁺
b) [Fe (SCN)]²⁺
c) [CO(NH₃)₆]³⁺
d) [Ag(CN)₂]^–
Answer:
c) [CO(NH₃)₆]³⁺

43. Phthalo blue – a bright blue pigment is a complex of
a) Ni²⁺
b) Co³⁺
c) Cu²⁺
d) Ag⁺
Ans:
c) Cu²⁺

44. For removing lead poisoning, the chelating ligand used is
a) DMG
b) EDTA
c) en
d)CO
Answer:
b) EDTA

45. The process in which coordination complexes are used in the extraction of silver and gold from their ores is
a) Mond’s process
b) Alumino thermic process
c) Mac – Arthur – Forrest cyanide process
d) Bessimerisation
Answer:
c) Mac- Arthur – Forrest cyanide process

46. Ni ions present in nickel chloride solution is estimated accurately using the ligand
a) DMG
b) EDTA
c) en
d) CO
Ans:
a) DMG

47. Catalyst used for hydrogenation of alkenes is
a) Ziegler – Natta Catalyst
b) Zeise’s Salt
c) Wilkinson’s catalyst
d) Magnus’s green Salt
Answer:
c) Wilkinson’s Catalyst

48. The catalyst used in the polymerization of ethene is
a) Ziegler – Natta Catalyst
b) Zeise’s Salt
c) Wilkinson’s catalyst
d) Magnus’s green Salt
Answer:
a. Ziegler – Natta Catalyst

49. The complex used as an antitumor drug in cancer treatment is
a) Ca – EDTA Chelate
b) Cisplatin
c) Trans – Platin
d) Cyano cobalamine
Answer:
b) Cis – Platin

50. RBC is composed of
a) Mg²⁺
b) Fe²⁺
c) Fe³⁺
d) Co⁺
Answer:
b) Fe²⁺

51. In RBC and chlorophyll, the ligand is
a) EDTA
b) DMG
c) Porphyrin
d) en
Answer:
c) Porphyrin

II. Match the following
Question 1.

Complex	Type
i) K₂[Ni(CN)₄l	a) Neutral complex
ii) [Co(NH₃)₅NO₂]Cl₂_	b) Metal carbonyl
iii) [Co(NH₃)₃C1₃]	c) anionic complex
iv) [Ni(CO₄]	d) Cationic complex

Answer:

Complex	Type
i) K₂[Ni(CN)₄l	c) anionic complex
ii) [Co(NH₃)₅NO₂]Cl₂_	d) Cationic complex
iii) [Co(NH₃)₃C1₃]	a) Neutral complex
iv) [Ni(CO₄]	b) Metal carbonyl

Question 2.

Isomer	Reason
i) Coordination isomers	a) Exchange of free solvent molecules with ligand
ii) Linkage isomers	b) Different ions in solution
iii) Ionisation isomers	c) Interchange of one or more ligands
iv) Solvate isomers	d) Bonding through different donor atoms

Answer:

Isomer	Reason
i) Coordination isomers	c) Interchange of one or more ligands
ii) Linkage isomers	d) Bonding through different donor atoms
iii) Ionisation isomers	b) Different ions in solution
iv) Solvate isomers	a) Exchange of free solvent molecules with ligand

Question 3.

Carbonyl	Type
i) [Ni(CO)₄]	a) Polynuclear carbonyl
ii) [Fe₃(CO)₁₂]	b) Bridged carbonyl
iii) Fe₂(CO)₉	c) Mononuclear carbonyl

Answer:

Carbonyl	Type
i) [Ni(CO)₄]	c) Mononuclear carbonyl
ii) [Fe₃(CO)₁₂]	a) Polynuclear carbonyl
iii) Fe₂(CO)₉	b) Bridged carbonyl

Question 4.

Carbonyl	Geometry
i) Chromium hexacarbonyl	a) Tetrahedral
ii) Iron pentacarbonyl	b) Octahedral
iii) Nickel tetra carbonyl	c) Trigonal bipyramidal

Answer:

Carbonyl	Geometry
i) Chromium hexacarbonyl	b) Octahedral
ii) Iron pentacarbonyl	c) Trigonal bipyramidal
iii) Nickel tetra carbonyl	a) Tetrahedral

Question 5.

Hybridisation	Geometry
i) sp	a) Square planar
ii) sp²	b) Octahedral
iii) sp³	c) Trigonal bipyramidal
iv) dsp²	d) Trigonal planar
v) dsp³	e) Lineal
vi) d²sp³	f) Tetrahedral

Answer:

Hybridisation	Geometry
i) sp	e) Lineal
ii) sp²	d) Trigonal planar
iii) sp³	f) Tetrahedral
iv) dsp²	a) Square planar
v) dsp³	c) Trigonal bipyramidal
vi) d²sp³	b) Octahedral

III. Pick the odd man out

1. Pick the odd man out w.r.t complexion
a) [CO(NH₃)₆] Cl₃
b) [Fe (H₂O)₆] Cl₂
c) [Cu(NH₃)₄]SO₄
d) K₄ [Fe(CN)₆]
Answer:
K₄ [Fe(CN)₆] – Anionic complex others are cationic complexes

2. Pick the odd man out w.r.t geometry
a) [CO(NH₃)₆] Cl₃
b) [Fe (H₂O)₆] Cl₂
c) [Cu(NH₃)₄]SO₄
d) K₄ [Fe(CN)₆]
Answer:
c) [Cu(NH₃)₄]SO₄

3. Pick the add man out w.r.t primary valence of the metal ion
a) [CO(NH₃)₆]Cl₃
b) K₃[Fe(CN)₆]
c) [Fe(CO)₅]
d) [CO(NH₃)₃ Cl₃]
Answer:
c) [Fe(CO)₅] – Primary valence is zero, in other complexes the primary valence is +3

4. Pick the odd man out w.r.t the nature of the ligand
a) aqua
b) carbonyl
c) nitrosyl
d) nitrato
Answer:
d) nitrato – negative ligand others are neutral ligand

IV. Pick the wrong statement

Question 1.
i) The secondary valence of a metal ion is satisfied by ligands.
ii) The secondary valence determines the geometry of the complex.
iii) The secondary valence is oxidation state of the metal ion.
iv) If the secondary valence is six, the geometry is tetrahedral.
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
d) (iii) & (iv)
Correct statement : (iii) The secondary valence is the coordination number of the metal ion (iv) If the secondary valence is six, the geometry is octahedral.

Question 2.
i) Central metal ion and ligands are bonded through ionic bond.
ii) Central metal ion accepts electron pairs donated by ligands.
iii) Ligands act as lewis acids
iv) Charge on the coordination sphere is the net change inside the coordination sphere.
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
b) (i) & (iii)
Correct statement: (i) Central metal ions and ligands are bonded through a coordinate covalent bond, (iii) Ligands act as Lewis bases.

Question 3.
i) Trans [Co Cl₂ (en)₂]⁺ is optically active.
ii) In octahedral complexes if nd orbitals are involved in hybridisation, they are called as low spin complexes.
iii) In a complex if all the electrons are paired, it is diamagnetic
iv) Ligands which cause the pairing of electrons are called strong field ligands.
a) (i) &(ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
a) (i) & (ii)
Correct statement:
i) Trans [Co Cl₂ (en)₂]⁺ is optically inactive.
ii) In octahedral complexes if nd orbitals are involved in hybridisation, they are called high spin complexes.

Question 4.
i) The ligands present on the right side of the spectrochemical series are called strong field ligands.
ii) The ligands present on the left side of the spectrochemical series are called weak field ligands.
iii) CFSE (∆E₀ = E_iso – E_LF
iv) \(\mu_{S}=\sqrt{n(n-2)}\)
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
c) (iii) & (iv)
Correct statement: (iii) CFSE (∆E₀) = E_LF – E_iso
iv) \(\mu_{s}=\sqrt{n(n+2)}\)

V. Pick the Correct statement

Question 1.
i) The primary valence of a metal ion is positive in most cases and zero in certain cases.
ii) The primary valences are directional
iii) The primary valence is satisfied by ligands in certain cases.
iv) The primary valence is co-ordination number.
a) (i) &(ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
b) (i) & (iii)
Correct statement: (ii) The primary valences are non – directional
(iv) The primary valence is oxidation state.

Question 2.
i) Coordination entity consists of central metal ion and ligands.
ii) In K₄[Fe(CN)₆], the coordination entity is [Fe(CN)₆]^4-
iii) The groups present inside the coordination entity can be ionised
iv) The coordination entity is not responsible for the geometry of the complex.
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
a) (i) & (ii)
Correct statement : (iii) The groups present inside the coordination entity can not be ionised.
(iv) The coordination entity is responsible for the geometry of the complex.

Question 3.
i) In thiocyanate ligand if sulphur forms a coordination bond with metal, the ligand is named thiocyanato-KN
ii) The ligand NH₃ is named amine.
iii) Ethylenediamine is a neutral ligand.
iv) The name of cationic ligand ends with – ium
a) (i) &(ii)
b) (i) & (iii)
c) (iii) & (iv)
d) (i) & (iii)
Answer:
c) (iii) & (iv)
Correct statement : (i) In thiocyanate ligand if sulphur forms a coordination bond with metal, the ligand is named thiocyanato-kS.
(ii) The ligand NH₃ is named as ammine.

Question 4.
i) Linkage isomer is not possible with nitrite ligand.
ii) Coordination isomerism is found in coordination compounds in which both cation and anion are complexions.
iii) Ionisation isomer arises when an ionisable counter ion itself can act as a ligand.
iv) In solvate isomer if the solvent molecule is alcohol it is called a hydrate isomer.
a) (i) &(ii)
b) (i) & (iii)
c) (iii) & (iv)
d) (i) & (iii)
Answer:
b) (ii) & (iii)
Correct statement : (i)Lingake isomer is possible with nitrite ligand (ii) In solvate isomer if the solvent molecule is water it is called as hydrate isomer.

VI. Assertion and Reason

i) Both A and R are correct, R explains A.
ii) A is correct, R is wrong
iii) A is wrong, R is correct
iv) Both A and R are correct, but R does not explain A.

Question 1.
Assertion (A) : [Co(NH₃)₄ Br₂] Cl and [CO(NH₃)₄ Cl Br] Br are ionisation isomers.
Reason (R) : In solution, they give free Cl^– and Br^– ions respectively
Answer:
i) Both A and R are correct, R explains A.

Question 2.
Assertion (A): Geometrical isomerism exists in homoleptic complexes.
Reason (R): Geometrical isomerism is due to different possible three-dimensional spatial arrangements of the ligands around the central metal atom.
Answer:
iii) A is wrong, R is correct.
Correct Assertion (A): Geometrical isomerism exists in heteroleptic complexes.

Question 3.
Assertion (A) : According to VBT the ligand → metal bond in a coordination complex, is covalent in nature.
Reason (R) : Mutually shared electrons are provided equally by the central metal atom and the ligand
Answer:
ii) A is correct, R is wrong Correct Reason: Mutually shared electrons are provided by ligands to the central metal atom.

VII. Two Mark Questions

Question 1.
Write any two medicinal uses of coordination compounds? (PTA – 6)
Answer:
Ca-EDTA chelate is used in the treatment of lead and radioactive poisoning. That is for removing lead and radioactive metal ions from the body.
cisplatin is used as an antitumor drug in cancer treatment.

Question 2.
What is a central metal ion?
Answer:
The metal ion which accepts electron pair donated by ligands is called the central metal ion of a complex.
Central metal ions are electron-pair acceptors.
Central metal ions are Lewis acids.

Question 3.
What are ligands?
Answer:
In a complex, the negative ions, neutral molecules positive ions which donate pair of electrons to the central metal ion are called ligands.
Ligands are electron-pair donors.
Ligands are Lewis bases.

