Category: CBSE

RS Aggarwal Solutions

Get Latest Edition of Maths RS Aggarwal Solutions Pdf Download on LearnInsta.com. It provides step by step RS Aggarwal Solutions Pdf Download. You can download the RS Aggarwal Maths Solutions with Free PDF download option, which contains chapter wise solutions. In RS Aggarwal Solutions all questions are solved and explained by expert Mathematic teachers as per CBSE board guidelines. By studying these RS Aggarwal Maths Solutions you can easily get good marks in CBSE Board Examinations.

• RS Aggarwal Solutions Class 10
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Online Education Maths Lab Manual Class 10: The student working in Maths Lab Manual Class 10 CBSE Pdf must rely to a large extent on methods of enquiry, trial and error in combination with his own curiosity, intuition, and ingenuity; nowhere else in any subject is rigorous proof is so often preceded by the patient, plodding experiments. If going occasionally becomes slow and difficult, one can take comfort in that a CBSE Maths Lab Manual Class 10 Free Download PDF available to increase his pace and make the subject enjoyable.

Online Education for CBSE Class 10 Maths Lab Manual Activities Solutions

Our treatment is structured with a revised list of activities and projects so as to facilitates students and teachers to go through maths activities Chapter wise.

Viva-voce Questions (Very Short Answer Type Questions) and Multiple Choke Questions (MCQ) are incorporated at the end of each activity to check the basics of the activity.

Maths lab manual class 10 activities will serve the needs of students and teachers alike by

• Encouraging the students to carry out activities in a systematic manner, and
• Helping the teachers to evaluate the student’s creative skills.

The activities contained in the book have been chosen from the child’s own environment and will make both the learning and teaching of mathematics not only more fruitful but also much more interesting and lively.

Lab Manual Class 10 Maths Activities Solutions

1. HCF of Two Numbers
2. Zeroes of Polynomials
3. System of Linear Equations
4. Basic Proportionality Theorem
5. Pythagoras Theorem
6. (A) A System of Similar Triangles
   (B) Areas of Two Similar Triangles
7. Median
8. Mode
9. Arithmetic Progression (I)
10. Arithmetic Progression (II)
11. Arithmetic Progression (III)
12. Verification of Lengths of Tangents
13. Area of a Circle
14. Areas of Sectors
15. Comparison of Surface Areas
16. Comparison of Volumes of Right Circular Cylinders
17. Volume of a Right Circular Cylinder
18. A Right Circular Cone
19. Surface Area of a Cone
20. Volume of a Cone
21. Surface Area of a Sphere
22. Volume of a Sphere
23. Centroid of a Triangle
24. (A) Idea of Probability
    (B) Determination of Experimental Probability
25. Measurement of Height Using Clinometer

Syllabus for Lab Manual Class 10 Maths

Subjects	80 Marks (Board Examination)	20 Marks (Internal Assessment)
	The student has to secure 33% marks out of 80 marks in each subject	The student has to secure 33% marks out of overall 20 marks earmarked in each subject
		Periodic Test (10 Marks)	Notebook Submission (5 Marks)	Subject Enrichment Activity (5 Marks)
		(i)	(ii)	(ill)
Mathematics	Class-X Examination will be conducted for 80 marks in each subject covering 100% syllabus of the subject of Class-X only. Marks and Grades both will be awarded for individual subjects. 9-point grading will be the same as followed by the Board.	Periodic written Test, restricted to three in each subject in an Academic Year. Average of the best two tests to be taken for final marks submission.	This will cover • Regularity • Assignment Completion • Neatness and upkeep of notebook	Maths Lab Practical

(i) Periodic Test (10 marks)
The school should conduct three periodic written tests in the entire academic year and the average of the best two will be taken. The schools have the autonomy to make their own schedule. However, for the purpose of gradient learning, three tests may be held as one being the mid-term test and the other the two being pre mid and post mid-term with a portion of syllabus cumulatively covered. The gradually increasing portion of contents would prepare students to acquire confidence for appearing in the Board examination with a 100% syllabus. The school will take the average of the best two tests for final marks submission.

(ii) Notebook Submission (5 marks)
Notebook submission as a part of internal assessment is aimed at enhancing the seriousness of students towards preparing notes for the topics being taught in the classroom as well as assignments. This also addresses the critical aspect of regularity, punctuality, neatness, and notebook upkeep.

(iii) Subject Enrichment Activities (5 marks)
These are subject-specific application activities aimed at enrichment of understanding and skill development. These activities are to be recorded internally by respective subject teachers.

In Online Education The students are required to have a thorough study and understanding of the given comprehension/Unseen passage which may consist of one or more than one paragraphs. Also, the Unseen passages with questions and answers following them constitute one of the highest weightage in the exam. The main purpose of this activity is to test the reading ability of the students and their intellectual skills.

You can refer to NCERT Solutions to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Online Education Unseen Passage for Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12

Short Unseen Passage with Questions and Answers in English

• Unseen Passage for Class 12
• Unseen Passage for Class 11
• Unseen Passage for Class 10
• Unseen Passage for Class 9
• Unseen Passage for Class 8
• Unseen Passage for Class 7
• Unseen Passage for Class 6
• Unseen Passage for Class 5
• Unseen Passage for Class 4
• Unseen Passage for Class 3
• Unseen Passage for Class 2
• Unseen Passage for Class 1

Tips for solving comprehension passages:

• Read the passage thoroughly. The reading should be quick.
• Focus on the relevant details and underline them with a pen or a pencil.
• Read the questions carefully and go back to the passage to find the answers.
• The answers are generally in a logical sequence.
• Try to write the answer in your own words.
• To find answers to the vocabulary-based questions like synonyms etc., replace the word with the meaning. If you find that it is the same in meaning, the answer is correct.
• To find the correct option in Multiple Choice Questions, go through all the options. Re-read the passage and then tick the correct option.

NCERT Solutions for Class 12 Chemistry Chapter 3 provides detailed solutions for the questions asked in the textbook. The solutions are provided by subject experts and the students can refer to these to prepare well for the exams. NCERT Solutions are a guide to the students appearing in different boards like MP board, UP board, CBSE, Gujarat board, etc.

Chemistry Class 12 Chapter 3 Electrochemistry is very important from the examination point of view. All the analytical and conceptual details are provided in the NCERT Solutions Class 12 Chapter 3 that will help the students to score well.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 3
Chapter Name	Electrochemistry
Number of Questions Solved	33
Category	NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

The production of electricity from chemical reactions and the use of electrical energy to bring non-spontaneous chemical transformations is known as electrochemistry. Both the theory and the practical portions are very important in this chapter.

Class 12 Chemistry chapter 3 Electrochemistry explains different types of cells and the differences between the two. Important concepts such as Gibb’s energy and equilibrium constant are also discussed here.

NCERT IN-TEXT QUESTIONS

Question 1.
How would you determine the standard electrode potential of the system ; Mg2+/Mg ?
Answer:
n order to determine E° value of Mg²⁺/Mg electrode, an electrochemical cell is set up in which a Mg electrode dipped in 1 M MgSO₄ solution acts as one half cell (oxidation half cell) while the standard hydrogen electrode acts as the other half cell (reduction half cell). The deflection of voltmeter placed in the cell circuit is towards the Mg^– electrode indicating the flow of current. The cell may be represented as :
Mg/Mg²⁺ (1 M) || H⁺(l M)/H₂(1 atm), Pt
The reading as given by voltmeter gives \({ E }_{ cell }^{ \circ }\)

The expected value of standard electrode potential (E°) = -2-36 V.

Question 2.
Can you store copper sulphate solution in a zinc pot ?
Answer:
No, it is not possible. The E° values of the copper and zinc electrodes are as follows :
Zn²⁺(aq) + 2e^– → Zn(s) ; E° = – 0·76 V
Cu²⁺(aq) + 2e^– → Cu(s) ; E° = + 0·34 V
This shows that zinc is a stronger reducing agent than copper. It will lose electrons to Cu²⁺ ions and redox reaction will immediately set in.
Zn(s) + Cu²⁺ (aq) → Zn²⁺(aq) + Cu(s)
Thus, copper sulphate solution cannot be stored in zinc pot.

Question 3.
Consultthe table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Answer:
Oxidation of Fe²⁺ converts it to Fe³⁺, i.e.,Fe²⁺ –>Fe³⁺ +e^– ; E°_ox= – 0.77 V Only those substances can oxidise Fe²⁺ to Fe³⁺ which are stronger oxidizing agents and have positive reduction potentials greater than 0.77 V, so that EMF of the cell reaction is positive. This is so for elements lying below Fe³⁺/Fe²⁺ in the series ex: Br₂, Cl₂ and F₂.

Question 4.
Calculate the potential of hydrogen electrode in contact with a solution with pH equal to 10.
Answer:
For hydrogen electrode, H⁺ + e^– → 1/2H₂
Applying Nernst equation,

Question 5.
Calculate e.m.f. of the cell in which the following reaction takes place
Ni(s) + 2Ag⁺(0·002M) → Ni²⁺(0·160M) + 2Ag(s) Given that \({ E }_{ cell }^{ \circ }\) = 1.05 V. (C.B.S.E. Outside Delhi 2015)
Answer:

Question 6.
The cell in which the following reaction occurs
2Fe³⁺(aq) + 2I^–(aq) → 2Fe²⁺(aq) + I₂(s) has \({ E }_{ cell }^{ \circ }\) = 0-236 V at 298 K.
Calculate standard Gibbs energy and equilibrium constant for the reaction.
Answer:
The two half reactions are :
2Fe³⁺ + 2e^– → 2Fe²⁺ and 2I^– → I₂ + 2e^–
For the above reaction, n = 2

Question 7.
Why does the conductivity of a solution decrease with dilution?
Answer:
Conductivity of a solution is the conductance of ions present in a unit volume of the solutions. On dilution, no. of ions per unit volume decreases. Hence, the conductivity decreases.

Question 8.
Suggest a way to determine the \({ A }_{ m }^{ \circ }\) for water.
Answer:
Water (H₂O) is a weak electrolyte. Its molar conductance at infinite dilution i.e., \({ A }_{ m }^{ \circ }\) can be determined in terms of \({ A }_{ m }^{ \circ }\) for strong electrolytes. This is in accordance with Kohlrausch’s Law.

Question 9.
The molar conductance of 0·025 mol L^-1of methanoic acid is 46·15 cm² mol^-1. Calculate its degree of dissociation and dissociation constant. Given λ°(H⁺) = 349·6 S cm² mol^-1 and λ°(HCOO^–) = 54·6 S cm² mol^-1.
Answer:

Question 10.
If a current of 0·5 ampere flows through a metallic wire for 2 hours, then how many electrons flow through the wire ?
Answer:
Quantity of charge (Q) passed = Current in amperes x Time in seconds = (0·5 A) X (2 x 60 x 60 s)
= 3600 As = 3600 C
No. of electrons flowing through the wire by passing a charge of one faraday (96500 C) = 6·022 x 10²³
No. of electrons flowing through the wire by passing a charge of 3600 C

Question 11.
Suggest a list of metals that are extracted electrolytically.
Answer:
Na, Ca, Mg and Al

Question 12.
Consider the reaction :
Cr₂\({ O }_{ 7 }^{ 2- }\) + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O.
What is the quantity of electricity in coulombs needed to reduce 1 mole of Cr₂\({ O }_{ 7 }^{ 2- }\) ions ?
Answer:
The quantity of electricity in coulombs is 6 F or 6 x 96500 C = 5·76 x 10⁵ C.