Question 4.
[Fe(CN₆)]^4- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions, why? (PTA – 2)
Answer:

[Fe(CN₆)]^4- and [Fe(H₂O)₆]²⁺ show different colours in dilute solution because
CN^– is a strong field ligand and H₂O is a weak ligand hence magnitude of CFSE is different
both CN^– and H₂O absorb same wavelength of energy.
complexes of weak field ligands are generally colourless
the sizes of CN^– and H₂O are different hence their colours are also different.

Question 5.
Define coordination number.
Answer:
The number of ligand donor atoms bonded to a central metal ion in a complex is called the coordination number of the metal.

Question 6.
What is the coordination sphere?
Answer:
The complexion containing the central metal atom/ion and the ligands attached to it is collectively called a coordination sphere or inner sphere.

Question 7.
What are inert and labile complexes? (PTA – 4)
Answer:
In some cases, complexes can undergo rapid ligand substitution, such complexes are called labile complexes. However, some complexes undergo ligand substitution very slowly (or sometimes no substitution), such complexes are called inert complexes.

Question 8.
What is d-d transition?
Answer:
The excitation of d – electrons of central metal ion from the lower energy t_2g level to the higher energy eg level by the absorption of light is known as d-d transition.

Question 9.
What are metallic carbonyls?
Answer:
Metal carbonyls are the transition metal complexes of carbon monoxide containing metal-carbon bond.

Question 10.
What is stability constant? (PTA – 5)
Answer:
The reciprocal of dissociation equilibrium constant or instability constant (∝) of a complex is called as the formation equilibrium constant or stability constant (β)

Question 11.
Calculate the magnetic moment and magnetic property of [CoF₆]^3- (MARCH 2020)
Answer:

Complex	[CoF₆]^3-
magnotic propertv	No. of unpaired electrons = 4 Hence paramagnetic.
magnetic moment	\(\mu_{\mathrm{s}}=\sqrt{n(n+2)}=\sqrt{4(4+2)}\) = 4.899 B.M.

VIII. Three Mark questions

Question 1.
Explain geometrical isomerism in square planar complexes.
Answer:

Geometrical isomerism exists in heteroleptic complexes due to different possible three-dimensional spatial arrangements of the ligands around the central metal atom.
In square planar complexes of the form [MA₂B₂]^n± and [MA₂BC]^n±, if similar groups present on the same side of the metal atom it is called Cis – isomer and if they are on the opposite sides of the metal atom it is called trans isomer.
Square planar complex of the type [M(xy)₂]^n± where XY is a bidentate ligand with two different coordinating atoms and of the type [MABCD]^n± by considering anyone ligands as a reference, the rest can be arranged in three different ways also show Cis – trans isomerism.

Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 24

Question 2.
Explain geometrical isomerism in octahedral complexes.
Answer:

Octahedral complexes of the type [MA₂B₄]^n±, [M(XX)₂B₂]^n± show Cis- trans isomerism Here A and B are monodentate ligands and XX is a bidentate ligand with two same kind of donor atoms.
In the above diagram the positions (1,2), (1.3) (1,4) (1,5), (2,3) (2,5), (2,6) (3,4) (3,6), (4,5), (4,6) and (5,6) are identical and if two similar groups are present in any one of these positions, the isomer is called as Cis isomer.
Similarly, positions (1, 6), (2,4) and (3, 5) are identical and if two similar groups are present in these positions, the isomer is called as trans-isomer.
(ex) [Co (NH₃)₄Cl₂]⁺

Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 25

Question 3.
Write about facial and meridional isomers.
Answer:

Octahedral complex of the type [MA₃B₃]^n± shows geometrical isomerism
If the three similar ligands (A) are present in the corners of one triangular face of the octahedron and the other three ligands (B) are present in the opposing triangular face, then the isomer is called as a facial isomer (fac isomer)
If the three similar ligands are present around the meridian which is an imaginary semicircle from one apex of the octahedron to the opposite apex, the isomer is called as a meridional isomer (mer isomer)
This is called meridional because each set of ligands can be regarded as lying on a meridian of an octahedron

Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 27

Question 4.
In a tetrahedral crystal field, draw the figure to show splitting of d-orbitals (PTA – 6)
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 28

Question 5.
Write the use of complex formation in photography.
Answer:
In photography, when the developed film is washed with sodium thiosulphate solution (hypo), the negative film gets fixed. Undecomposed AgBr forms a soluble complex called
sodiumdithiosulphatoargentate(I) which can be easily removed by washing the film with water.
AgBr + 2Na₂S₂O₃ → Na₃[Ag (S₂O₃)₂] + 2NaBr

Question 6.
Write the use of metal complexes in biological systems.
Answer:

RBC is composed of heme group which is a complex of Fe²⁺ and Porphyrin ligand. It carries oxygen from lungs to tissues.
Chlorophyll, a green pigment present in green plants, algae is a complex of Mg²⁺ and corrin ring, a modified porphyrin ligand. It is responsible for the photosynthesis and the conversion of CO₂ and water into carbohydrates and oxygen.
The only vitamin with a metal ion Vitamin B₁₂ (Cyanocobalamine) is a complex of Co⁺ and porphyrin ligand.
Many enzymes which regulate biological processes are metal complexes.
(ex) Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in digestion is a zinc complex with protein-ligand.

Question 7.
How is Cisplatin used as an antitumour drug in cancer treatment?
Answer:

C is platin is a square planar coordination complex Cis – [Pt (NH₃)₂ Cl₂]
It is a platinum-based anti-cancer drug
This drug undergoes hydrolysis and reacts with DNA to produce various cross-links.
These cross-links hinder the DNA replication and transcription, which results in cell growth inhibition and ultimately cell death.
It also cross-links with cellular proteins and inhibits mitosis.

IX. Five Mark Questions

Question 1.
How are coordination complexes classified?
Answer:
Coordination complexes are classified into two types.
I. Classification based on the net charge on the complex
A coordination compound in which the complexion
(i) Carries a net positive charge is called a cationic complex. (Ex). [Ag(NH₃)2]⁺
(ii) Carries a net negative charge is called an anionic complex. (Ex). [Ag(CN)₂]^–
(iii) Bears with no net charge is called a neutral complex. (Ex). [Ni(CO)₄]

II. Classification based on kind of ligands.
A coordination compound in which
(i) The central metal ion/ atom is coordinated to only one kind of ligand is called a homoleptic complex. (Ex). [Ag(NH₃)₂]⁺
(ii) The central metal ion/ atom is coordinated to more than one kind of ligands is called a heteroleptic complex. (Ex). [Co (NH₃)₅ Cl]²⁺

Question 2.
Write the steps involved in the IUPAC nomenclature of coordination compounds.
Answer:
i) The cation is named first, followed by the anion regardless of whether the ion is simple or complex.
ii) The simple ions are named as in other ionic compounds.
iii) To name a complexion, the ligands are named first followed by the central metal atom/ ion. If more than one ligand is present they are named in alphabetical order.

(a) Naming the ligands:

Name of anionic ligands end with the letter ‘O’.
Name of cationic ligands end with ‘ium’.
Neutral ligands are usually called with their molecular names with fewer exception namely H₂O (aqua), NH₃ (ammine) etc.,
K-term is used to denote an ambidentate ligand in which more than one coordination mode is possible.
If the coordination entity contains more than one ligand of a particular type, the multiples of ligand (2,3,4, …..) is indicated by adding Greek prefixes (di,tri,tetra ….) to the name of the ligand.
If the name of the ligand contains a Greek prefix (eg. ethylene diamine) alternate prefixes (bis, tris, tetrakis ) can be used.
These numerical prefixes are not taken into account for alphabetising the name of ligands.

(b) Naming the central metal atom/ion.

In cationic/neutral complexes, the element name is used as such for naming the central metal atom/ion.
In anionic complexes a suffix ‘ate’ is used along with the element name or its Greek or Latin name if any.

Question 3.
Explain structural isomerism exhibited by coordination compounds.
Answer:

The coordination compounds with same formula, but have different connections among their constituent atoms are called structural isomers or constitutional isomers.
It is of four types.

i) Linkage isomers:
This type arises when an ambidentate ligand is bonded to the central metal atom/ion through either of its two different donor atoms.
(ex). [Co (NH₃)₅ NO₂]²⁺ – nitritoK-N ligand N – attached.
[Co (NH₃)₅ ONO]²⁺ – nitrito K-O ligand O – attached.

ii) Coordination isomers:
This type arises in the coordination compounds having both the cation and anion as complex ion.

The interchange of one or more ligands between the cationic and the anionic coordination entities result in different isomers.
(ex). [Co (NH₃)₆] [Cr (CN)₆] &
[Cr (NH₃)₆] [CO (CN)₆]

iii) Ionisation isomers:

This type arises when an ionisable counter ion itself can act as a ligand.
The exchange of such counterions with one or more ligands in the coordination entity will result in ionisation isomers.
These isomers will give different ions in solution.
(ex). [Co (NH₃)₄ Br₂] Cl &
[Co (NH₃)₄ Cl Br] Br.

iv) Solvate isomers:
This arises by the exchange of free solvent molecules such as water, ammonia, alcohol etc., in the crystal lattice with a ligand in the coordination entity.

If the solvent molecule is water, these isomers are called hydrate isomers. (ex).[Cr(H20)6]Cl3;
[Cr (H₂O)₅ Cl] Cl₂ H₂O;
[Cr (H₂O)₄ Cl₂] Cl 2H₂O.

Question 4.
Explain the main assumption of VBT of coordination compounds (PTA – 1)
Answer:

The ligand → Metal bond in coordination complex is an aicnl in nature, it is formed by sharing ol electrons between theienlrul metal atom and the ligand.
Each ligand should have at least one filled orbital containing a lone pair of electrons.
To accept the electron pairs donated by the ligands, the central metal ion must contain vacant orbitals equal to its coordination number.
This vacant orbital of central metal atom undergo hybridisation.
The vacant hybridized orbitals of central metal ion linearly overlap with filled orbital of the ligands forming coordinate covalent sigma bonds.
The hybridized orbitals are directional and give definite geometry to the complexion.
In octahedral complexes, if (n-1) d orbitals are involved in hybridisation, they are called inner orbital complexes or low spin complexes or spin paired complexes. If nd orbitals are involved in hybridisation, they are called as outer orbital complexes, or high spin complexes or spin-free complexes.
If the metal ion contains unpaired electrons the complex is paramagnetic, if the electrons are paired, the complex is diamagnetic.
Ligands such as CO, CN^–, en and NH₃ which cause pairing of electrons are called strong field ligands.
Ligands such as, F^–, Cl^–, Br^– which do not cause pairing of electrons are called weak field ligands.
Greater the overlapping between the ligand orbitals and the hybridised metal orbital, greater is the bond strength.
The relation between crystal field splitting energies of octahedral and tetrahedral field is \(\Delta t=\frac{4}{9} \Delta 0\)

Question 5.
Write the salient features of Crystal Field Theory.
Answer:

Crystal field theory assumes that the bond between the ligand and the central metal atom is purely ionic, ie. the bond is formed due to the electrostatic attraction between the electron-rich ligand and the electron-deficient metal.
In case of charged metal ions or ligands, they are considered as point charges and in case of neutral metal atoms or ligands they are considered as electric dipoles.
The complex formation is considered as the following series of hypothetical steps.

Step-1:
In an isolated gaseous state, all the five ‘d’ orbitals of the central metal ion are degenerate. Initially, the ligands form a spherical field of negative charge around the metal. In this field the energies of all the five d orbitals will increase due to the repulsion between the electrons of the metal and the ligand.