Question 13.
Write the chemistry of recharging the lead storage battery highlighting all the materials that are involved during recharging.
Answer:
Chemical reactions while recharging :
Chemical reactions while recharging :
2PbSO₄ + 2H₂O → PbO₂ + Pb + 2H₂SO₄
Electricity is passed through the electrolyte PbSO₄ which is converted into PbO₂ and Pb.
Recharging is possible in this case because the PbSO₄ formed during discharging is a solid and sticks to the electrodes. Therefore, it can either take up or give electrons during recharging.

Question 14.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Answer:
Methane and Methanol.

Question 15.
Explain how rusting of iron can be envisaged as the setting up of an electrochemical cell.
Answer:
Iron (Fe) is involved in the redox-reaction that is carried in the electrochemical cell which is set up. As a result, it slowly dissolves and the metal surface gets rusted or corroded.
The redox-reaction may be described as follows :
At anode: Fe (s) undergoes oxidation to release electrons
F2(s) → Fe²⁺(aq) + 2e^–              ….(oxidation)
At cathode: The electrons which are released participate in the reduction reaction and combine with H⁺ ions released from carbonic acid (H₂CO₃) formed by the combination of CO₂ and H₂O present.

NCERT EXERCISE

Question 1.
Arrange the following metals in the order in which they displace each other from the solution of their salts: Al, Cu, Fe, Mg and Zn.
Answer:
Mg, Al, Zn, Fe, Cu, Ag.

Question 2.
Given the standard electrode potentials
K⁺/K = – 2·93 V, Ag⁺/Ag = 0·80 V
Hg²⁺/Hg = 0·79 V ; Mg²⁺/Mg = – 2·37V, Cr³⁺/Cr = – 0·74 V
Arrange these metals in increasing order of their reducing power.
Answer:
Less the electrode potential more will be the reducing power.
Ag < Hg < Cr < Mg < K.

Question 3.
Depict the galvanic cell in which the reaction
Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s) takes place. Further show :
(i) which electrode is negatively charged ?
(ii) the carriers of the current in the cell.
(iii) individual reaction at each electrode. (C.B.S.E. Delhi 2008)
Answer:
The galvanic cell in which the given reaction takes place is depicted as:
Zn(s) | Zn²⁺ (aq) || Ag⁺ (aq) | Ag(s)
(i) Zn electrode (anode) is negatively charged
(ii) Tons are carriers of current in the cell and in the external circuit, current from silver to Zinc.
(iii) The reaction taking place at the anode is given by,
Zn(s) -H → Zn²⁺(aq) + 2e^–
The reaction taking place at the cathode is given
Ag⁺+ e^– → Ag(s)

Question 4.
Calculate the standard cell potentials of galvanic cell in which the following reactions take place

Answer:

Question 5.
Write the Nernst Equation and calculate e.m.f. of the following cells at 298 K :
(i) Mg(s) | Mg²⁺ (0·001 M) || Cu²⁺ (0·0001 M) | Cu(s) (C.B.S.E. Delhi 2008, 2013)
(ii) Fe(s) | Fe²⁺ (0·001 M) || H⁺ (1 M) | H₂(g) (1 bar) | Pt(s)
(iii) Sn(s) | Sn²⁺ (0·050 M) || H⁺ (0·02 M) | H₂(g) (1 bar) | Pt(s) (C.B.S.E. Outside Delhi 2013, 2015)
(iv) Pt(s) | Br₂(l) | Br^– (0·010 M) || H⁺ (0·030 M) | H₂(g) (1 bar) | Pt(s)
Answer:

Question 6.
In the button cell widely used in watches and in other devices, the following reaction takes place:
Zn (s) + Ag₂O (s) + H₂O (l) → Zn²⁺ (aq) + 2Ag (s) + 2OH^– (aq)
Determine E° and ∆G° for the reaction. (C.B.S.B. Delhi 2005, Outside Delhi 2006 Supp., 2008, C.B.S.E. Sample Paper 2010)
Answer:

Question 7.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer:
The conductivity of a solution is defined as the conductance of a solution 1 cm in length and the area of cross-section cm2.1 is represented by K.

Conductivity always decreases with a decrease in concentration both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.

Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.

Molar conductivity increases with a decrease in concentration. This is because the total volume of the solution containing one mole of the electrolyte increases on dilution.

Question 8.
The conductivity of 0·20 M solution of KCl at 298 K is 0·0248 S cm^-1. Calculate its molar conductivity.
(C.B.S.E. Delhi 2008, 2013)
Answer:

Question 9.
The resistance of a conductivity cell containing 0·001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0·001M KCl solution at 298 K is 0·146 x 10^-3 S cm^-2? (C.B.S.E. Outside Delhi 2007, 2008, 2013)
Answer:

Question 10.
The conductivity of sodium chloride solution at 298 K has been determined at different concentrations and results are given below :

Calculate molar conductivity for all the concentrations and draw a plot between \({ A }_{ m }^{ c }\) and \(\sqrt { c } \). Find the value \({ A }_{ m }^{ \circ }\) from the graph.
Answer:
\(\frac { 1S{ cm }^{ -1 } }{ 100S{ m }^{ -1 } } =1\) (unit conversion factor)


\({ A }_{ m }^{ \circ }\) can be obtained on extrapolation to zero concentration along Y-axis. It is 124·0Scm²mol^-1.

Question 11.
The conductivity of 0·00241 M acetic acid is 7·896 x 10^-5 S cm^-1. Calculate the molar conductivity. If A° for acetic acid is 390·5 S cm² mol^-1, what is its dissociation constant? (C.B.S.E. Delhi 2008, C.B.S.E. Outside Delhi 2016)
Answer:

Question 12.
How much charge is required for the reduction of :
(i) 1 mol of Al³⁺ to Al
(ii) 1 mol of Cu²⁺ to Cu
(iii) 1 mol of Mn\({ O }_{ 4 }^{ – }\) to Mn²⁺.
Answer:

Question 13.
How much electricity in terms of Faraday is required to produce.
(i) 20.0 g of Ca from molten CaCl₂?
(ii) 40.0 g of Al from molten Al₂0₃?
Answer:

Question 14.
How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H₂O to O₂
(ii) 1 mol of FeO to Fe₂O₃.
Answer:

For the oxidation of two moles of FeO, charge required = 2 F
For the oxidation of one mole of FeO, charge required = 1 F = 96500 C.

Question 15.
A solution of Ni(NO₃)₂ is electrolyzed between platinum electrodes using a current of 5·0 ampere for 20 minutes. What weight of Ni will be produced at the cathode? (Atomic mass of Ni = 58·7). (Jharkhand Board 2009)
Answer:

Question 16.
Three electrolytic cells A, B, and C containing electrolytes zinc sulphate, silver nitrate, and copper sulphate respectively, were connected in series. A steady current of 1·50 ampere was passed through them until 1·45 g of silver was deposited at the cathode of cell B. How long did the current flow? What weight of copper and of zinc were deposited? (Atomic mass of Cu = 63·5 ; Zn = 65·3; Ag = 108) (C.B.S.E. Outside Delhi 2008, Jharkhand Board 2010)
Answer:

Question 17.
Predict if the reaction between the following is feasible:


Answer:
The reaction is feasible if the EMF of the cell reaction is positive.

Question 18.
Predict the products of electrolysis of each of the following :
(i) An aqueous solution of AgNO₃ using silver electrodes.
(ii) An aqueous solution of AgNO₃ using platinum electrodes.
(iii) A dilute solution of H₂SO₄ using platinum electrodes. (C.B.S.E. Outside Delhi 2007)
(iv) An aqueous solution of CuCl₂ using platinum electrodes. (C. B. S. E. Sample Paper 2010)
Answer:
(i) An aqueous solution of AgNO₃ using silver electrodes :
Both AgNO₃ and water will ionise in aqueous solution

At cathode: Ag⁺ ions with less discharge potential are reduced in preference to H⁺ ions which will remain in solution. As a result, silver will be deposited at cathode.
Ag⁺ (aq) + e^– → Ag (deposited)
At anode: An equivalent amount of silver will be oxidised to Ag⁺ ions by releasing electrons.

(ii) An aqueous solution of AgNO₃ using platinum electrodes:
In this case, the platinum electrodes are the non-attackable electrodes. On passing current the following changes will occur at the electrodes.
At cathode: Ag⁺ ions will be reduced to Ag which will get deposited at the cathode.
At anode: Both \({ NO }_{ 3 }^{ – }\) and OH^– ions will migrate. But OH^– ions with less discharge potential will be oxidised in preference to \({ NO }_{ 3 }^{ – }\) ions which will remain in solution.

Thus, as a result of electrolysis, silver is deposited on the cathode while O₂ is evolved at the anode. The solution will be acidic due to the presence of HNO₃.
(iii) A dilute solution of H₂SO₄ using platinum electrodes:
On passing current, both acid and water will ionise as follows:

At cathode: H⁺ (aq) ions will migrate to the cathode and will be reduced to H₂.

Thus, H₂ (g) will be evolved at the cathode.
At anode: OH^– ions will be released in preference to \({ SO }_{ 4 }^{ 2- }\) ions because their discharge potential is less. They will be oxidized as follows:

Thus, O₂ (g) will be evolved at the anode. The solution will be acidic and will contain H₂SO₄.
(iv) An aqueous solution of CuCl₂ using platinum electrodes :
The electrolysis proceeds in the same manner as discussed in the case of AgNO₃ solution. Both CuCl₂ and H₂O will ionise as follows :

At cathode: Cu²⁺ ions will be reduced in preference to H⁺ ions and copper will be deposited at the cathode

Cu²⁺ (aq) + 2e^– → Cu (deposited)

At anode: C^– ions will be discharged in preference to OH^– ions which will remain in solution.

Cl^– → Cl^– + e^–; Cl + Cl → Cl₂ (g)

Thus, Cl₂ will be evolved at the anode.

We hope the NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry, drop a comment below and we will get back to you at the earliest.

NCERT Class 12 Chemistry Solutions for Chapter 9 Coordination Compounds provides in-depth knowledge of the concepts given in the chapter. It helps the students prepare well for boards as well as competitive exams. It also helps the students strengthen their basics for advanced concepts.

The students appearing for UP board, MP board, CBSE, Gujarat board, Maharashtra board, etc. can refer to these NCERT Solutions and score well in the examination.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 9
Chapter Name	Coordination Compounds
Number of Questions Solved	43
Category	NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds

Complex compounds are known as coordination compounds. It is an important chapter from the examination perspective. The students need to be thorough with the concepts such as properties, examples and applications of coordination compounds.

A thorough knowledge of the basic concepts helps the students while studying the advanced concepts. This not only helps them during boards and home exams but also during competitive exams.