Step-2:
The ligands are approaching the metal atom in actual bond directions. In an octahedral field the central metal ion is located at the origin and the six ligands are coming from +x, -x, +y, -y, +z and -z directions.
Orbitals lying along the axes d_x²-y² and d_z² (e_g) will experience strong repulsion and raise in energy to a greater extent than the orbitals with lobes directed between the axes (d_xy/ d_yz, d_zx) (t_2g). This splitting of degenerated orbitals into two sets is called crystal field splitting.

Step-3:
Upto this point the complex formation would not be favoured. When the ligands approach further, there will be an attraction between the negatively charged electron and the positively charged metal ion resulting in a net decrease in energy. This decrease in energy is the driving force for the complex formation.

In the case of the tetrahedral complex, none of these orbital is directly pointing to the ligands. The d_xy, d_yz, d_zx (t_2g) orbitals are nearer to the ligands and hence they interact to a greater extent with the ligand orbitals, therefore d_xy, d_yz, d_zx (t_2g) orbitals become high energy orbitals than d_x²-y² and d_z² (e_g) orbitals in a tetrahedral complex.
Thus the mode of splitting of d orbitals in an octahedral complex is just the reverse of that observed in a tetrahedral complex.

Question 6.
Explain the classification of metallic carbonyls.
Answer:
Metallic carbonyls are classified in two different ways.

I. Classification based on the number of metal atoms present
a) Mononuclear carbonyls:
These contain only one metal atom.
(ex). [Ni (CO)₄]

b) Polynuclear carbonyls
These contain two or more metal atoms. They may be homonuclear or heteronuclear.
(ex). [Co₂ (CO)₈]; [MnCo (CO)₉]

II. Classification based on the structure:
a) Non bridged metal carbonyls.
These do not contain any bridging carbonyl ligands.
Those contain only terminal carbonyls.
(ex) Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 29
Those contain terminal carbonyls and metal-metal bonds

b) Bridged carbonyls:
These contain one or more bridging carbonyls along with terminal carbonyl ligands and one or more metal-metal bonds, (ex).
Samacheer Kalvi 12th Chemistry Guide Chapter 5 Coordination Chemistry 31

Question 7.
Write the IUPAC names for the following complexes. (MARCH 2020)

[Co (NH₃)₅Cl]²⁺
K₃ [Fe (C₂O₄)₃]
[Co <NH₃)₆] [Cr(CN)₆]
[Cr (H₂O)₄ Cl₂] Cl 2H₂O
[Ft (Py)₄] [Pt Cl₄]
[Zn(NCS)₄]^2-
[Ag(NH₃)₂]²⁺

Answer:

Pentaamminechloridocobalt (III)ion
Potassium trioxalato ferrate (III)
Hexaammine Cobalt (III) hexa cyanido K-C chromate (III)
Tetra aqua dichlorido chromium (III) chloride dihydrate
Tetrapyridine platinum (II) tetra chloridoplatinate (II)
Tetrathiocyanato – kN Zincate (II) ion
Diammine Silver (I) ion

Question 8.
Write the formula for the following coordination compounds.

Tris (ethylenediamine) Chromium (III) Chloride
Potassium tetracyanido K-C nickelate (II)
Ammine bromido chlorido nitrito – kN platinate (II) ion
Dichlorido bis (ethane -1, 2 – diamine) Platinum (IV) nitrate
Hexa aqua manganese (II) Phosphate

Answer:

[Cr (en)₃] Cl₃
K₂[Ni(CN)₄]
[Pt (NH₃) BrCl No₂]^–
[Pt (en)₂ Cl₂] (N0₃)₂
[Mn (H₂O)₆]₃ (PO₄)₂

Question 9.
What will be the correct order for the wavelengths of absorption in the visible region and explain for the following (PTA – 3)
[Ni(NO₂)₆]^4-; [Ni(NH₃)₆]²⁺; [Ni(H₂O)₆]²⁺
Answer:
The central metal ion in all three complexes is the same. Therefore, absorption in the visible region depends on the ligands. The order in which the CFSE values of the ligands increases in the spectrochemical series is as follows:
H₂O < NH₃ < NO₂^–
Thus, the amount of crystal-field splitting observed will be in the following order.
∆_0(H₂O) < ∆_0(NH₃)< ∆_0(NO₂^–)
Hence, the wavelengths of absorption in the visible region will be in the order.
[Ni(H₂O)₆]²⁺ >[Ni(NH₃)₆]²⁺ > [Ni(NO₂)₆]^4-

Question 10.
Answer all the questions for the complex [Fe(en)₂Cl₂]Cl₂ (PTA -6)

Oxidation number of Fe
Hybridisation and shape
Magnetic behaviour
Number of geometric isomers
Whether there may be optical isomer also?
IUPAC name

Answer:

Oxidation number of Fe is +3
Hybridisation and shape is -d²sp³
Paramagnetic due to presence of three unpaired electrons
Two cis and trans isomers
Yes, cis isomer will also show optical isomerism
Dichlorido bis (ethane 1, 2 diamine) iron (III) chloride (or)
Dichloro bis (ethylenediamine) iron (III) chloride.

Samacheer Kalvi 12th Chemistry Guide Chapter 4 Transition and Inner Transition Elements

January 1, 2021

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 4 Transition and Inner Transition Elements Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements

12th Chemistry Guide Transition and Inner Transition Elements Text Book Questions and Answers

Part – I – Text Book Evaluation

I. Choose the correct answer

1. Sc (Z = 21) is a transition element but Zinc (Z = 30) is not because
a) both Sc³⁺ and Zn²⁺ ions are colourless and form white compounds
b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled
c) last electron as assumed to be added to 4s level in case of zinc
d) both Sc and Zn do not exhibit variable oxidation states
Answer:
b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled

2. Which of the following d block element has half-filled penultimate d sub shell as well as half-filled valence sub shell?
a) Cr
b) Pd
c) Pt
d) none of these
Answer:
a) Cr

3. Among the transition metals of 3d series, the one that has highest negative \(\left(\mathrm{M}^{2+} / \mathrm{M}\right)\) standard electrode potential is
a) Ti
b) Cu
c) Mn
d) Zn
Answer:
a) Ti

4. Which one of the following ions has the same number of unpaired electrons as present in V³⁺?
a) Ti³⁺
b) Fe³⁺
c) Ni²⁺
d) Cr³⁺
Answer:
c) Ni²⁺

5. The magnetic moment of Mn²⁺ ion is
a) 5.92BM
b) 2.80BM
c) 8.95BM
d) 3.90BM
Answer:
a) 5.92BM

6. The catalytic behaviour of transition metals and their compounds is ascribed mainly due to
a) their magnetic behaviour
b) their unfilled d orbitals
c) their ability to adopt variable oxidation states
d) their chemical reactivity
Answer:
c) their ability to adopt variable oxidation states

7. The correct order of increasing oxidizing power in the series
Samacheer Kalvi 12th Chemistry Guide Chapter 4 Transition and Inner Transition Elements 1
Answer:
a)VO₂⁺ < Cr₂O₇^2- < MnO₄^–

8. In acid medium, potassium permanganate oxidizes oxalic acid to
a) oxalate
b) Carbon dioxide
c) acetate
d) acetic acid
Answer:
b) Carbon dioxide

9. Which of the following statements is not true?
a) on passing H₂S, through acidified K₂Cr₂O₇ solution, a milky colour is observed
b) Na₂Cr₂O₇ is preferred over K₂Cr₂O₇ in volumetric analysis
c) K₂Cr₂O₇ solution in acidic medium is orange in colour
d) K₂Cr₂O₇ solution becomes yellow on increasing the pH beyond 7
Answer:
b) Na₂Cr₂O₇ is preferred over K₂Cr₂O₇ in volumetric analysis

10. Permanganate ion changes to ________ in acidic medium
a) MnO₄^2-
b) Mn²⁺
c) Mn³⁺
d) MnO₂
Answer:
b) Mn²⁺

11. How many moles of I₂ are liberated when 1 mole of potassium dichromate react with potassium iodide?
a) 1
b) 2
c) 3
d) 4
Answer:
c) 3

12. The number of moles of acidified KMnO₄ required to oxidize 1 mole of ferrous oxalate (FeC₂O₄) is
a) 5
b) 3
c) 0.6
d) 1.5
Answer:
c) 0.6

13. Which one of the following statements related to lanthanons is incorrect?
a) Europium shows +2 oxidation state
b) The basicity decreases as the ionic radius decreases from Pr to Lu.
c) All the lanthanons are much more reactive than aluminium.
d) Ce⁴⁺ solutions are widely used as oxidising agents in volumetric analysis.
Answer:
c) All the lanthanons are much more reactive than aluminium.

14. Which of the following lanthanoid ions is diamagnetic?
a) Eu²⁺
b) Yb²⁺
c) Ce²⁺
d) Sm²⁺
Answer:
b) Yb²⁺

15. Which of the following oxidation states is most common among the lanthanoids?
a) 4
b) 2
c) 5
d) 3
Answer:
d) 3

16. Assertion : Ce⁴⁺ is used as an oxidizing agent in volumetric analysis
Reason : Ce⁴⁺ has the tendency of attaining +3 oxidation state.
a) Both assertion and reason are true and reason is the correct explanation of assertion.
b) Both assertion and reason are true but reason is not the correct explanation of assertion.
c) Assertion is true but reason is false
d) Both assertion and reason are false.
Answer:
a) Both assertion and reason are true but reason is the correct explanation of assertion.

17. The most common oxidation state of actinoids is (PTA -2)
a) +2
b) +3
c) +4
d) +6
Answer:
b) +3

18. The actinoid elements which show the highest oxidation state of +7 are
a) Np, Pu, Am
b) Li, Fm, Th
c) U, Th, Md
d) Es, No, Lr
Ans : a) Np, Pu, Am

19. Which one of the following is not correct? (PTA -5)
a) La (OH)₃, is less basic than Lu (OH)₃
b) In lanthanoid series ionic radius of Ln³⁺ ions decreases
c) La is actually an element of transition metal series rather than lanthanide series
d) Atomic radii of Zr and Hf are same because of lanthanide contraction,
Answer:
a) La(OH)₃, is less basic than Lu(OH)₃

II. Answer the following questions

Question 1.
What are transition metals? Give four examples (PTA – 3)
Answer:

Transition metal is an element whose atom has an incomplete d sub shell or which can give rise to cations with an incomplete d sub shell.
They occupy the central position of the periodic table, between s-block and p-block elements.
Their properties are transitional between highly reactive metals of s-block and elements of p-block which are mostly non-metals.
Example – Iron, Copper, Tungsten, Titanium.

Question 2.
Explain the oxidation states of 4d series elements.
Answer:

The oxidation states of 4d series elements vary from +3 for Y to +8 for Ru.
The highest oxidation state of 4d elements are found in their compounds with higher electronegative elements like O, F and Cl.
Ex: In RuO₄, the oxidation state of Ru is +8.

Question 3.
What are inner transition elements?
Answer:

1. The inner transition elements have two series of elements.

Lanthanoids
Actinoids

2. Lanthanoid series consists of 14 elements from Cerium (₅₈Ce) to Lutetium (₇₁Lu) following Lanthanum (₅₇La). These elements are characterised by the preferential filling of 4f orbitals. .

3. Actinoids consists of 14 elements from Thorium (₉₀Th) to Lawrencium (₁₀₃Lr) following Actinium (₈₉Ac). These elements are characterized by the preferential filling of 5f orbital.

The elements which in their elemental or ionic form have partly filled f orbitals are called f block elements.
As the f orbitals lie inner to the penultimate shell, therefore these elements having partially filled f orbitals, are also called inner transition elements.

Question 4.
Justify the position of lanthanides and actinides in the periodic table. (PTA – 1)
Answer:
Lanthanides:

The actual position of lanthanides in the periodic table is at group number 3 and period number 6.
In the sixth period, the electrons are preferentially filled in inner 4f-sub shell.
The fourteen elements following lanthanum (Ce to Lu) show similar chemical properties.
Hence they are grouped together and placed at the bottom of the periodic table.
This position is justified as follows:
(1) General electronic configuration:
[ Xe] 4f^1-145d^0,16S²
(2) Common Oxidation state: +3
(3) They have similar physical and chemical properties.

Actinides:

The actual position of actinides in the periodic table is at group number 3 and period number 7.
In the seventh period, the electrons are perferentially filled in inner 5f-Sub shell.
The fourteen elements following actinides (Th to Lr) show similar chemical properties.
Hence they are grouped together and placed at the bottom of the periodic table.
This position is justified as follows:
(1) General electronic configuration:
[Rn] 5f^2-146d^0-27S²
(2) Common oxidation state: +3
(3) They have similar physical and chemical properties.

Question 5.
What are actinides? Give three examples.
Answer:
The fourteen elements following actinium is from thorium to lawrencium are called actinoids.
Examples: Thorium, uranium, Neptunium.

Question 6.
Describe the preparation of potassium dichromate. (PTA – 2)
Answer:

The fourteen elements following actinium ,i.e., from thorium (Th) to lawrentium (Lr) are called actinoids.
All the actinoids are radioactive and most of them have short half lives.
The heavier members being extremely unstable and not of natural occurrence. They are produced synthetically by the artificial transformation of naturally occuring elements by nuclear reactions.
Example – Thorium, Uranium, Plutonium, Californium.

Question 7.
What is lanthanide contraction and what are the effects of lanthanide contraction? (PTA – 3)
Answer:
As we move across 4f series, the atomic and ionic radii of lanthanoids show gradual decrease with increase in atomic number. This decrease in ionic size is called lanthanide contraction.

Consequences:
Basicity differences:
As we from Ce³⁺ to Lu³⁺, basic character of LU³⁺ ions decrease. Due to the decrease in the size of LU³⁺ ions, the ionic character of Ln-OH bond decreases (covalent character increases)

Similarities among lanthanoids:
In the complete f-series very small change in radii of Lanthanoids are observed and their chemical properties a requite similar.

Second and third transition series resemble each other due to lanthanide contraction.

Question 8.
Complete the following.
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 4 Transition and Inner Transition Elements 5

Question 9.
What are interstitial compounds?
Answer:
An interstitial is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice. They are non-stoichiometric compounds.
Ex: Tic, ZrH_1.92

Question 10.
Calculate the number of unpaired electrons in Ti³⁺, Mn^2+, and calculate the spin-only magnetic moment.
Answer:
Electronic configuration of Ti³⁺ = [Ar]3d¹
It has only one unpaired electron, ie. n = 1
Spin only magnetic moment
\(=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{1(1+2)}=\sqrt{3}\) = 1.732BM

Electronic configuration of Mn²⁺ = [Ar] 3d⁵
It has five unpaired electrons ie. n = 5
Spin only magnetic moment
\(=\sqrt{n(n+2)}=\sqrt{5(5+2)}=\sqrt{35}\) = 5.92BM

Question 11.
Write the electronic configuration of Ce⁴⁺ and Co²⁺
Answer:
Electronic cofiguration of Ce⁴⁺ = [Xe] 4f⁰5d⁰6s⁰
Electronic configuration of Co²⁺ = [Ar] 3d⁷.

Question 12.
Explain briefly +2 states become more and more stable in the first half of the first-row transition elements with increasing atomic number.
Answer:
In the elements of the first half of the first row, with the removal of valance 4s electrons (+2 oxidation state) the 3d orbital get gradually occupied. Since the number of empty d orbital decreases, the stability of the cations (M²⁺) increases from Sc²⁺ to Mn²⁺.

Question 13.
Which is more stable? Fe³⁺ or Fe²⁺ – explain.
Answer:

Electronic configuration of Fe³⁺ = [Ar] 3d⁵. It consists of 5 unpaired electrons. Half filled and stable.
Electronic configuration of Fe²⁺ = [Ar]3d⁶. It consists of 4 unpaired electrons, partially filled d subshell.
Hence Fe³⁺ is more stable than Fe²⁺.

Question 14.
Explain the variation in E°_{M³⁺/M²⁺} 3d series.
Answer:
Transition metals in their high oxidation states tend to be oxidising. The standard reduction potential for the M³⁺/M²⁺ half cell gives the relative stability between M³⁺ and M²⁺.

Reaction	Standard reduction potential (V)
Ti³⁺ +e^– → Ti²⁺	-0.37
y³⁺ + e^– → V²⁺	-0.26
Cr³⁺ +e^– → Cr²⁺	-0.41
Mn³⁺ + e^– → Mn²⁺	+1.51
Fe³⁺ + e^– → Fe²⁺	+0.77
Co³⁺ + e^– → Co²⁺	+1.81

The negative values for titanium, vanadium and chromium indicate that the higher oxidation state is preferred.

The high reduction potential of M³⁺/M²⁺ indicates Mn²⁺ is more stable than Mn³⁺

Question 15.
Compare lanthanides and actinides.
Answer:

Lanthanides	Actinides
1. Differentiating electron enters in 4f orbital	The differentiating electron enters in 5f orbital
2. Binding energy of 4f orbitals are higher	Binding energy of 5f orbitals are lower
3. Do not form complexes	Form complex compounds
4. Most of the lanthanoids are colourless	Must of the actinoids are coloured . U³⁺red, U⁴⁺ Green, UO₂²⁺ Yellow
5. They do not form oxocations.	They form oxocations Ex. UO₂²⁺
6. Maximum oxidation state is +4	Maximum oxidation state is +7

Question 16.
Explain why Cr²⁺ is slrongly reducing while Mn³⁺ is strongly oxidizing. (PTA – 5)
Answer:
E° value for Cr³⁺/Cr²⁺ is negative (-0.41 V), where as E° Value for Mn³⁺/Mn²⁺ is positive (+1.57V).

Hence Cr²⁺ ions can easily undergo oxidation to give Cr³⁺ ions and therefore, acts as strong reducing agent. On the otherhand, Mn³⁺ can easily undergo reduction to give Mn²⁺ and hence acts as oxidising agent.

Question 17.
Compare the ionization enthalpies of first series of the transition elements.
Answer:

Ionisation energy of transition element is intermediate between those of s and p block elements.
As we move from left to right in a transition metal series, the ionisation enthalpy increase.
This is due to the increase in nuclear charge corresponding to the filling of d electrons.
The following figure shows the trends in ionisation enthaply of first series of transiton elements.

In 3d series moving from	Ionisation energy
Sc to Ti	increases
Ti to Cr	no change
Cr to Mn and Fe	increases
Fe to Co, Ni and Cu	not much change
Cu to Zn	increases

Samacheer Kalvi 12th Chemistry Guide Chapter 4 Transition and Inner Transition Elements 7

Question 18.
Actinoid contraction is greater from element to element than the lanthanoid contraction why?
Answer:
This is due to poor shielding by 5f electrons in actinoids as compared to that by 4f electrons in lanthanoids.

Question 19.
Out of Lu(OH)₃ and La(OH)₃ which is more basic and why?
Answer:

La(OH)₃	Lu(OH)₃
1. Size of La³⁺ ion is larger	Size Lu³⁺ ion is smaller
2. La-OH is more ionic	Lu-OH is less ionic
3. La-OH bond is less covalent	Lu-OH bond is more covalent
4. Hence La(OH)₃ is. more basic.	Hence Lu(OH)₃ is. less basic.

Question 20.
Why europium (II) is more stable than cerium (II)
Answer:

Ion	Cerium (II)	Europium (II)
Electronic configuration	[Xe] 4f¹5d¹ 6S^o	[Xe] 4f⁷5d°6S°
Nature of Orbitals	PartialK filled unstable d orbitals and f – orbitals	Mall filled stable f-orbitals
Stability of ion	Less stable	More stable

Question 21.
Why do zirconium and Hafnium exhibit similar properties?
Answer:
This is because Zr and Hf have similar atomic sizes which is due to lanthanoid contraction.

Question 22.
Which is stronger reducing agent Cr²⁺ or Fe²⁺?
Answer:
Cr²⁺ is a stronger reducing agent than Fe.
Reason :
E°(Cr³⁺/Cr²⁺) is negative (-0.41 V) where as E°(Fe³⁺/Fe²⁺) is positive (+0.77V). Thus Cr³⁺ is easily oxidised to Cr³⁺ but Fe²⁺ cannot be easily oxidised to Fe³⁺. Hence Cr²⁺ is a stronger reducing agent than Fe²⁺.

Question 23.
The E°_M⁺/M value for copper is positive. Suggest a possible reason for this.
Answer:
E°_M⁺/M for any metal depends upon the sum of the enthalpy changes taking place in the following steps.
M_(s) + ∆H_(a) → M_(g) (∆H_(a) = enthalpy of atomisation)
M_(g) + ∆H_(i) → M²⁺_(g) (∆H_(i) = ionisation enthalpy)
M²⁺_(g) + ∆_aq → M²⁺_(aq) (∆H_(Hyd) = hydration enthalpy)

Coppes possesses a high enthalpy of atomisation and low enthalpy of hydration. Hence E° Cu²⁺/ Cu is positive.

Question 24.
Describe the variable oxidation on state of 3d series elements.
Answer:

First transition metal Sc exhibits only +3 oxidation state
All other transition elements exhibit variable oxidation states by losing electrons from (n-l)d orbital and ns orbital
Because the energy difference between these two orbitals is very small.

(ex)

atom/ion	Electronic configuration
Fe	3d⁶4S²
Fe²⁺	3d⁶4S⁰
Fe³⁺	3d⁵4S⁰

At the beginning of the series, +3 oxidation is stable.
Number of oxidation states increases with the number of electrons available.
Number of oxidation states decreases as the number of paired electrons increases.
Hence the first and last elements show less number of oxidation states and the middle elements show more number of oxidation states.
(Ex): First element Sc has only one oxidation state +3. Middle elements Mn has Six different oxidation states from +2 to +7. Last element Cu has +1 and +2 oxidation states.
Relative stability of different oxidation states of 3d-metals is correlated with the extra stability of half filled and fully filled electronic configurations.
(Ex): Mn²⁺(3d⁵) is more stable than Mn⁴⁺(3d³)

Question 25.
Which metal in the 3d series exhibits +1 oxidation state most frequently and why?
Answer:
Copper has electronic configuration 3d¹⁰4s¹. It can easily lose 4s¹ electron to give the stable 3d¹⁰ configuration. Hence, it shows +1 oxidation state.

Question 26.
Why first ionization enthalpy of chromium is lower than that of Zinc.
Answer:

Chromium	Zinc
i) Electronic configuration is 3d⁵s¹	Electronic Configuration is 3d¹⁰4s²
ii) Easily loses one electron from 4s orbital to attain stable half filled d-orbital configuration	Does not easily lose one electron from completely filled 4s orbital.
iii) Hence first ionisation energy of chromium is less.	Hence first ionisation energy of zinc is more.

Question 27.
Transition metals show high melting points, why? (PTA -6)
Answer:
Transition metals show high melting points due to the presence of unpaired d-electrons available for metallic bonding.