NCERT IN-TEXT QUESTIONS

Question 1.
Write the formulas of the following coordination compounds:
(a) tetraamminediaquacobalt (III) chloride
(b) potassium tetracyanonickelate (II)
(c) tris (ethane-1, 2-diamine) chromium (III) chloride
(d) amminebromidochloridonitrito-N-platinate (II)
(e) dichlorobis (ethane-1, 2-diamine) platinum (TV) nitrate
(f) iron (III) hexacyanoferrate (II).
Answer:
(a) [Co(NH₃)₄(H₂O)₂]Cl₃
(b) K₂[Ni(CN)₄]
(c) [Cr(en)₃]Cl₃
(d) [Pt(NH₃)BrCl(NO₂)]^–
(e) [PtCl₂(en)₂](NO₃)₂
(f) Fe₄[Fe(CN)₆]₃.

Question 2.
Write the IUPAC names of the following coordination compounds:

1. [Co(NH₃)₆]Cl₃
2. [Co(NH₃)₅Cl]Cl₂
3. K₃[Fe(CN)₆l
4. K₃lFe(C₂O₄)₃]
5. K₂[PdCl₄]
6. [Pt(NH₃)₂Cl(NH₂CH₃)]Cl

Answer:

1. Hexaamminecobalt (III) chloride
2. Pentaamminechloridecobalt (III) chloride
3. Potassium hexacyanoferrate (III)
4. Potassiumtrioxalatoferrate(III)
5. Potassium tetrachloridopalladate (II)
6. Diamminechloride (methylamine) platinum (II) chloride

Question 3.
Give the types of isomerism exhibited by the following complexes and draw the structures of these isomers:
(a) K[Cr(H₂O)₂(C₂O₄)₂]
(b) [Co(en)₃]Cl₃
(c) [CO(NH₃)₅(NO₂)] (NO₃)₂
(d) [Pt(NH₃) (H₂O)Cl₂] (C.B.S.E. Outside Delhi 2013)
Answer:
(a) K[Cr(H₂O)₂(C₂O₄)₂] or K[Cr(H₂O)₂(OX)₂]
(i) It exists as geometrical isomers :

(ii) The cis isomer can also exist as pair of optical isomers.

(b) The complex can exist as optical isomers. For structure, consult section 9.7.
(c) The complex can exist as pair of ionisation isomers as well as linkage isomers.
Ionisation isomers: [Co(NH₃)₅(NO₂)] (NO₃)₂ and [Co(NH₃)₅(NO₃)](NO₂)(NO₃)
Linkage isomers: [Co(NH₃)₅(NO₂)](NO₃)₂ and [Co(NH₃)₅ONO](NO₃)₂
(d) The complex can exist as pair of geometrical isomers.

Question 4.
Give evidence to show that [Co(NH₃)₅Cl]SO₄ and [CO(NH₃)₅SO₄]Cl exist as ionisation isomers.
Answer:
Dissolve both the complexes separately in water. First add a few drops of BaCl₂ solution to both these complexes. Only one of these will give white precipitate with the solution indicating that SO₄ is not a part of complex entity. If exists as an anion.

Now again add a few drops of AgN03 solution to both these complexes. Only one of these will give white precipitate with the solution indicating that in this case Cl is not a part of complex entity. It exists as anion.

This shows that the complexes exist as pair of ionisation isomers.

Question 5.
Explain on the basis of valence bond theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and the[Ni(CN)₄]^2- ion with tetrahedral geometry is paramagnetic.
Answer:

Question 6.
[MCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why ? (C.B.S.E. Outside Delhi2012)
Answer:
In the complex [NiCl₄]^2-, Ni is in + 2 oxidation state and has the configuration 3d⁸4⁰. The CL ion being a weak ligand cannot pair the two unpaired electrons present in 3d orbitals. This means that 3d orbitals are not involved in hybridisation. The complex is sp³ hybridised (tetrahedral) and is paramagnetic in nature. In the other complex [Ni(CO)₄], the oxidation state of Ni is zero and electronic configuration is 3d⁸4s². In the presence of the ligand CO, the 4s electrons shift to the two half filled 3d orbitals and make all the electrons paired. The valence 4s and 3p orbitals are involved in hybridisation. The complex is tetrahedral but diamagnetic. For more details, consult text part.

Question 7.
[Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly paramagnetic. Explain.
Answer:
Outer electronic configuration of iron (Z = 26) in-ground state is 3d⁶4s². Iron in this complex is in +3 oxidation state. Iron achieves +3 oxidation state by the loss of two 4s electrons and one 3d electron. The resulting Fe³⁺ ion has outer electronic configuration of 3d⁵. Since H₂O is not a strong field ligand, it is unable to cause electron pairing.

Outer electronic configuration of iron (Z=26) in ground state is 3d⁶4s². Iron in this complex is in a +3 oxidation state. Iron achieves +3 oxidation state by the loss of two 4s electrons and one 3d electron. The resulting Fe³⁺ ion has an outer electronic configuration as 3d⁵. CN^– ion is a strong field ligand

Question 8.
Explain [Co(NH₃)₆]³⁺ is an inner orbital complex while [Ni(NH₃)₆]²⁺ is an outer orbital complex.
Answer:
In the complex [Co(NH₃)₆]³⁺, the oxidation state of cobalt is +3 and has 3d⁶ configuration. In the presence of NH₃ molecules (ligands), two 3d electrons pair up and two 3d orbitals remain empty. Since six ligands are to be accommodated the hybridisation of the metal ion is d²sp³.

As inner d-electrons are involved, the complex is inner orbital complex and is diamagnetic in nature.
In the complex [Ni(NH₃)₆]^{3 +} , the oxidation state of Ni is +2 and has 3d⁸ configuration. Since six NH₃ molecules (ligands) are to be accomodated, the hybridisation of metal ion is sp³d². This means that 4d orbitals are involved in the hybridisation.

The complex is paramagnetic as well as outer orbital complex since outer (4d) electrons are involved in the hybridisation.

Question 9.
Predict the number of unpaired electrons in the square planar [Pt(CN)₄]^2- ion.
Answer:
The element Pt(Z = 78) is present in group 10 with electronic configuration 5d⁹6s¹. The divalent cation Pt²⁺ has 5d⁸ configuration.

For square planar complex, Pt (II) is in dsp² hybridisation state. To achieve this, the two unpaired electrons present in 5d orbitals get paired. The complex has, therefore, no unpaired electrons.

Question 10.
The hexaquomanganese(II) ion contains five unpaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.
Answer:

Question 11.
Calculate the overall complex dissociation equilibrium constant for [Cu(NH₃)₄]²⁺ ion, given that p4 for the complex is 2·1 x 10¹³.
Answer:
The dissociation constant is the reciprocal of overall stability constant (β₄)

NCERT EXERCISE

Question 1.
Explain bonding in co-ordination compounds in terms of Werner’s postulates.
Answer:
Werner’s coordination theory: Alfred Werner gave his co-ordination theory in 1893. The important postulates of this theory are:
(i) All metals in atomic or ionic form exhibit two types of valencies in coordination compounds :
(a) Primary or principal or ionic valency (—–),
(b) Secondary or auxiliary or nonionic valency (—).
The primary valency is ionizable and it is shown by dotted lines. The secondary valency is non-ionizable and is shown by a continuous line.
(ii) Primary valency represents oxidation states of a metal atom or ion and secondary valency represents the co-ordination number of metal ion which is fixed for a particular atom.
(iii) The primary valencies are satisfied by negative ions whereas the secondary valencies may be satisfied either by negative ions (e.g., Cl^–, Br^–, CN^– etc.) or neutral molecules (ag., H₂O).
(iv) Secondary valencies are directed towards a fixed position in space.
(v) Every element tends to satisfy both its primary and secondary valencies. For this purpose a negative ion may often act a dual behaviour i.e., it may satisfy primary as well as secondary valency (—–,-).
Example : (i) Luteo cobaltic chloride CoCl₃.6NH₃ or [Co(NH₃)₆]Cl₃.
(ii) Purpureo cobaltic chloride COC1₃.5NH₃ or [Co(NH₃)₅Cl]Cl₂

Question 2.
FeSO₄ solution mixed with (NH₄)₂SO₄solution in 1:1 molar ratio gives the test of, Fe²⁺ion but CuSO₄solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu²⁺ ion. Explain why.
Answer:
FeSO₄ solution mixed with (NH₄),SO₄ solution in 1 : 1 molar ratio forms a double salt, FeS04 (NH₄)₂SO₄-6H₂O (Mohr’s salt) which ionizes in the solution to give Fe²⁺ions. Hence it gives the tests of Fe²⁺ ions. CuSO₄ solution mixed with aqueous ammonia in 1:4 molar ratio forms a complex salt, with the formula [CU(NH₃)₄]SO₄. The complex ion [Cu(NH₃)₄]²⁺ does not ionize to give Cu²⁺ ions. Hence, it does not give the tests of Cu²⁺ ion.

Question 3.
Explain with two examples each of the following:
coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heterolepic.
Answer:
(a) Co-ordination Entity: A coordination complex or entity is the species enclosed in square bracket. It is also called co-ordinate sphere. It contains in it a certain metal atom or ion to which a fixed number of neutral molecules or ions capable of donating electron pairs are linked with co-ordinate bonds. e.g. [COCl₃(NH₃)₃]

(b) Ligand: Ligands are the electron donor molecules or ions which may be either neutral, or anionic (sometimes cationic as well) and are linked to the central metal atom or ion by co-ordinate bonds also called dative bonds. In fact, the electrons needed for the bond are provided by the ligands. e.g. H₂NCH₂CH₂NH₂ or N(CH₂CH₂NH₂)₃

(c) Co-ordination Number: The coordination number [C.N.] of a metal ion in a complex can be defined as the number of ligand or donor atoms to which the metal is directly bonded. For example, in the complex ions, [PtCl₆]^2- and [Ni(NH₃)₄]²⁺, the coordination number of Pt and Ni are 6 and 4 respectively. Similarly, in the complex ions, [Fe(C₂O₄)₃]^3- and [CO(en)₃]³⁺,the coordination number of both, Fe and Co, is 6 because C₂O₄^2-and en (ethane-1,2-diamine) are bidentate ligands.

(d) Coordination polyhedron : The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom. The most common coordination polyhedra are octahedral, square planar and tetrahedral. For example, [CO(NH₃)₆]³⁺ is octahedral, [Ni(CO)₄] is tetrahedral and [PtCl₄]^2- is square planar.

(e) Homoleptic and heteroleptic complexe : Complexes in which a metal is bound to only one kind of donor groups, e.g., [CO(NH₃)₆]³⁺, are known as homoleptic. Complexes in which a metal is bound to more than one kind of donor group, e.g., [CO(NH₃)₄Cl₂]⁺, are known as heteroleptic.

Question 4.
What is meant by unidentate, didentate and ambidentate ligands ? Give two examples for each.
Answer:
Unidentate ligands are those which bind to the metal ion through a single donor atom, e.g., Cl^– , H₂O.

Bidentate ligands are those which bind to the metal ion through two donor atoms. e.g., ethane-1,2-diamine (H₂NCH₂CH₂NH₂), oxalate (C₂O₄^2-) ion.

Ambidentate ligands are those which can bind to metal ion through two different donor atoms, e.g., NO₂^– and SCN^– ion.

Question 5.
Specify the oxidation numbers of metals in the following co-ordination entities :
(a) [Co(H₂O)(CN)(en₂)]²⁺
(b) [PtCl₄]^2-
(c) [Cr(NH₃)₃Cl₃]
(d) [CoBr₂(en)₂]⁺
(e) K₃[Fe(CN)₆].
Answer:
(a) O.N. of Co : x + 0 + (-1) + 2(0)= + 2 or x = + 2+ 1 = + 3
(b) O.N. of Pt : x + 4 (-1) =-2 or x =-2 + 4 = + 2
(c) O.N. of Cr : x + 3(0) + 3(- 1) = 0 or x = + 3
(d) O.N. of Co : x + 2(- 1) + 2(0) = + 1 or x = + 1 + 2 = + 3
(e) O.N. of Fe : x + 6 (- 1) = – 3 or x = – 3 + 6 = + 3

Question 6.
Using IUPAC norms write the formulas for the following:
(i)Tetrahydroxozincate(Il)
(ii)Potassium tetrachloridopalladate (II)
(iii)Diamminedichlorido platinum (II)
(iv)Potassium tetracyanonickelate (II)
(v)Pentaamminenitrito-O-cobalt(III)
(vi)Ilexaamminccobalt (III) sulphate
(vii)Potassium tri(oxalato) chromate (III)
(yiii)Hexaammineplatinum (IV)
(ix)Tetrabromidocuprate(II)
(x) Pentaamminenitrito-N-cobalt (III)
Answer:
(i)[Zn(OH)₄]^2- (ii)K₂[PdCl₄]
(iii)[Pt(NH₃)₂Cl₂]
(iv)K₂[Ni(CN)₄]
(v)[Co(NH₃)₅(ONO)]²⁺
(vi)[Co(NH₃)₆]²(SO₄)₃
(vii)K₃[Cr(C₂O₄)₃]
(viii)[Pt(NH₃)₆]4+
(ix)[CuBr₄]^2-
(x)[Co(NH₃)₅(N0₂)]²⁺

Question 7.
Using IUPAC norms write the systematic names of the following :
(a) [CO(NH₃)₆]Cl₃
(b) [CO(NH₃)₄Cl(NO₂)]Cl
(c) [Ni(NH₃)₆]Cl₂
(d) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
(e) [Mn(H₂O)₆]²⁺
(f) [NiCl₄]^2-
(g) [Co(en)₃]³⁺
(h) [Ti(H₂O)₆]³⁺
(i) [Ni(CO)₄]. (Jharkhand Board 2015)
Answer:
(a) hexamminecobalt(III) chloride
(b) tetramminechloriodonitrito-N-cobalt(III) chloride
(c) hexaamminenickel(II) chloride
(d) diamminechlorido (methaneamine) platinum(II) chloride
(e) hexaaquamanganese(II) ion
(f) tetrachloriodonickelate(II) ion
(g) tris(ethane-l, 2-diammine) cobalt(III) ion
(h) hexaaquatitanium(III) ion
(i) tetracarbonylnickel (0).

Question 8.
List various types of isomerism possible for coordination compounds, giving an example of each.
Answer:
(i) Geometrical isomerism: The isomer in which similar ligands occupy adjacent positions is referred to as cis isomer and the isomer in which similar ligands occupy opposite positions is referred to as trans isomer. Therefore, this type of isomerism is also known as cis-trans isomerism.

(ii) Optical isomerism: The isomer which rotates the plane of polarised light to the right is called dextro rotatory and designated as d- and the one which rotates the plane of polarised light to the left is called laevo rotatory and designated as l. These optical isomers have identical physical and chemical properties except their behaviour towards the plane polarised light.

(iii) Linkage isomerism: Linkage isomerism arises in a coordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocyanate ligand, NCS^–, which may bind through the nitrogen to give M-N CS or through sulphur to give M-SCN. This behaviour was seen in the complex [CO(NH₃)₅(NO₂)]Cl₂, which is obtained as the red form, in which the nitrite ligand is bound through oxygen ( ONO), and as the yellow form, in which the nitrite ligand is bound through nitrogen (-NO₂).

(iv) Coordination isomerism : This type of isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex. An example is provided by [Co(NH₃)₆] [Cr(CN)₆], in which the NH₃ ligands are bound to CO³⁺ and the CN^– ligands to Cr³⁺. In its coordination isomer [Cr(NH₃)₆][CO(CN)₆], the NH₃ ligands are bound to Cr³⁺ and the CN^– ligands to CO³⁺.

(v) Ionisation isomerism : This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion. An example is provided by the ionisation isomers [CO(NH₃)₅SO₄]Br and [CO(NH₃)₅Br]SO₄.

(vi) Solvate isomerism : This form of isomerism is known as ‘hydrate isomerism in case where water is involved as a solvent. This is similar to ionisation isomerism. Solvate isomers differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice. An example is provided by the aqua complex [Cr(H₂O)₆]Cl₃ (violet) and its solvate isomer [Cr(H₂O)₅Cl]Cl₂ H₂O (grey green).

Question 9.
How many geometrical isomers are possible in the following coordination entities ?
(a) [Cr(C₂O₄)₃]^3-
(b) [Co(NH₃)₃Cl₃]
Answer:
(a) [Cr(C₂O₄)₂]^3- : No geometrical isomerism is possible.
(b) [CO(NH₃)₃Cl₃] : Two geometrical isomers : fac and mer

Question 10.
Draw the structures of optical isomers of :
(a) [Cr(C₂O₄)₃]^3-
(b) [PtCl₂(en)₂]²⁺
(c) [Cr(NH₃)₂Cl₂(en)]⁺ (C.B.S.E. Foreign 2015)
Answer:

Question 11.
Draw all the isomers (geometrical and optical) of :
(a) [CoCl₂(en)₂]⁺
(b) [CoNH₃Cl(en)₂]²⁺
(c) [Co(NH₃)₂Cl₂(en)]⁺
Answer:
(a)

(b)

(c)

Question 12.
Write all the geometrical isomers of [Pt(NH₃)(Br)(Cl)(py)]. How many of these will exhibit optical isomerism ?
Answer:
There are three geometrical isomers.

Optical isomerism is generally not shown by the square planar complexes with CN = 4.

Question 13.
Aqueous copper sulphate solution (blue in colour) gives (a) green precipitate with aqueous potassium fluoride and (b) a bright green solution with aqueous potassium chloride solutions. Explain these experimental results.
Answer:
Aqueous solution of copper sulphate which is blue in colour exists as [Cu(H₂O)₄]SO₄ and gives [Cu(H₂O)₄]²⁺ in solution. It is a labile complex entity in which the ligands H₂O get easily replaced by F^– ions of KF and by Cl^– ions of KCl.

Question 14.
What is the coordination entity formed when excess aqueous KCN is added to an aqueous solution of copper sulphate? Why is that no precipitate of copper sulphide is obtained when H₂S(g) is passed through the solution?
Answer:
When an excess aqueous KCN is added to an aqueous solution of CuSO₄, Potassiumtetra-cyanocuprate (II) is formed. When H₂S_(g) is passed through the above solution, no precipitate of copper sulphide is obtained because CN^– ions are strong ligands so the complex [Cu(CN)₄]^2- is very stable. As Cu²⁺ ions are not available so CuS precipitate is not formed.

Question 15.
Discuss the nature of bonding in the following co-ordination complexes on the basis of valence bond theory :
(a) [Fe(CN)₆]^4-
(b) [FeF₆]^3-
(c) [Co(C₂O₄)₃]^3-
(d) [CoF₆]^3-
Answer:
(a) Hexacyanoferrate(II) ion [Fe(CN)₆]^4-: Iron in this complex is in +2 oxidation state. Iron achieves + 2 oxidation state by the loss of two 4s electrons. The resulting Fe²⁺ ion has outer electronic configuration of 3d⁶.

(b) Hexafluoriodoferrate(II) ion [FeF₆]^3-: Iron in this complex is in +2 oxidation state. Iron achieves + 2 oxidation state by the loss of two 4s electrons. The resulting Fe²⁺ ion has outer electronic configuration of 3d⁶.

(c)

(d) Hexafluorocobaltate(III) [CoF₆]^3-: Cobalt ion in the complex is in + 3 oxidation state. Cobalt achieves + 3 oxidation state by the loss of two 4s electrons and one 3d electron. The resulting Co^{3 +} ion has outer electronic configuration of 3d⁶.

Question 16.
Draw figure to show the splitting of d-orbitals in octahedral crystal field.
Answer:
Let us assume that the six ligands are positioned symmetrically along the Cartesian axes, with a metal atom at the origin.

As the ligands approach, first there is an increase in energy of d-orbitals relative to that of the free ion just as would be the case in a spherical field.

The orbitals lying along the axes (d_z2 and d_x2 – y₂) get repelled more strongly than d_xy, d_yz and d_zx orbitals which have lobes directed between the axes.

The d_z2 and d_x2-y2 orbitals get raised in energy and d_xy d_yz, d_xz orbitals are lowered in energy relative to the average energy in the spherical crystal field. Thus, the degenerate set of d-orbitals get split into two sets: the lower energy orbitals set, t_2g, and the higher energy orbitals set, e_g. The energy is separated by ∆₀

Question 17.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer:
Spectrochemical series: The arrangement of ligands in order of their increasing field strengths i.e. increasing crystal field splitting energy (CFSE) values is called spectrochemical series, which is as follows:
I < Br < SCN < Cl < S^2- < F^– < OH^– < C₂O^2-₄ < H₂O < NCS^– < edta^-4 < NH₃ < en < CN^– < CO.
Difference between weak field ligand and a strong field ligand: The ligand with a small value of CFSE (∆₀) are called weak field ligands whereas those with a large value of CFSE are called strong field ligands.

Question 18.
What is crystal field splitting energy ? How does the magnitude of ∆₀ decide the actual configuration of d- orbitals in a coordination entity ?
Answer:
The degenerate d-orbitals (in a spherical field environment) split two-level i.e. e_g and t₂g in the presence of ligands. The splitting of the degenerate orbitals in the presence of ligands is called crystal field splitting and the energy difference between the two levels (e and t₂g) is called the crystal field splitting energy. It is denoted by ∆_o. After the orbitals have split, the filling of the electrons takes place. After 1 electron (each) has filled in the three t₂g orbitals, the filling of the electrons takes place in 2 ways.

It can enter the orbital (giving) rise to t³g e_g, like electronic configuration on the pairing of the electrons can take place in the t₂g orbitals (giving rise to t⁴₂g eg⁰ like electronic configuration). If the ∆_o value of a ligand is less than the pairing energy, then the electrons enter the eg orbital. On the other hand, if the ∆_o value of a ligand is more than the pairing energy, then the electrons enter the t₂g orbitals.