III. Evaluate Yourself

Question 1.
Compare the stability of Ni⁴⁺ and Pt⁴⁺ from their ions enthalpy value.
Answer:

IE	Ni	Pt
I	737	864
II	1753	1791
III	3395	2800
IV	5297	4150

The value of the ionisation enthalpies can be used is estimating the relative stability.
Pl⁴⁺ relative to Ni⁴⁺ can be explained as follows:
Ni⁴⁺ – IE₃ +IE₄ = (3395 +5297) = 8692 KJmol^-1
Pt⁴⁺ – IE₃ +IE₄ = (2800 +4150) = 6950 KJmol^-1
Pt⁴⁺ requires lesser energy as compared to the formation of Ni⁴⁺
Therefore Pt⁴⁺ compounds are more stable than Ni⁴⁺ compounds.

Question 2.
Why ion is more stable in +3 oxidation state than in +2 and the reverse is true for manganese?
Answer:
(i) Atomic number of Fe=26
(ii) Electric configuration of Fe = [Ar] 3d⁶4S²
(iii) After removal of two electrons [Fe²⁺]=[Ar]3d⁶
(iv) Further, with removal of one more electron it becomes Fe³⁺ and d⁵ (half filled) that is most stable.
(v) Atomic number of Mn=25
(vi) Electronic configuration of Mn= [Ar] 3d⁵4S²
(vii) After removal of two outer electrons it will become Mn²⁺ and remaining d will be half filled d⁵ and that is most stable.
(viii) Removal of one electron Mn³⁺ having d⁴ configuration which is incompletely filled.

12th Chemistry Guide Transition and Inner Transition Elements Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

1. Transition elements are good conductors because
a) They are metals
b) They are all solids
c) They have free electrons in outer energy orbits
d) All of these
Answer:
d) All of these

2. Transition elements are
a) All metals
b) All non metals
c) Metals and non metals
d) Gases
Answer:
a) All metals

3. Transition elements form complexes very readily because of
a) Small cation size
b) Vacant – d – orbitals
c) Large ionic charge
d) All are correct
Answer:
d) All are correct

4. The transition metal present in vitamin B12 is
a) Fe
b) Co
c) Ni
d) Na
Answer:
b) Co

5. Transition elements are frequently used as catalysts, because of
a) Large ionic charge
b) Large surface area for the reactions to be absorbed
c) Unpaired d electrons
d) Both (b) and (c) are correct
Answer:
d) Both (b) and (c) are correct

6. d – Block elements are arranged in
a) Three series
b) Six series
c) Two series
d) Four series
Ans : d) Four series

7. d – Block element generally for
a) Covalent hydrides
b) Metallic hydrides
c) Interstitial hydrides
d) Salt like hydrides
Answer:
c) Interstitial hydrides

8. Metals which are hard and lustrous substances with high melting points form highly coloured compounds are known as
a) Alkali metals
b) Alkaline earth metals
c) Transition metals
d) None of these
Answer:
c) Transition metals

9. In the first transition series, the incoming electrons enters
a) 5 d orbitals
b) 4 d orbitals
c) 3 d orbitals
d) 2 d orbitals
Answer:
c) 3 d orbitals

10. Transition elements form alloys easily because they have
a) Same atomic number
b) Same electronic configuration
c) Nearly same atomic size
d) None
Answer:
c) Nearly same atomic size

11. Transition elements that show anaomalous electronic configuration in first series are.
a) Cr and Ni
b) Cu and Co
c) Fe and Ni
d) Cr and Cu
Answer:
d) Cr and Cu

12. Lightest transition element is
a) Fe
b) Sc
c) Os
d) None
Answer:
b) Sc

13. Densest transition element is
a) Fe
b) Sc
c) Os
d) Mn
Answer:
b) Sc

14. Variable valencies of transition elements is due to
a) Different energies of (n -1) d electrons
b) Different energies of ns electrons
c) Similar energies of (n – 1)d electrons
d) Similar energies of (n – 1)d and ns electrons
Answer:
d) Similar energies of (n-l)d and ns electrons

15. Which of the following ions has minimum ionic radius
a) Ni²⁺
b) Co²⁺
c) Cr²⁺
d) V²⁺
Answer:
a) Ni²⁺

16. Ionic radii of zirconium and hofminium become almost identical because
a) They are d block elements
b) They belong to the same group
c) Of increased nuclear charge
d) Of lanthanoids contraction
Answer:
d) Of lanthanoids contraction

17. The colour of Fe ions is
a) Blue
b) Light green
c) Very dark green
d) Yellow
Answer:
b) Light green

18. Magnetic property of transition metal is due to
a) Spin of electron
b) Orbital moment
c) Both
d) Neither of the two
Answer:
a) Spin of electron

19. All ferromagnetic substances
a) Can be magnetised
b) Can be electrolysed
c) Have completely filled d – orbitals
d) None
Answer:
a) Can be magnetised

20. Maxmium paramagnetism in 3d series shown by
a) Mn
b) Co
c) Ni
d) Fe
Answer:
a) Mn

21. Which of the following has the minimum magnetic moment
a) Mn²⁺
b) Fe²⁺
c) Cr²⁺
d) V³⁺
Answer:
d) V³⁺

22. Paramagnetism in the substance increases as
a) The number of paired electrons increases
b) The number of unpaired electrons increases
c) The number of unpaired electrons decreases
d) The number of paired electrons decreases
Answer:
b) The number of unpaired electrons increases

23. The number of unpaired electrons present in Cr³⁺ ion is
a) 1
b) 2
c) 3
d) 4
Answer:
c) 3

24. Which of the following has more unpaired d electrons
a) Zn²⁺
b) Fe²⁺
c) N^3-
d) Cu⁺
Answer:
b) Fe²⁺

25. The lanthanoids contraction is responsible for the fact that
a) Zr and Y have the same radius
b) Zr and Nb have similar oxidation state
c) Zr and Hr have almost the same radius
d) Zr and Zn have same oxidation state
Answer:
c) Zr and Hr have almost the same radius

26. Which of the following is coloured
a) Cu⁺
b) Cu²⁺
c) Ti⁴⁺
d) V⁵⁺
Answer:
Cu⁺

27. In the four successive transition elements (Cr, Mn, Fe & Co) the stability of +2 oxidation state will be are in which of the following order? (PTA – 5)
a) Fe>Mn>Co>Cr
b) Co>Mn>Fe>Cr
c) Cr>Mn>Co>Fe
d) Mn>Fe>Cr>Co
Answer:
d) Mn>Fe>Cr>Co

28. Chromyl chloride when dissolved in NaOH solution gives yellow solution. The yellow solution contains (PTA – 3)
a) Cr₂O₇^2-
b) CrO₄^2-
c) CrO₅
d) Cr₂O₃
Answer:
b) CrO₄^2-

29. In the dichromate anion (Cr₂O₇)^2- (PTA – 1)
a) 4Cr – O bonds are equivalent
b) 6 Cr – O bonds are equivalent
c) All Cr – O bonds are equivalent
d) All Cr-O bonds are non-equivalent
Answer:
b) 6 Cr – O bonds are equivalent

30. Which of the following forms colourless compound?
a) Sc³⁺
b) V³⁺
c) Ti³⁺
d) Cr³⁺
Answer:
a) Sc³⁺

31. The basic character of the transition metal monoxides follows the order (Atomic number, Ti = 22; V = 23; Cr = 24)
a) VO > CrO > TiO > FeO
b) Cro > VO > FeO > TiO
c) TiO > FeO > VO > CrO
d) TiO > VO > CrO > FeO
Answer:
d) TiO > VO > CrO > FeO

32. The correct order of ionic radii of Y³⁺, La³⁺, EU³⁺, and Lu³⁺ is (Atomic number Y = 39; La = 57; Eu = 60; Lu = 71)
a) Y³⁺ < La³⁺ < Eu³⁺ < Lu³⁺
b) Y³⁺ < Lu³⁺ < Eu³⁺ < La³⁺
c) Lu³⁺ < Eu³⁺ < La³⁺ < Y³⁺
d) La³⁺ < Eu³⁺ < Lu³⁺ < Y³⁺
Answer:
b) Y³⁺ < Lu³⁺ < Eu³⁺ < La³⁺

33. During the smelting process silica is added to roasted copper to remove
a) Copper sulphide
b) Ferrous sulphide
c) Ferrous oxide
d) Ferrous Chloride
Answer:
c) Ferrous oxide

34. The ore which contains copper and iron both
a) Malachite
b) Chalcopyrite
c) Chalocite
d) Azurite
Answer:
b) Chalcopyrite

35. According to Ellingham diagram, the oxidation reaction of carbon to carbon monoxide may be used to reduce which one of the following oxides at lowest temperature?
a) Al₂O₃
b) Cu₂O
c) MgO
d) ZnO
Answer:
a) Al₂O₃

36. In electro chemical process (Electrolysis of fused salt) is used to extract
a) Iron
b) Lead
c) Sodium
d) Silver
Answer:
c) Sodium

37. In metallurgy, flux is a substance used to convert
a) Mineral into silicate
b) Fusible impurities to soluble impurities
c) Infusible impurities to soluble impurities
d) none of these
Answer:
c) Influsible impurities to soluble impurities

38. Heating of Iron pyrites in air to remove sulphur is called
a) Fusion
b) Calcination
c) Smelting
d) Roasting
Answer:
d) Roasting

39. Which of the following metal is leached by cyanide process?
a) Silver
b) Sodium
c) Aluminium
d) Copper
Answer:
a) Silver

40. The elements in which extra electrons enter into (n – 2) orbitals are called
a) f – block elements
b) p – block elements
c) d – block elements
d) s – block elements
Answer:
a) f – block elements

41. The most common oxidation state of lanthanoides
a) +4
b) +2
c) +6
d) +3
Answer:
d) +3

42. ………………. form complexes.
a) lanthanoides
b) actinides
c) Thorium
d) Cerium
Answer:
b) Actnides

43. ……………. form oxocations.
a) Actnides
b) lanthanoides
c) s – block
d) p – block
Answer:
a) Actnides

44. The correct electronic configuration of Gd³⁺ is
a) [Xe]4f¹⁴
b) [Xe]4f⁷
c) [Xe]4f⁰
d) [Xe]4f⁷
Answer:
b) [Xe]4f⁷

45. The general electronic configuration of lanthanoides
a) [Xe]4f^0-14
b) [Xe]5d^0-1
c) [Xe]4f^2-145d^0-1bs²
d) [Xe]4f^0-145d^1-10 6s²
Answer:
c) [Xe]4f^2-145d^0-1bs²

46. As we move from lanthanum to Lutetium, basic character of hydroxides.
a) decreases
b) increases
c) decrease and then increases
d) none of these
Answer:
a) decreases

47. The colour of U³⁺ is
a) red
b) green
c) yellow
d) pink
Answer:
a) red

II. Assertion and Reason

Question 1.
Assertion : For chromium, the ground state electronic configuration is 3d⁵4s¹ rather than 3d⁴4s².
Reason : The energy of the Cr atom is lower, when the six valence electrons are in different atomic orbitals with parallel spins
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason not is the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false but Reason is true
Answer:
A) Both Assertion and Reason are true and Reason is the correct explanation of A.

Question 2.
Assertion : The energies of the 5s and 4d orbitals are very close.
Reason: The relative energies of the 4d and 5s orbitals vary with the nuclear charge and the electronic distribution.
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false but Reason is true
Answer:
B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 3.
Assertion : The densities, melting and boiling points of the transition elements are generally very high
Reason: Zn, Hg and Cd have low melting and boiling points as the d – block is complete.
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false bu t Reason is true
Answer:
B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 4.
Assertion: The transition metals more similar to one another than are representative metals of group 1 and group 2
Reason: Inner d orbitals are being filled
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason is the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false but Reason is true
Answer:
A) Both Assertion and Reason are true and Reason is the correct explanation of A.

Question 5.
Assertion : All the transition elements are metals and good conductors of heat and electricity.
Reason : The penultimate shell of electrons of all these elements is expanding and they are, therefore, expected to have physical and chemical properties in common.
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason is the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false but Reason is true
Answer:
A) Both Assertion and Reason are true and Reason is the correct explanation of A.