Question 19.
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Explain why?
Answer:
(i) Hexaamminechromium(III) ion [Cr(NH₃)₆]³⁺: Outer electronic configuration of chromium (Z=24) in ground state is 3d²4s¹ and in this complex, it is in the +3 oxidation state. Chromium achieves +3 oxidation state by the loss of one 4s electron and two 3d-electrons. The resulting Cr³⁺ ion has outer electronic configuration of 3d³.

(ii) Tetracyanonickelate (II) ion [Ni(CN)₄]^2-: Outer electronic configuration of nickel (Z = 28) in ground state is 3d⁸4s². Nickel in this complex is in + 2 oxidation state. It achieves + 2 oxidation state by the loss of the two 4s-electrons. The resulting Ni²⁺ ion has outer electronic configuration of 3d⁸.

Question 20.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Explain. (C.B.S.E. Delhi 2017)
Answer:
Formation of [Ni(H₂O)₆]²⁺

Since H₂P molecules represent weak field ligands, they do not cause any electron pairing. As a result the complex has two unpaired electrons and is coloured. The d-d transitions absorb radiations corresponding to a red light and the complementary colour emitted is green.
Formation of [Ni(CN)₄]^2-. For the details of the structure of the complex,

Since the complex has no unpaired electrons there is no scope for any d-d transition. The complex is therefore, colourless.

Question 21.
[Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺ are of different cdlours in dilute solutions. Why?
Answer:
In both the complexes, Fe is in +2 state with the configuration 3d⁶ i.e., it has four unpaired electrons. As the ligand H₂O and CN^– possess different crystal field splitting energy (∆₀), they absorb different components of the visible light (VIBGYOR) for the transition. Hence, the transmitted colours are different.

Question 22.
Discuss the nature of bonding in metal carbonyls.
Answer:
The metal-carbon bond in metal carbonyls possess both s and p character. The M-C σ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The M-C π bond is formed by the donation of a pair of electrons from a filled d-orbital of metal into the vacant antibonding π* orbital of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal.

Question 23.
Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes :
(a) K₃[CO(C₂O₄)₃]
(b) (NH₄)₂[CoF₄]
(C) Cis – [CrCl₂2(en)₂]Cl
(d) [Mn(H₂O)₆]SO₄
Answer:
(a) OS = + 3, CN = 6, d-orbital occupation is 3d⁶ \({ t }_{ 2g }^{ 6 }{ e }_{ g }^{ 0 }\),
(b) OS = + 2, CN = 4, 3d⁷ (\({ t }_{ 2g }^{ 5 }{ e }_{ g }^{ 2 }\)),
(c) OS = + 3, CN = 6, 3d³ (\({ t }_{ 2g }^{ 3 }\)),
(d) OS = + 2, CN = 6, 3d⁶ (\({ t }_{ 2g }^{ 3 }{ e }_{ g }^{ 2 }\)).

Question 24.
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex :
(a) K[Cr(H₂O)₂(C₂O₄)₂]3H₂O
(b) [CrCl₃(py)₃]
(c) K₄[Mn(CN)₆]
(d) [Co(NH₃)₅Cl]Cl₂
(e) Cs[FeCl₄]
Answer:
(a) IUPAC name : potassium diaquadioxalatochromate (III) hydrate.
O.S. of Cr = + 3 ; 3d³ (\({ t }_{ 2g }^{ 3 }{ e }_{ g }^{ 0 }\)) CN = 6 ; shape = octahedral, three unpaired electrons.
Magnetic moment (μ) = \(\sqrt { n\left( n+2 \right) } =\sqrt { 3\times 5 } =\sqrt { 15 } =3\cdot 87BM\)
(b) IUPAC name: trichloridotripyridinechromium (III) O.S. of Cr = + 3; 3 d³ (\({ t }_{ 2g }^{ 3 }{ e }_{ g }^{ 0 }\)) CN = 6
shape = octahedral ; three unpaired electrons.
Magnetic moment (μ) = \(\sqrt { n\left( n+2 \right) } =\sqrt { 3\times 5 } =3\cdot 87BM\)
(c) IUPAC name : potassiumhexacyanomanganate (II) O.S. of Mn = + 2 ; 3d⁵ (\({ t }_{ 2g }^{ 5 }{ e }_{ g }^{ 0 }\)), CN = 6, shape = octahedral; one unpaired electron.
Magnetic moment (μ) = \(\sqrt { n\left( n+2 \right) } =\sqrt { 1\times 3 } =\sqrt { 3 } =1\cdot 73BM\)
(d) IUPAC name : pentaamminechloridocobalt (III) chloride
O.S. of Co = + 3 ; 3d⁶ (\({ t }_{ 2g }^{ 6 }{ e }_{ g }^{ 0 }\)), CN = 6
shape = octahedral; zero unpaired electron. Magnetic moment (μ) = 0
(e) IUPAC name : cesium tetrachloridoferrate (III)
O.S. of Fe = + 3 ; 3d⁵ (\({ e }^{ 2 }{ t }_{ 2 }^{ 3 }\)), CN = 4.
shape = tetrahedral ; five unpaired electrons.
Magnetic moment (μ) = \(\sqrt { n\left( n+2 \right) } =\sqrt { 5\times 7 } =\sqrt { 35 } =5\cdot 92BM\)

Question 25.
What is meant by stability of a coordination compound in solution ? State the factors which govern stability of complexes.
Answer:
The stability of a complex in solution refers to the degree of association between the two species involved in the state of equilibrium. The magnitude of the (stability or formation) equilibrium constant for the association, quantitatively expresses the stability. Thus, if we have a reaction of the type :

then, the larger the stability constant, the higher is the proportion of ML₄ that exists in the solution. Free metal ions rarely exist in the solution so that M will usually be surrounded by solvent molecules which will compete with the ligand molecules, L, and be successively replaced by them. For simplicity, we generally ignore these solvent molecules and write four stability constants as follows :

Factors affecting stability of complexes :

1. The smaller the size of the cation, the greater will be the stability of the complex e.g., Fe³⁺ forms a more stable complex than Fe²⁺.
2. The greater the charge on the central metal ion, the more stable will be the complex e.g., Pt⁴⁺ forms a more stable complex than Pt²⁺.
3. Stronger the ligand, more stable will be the complex formed e.g., CN forms more stable complex then NH₃.

Question 26.
What is meant by chelate effect ? Give an example.
Answer:
When a ligand attaches to the metal ion in a manner that form’s a ring, then the metal-ligand association is found to be more stable. In other words, we can say that complexes containing chelate ring more stable than complexes without rings. This is known as the chelate effect.
Examples: EDTA, DMG, etc.

Question 27.
Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems
(ii) medicinal chemistry
(iii) analytical chemistry and
(iv) extraction/ metallurgy of metals.
Answer:
(i) Role of coordination compounds in biological systems:
We know that photosynthesis is possible by the presence of chlorophyll pigment. This pigment is a coordination compound of magnesium. In the human biological system, several coordination compounds play important roles. For example, the oxygen – carrier of blood, i.e hemoglobin is a coordination compound of iron.

(ii) Role of coordination compounds in Medicinal chemistry: Certain coordination compounds of platinum (for example cis-platin) are used for inhibiting the growth of tumors.

(iii) Role of coordination compounds in analytical chemistry: During salt analysis, a number of basic radicals are detected with the help of the colour changes they exhibit with different reagents. These colour changes are a result of the coordination compounds or complexes that the basic radicals form with different ligands.

(iv) Role of coordination compounds in interaction or metallurgy of metals: The process of extraction of some of the metals from their ores involves the formation of complexes. For example in an aqueous solution, gold combines with cyanide ions to form [Au (CN)₂]. From this solution, gold is later extracted by the addition of Zn metal.

Question 28.
How many ions are produced from the complex Co(NH₃)₆Cl₂ in solution?
(a) 6
(b) 4
(c) 3
(d) 2.
Answer:
The complex will dissociate in aqueous solution to give three ions

Therefore, (c) is the correct answer.

Question 29.
Amongst the following ions which one has the highest magnetic movement value?
(a) [Cr(H₂O)₆]³⁺
(b) [Fe(H₂O)₆]²⁺
(c) [Zn(H₂O)₆]²⁺
Answer:
The oxidation states of the metals in the complexes along with the electronic configuration are given:
(a) Cr³⁺ : 3d³ configuration ; unpaired electrons = 3
(b) Fe²⁺ : 3d⁶ configuration ; unpaired electrons = 4
(c) Zn²⁺ : 3d¹⁰ configuration ; unpaired electrons = 0
The complex (b) with maximum number of unpaired electrons has the highest magnetic moment. Therefore, (b) is the correct answer.

Question 30.
The oxidation number of cobalt in K[Co(CO)⁴] is
(a) + 1
(b) + 3
(c) – 1
(d) – 3.
Answer:
O.N. of Co : x + 4(0) = -1 or x = -1. Therefore, (c) is the correct answer.

Question 31.
Amongst the following, the most stable complex is :
(a) [Fe(H₂O)₆]³⁺
(b) [Fe(NH₃)₆]³⁺
(c) [Fe(C₂O₄)₃]^3-
(d) [FeCl₆]^3-.
Answer:
In all the complexes, Fe is in + 3 oxidation state. However, the complex (c) is a chelate because three \({ C }_{ 2 }{ O }_{ 4 }^{ 2- }\) ions act as the chelating ligands. Thus, the most stable complex is (c).

Question 32.
What will be the correct order for the wavelengths of absorption in the visible region for the following : [Ni(NO₂)₆]^4-,[Ni(NH₃)₆]²⁺,[Ni(H₂O)₆]²⁺.
Answer:
In all the complexes, the metal ion is the same (Ni²⁺). The increasing field strengths of the ligands present as per electrochemical series are in the order :
H₂O < NH₃ < \({ NO }_{ 2 }^{ – }\)
The energies absorbed for excitation will be in the order :
[Ni(H₂O)₆]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(NO₂)₆]^4-
As E = hc/λ i.e., E ∝ 1/λ; the wavelengths absorbed will be in the opposite order.

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NCERT Solutions Class 12 Chemistry Chapter 1 contains the solved questions and answers provided in the textbook. The answers are provided by subject experts and hence the students can refer to these solutions for better preparations. The explanations are provided in an easy language which the students find easy to understand. The diagrammatic representations make the explanations even clearer.

The NCERT Solutions for Class 12 Chemistry Chapter 1 not only helps the students prepare for their board exams, but also prepares them for competitive exams. The solutions strengthen the conceptual knowledge of the students that clarifies even the minute doubts.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 1
Chapter Name	The Solid State
Number of Questions Solved	50
Category	NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State

“The Solid State” is an important chapter in Chemistry from the examination perspective. We are surrounded by different states of matter. Most of the matter around us is in solid state. The NCERT Solutions for Class 12 Chemistry Chapter 1 explain the structure, classification and properties of solids. The explanation of the structure of solids states the correlation between the structure and properties of solids.

NCERT IN-TEXT QUESTIONS

Question 1.
Why are solids rigid?
Answer:
Solids are rigid because the constituent particles are very closely packed. They don’t have any translatory movement
and can only oscillate about their mean positions.

Question 2.
Why do solids have a definite volume?
Answer:
The constituent particles of a solid have fixed positions and are not free to move about, i.e., they possess rigidity. That is why they have definite volume.

Question 3.
Classify the following as amorphous and crystalline solids; polyurethane, naphthalene, benzoic acid, Teflon,
potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.
Answer:
Amorphous solids: Polyurethane, naphthalene, Teflon, cellophane, polyvinyl chloride, fiberglass. Crystalline solids: Benzoic acid, potassium nitrate, copper.

Question 4.
Why is glass considered a supercooled liquid?
Answer:
Glass is an amorphous solid. Like liquids, it has a tendency to flow, though very slowly. This can be seen from the glass panes of windows or doors of very old buildings which are thicker at the bottom than at the top. Therefore, glass is considered as a supercooled liquid.

Question 5.
The Refractive index of a solid is observed to have the same value along with all the directions. Comment on the nature of the solid. Would it show cleavage property?
Answer:
As the solid has the same refractive index along with all the directions, it is isotropic in nature and is, therefore, an amorphous solid. It is not expected to show a clean cleavage when cut with a special type of knife. It will break into pieces with irregular surfaces.

Question 6.
Classify .the following solids in different categories based on the nature of intermolecular forces operating in them: Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide
Answer:
Potassium sulphate = Ionic Tin=Metallic.
Benzene = Molecular (non-polar)
Urea=Molecular (polar).
Ammonia=Molecular (H-bonded)
Water = Molecular (H-bonded)
Zinc sulphide = Ionic Graphite=Covalent Rubidipm Metallic Argon = Molecular (non-polar)
Silicon Carbide=Covalent

Question 7.
A solid substance ‘A’ is a very hard and electrical insulator both in the solid-state as well as in the molten state. It has also the very high melting point. Is the solid metal like silver or network solid like silicon carbide (SiC)?
Answer:
Since the solid behaves as an insulator even in the molten state, it cannot be metal like silver. Therefore, it is a covalent or network solid like SiC.

Question 8.
Ionic solids conduct electricity in molten state but not in solid state. Explain
Answer:
In a solid-state, the ions cannot move, they are held by strong electrostatic forces of attraction. So, ionic solids do not conduct electricity in the solid state. However, in the molten state, they dissociate to give tree ions and hence conduct electricity.

Question 9.
What types of solids are electrical conductors, malleable and ductile? (C.B.S.E. Outside Delhi 2013)
Answer:
Metallic solids exhibit these characteristics. Their atoms are linked to one another by metallic bonds.

Question 10.
Give the significance of a ‘lattice point’.
Answer:
Each lattice point represents one constituent particle of the solid. This constituent particle may be an atom, a molecule or an ion.

Question 11.
Name the parameters which characterize a unit cell.
Answer:
A unit cell is characterized by two types of parameters. These are edges (a, b, c) which may or may not be mutually perpendicular, and angles between the edges (α, β, and γ).
(i) Edges or edge lengths. The edges a, b and c represent the dimensions of the unit cell in space along the three axes. The edges may or may not be mutually perpendicular.
(ii) Angles between the edges. There are three angles between the edges. These are denoted as α (between b and c), β (between a and c), and γ (between a and b). Thus, a unit cell may be characterized by six parameters as shown in Fig. 1.12. The various types of crystal systems differ with respect to edge lengths as well as angles between the edges.

Question 12.
Distinguish between
(i) Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centered unit cells.

Answer:
(i)

Hexagonal unit cell	Monoclinic unit cell
a=b≠c	a≠b≠c
α = β = 90°	α = γ = 90°
γ = 120°	β ≠ 90°

(ii)

Face-centred unit cell	End-centred unit cell
A Face-centred unit cell the constituent particles are present at the corners and one at the centre of each face.	An End-centred unit cell contains particles at the corners and one at the centre of any two opposite faces.
Total no of particles in a face centered unit cell= 4	Total no. of particles in an end centered unit cell = 2

Question 13.
Explain how many portions of an atom located at the
(i) corner and
(ii) body centre of a cubic unit cell is a part of the neighbouring unit cell.
Answer:
(i) An atom located at the corner is shared by eight unit cells. Therefore, its contribution to a particular unit cell is 1/8.
(ii) An atom located at the body of the unit cell is not shared by any unit cell. It belongs to one particular unit cell only.

Question 14.
What is the two-dimensional coordination number of a molecule in a square close-packed layer?
Answer:
In 2D, square close-packed layer, an atom touches 4 nearest neighbouring atoms. Hence, its CN=4

Question 15.
A compound forms a hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
Answer:
No. of atoms in 0.5 mole of the compound = 0.5 x N₀ = 0.5 x 6.022 x 10²³ = 3.011 x 10²³
No. of octahedral voids = No. of atoms = 3.011 x 10²³
No. of tetrahedral voids = 2 x 3.011 x 10²³ = 6.022 x 10²³
Total no. of voids = (3.011 + 6.022) x 102²³ = 9.033 x 10²³

Question 16.
A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy l/3rd of tetrahedral voids. What is the formula of the compound?
Answer:
Atoms of N from ccp, therefore, if the lattice points are n, then
No.of atoms of N = n
No. of oct voids = n
No. of td voids = 2n= 2 x 1n/3 = 2n/3
∴ Formula of the compound is: M : N
2/3 n : n
2n: 3n
2: 3
i.e., M₂N₃

Question 17.
Which of the following lattices has the highest packing efficiency :
(i) simple cubic
(ii) body-centered cubic and
(iii) hexagonal close-packed lattice?
Answer:
The packing efficiency of the different types of arrangement is :
(i) Simple cubic = 52.4%
(ii) Body-centred cubic = 68%
(iii) Hexagonal close-packed = 74%
his means that hexagonal close-packed arrangement has the maximum packing efficiency (74%).

Question 18.
An element with a molar mass 2.7 x 10^-2 kg  mol^-1 forms a cubic unit cell with an edge length of 405 pm. If its density is 2.7 x 10³ kg m^-3, what is the nature of the cubic unit cell?
Answer:

Question 19.
What type of defect can arise when a solid is heated? Which physical property is affected by it and in what
way?
Answer:
When a solid is heated, some atoms or ions may leave the crystal lattice. As a result, vacancies are created and this leads to vacancy defects in the crystalline solid. Since the number of atoms/ions per unit volume decreases, the vacancy defects lead to a decrease in the density.

Question 20.
What type of stoichiometric defect is shown by:
(i) ZnS
(ii) AgBr
Answer:
(i) ZnS shows Frenkel defect
(ii) AgBr shows Frenkel as well as Schottky defect.

Question 21.
Explain how vacancies are introduced in the ionic solid when a cation of higher valence is added as an impurity to it.
Answer:
Let us consider an ionic solid sodium chloride (Na⁺Cl^–) to which a small amount of strontium chloride (SrCl₂) has been added to act as an impurity. Since the crystal as a whole is to remain electrically neutral, two Na+ ions have to leave their sites to create two vacancies. Out of these, one will be occupied by Sr²⁺ ion while the other will be vacant. Thus, vacancies will be created in the ionic solid. When a cation of higher valency is added as an impurity in the ionic solid, some of the sites of the original cations are occupied by the cations of higher valency. Each cation of higher valency replaces two or more original cations and occupies the site of one original cation and the other site(s) remains vacant.

Question 22.
Ionic solids which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Answer:
Let us illustrate by sodium chloride (Na⁺Cl^–) crystals. Upon heating in the atmosphere of sodium (Na) vapours, sodium atoms get deposited on the surface of the crystals. The Cl^– ions from the crystal lattice leave their sites and diffuse into the surface. They tend to combine with sodium atoms present in the vapours which in turn get ionised to form Na⁺ ions by releasing electrons. The latter is trapped by the anionic vacancies created by Cl^–ions in order to maintain the crystals electrically neutral. Now, the electrons absorb radiations corresponding to a certain colour from white light and start vibrating. They emit radiations corresponding to yellow colour. That is how the crystals of sodium chloride develop yellow colour. These electrons are called F-centres because these are responsible for colours (In German, F = Farbe meaning colour).

Question 23.
A group 14 element is to be converted into an n-type semiconductor by doping it with a suitable impurity. To which group should the impurity element belong?
Answer:
n-type semiconductors are conducting due to the presence of excess negatively charged electrons. In order to convert group 14 elements (e.g. Si, Ge) into n-type semi-conductors, doping is done with some elements of group 15 (e.g. P, As)

Question 24.
What type of substances would make better permanent magnets; ferromagnetic or ferrimagnetic? Justify your answer. (C.B.S.E. Outside Delhi 2013)
Answer:
Ferromagnetic substances make better permanent magnets than ferrimagnetic substances. The metal ions of a ferromagnetic substance are grouped into small regions known as domains and these are randomly oriented. Under the influence of the applied magnetic field, all domains are oriented in the direction of the magnetic field and as a result, a strong magnetic field is produced. The ferromagnetic substance behaves as a magnet. This characteristic of the domains persists even when the external magnetic field is removed. This imparts permanent magnetic character to these substances. However, this property is lacking in ferrimagnetic substances. Therefore ferromagnetic substances are better magnets.

NCERT EXERCISE

Question 1.
Define the term ‘amorphous’. Give a few examples of amorphous solids.
Answer:
Amorphous solids are those solids in which the constituent particles may have short-range order but do not have a long-range order. They have irregular shapes and are isotropic in nature. They do not undergo a clean cleavage. They do not have sharp melting points or definite heat of fusion. E.g.: Glass, rubber, and plastics.

Question 2.
What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Answer:
Glass is made up of Si04 tetrahedral units. These constituent particles have short-range order only. Quartz is also made up of Si04 tetrahedral units. On heating it softens and melts over a wide range of temperature. It is a crystalline solid having long-range ordered structure. It has a sharp melting point. Quartz can be converted into glass by first melting and then rapidly cooling it.

Question 3.
Classify each of the following solids as ionic, metallic, molecular, network (covalent), or amorphous:
(a) Tetra phosphorus decoxide (P₄O₁₀)
(b) Graphite
(c) Brass
(d) Ammonium phosphate (NH₄)₃PO₄
(e) SiC
(f) Rb
(g)l₂
(h) LiBr
(i) P₄
(j) Si
(k) Plastic.
Answer:
(a) Molecular solid
(b) Covalent (Net-work) solid
(c) Metallic solid
(d) Ionic solid
(e) Covalent solid (Network)
(f) Metallic solid
(g) Molecular solid
(h) lonic solid
(i) Molecular solid
(j) Covalent solid
(k) Amorphous solid.

Question 4.
(a) What is meant by the term coordination number?
(b) What is the coordination number of atoms
(i) in a cubic close-packed structure
(ii) in a body-centered cubic structure?
Answer:
(i) The number of nearest neighbours of a particle in its close packing is called its coordination number.
(ii) (a) 12, (b) 8.