III. Match the following

1. Match the catalysts given in column I with the processes given in column II

Column I (catalyst))	Column II process
a. Ni in the presence of hydrogen	i) Ziegler natta catalyst
b. CuCl₂	ii) Contact process
c. V₂O_s	iii) Vegetable oil to ghee
d. Finely divided iron	iv) Sandmeyer reaction
e. TiCl₄ + Al(C₂H₅)₃	v) Haber’s process
	vi) Decomposition of KCl0₃

Answer:

Column I (catalyst))	Column II process
a. Ni in the presence of hydrogen	iii) Vegetable oil to ghee
b. CuCl₂	iv) Sandmeyer reaction
c. V₂O_s	ii) Contact process
d. Finely divided iron	v) Haber’s process
e. TiCl₄ + Al(C₂H₅)₃	i) Ziegler natta catalyst

2. Match the properties given in column I with the metals in colum II

Property	Metal
a. An element shows + 8 oxidation state	i) Zero
b. 3d series element shows + 70xidation state	ii) Osmium
c. 3d series element shows high melting point	iii) Manganese
d. oxidation state metal in metal carbonyls	iv) Chromium

Answer:

Property	Metal
a. An element shows + 8 oxidation state	ii) Osmium
b. 3d series element shows + 70xidation state	iii) Manganese
c. 3d series element shows high melting point	iv) Chromium
d. oxidation state metal in metal carbonyls	i) Zero

3. Match the statements given in column I with oxidation states given in column II

a. oxidation state of Mn in MnO₂	i) +3
b. Most stable O.S of Mn	ii) +7
c. Most stable O.S Mn in oxides	iii) +2
d. Common O.S of lanthanoids	iv) +4

Answer:

a. oxidation state of Mn in MnO₂	iv) +4
b. Most stable O.S of Mn	iii) +2
c. Most stable O.S Mn in oxides	ii) +7
d. Common O.S of lanthanoids	i) +3

4. Match the following column I with column II

Column I	Column II
a) Permagnaic acid	i) CrO
b) Chromic acid	ii) Cr₂O₃
c) Dichromic acid	iii) H₂CrO₄
d) Dichromic acid	iv) CrO₃
	v) HMnO₄

Answer:

Column I	Column II
a) Permagnaic acid	v) HMnO₄
b) Chromic acid	iv) CrO₃
c) Dichromic acid	iii) H₂CrO₄
d) Dichromic acid	ii) Cr₂O₃

5. Match the following

a) Lanthanoides which shows + 4 O.S	i) Pm
b) Lanthanoides which shows +2 O.S	ii) Ce
c) Radio active lanthanide	iii) Gd
d) Lanthanoide has 4f⁷electronic configuration	iv) Eu

Answer:

a) Lanthanoides which shows + 4 O.S	ii) Ce
b) Lanthanoides which shows +2 O.S	iv) Eu
c) Radio active lanthanide	i) Pm
d) Lanthanoide has 4f⁷electronic configuration	iii) Gd

IV. Choose the correct statements

Question 1.
Consider the following statements which is/are correct.
I. Sc ion has no unpaired electrons and the expected magnetic moemnt value is zero B.M
II. The oxidation state of Cr in Cr₂O₇^2- is +4
III. The spin only magnetic moment is given by \(\sqrt{n(n+2)} B M\)
a) I, III
b) II, I
c) III, II
d) II
Answer:
a) I, III

Question 2.
Consider the following statements which is / are correct
I. All Lanthanoids displaces hydrogen from acidified water.
II. The colour of tripositive Lanthanides becomes darker as one goes from Cr³⁺ to LU³⁺
III. The ionic size of tripositive Lanthanoid ions increases with atomic number
a) I
b) II
c) II, III
d) I, III
Answer:
a) I

Question 3.
Consider the following statement which is / are correct.
I. The maximum oxidation state of Osmium is +8
II. The highest oxidation state of a transition metal is given by outermost ‘s’ plus ‘d’ electrons.
III. The maximum magnetic moment is shown by the ion having d orbital electronic configuration is 3d⁵
a) I
b) II
c) I, III
d) I, II, III
Answer:
d) I, II, III

Question 4.
Consider the following statements which is / are correct.
I. Compared to Cu²⁺ salts Cu⁺ salts are less stable.
II. Iron does not form interstitial compounds
III. Ag⁺ is isoelectronic with Cd²⁺
a) I, III
b) II, III
c) III
d) II
Answer:
a) I, III

V. Choose the incorrect statements

Question 1.
Consider the following statements which is / are incorrect.
I. The common oxidation state of lanthanides is +3
II. All lanthanides form oxocations
III. All lanthanides are non – radioactive
a) I
b) II
c) II, III
d) II, III
Answer:
d) II, III

Question 2.
Consider the following statements which is / are incorrect.
I. The common oxidation state of Lanthanum is +4
II. UO₂²⁺ is colourless.
III. Actindes forms oxocations.
a) I, II
b) II, III
c) III, I
d) I, II, III
Answer:
a) I, II

Question 3.
Consider the following statements which is incorrect regarding potassium dichromate.
a) It oxidises ferric salt to ferrous salts.
b) It oxidises KI to I₂
c) It oxidises H₂S to S
d) None of the above
Answer:
a) It oxidises ferric salt to ferrous salts

Question 4.
Consider the following statements which is / are correct.
I. Potassium permanganate is extracted from chromite iron ore.
II. Potassium dichromate is extracted from Pyrolusite ore.
III. Potassium permangate is mainly used in Tanning of leather.
a) III
b) II
c) I, II
d) I, III
Answer:
c) I, II

VI. Fill in the blanks

1. The electronic configuration of scandium is ……………….. .
Answer:
[Ar] 3d¹4s²

2. The metal with highest electrical conductivity at room temperature is ……………….. .
Answer:
Silver

3. ……………….. values are used to predict the thermodynamic stability of the compounds
Answer:
Ionisation energy

4. The common oxidation state of scandium is ………………… .(MARCH 2020)
Answer:
+3

5. Mn²⁺ is ………………… than Mn⁴⁺
Answer:
More stable than

6. The oxidation state of W in Wcl₆ is
Answer:
+6

7. ………………… metal is unique in 3d series
Answer:
Copper

8. The metal shows maximum ferromagnetic character is
Answer:
Iron

9. Paramagnetism is the property of ………………… .
Answer:
Unpaired electrons

1o. Paramagnetism is common in
Answer:
Transition elements

11. Hydroformylation catalyst is
Answer:
(Co₂(CO)₈)

12. Potassium dichromate is in colour
Answer:
Orange red

13. The oxidation state of chromium in potassium dichromate is
Answer:
+6

14. The colour of potassium permanganate is
Answer:
dark purple colour

15. The oxidation state uranium in UP6 is
Answer:
+6

VII. Choose the correct Pair

1. a) Lead – Lead pencil
b) Cerusite – Zinc
c) Zinc sulphate – antiseptic
d) Lead pipes – softwater
Answer:
a) Lead – Lead pencil

2. a) Ag + hot NaOH – Products
b) Zn + hot NaOH – Products
c) Au + NaOH – Products
d) Cr + NaOH – Products
Answer:
b) Zn + hot NaOH – Products

3. a) Cu²⁺ – Diamagnetic
b) Cu²⁺ – Colourless
c) Cu²⁺ – Zero magnetic moment
d) Cu²⁺ – One unpaired electron
Answer:
d) Cu²⁺ – One unpaired electron

4. a) TiCl₄ – Polymerisation catalyst
b) Ni – Hydrogenation catalyst
c) Fe – Haber’s process
d) All the above pairs are correct
Answer:
All the above pairs are correct

VIII. Choose the incorrect Pair

1. a) Sc – 3d series
b) Zn – 3d series
c) Cr – 3d series
d) Cu – 4d series
Answer:
d) Cu – 4d series

2. a) Mn⁴⁺ – 3d³
b) Mn³⁺ – 3d⁴
c) Mn⁵⁺ – 3d⁵
d) Mn⁵⁺ – 3d²
Answer:
c) Mn⁵⁺ – 3d⁵

3. a) Cu⁺, Zn²⁺ – Diamagnetic
b) Sc³⁺, Ti⁴⁺, V⁵⁺ – Paramagnetic
c) Co⁵⁺, Fe²⁺ – Paramagnetic
d) Cu²⁺ – Paramagnetic
Answer:
b) Sc³⁺, Ti⁴⁺, V⁵⁺ – Paramagnetic

4.a) Potassium dichromate – +6
b) Potassium dichromate – oxidising
c) Potassium dichromate – Chrome tanning
d) Potassium dichromate – Bayer’s reagent
Answer:
d) Potassium dichromate – Bayer’s reagent

IX. Choose the odd man outer

1. Scandium, Titanium, Vanadium, Cadmium
Answer:
Cadmium
Hint : Cadmium belongs to 4d series. All others are 3d series.

2. Zirconium, Niobium, Technicium,
Ruthenium
Answer:
Technicium
Hint: Technicium is radio active. All others are non radio active.

3. U³⁺, UO₂²⁺, U⁴⁺, Ce³⁺
Answer:
Ce⁴⁺
Hint: Ce⁴⁺ is colourless. All others are coloured.

X. Additional Questions – 2 Mark

Question 1.
Write the classfication of transition elements.
Answer:

d – Block elements composed of 3d series (4th period) Scandium to Zinc (10 elements),
4d series (5th period) Yttrium to Cadmium (10 elements)
5d series (6th period) Lanthanum, Hafinium to mercury (10 elements).
6d series (7th period) Actinium, Ruther Fordium to copernicium (10 mercury)

Question 2.
Give the electronic configuration of copper and chromium.
Answer:
The electronic configurations of
Chromium Cr [Ar] 3d⁵ 4S¹
Copper, Cu [Ar] 3d¹⁰ 4S¹

Question 3.
What are the type of packing possible in transition metals?
Answer:

Most of the transition elements are hexagonal close packed,
cubic close packed
body centrered cubic which are the characteristics of true metals.

Question 4.
Why transition elements show variable oxidation states? (PTA – 4)
Answer:
Transition elements exhibit variable oxidation states due to

loosing electrons from (n-1) d orbital
ns orbital as the energy difference between them is very small.

Question 5.
Explain why Mn²⁺ is more stable than Mn³⁺?
Answer:

Mn²⁺ has outer electronic configuration as 3d⁵
Mn³⁺ has outer electronic configuration as 3d⁴
Half filled orbitals and completely filled orbital are more stable than the partially filled orbitals so Mn²⁺ with half filled d orbital is more stable than Mn³⁺

Question 6.
Explain about ferromagnetic materials.
Answer:

Ferromagnetic materials have domain structure.
In each domain the magnetic dipoles are arranged.
But the spin dipoles of the adjacent domains are randomly oriented.
Some transition elements or ions with unpaired d electrons show ferromagnetism.

Question 7.
Explain why the melting and boiling points of Cd, Hg and Zn are low?
Answer:
The elements of group 12 [Zn, Hg, Cd] are quite soft with low melting points. Mercury is a liquid and melts at – 38°C. These elements has no unpaired electrons available for metallic bonding and therefore their melting and boiling points are low.

Question 8.
Which of the following ions would from colourless complexes?
Answer:
Cu²⁺, Zn²⁺, Ti , Ti⁴⁺, Cd²⁺
Zn²⁺ – 3d¹⁰ > Ti⁴⁺ – 3d° > Cd²⁺ – 4d¹⁰

These elements from colourless complexes because they have fully filled d – orbitals and do not possess impaired electrons and so no d – d transition is possible.