Question 5.
How can you determine the atomic mass of an unknown metal if you know its density and the dimensions of its unit cell? Explain.
Answer:
Let the edge length of a unit cell = a
Density = d
Molar mass = M
The volume of the unit cell = a3
Mass of the unit cell = No. of atoms in unit cell x Mass of each atom = Z × m

Question 6.
(a) Stability of a crystal is reflected in the magnitude of the melting point. Comment.
(b) Collect the melting point of

• Ice
• ethyl alcohol
• diethyl ether
• methane from a data book. What can you say about intermolecular forces between the molecules?

Answer:
(a) Higher the melting point, greater are tire “forces holding the constituent particles together and hence greater is the stability.

(b) The intermolecular forces in water and ethyl alcohol are mainly hydrogen bonding. The higher melting point of water than alcohol shows that hydrogen bonding in ethyl alcohol molecules is not as strong as in water molecules. Diethyl ether is a polar molecule. The intermolecular forces present in them are the dipole-dipole attraction. Methane is a non-polar molecule. The only forces present in them are the weak Vander Waal’s forces (London / dispersion forces).

Question 7.
How will you distinguish between the following pairs of terms:
(i) Cubic close packing and hexagonal close packing?
(ii) Crystal lattice and unit cell?
(iii) Tetrahedral void and octahedral void?
Answer:
(i) Cubic close packing: When the third layer is placed over the second layer in such a way that the spheres cover the octahedral voids, a layer different from first (A) and second (B) is produced. If we continue packing in this manner, then packing is obtained where the spheres in every fourth layer will vertically aligned. This pattern of packing spheres is called the ABCABC pattern or cubic close packing.

Hexagonal close-packing: When a third layer is placed over the second layer in such a manner that the spheres cover the tetrahedral void, a three-dimensional close packing is obtained where the spheres in every third or alternate layer are vertically aligned. If we continue packing in this manner, then the packing obtained would be called ABAB pattern or hexagonal close packing.

(ii) Crystal lattice: It is a regular arrangement of the constituent particles (i?.e., ions, atoms or molecules) of a crystal in three-dimensional space.
Unit cell: The smallest three-dimensional portion of a complete space lattice which when repeated over and over again in different directions produces the complete crystal lattice is called the unit cell.

(iii) Tetrahedral void: A simple) the triangular void is a crystal is surrounded by four spheres and is called a tetrahedral void.
Octahedral void: A double triangular void is surrounded by six spheres and is called an octahedral void.

Question 8.
How many lattice points are there in one unit cell of each of the following lattices
(a) face-centered-cubic
(b) face centred tetragonal
(c) body-centered cubic?
Answer:
(i) In the face-centered cubic arrangement, a number of lattice points are = 8 (at comers) + 6 (at face centres)
Lattice points per unit cell = 8 × \(\frac { 1 }{ 8 } \) + 6 × \(\frac { 1 }{ 2 } \) = 4.

(ii) In face centred tetragonal, number of lattice points are = 8 (at comers) + 6 (at face centres)
Lattice points per unit cell = 8 × \(\frac { 1 }{ 8 } \) + 6 × \(\frac { 1 }{ 2 } \) = 4

(iii) In body centred cubic arrangement, number of lattice points are = 8 (at comers) + 1 (at body centres)
Lattice points per unit cell = 8 × \(\frac { 1 }{ 8 } \) + 1 = 2

Question 9.
Explain:
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
Answer:
(i) Basis of similarities. The basis of similarities between the metallic and ionic crystals are the presence of strong electrostatic forces of attraction. These are present among the ions in the ionic crystals and among the kernels and
valence electrons in the metallic crystals. That is why both metals and ionic compounds are good conductors of electricity and have high melting points.
Basis of differences. The basis of differences in the absence of mobility of ions in the ionic crystals while the same is present in the valence electrons and kernels in the case of metallic crystals. As a consequence, the ionic compounds conduct electricity only in the molten state while the metals can do so even in the solid-state.
(ii) The ionic solids are hard and brittle because of strong electrostatic forces of attraction which are present in the oppositely charged ions.
The ionic solids are hard because of the presence of strong inter-ionic forces of attraction in the oppositely charged ions. These ions are arranged in three-dimensional space. The ionic solids are brittle because the ionic bond
is non-directional.

Question 10.
Calculate the efficiency of packing in the case of metal crystal for:
(i) Simple cubic
(ii) Body centered cubic
(iii) Face centered cubic (with the assumption that the atoms are touching each other). (C.B.S.E. Outside Delhi 2011) Answer:
(i) Simple cubic: We know that in a simple cubic unit cell, there is one atom (or one sphere) per unit cell. If r is the radius of the sphere, the volume occupied by one sphere present in the unit cell = 4/3πr³.


(ii) Body-centred cubic: We know that a body-centered cubic unit cell has 2 spheres (atoms) per unit cell. If r is the radius of the sphere Volume of one sphere = 4/3πrsup>3


(iii) Face centred cubic: We know that a face centered cubic unit cell (fcc) contains four spheres (or atoms) per unit cell.

Question 11.
Silver crystallizes in a face centered cubic lattice with all the atoms at the lattice points. The length of the edge of
the unit cell as determined by X-ray diffraction studies is found to be 4.077 x 10^-8 cm. The density of silver is 10.5 g cm^-3. Calculate the atomic mass of silver. (C.B.S.E. Sample Paper 2012)(Uttarakhand Board 2015)
Answer:

Question 12.
A cubic solid is made up of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body centre. What is the formula of the compound? What are the coordination numbers of P and Q?
Answer:
As atom Q are present at the 8 comers of the cube, therefore, number of atoms of Q in the unit cell = 8 × \(\frac { 1 }{ 8 } \) = 1.
As atoms P are present at the body centre, therefore a number of atoms P in the unit cell = 1.
∴ The formula of the compound = PQ
Co-ordination number of each P and Q = 8.

Question 13.
Niobium crystallizes in a body-centered cubic structure. If the density is 8.55 g cm-1, calculate the atomic radius of niobium given that the atomic mass of niobium is 93 g mol^-1. (C.B.S.E. Delhi 2008)
Answer:
Step I. Calculation of edge length of unit cell.
No. of particles in b.c.c. type unit cell (Z) = 2
Atomic mass of the element (M) = 93 g mol^-1

Step II. Calculation of radius of unit cell

Question 14.
If the radius of octahedral void is r and the radius of the atom in close packing is R, derive the relation between r and R. (C.B.S.E. Sample Paper 2017)
Answer:
Let length of each side of the square is a and the radii of the void and the sphere are r and R respectively. Consider the right angled triangle ABC.


∴ Radius of octahedral void is 0.414R(or 41.4% as compared to that of the sphere).

Question 15.
Copper crystallises into a foc lattice with edge length 3•61 x 10-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm. (C.B.S.E. Delhi 2009 Comptt.)
Answer:

The calculated value of the density is nearly the same as the measured value.

Question 16.
Analysis shows that nickel oxide has formula Nin.os 01.00. What fraction of nickel exists as Ni²⁺ and as Ni³⁺ ions ?
Answer:
The ratio of Ni and O atoms in pure nickel oxide (NiO) = 1:1
Let x be the no. of Ni (II) atoms replaced by Ni (III) atoms in the oxide.
∴ No. of Ni (II) atoms present = (0.98 – x)
Since the oxide is neutral in nature,
Charge on Ni atoms = Charge on oxygen atoms
2(0.98 – x) + 3x = 2
1.96 – 2x + 3x = 2
x = 2 – 1.96 = 0.04

% of Ni (II) atoms in nickel oxide = 100 – 4:01 = 95.99%

Question 17.
What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
Answer:
Those solids which have intermediate conductivities ranging from 10^-6 to 10⁴ ohm^-1 m^-1 are classified as semiconductors. As the temperature rises, there is a rise in conductivity value because electrons from the valence band jump to the conduction band.

(i) n-type semiconductor: When a silicon or germanium crystal is doped with group 15 elements like P or As, the dopant atom forms four covalent bonds like Si or Ge atom but the fifth electron, not used in bonding, becomes delocalized and continues its share towards electrical conduction. Thus silicon or germanium doped with P or As is called an H-type semiconductor, a-indicative of negative since it is the electron that conducts electricity.

(ii) p-type semiconductor: When silicon or germanium is doped with group 13 elements like B or Al, the dopant is present only with three valence electrons. An electron vacancy or a hole is created at the place of the missing fourth electron. Here, this hole moves throughout the crystal-like a positive charge giving rise to electrical conductivity. Thus Si or Ge doped with B or Al is called p-tvpe semiconductor, p stands for the positive hole since it is the positive hole that is responsible for conduction.

Question 18.
Non-stoichiometric cuprous oxide (Cu₂O) can be prepared in the laboratory. In this oxide, the copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?
Answer:
The ratio less than 2: 1 in Cu₂O shows that some cuprous (Cu⁺) ions have been replaced by cupric (Cu²⁺) ions. To maintain electrical neutrality, every two Cu⁺ ions will be replaced by one Cu²⁺ ion thereby creating a hole. As conduction will be due to the presence of this positive hole, hence it is a p-type semiconductor.

Question 19.
Ferric oxide crystallizes in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of ferric oxide.
Answer:
There is one octahedral hole for each atom in the hexagonal close-packed arrangement.
If the number of oxide ions (O^2-) per unit cell is 1, then the number of Fe³⁺ ions = 2/3 x octahedral holes = 2/3 x 1 = 2/3.
Thus, the formula of the compound = Fe^2/3O¹, or Fe²O³.

Question 20.
Classify each of the following as being either a p-type or an n-type semiconductor.
(i) Ge doped with In
(ii) B doped with Si.
Answer:
(i) Ge belongs to group 14 and In belongs to group 13, therefore an electron-deficient hole is created and hence it is an n-type semiconductor.
(ii) B belongs to group 13 and Si belongs to group 14, therefore there will be a free electron and hence it is an n-type semiconductor.

Question 21.
Gold (atomic radius = 0.144 nm) crystallizes in a face-centered unit cell. What is the length of a side of the cell?
Answer:
For fee lattice, edge length,
a = 2 √2 x 0.144 nm = 0.407 nm

Question 22.
In terms of Band Theory, what is the difference
(i) between a conductor and an insulator
(ii) between a conductor and semi-conductor?
Answer:
The variation in the electrical conductivity of the solids can be explained with the help of the band theory.

(i) In insulators, the energy gaps are very large and the no electron jump is feasible from the valence band to the conduction band. The energy gaps also called forbidden zones. The insulators, therefore, do not conduct electricity.
(ii) In semi-conductors, there is a small energy gap between the valence band and conduction band. However, some electrons may jump to the conduction band and these semiconductors can exhibit a little electrical conductivity.

Question 23.
Explain the following terms with suitable examples.
(i) Schottky defect,
(ii) Frenkel defect,
(iii) Interstitials,
(iv) F-centres.
Answer:
(i) Schottky defect: This rises because certain ions are missing from the crystal lattice and vacancies or holes are created at their respective positions. Since a crystal is electrically neutral, the number of such missing cations (A⁺) and anions (B^–) must be the same. e.g., KCl, NaCl, KBr, etc.