Question 9.
State Hume – Rothery rule for alloy formation.
Answer:
According to Hume – Rothery rule to form a substitute alloy the difference between the atomic radii of solvent and solute is less than 15%.

Both the solvent and solute must have the same crystal structure and valence and their electronegativity difference must be dose to zero.

Question 10.
Explain why transition metals form alloys? Transition metals form more alloys themselves because of their
Answer:
a tomic sizes are similar

one metal atom can be easily replaced by another metal atom from its crystal lattice to form an alloy.

Question 11.
Silver atom has completely filled d-orbitals (4d¹⁰) in its ground state, How can you say that it is a transition element?
Answer:
The outer electronic configuration of Ag (Z = 47) is 4d¹⁰ 5S¹, In addition to +1, it shows an oxidation state of +2 also. In +2 oxidation state, the configuration is d⁹
ie. the d – subshell is incompletely filled. Hence, it is transition element.

Question 12.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer:
Oxygen and fluorine have small size high electro negativity. Hence, they can oxidise the metal to the highest oxidation state by prompting all the valance electrons to participate in bonding.

Question 13.
Calculate the ‘Spin only’ magnetic moment of M²⁺ (aq) ion (z=27)
Answer:
Electronic configuration of M with z = 27 is [Ar] 3d⁷4S²
Thus, electronic configuation of M²⁺ will be [Ar] 3d⁷4S°
Samacheer Kalvi 12th Chemistry Guide Chapter 4 Transition and Inner Transition Elements 8

Question 14.
In what way is the electronic configuration of the transition elements different from that of the non – transition elements?
Answer:
Transition elements contain incompletely filled d-sub shell i.e their electronic configuration is (n-l)d^”1( ns0’1, , whereas non – transition elements have either no d- subshell or their

d-subshell is completely filled and have ns^1,2 or n² np^1-6 in their outermost shell.

Question 15.
Name the oxometa! anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
Answer:
Cr²O₇^2- and Cr²O₄^2-
(Group number = Oxidation state of Cr = 6 MnO₄^–
(Group = oxidation stale of Mn=7)

Question 16.
What is the effect of increasing pH of a solution of Potassium dichromate?
Answer:
On increasing pH of potassium dichromate solution (i.e. on adding alkali) it changes to potassium chromate.

Question 17.
Write chemical equations for the reactions involved in the manufacture of potassium permanganate from Pyrolusite. (PTA- 5)
Answer:
Relevant equations for the manufacture of KMn04 from pyrolusite ore are given below:
Samacheer Kalvi 12th Chemistry Guide Chapter 4 Transition and Inner Transition Elements 9

Question 18.
Classify the following elementsintod-block and f-block elements. (MARCH 2020)
Answer:
(i) Tungsten
(ii) Ruthenium
(iii) Promethium
(iv) Einstenium
d-block elements – Tungsten, Ruthenium
f-block elements- promethium, Einstenium

Question 19.
The halides of transition elements become more covalent with increasing oxidation state of the metal why?
Answer:
As the oxidation slate of the metal increases, its charge increases. According to Eajan’s Rules, as the charge of the metal ion increases covalent character increases because the positively charged cation attracts the electron could on the anion towards itself.

Question 20.
Although Cr³⁺ and Co²⁺ ions have the same number of unpaired electrons but the magnetic moment of Cr³⁺ is 3.87 B.M. and that of Co²⁺ is 4.87 B.M. why?
Answer:
Cr³⁺ ion has symmetrical electronic configuration in the outermost orbit i.e. 3d³. In such ions there is no orbital configuration to magnetic moment. However appreciable orbital contribution takes place in Co²⁺ with 3d⁷ configuration.

Question 21.
Why E° value for Mn, Ni and Zn are more megative than expected?
Answer:
Negative values for Mn²⁺ and Zn²⁺ are related to the stabilities of half – filled and fully filled configurations respectively.

Question 22.
Calculate the spin only magnetic moment of Mn²⁺
Answer:
E.C of Mn²⁺ = [Ar] 3d⁵

Question 23.
Why Ni^II complexes are thermodynamically more stable than Pt^II complexes?
Answer:
The ionisation energy of Ni²⁺ less than that of Pt²⁺. Hence Ni²⁺ complexes are more stable than Pt²⁺ complexes.

Question 24.
Why do transition elements and its compounds act as catalyst? (PTA – 5)
Answer:
Transition metals has energetically available d orbitals that can accept electrons from reactant molecule or metal can form bond with reactant molecule using it’s d electrons.

Question 25.
WhatisZeigler-Natta Polymerisation
Answer:
A mixture of TiCl₄ and Tri alkyl aluminium is used for polymerisation.
Samacheer Kalvi 12th Chemistry Guide Chapter 4 Transition and Inner Transition Elements 11

Question 26.
Actinoid atoms are generally coloured? Justify your answer.
Answer:
The actinoid ions are generally coloured. This can be explained interms of unpaired
electrons u nde rgoi ng f – f Transi t ion.

Question 27.
How many unpaired electrons are present in Mn²⁺ ion. (Z=25) How does it influences magnetic behaviour of Mn²⁺ ion.
Answer:
Mn²⁺ :3d⁵ has 5 unpaired electorns. It is highly paramagnetic and it is attracted by magnet.

Question 28.
Transition metal atoms/ions are usually coloured. Justify.
Answer:
Transition metal ions have unpaired electrons. They can undergo d-d transition by absorbing light from visible region and radiating complementary colour.

Question 29.
Give the disproportionation of manganese (VI) in acidic medium.
Answer:
3MnO₄^2- + 4H⁺ → 2MnO^– + MnO₂ + 2H₂O

Question 30.
Oxoanions of a metal show higher oxidation state. Give reason.
Answer:
Oxo anions of a metal show higher oxidation state due to the ability of oxygen to form multiple bonds and its high electronegativity.

XI. Additional Questions – 3 Mark

Question 1.
In the series Sc (Z = 21) to Zn (Z = 30) the enthalpy of atomisation of Zinc is the lowest i.e. 126 KJ mol¹. why?
Answer:
In the series Sc to Zn, all elements have one or more unpaired electrons except zinc which has no unpaired electrons, its outer electronic configuration being 3d¹⁰ 4s². Lower the number of unpaired electrons, lower is the metal – metal bonding. Hence, metal – metal bonding is weakest in zinc. Therefore, enthalpy of atomisation is lowest.

Question 2.
Explain why Cu⁺ ion is not stable in aqueous solutions.
Answer:
Cu²⁺ (aq) is much more stable than Cu⁺ (aq). Although, second ionisation enthalpy of copper is large but ∆H_hyd for Cu²⁺ (aq) is much more negative than thatfor Cu⁺ (aq) and hence it more than compensates for second ionisation enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproport¬ionation.

Question 3.
Write down the electronic configuration of :
Answer:
i) Cr³⁺
ii) Cu⁺
iii) Co²⁺
iv) Mn²⁺
v) Pm³⁺
vi) Ce⁴⁺
vii) Lu²⁺
viii) Th⁴⁺
Answer:

ion	electronic configuration
i) Cr³⁺	[Ar]3d³
ii) Cu⁺	[Ar]3d¹⁰
iii) Co²⁺	[Ar]3d⁷
iv) Mn²⁺	[Ar]3d⁵
v) Pm³⁺	[Xe]4f⁴
vi) Ce⁴⁺	[Xe]4f⁰
vii) Lu²⁺	[Xe]4f¹⁴ 5d¹
viii) Th⁴⁺	[Rn]5f⁰

Question 4.
What are the different oxidation states exhibited by lanthanoids?
Answer:
The most common oxidation state of lanthanoids is +3. However, some lanthanoids also show an oxidation state of +2 and +4. For example, Eu shows an oxidation state of +2 and Ce shows on oxidation state of +4.

Question 5.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reactions with (i) iodide (ii) iron II solution and (iii) H₂S (PTA – 5)
Answer:
i) Cr₂O₇^2- + 14H⁺ + 6I^– → 2Cr³⁺ + 7H₂O + 3I₂
ii) Cr₂O₇^2- + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺+ 7H₂O
iii) Cr₂O₇^2- + 8H⁺ + 3H₂S → 2Cr³⁺ + 7H₂O + 3S

Question 6.
What is meant by disproportionations’? Give two examples of disporportionation reaction in aqueous solution.
Answer:
When the oxidation state of an element in a reactant increases in one of the products and decreases in other product, the phenomenon is called disproportionation. For example, Mn(VI) in MnO₄² changes to Mn (VII) in the product MnO₄ and Mn(IV) in the product MnO₂ as shown by the reaction.
3MnO₄^2- + 4H⁺ → 2MnO_4-+ MnO₂ + 2H₂O
similarly, Cr (V) undergoes disproportionation in acidic medium as follows.
CrQ₄^3-+ 8H⁺ → 2CrO₄^2- + Cr³⁺ + 4H₂O

Question 7.
Calculate the no. of unpaired electrons in the following gaseous ions :
Answer:
Mn³⁺, Cr³⁺, V³⁺ and Ti³⁺ which one of these is the most stable in aqueous solution?
Mn³⁺ : 3d⁴ has 4 unpaired electrons,
Cr³⁺ : 3d³ has 3 unpaired electrons,
V³⁺ : 3d² has 2 unpaired electrons,
Ti³⁺ : 3d¹ has 1 unpaired electrons.
Cr³⁺ is most stable out of these in aqueous solution because it has half filled t₂g level (i.e.t³_2g)

Question 8.
Why is the +2 oxidation state of manganese quite stable, while the same is not true for iron?
Answer:
Half filled configuration is more stable than others. Mn²⁺ is more stable due to half-filled d – orbitals.
Fe²⁺ is not stable because it does not have half filled d – orbitals. Configurations of Mn²⁺ and Fe²⁺ are as follows :
Mn²⁺ : 3d⁵ 4s⁰
Fe²⁺ : 3d⁶ 4s⁰

Question 9.
Write two characteristics of the transition elements. (PTA – 3)
Answer:
All the transition elements are metals.

All the transition metals are good conductors of heat and electricity.

Question 10.
Explain about diamagnetic materials.
Answer:

Materials with no elementary magnetic dipoles are diamagnetic or a species with all paired electrons exhibits diamagnetism.
This kind of materials are repelled by the magnetic field.
Because the presence of external magnetic field, a magnetic induction is introduced to the material which generates weak magnetic field that oppose the applied field.

Question 11.
Explain about paramagnetic materials.
Answer:

Paramagnetic solids having unpaired electrons.
In the absence of external magnetic field, the dipoles are arranged at random and hence the solid shows no net magnetism.
In the presence of magnetic field, the dipoles are aligned parallel to the direction of the applied field and therefore, they are attracted by an external magnetic field.

Question 12.
A substance is found to have a magnetic moment of 3.9 BM. How many number of unpaired electrons does it certain?
Answer:
Magnetic moment µ = 3.9
\(\sqrt{\mathrm{n}(\mathrm{n}+2)}\) = 3.9B.M
n(n + 2) = (3.9)² = 15
n (n + 2 ) = 3(3+ 2)
n = 3
The number of unpaired electrons is 3.

Question 13.
Why Zn²⁺ salts are white while Ni²⁺ salts are coloured? (PTA – 3)
Answer:

Zn²⁺ has the configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰
It has completely filled d orbital and there is no unpaired electrons.
Hence Zn²⁺ is colourless.
Ni²⁺ has the configurations 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸
It has two unpaired electrons and d – d transitions are possible.
Hence Ni²⁺ is coloured.

Question 14.
[Ti(H₂O)₆]³⁺ is coloured while
[Sc(H₂O)₆]³⁺ is colourless. Explain.
Answer:

In [Ti(H₂O)₅]¹⁺ ,Tr³⁺ has outer electronic configuration as 3d¹
In this case d-d transition is possible by the absorption of energy from the visble light and produce purple colour.
But is [Sc(H₂O)₆]³⁺, Sc³⁺ has outer electronic configuration as 3d°.
In this case d-d transition is not possible and it is colourless.

Question 15.
What are the characteristics of interstitial compounds?
Answer:

They are hard and show electrical and thermal conductivity
They have high melting points higher than those of pure metals
Transition metal hydrides are used as powerful reducing agents
Metallic carbides are chemically inert.

Question 16.
Explain why d block elements form more complexes?
Answer:

Transition metal ions are small and highly charged
They have vacant low energy orbitals to accept an electron pair donated by other groups.
Examples : [Fe(CN)₆]^4-, [Co(NH₃)₆]³⁺

Question 17.
Explain chromyl chloride test. (MARCH 2020)
Answer:
When potassium dichromate is heated with any chloride salt in the presence of Cone H₂SO₄, orange red vapours of chromyl chloride (CrO₂Cl₂) is evolved.

This reaction is used in the detection of chloride ions in qualitative analysis.

Question 18.
Write the uses of potassium dichromate.
Answer:

It is used as a strong oxidizing agent.
It is used in dyeing and printing.
It used in leather tanneries for chrome tanning.
It is used in quantitiative analysis for the estimation of iron compounds and iodides.

Question 19.
Write the uses of potassium permanganate.
Answer:

It is used as a strong oxidizing agent.
It is used for the treatment of various skin infections and fungal infections of the foot.
It used in water treatment industries to remove iron and hydrogen sulphide from well water.
It is used as a Bayer’s reagent for detecting unsaturation in an organic compound.
it is used in quantitative analysis for the estimation of ferroussalts, oxalates, hydrogen peroxide and iodides.

Question 20.
What is the action of heat on K₂Cr₂O₇ ?
Answer:
Potassium dichromate on heating gives potassium chromate.

XII. Additional Questions – 5 Mark

Question 1.
For M²⁺ /M and M³⁺ / M systems, the E° values for some metals are as follows.
Cr²⁺ /Cr = 0.9 V ; Cr³⁺ /Cr2+ = – 0.4 V
Mn²⁺ /Mn = -1.2 V ; Mn³⁺/Mn²⁺= +1.5 V
Fe²⁺ /Fe = +0.4 V ;Fe³⁺/Fe²⁺ = +0.8 V
Use this data to comment upon
(a) the stability of Fe³⁺ in acid solution as composed to that of Cr³⁺ or Mn²⁺ and
(b) the ease with which iron can be oxidised as compared to the similar process for either chromium or manganese metals.
Answer:
(a) Cr³⁺ / Cr²⁺ has a negative reduction potential Hence, Cr³⁺ cannot be reduced to Cr²⁺ i.e. Cr³⁺ is most stable. Mn³⁺/ Mn²⁺ has large positive E° value. Hence, Mn³⁺ can be easily reduced to Mn²⁺, i.e. Mn³⁺ is least stable. Eo value for Fe³⁺ / Fe²⁺ is positive but small. Hence, Fe³⁺ is more stable than Mn³⁺ but less stable than Cr³⁺. Thus, the stability follows the order.
Cr³⁺>Fe³⁺>Mn³⁺

(b) Oxidation potentials for the given pairs will be +0.9V, +1.2V, and +0.4V. Thus the order of getting oxidised will be
Mn > Cr > Fe

Question 2.
(i) Why do transition elements show variable oxidation states?
(ii) Why do most transition metal ions exhibit para magnetism?
(iii) How is the magnetic moment of a species related to the number of unpaired electrons?
Answer:
(i) This is because ns and (n-1) d orbitals do not differ much in energy. Electrons from both may take part in bonding. Hence, they show variable valency.
(ii) This is because of presence of unpaired electrons in most of the transition metal ions.
(iii) \(\mu_{\mathrm{s}}=\sqrt{\mathrm{n}(\mathrm{n}+2)}\) BM, where n stands for the number of unpaired electrons.

Question 3.
(i) Why is copper (z=29) considered as transition element?
(ii) K₂PtCl₆ is well – known compound while corresponding Ni compound is not known.
(iii) Why is radius of Fe²⁺ less than that of Mn²⁺?
(iv) Why is electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴ not correct for the ground state of Cr (Z-24)?
Answer:
(i) This is because Cu²⁺ ion has incomplete d – orbital
(ii) This is because Pt⁴⁺ is more stable than Ni⁴⁺ Energy required to remove 4 electrons from pt is less and that in Ni than.
(iii) This is because effective nuclear charge in Fe²⁺ is more compared to that in Mn²⁺
(iv) Correct electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵ because half – filled configuration

Question 4.
(i) Which trivalent cation is the largest in lanthanoid series?
(ii) One unpaired electron in atom contributes a magnetic moment of 1.1 B.M. Calculate the magnetic moment of Cr. (Atomic number = 24)
(iii) In a paramagnet ion, all the bonds formed between Mn and O are covalent. Give reasons.
Answer:
(i) La³⁺ is the largest ion
(ii) As Cr has 6 unpaired electrons, its magnetic moment is 6 x 1.1 = 6.6 B.M.
(iii) Oxidation state of Mn is MnO₄ is +7. It is energitacally not possible to lose 7 electrons to give ionic species. It forms bonds by sharing of elctrons. Hence, covalent bonds are formed.

Question 5.
There is only a marginal difference in decrease in ionisation elthalpy from Aluminium to Thallium. Explain why? (MARCH 2020)
Answer:
This is due to the presence of inner d-electrons and f-electrons which has poor shielding effect compared to s and p electrons.

Question 6.
(i) Name the metal with tripositive charge represented by the following electronic configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³
(ii) Why is K₂Cr₂O₇ generally preferred over Na₂Cr₂O₇ in volumetric analysis though both are oxidising agents?
(iii) Why does V₂O₅, act as catalyst?
Answer:
(i) Cr³⁺ is represented by Ihe configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³
(ii) Na₂Cr₂O₇ absorbs moisture from, the atmosphere.
(iii) It can form unstable intermediates with the reactants which readilv change into products

Question 7.
Explain the following observations.
(a) The elements of the d – series exhibit a large number of oxidation states than the elements of f – series.
(b) The Cu⁺ salts are colourless while Cu²⁺ salts are coloured (Atomic number of [Cu=29]
Answer:
(a) In d – series there are large number of unpaired electrons which take part in bond formation due to less effective nuclear charge, therefore, number of oxidation states are more. In f – block, there is more effective nuclear charge due to poor shielding effect of f – orbitals. Therefore, less number of electrons take part in bond formation.

(b) Cu⁺ does not have unpaired electtron, therefore the electrons cannot undergo d-d transitions. That is why, Cu⁺ salts are colourless. Cu²⁺ salts are coloured due to presence of one unpaired electron, it can undergo d-d transition by absorbing light from visible region and radiating blue colour.

Question 8.
How do you account for the following?
(i) All Scandium salts are white [Atomic number of Sc=21]
(ii) The first ionisation energies of the 5d transition elements are higher than those of the 3d and 4d transition elements in respective groups.
Answer:
(i) In Scandium Salts, Scandium has +3 oxidation state. Sc³⁺ does not have unpaired electrons and has empty d – orbitals. Therefore, there is no d-d transition, Hence its salts are white.

(ii) Due to poor shielding effect of 5d and 4f – electrons, effective nuclear charge increases, Hence, ionisation energy of 5d transition elements is more than that of 3d and 4d transition elements in respective groups.

Question 9.
How would you account for the following Situations?
(i) The transition metals generally form coloured compounds.
(ii) With 3d⁴ configuration, Cr²⁺ acts as a reducing agent, actinoids exhibit a larger number of oxidation agent.
(iii) The actinoids exhibit a larger number of oxidation states than the corresponding lanthanoids.
Answer:
(i) Transition metals contain unpaired d- electrons. The d – electrons absorb light from the visible range and are excited to higher energy d – orbitals (from t₂g to e_g orbitals). Transmitted light is the colour shown by the transition metals.

(ii) Cr²⁺ has d⁴ configuration while Cr³⁺ has more stable d⁵ configuration. Thus Cr, has a tendency to change from Cr²⁺ to Cr³⁺ or Cr²⁺ acts as reducing agent. Mn³⁺ has a d⁴ configuration. It has a tendency to change into Mn²⁺ with more stable d⁵ configuation. Thus, Mn³⁺ acts as an oxidising agent.

(iii) Actinoids exhibit a larger number of oxidation states than the corresponding lanthanoids because 5f, 6d and 7s levels are comparable energies.

Question 10.
A Violet compound of manganese (A) decomposes on heating to liberate oxygen and compound (B) and (C) of manganese are formed. Compound (C) reacts with KOH in presence of potassium nitrate to give compound.
(B) On heating compound (C) with Cone. H₂SO₄ and NaCl, Chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds (A) to (D) and also explain the reactions involved.
Answer:
The compounds A,B,C and D are given as under : A = KMnO₄, B = K₂MnO₄, C = MnO₂, D = MnCl₂
The reactions are explained as under:
Samacheer Kalvi 12th Chemistry Guide Chapter 4 Transition and Inner Transition Elements 14

2MnO₂ + 4KOH+ O₂ → 2K₂MnO₄ + 2H₂O
MnO₂ + 4NaCl + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + Cl₂ + 2H₂O

Question 11.
When a chromite ore (A) is fused with Sodium Carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallized from the solution. When compound (C) is treated with KC1, orange crystals of compound (D) crystallises out. Identify (A) to (D) and also explain the reactions.
Answer:
The compounds A,B,C and Dare given as under
A = FeCr₂O₄
B = Na₂CrO₄
C = Na₂Cr₂O₇-2H₂O
D = K₂Cr₂O₇
The reactions are explained as under :
Samacheer Kalvi 12th Chemistry Guide Chapter 4 Transition and Inner Transition Elements 15

Question 12.
Explain the following facts:
(a) Transition metals acts as catalysts.
(b) Chromium group elements have the highest melting points in their respective series.
(c) Transition metals form coloured complexes.
Answer:
(a) Transition metals have incomplete d-orbitals. They combine with the reactants to form intermediate products which change into the final products.
(b) Chromium group elements have the greatest number of unpaired electrons in d-orbitals. They ae capable of forming maximum interatomic metallic bonds. This raises the melting point of the chromium group elements.
c) Transition metals have incompletely filled d-orbitals and have unpaired electrons which are excited to higher energy d-orbitals in the same sub-shell (from t2g to eg). These are called d-d transitions. During this process, the molecules absorb energy from the visible region. The transmitted light is the colour shown by the substance.

Question 13.
Discuss the structure of dichromate ion:
Answer:
(i) Both dichromate and chromate ions are oxo anions of chromium and they are oxidising agents.
(ii) In these ions Cr is in +6 oxidation state.
(iii) In aqueous solution these ions are interconvertible.
(iv) In alkaline solution chromate ion is predominant.
(v) In acidic solution dichromate ion is predominant.

Question 14.
Discuss the oxidising power of KMnO₄ in
a) Acidic medium
b) neutral medium
c) alkaline medium
Answer:
a) Acidic medium
It oxidises ferrous salt to ferric salt
2MnO₄^– + 10Fe²⁺ + 16H⁺ → 2Mn²⁺ + 10Fe³⁺ + 8H₂O

b) Neutral medium:
It oxidises H2S to sulphur
2MnO₄^– + 3H₂S → 2MnO + 3S + 2OH^– + 2H₂O

c) Alkaline medium:
In the presence alkali the permanganate ion is converted into manganate ion.
MnO₄^– + 2H₂O + 3e^– → MnO₂ + 40H^–