(ii) Frenkel defect: It results when certain ions leave their normal sites and occupy positions elsewhere in the crystal lattice. Holes are created at their respective positions. Since cations are smaller in size as compared to anions normally these are involved in Frenkel defect. e.g., AgBr, ZnS, etc.

(iii) Interstitials: This defect is noticed when constituent particles (atoms or molecules) occupy the interstitial sites in the crystal lattice. As a result, the number of particles per unit volume increases and so the density of the solid.

(iv) F-centres: These are the anionic sites occupied by unpaired electrons. F-centres impart colour to crystals. The colour results by the excitation of electrons when they absorb energy from the visible light falling on the crystal.

Question 24.
Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.
(a) What is the length of the side of the unit cell?
(b) How many unit cells are there in 1.00 cm? of aluminum? (C.B.S.E. Outside Delhi 2013)
Answer:
Step I. Calculation of length of side of the unit cell
For f.c.c. unit cell, a = \(2\sqrt { 2 } r\) = \(2\sqrt { 2 } \) (125pm) = 2 x 1.4142 x (125 pm) = 354 pm.
Step II. Calculation of no. of unit cells in 1:00 cm³ of aluminium.
Volume of one unit cell = (354 pm)3 = (354 x 10^-10 cm)³ = 44174155 x 10^-30 cc.

Question 25.
If NaCl is doped with 10^-3 mol % of SrCl₂, what is the concentration of cation vacancies?
Answer:

Question 26.
Explain the following with suitable examples.
(a) Ferromagnetism
(b) Piezoelectric effect
(c) Paramagnetism
(d) Ferrimagnetism
(e) Antifluoride structure
(f) 12 – 16 and 13 – 15 compounds.
Answer:
(a) Ferromagnetism: A few solids like iron. cobalt, nickel, gadolinium and CeO₂ are attracted very strongly by magnetic fields. These are known as ferromagnetic solids. Apart from that, they can be even permanently magnetised or become permanent magnet. e.g., Fe, Ni, Co and CrO₂

(b) Piezoelectric effect: A dielectric crystal which has a resultant dipole moment can produce electricity or show the electrical property when external pressure is applied. Such a crystal is known as piezoelectric crystal and this property is called piezoelectricity or pressure electricity. e.g., PbZrO₂, Nh₄H₂PO₄ etc.

(c) Paramagnetism: These are the solids attracted by a magnet. Actually, the atoms of the elements present have certain unpaired electrons. Their spins or magnetic moments may lead to magnetic characters. Many transition metals such as Co, Ni, Fe, Cu, etc. and their ions are paramagnetic. e.g., O₂, Cu²⁺, Fe³⁺, etc

(d) Ferrimagnetism: They have certain resultant magnetic moment or magnetic character which is of permanent nature. However, ferrimanetic solids are less magnetic than ferromagnetic solids. For example, magnetic oxide of iron (Fe₃O₄) and ferrites with general formula MFe₂O₄. e.g., Fe₃O₄

(e) Antifluoride structure: In this structure, the positions of the cations and anions as compared to fluorite structure get reversed i.e. the smaller cations occupy the position of fluoride ions while the anions with bigger size occupy the positions of calcium ions. e.g., Li₂O, K₂O, Rb₂O and Rb₂S.

(f) 12 – 16 and 13 – 15 compounds: A large variety of solid-state materials have been prepared by the combination of elements belonging to groups 13 and 15 or group 12 and 16. A few examples of compounds 13-15 combinations are InSb, Alp and GaAs. Similarly, compounds resulting from 12 – 16 combinations are AdS, CdSe, HgTe.

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Class 12 Chemistry NCERT Solutions for Chapter 7 The p-Block Elements is the best guide for the students appearing for boards and competitive exams. The solutions contain answers to the questions provided in the textbook. These help in strengthening all the concepts related to chapter 7 that help in better preparations.

NCERT Solutions not only help the students appearing in different boards but also the one’s appearing in the competitive exams. The students can practice the solutions provided by the subject experts.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 7
Chapter Name	The p-Block Elements
Number of Questions Solved	74
Category	NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 7 The p-Block Elements

Class 12 Chemistry Chapter 7 explains the p block elements and their properties, The elements in group 15, 16 and 17 are discussed along with their properties. Various concepts such as electronegativity, chemical and physical properties, ionization therapy, etc. are discussed in detail.

This chapter is important from examination perspective and the students are advised to go through the NCERT Solutions for better practice. The solutions are provided along with the diagrams for better understanding.

NCERT IN-TEXT QUESTIONS

Question 1.
Why are pentahalides more covalent than trihalides in the members of the nitrogen family?
Answer:
The electronic configuration of the elements of nitrogen family (group 15) is ns²p³. Because of the inert pair effect, the valence s-electrons cannot be released easily for the bond formation. This means that the elements can form trivalent cation (E³⁺) by releasing valence p-electrons while it is difficult to form pentavalent cation (E⁵⁺). Under the circumstances, if all the five valence electrons are to be involved in the bond formation, the compounds showing pentavalency or +5 oxidation state must be of covalent nature. This is particularly the case in the higher members (Sb, Bi) of the family where the inert pair effect is quite prominent.

Question 2.
Why is BiH₃ the strongest reducing agent amongst all the hydrides of Group 15 elements?
Answer:
This is because as we move down the group, the size increases, as a result, length of E-H bond increases and its strength decreases, so that the bond can be broken easily to release H₂ gas. Hence, BiH₃ is the strongest reducing agent.

Question 3.
Why is N₂ less reactive at room temperature?
Answer:
In the nitrogen molecule (N₂), two atoms of nitrogen are linked by triple bond (N = N). Due to small atomic size of the element (atomic radius = 70
pm), the bond dissociation enthalpy is very high (946 kJ mol¹). This means that it is quite difficult to cleave or break the triple bond at room temperature. As a result, N₂ is less reactive at room temperature.

Question 4.
Mention the conditions required for the maximum yield of ammonia.
Answer:
In Haber’s process, ammonia is formed by the following reaction.

According to Le-chatelier’s principle, the favourable conditions for the maximum yield of ammonia are :
(i) Low temperature : But optimum temperature of 700 K is necessary to keep the forward reaction in progress.
(ii) High pressure : Pressure to the extent of about 200 atm is required.
(iii) Catalyst & promoter : In order to achieve the early attainment of equilibrium, iron oxide acts as catalyst. Along with that; K₂O, Al₂O₃ or Mo metal may act as the promoter to increase the efficiency of the catalyst.

Question 5.
How does ammonia react with blue solution having Cu²⁺ ions ?
Answer:
When ammonia gas is passed through blue solution containing Cu²⁺ ions to give a soluble complex with deep blue colour.

Question 6.
What is the covalency of nitrogen in N₂O₅?
Answer:
The covalent structure of nitrogen pentoxide (N₂O₅) is given. Since the nitrogen atom has shared four electron pairs, its covalency is four in the molecule of N₂O₅.

Question 7.
Bond angle in PH₄⁺ is higher than that in PH₃. Why?
Answer:
P in PH₃ is sp³-hybridized with 3 bond pairs and one lone pair around P. Due to stronger lp-bp repulsions than bp-bp repulsions, tetrahedral angle decreases from 109°28′ to 93.6°. As a result, PH₃ is pyramidal. In PH₄⁺, there are 4 bp’s and no lone pair. As a result, there are only identical bp-bp repulsions so that PH₄⁺ assumes tetrahedral geometry and the bond angle is 109°28′.Hence, bond angle of PH₄⁺ > bond angle of PH₃

Question 8.
What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂?
Answer:
A mixture of sodium hypophosphite and phosphine gas is formed upon heating the reaction mixture in an inert atmosphere.

Question 9.
What happens when PCl₅ is heated?
Answer:
Upon heating, PCl₅ dissociates to give molecules of PCl₃ and Cl₂. Actually, the two P—Cl (a) bonds with more bond length break away from the molecule leaving three P— Cl(e) bonds attached to the central P atom since these are more firmly linked
PCl₅ \(\underrightarrow { heat } \) PCl₃ + Cl₂

Question 10.
Write the balanced equation for the hydrolytic reaction of PCl₅ in heavy water.
Answer:
With heavy water (D₂O) ; phosphorus pentachloride (PCl₅ reacts as follows :

Question 11.
What is the basicity of H₃PO₄ ?
Answer:
The acid is tribasic since it has three P—OH bonds which can release H⁺ ions.

Question 12.
What happens when phosphorus acid (H₃PO₃) is heated ? (C.B.S.E. 2008)
Answer:
In phosphorus acid (H₃PO₃), central atom P is in +3 oxidation state. Upon heating, it gives a mixture PH ₃ (P in -3 oxidation state) and H₃PO₄ (P in +5 oxidation state). This means that phosphorus acid undergoes disproportionation reaction.

Question 13.
List the important sources of sulphur.
Answer:
Combined sulphur exists as sulphates, such as gypsum, epsom, baryte and sulphides such as galena, zinc blende, copper pyrites, etc. Traces of sulphur occur as hydrogen sulphide in volcanoes. Few organic materials like eggs, proteins, garlic, onion, mustard, hair and wool contain sulphur. 0.03 – 0.1% sulphur is present in the earth’s crust.

Question 14.
Write the order of the thermal stability of the hydrides of group 16 elements.
Answer:
The order of thermal stability of hydride is :
H₂O > H₂S > H₂Se > H₂Te
This is related to the bond dissociation enthalpies of the E—H bonds where E stands for the element.
E—H bond :                                           O—H S—H Se—H Te—H
Bond dissociation enthalpy : (kJ mol^-1) 463    347      276  238
Based on bond dissociation enthalpy, H₂O is maximum stable thermally while H₂Te is the least stable.

Question 15.
Why is H₂O a liquid and H₂S a gas?
Answer:

In H₂O, the electronegativity difference between 0(3·5) and H(2·1) is more than difference between S(2·5) and H(21) in H₂S. As a result, O—H bond is more polar than S—H bond. This leads to inter molecular hydrogen bonding in H₂0 molecules while it is almost absent in the molecules of H₂S. The H₂O molecules get associated and consequently exist as liquid (water). The association in H₂S molecules is negligible and it exists as a gas at room temperature.

Question 16.
Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe
Answer:
Pt being a noble metal does not react with oxygen directly. In contrast, Zn, Ti and Fe are active metals and hence they react with oxygen directly to form their oxides.

Question 17.
Complete the following reactions :
(i) C₂H₄ + O₂ →
(ii) Al + O₂ →
Answer:
(i) C₂H₄ + 3O₂ \(\underrightarrow { heat } \) 2CO₂ + 2H₂O
(ii) 4Al + 3O₂ \(\underrightarrow { heat } \) 2Al₂O₃

Question 18.
Why does O₃ act as a powerful oxidising agent ? (C.B.S.E. 2013)
Answer:
Upon heating, ozone (O₃) readily decomposes to give molecular oxygen (O₂) which is more stable along with nascent oxygen (O). The released nascent oxygen readily takes part in oxidation reactions. Therefore, ozone acts as a powerful oxidising agent.
O₃ \(\underrightarrow { heat } \) O₂ + O (Nascent)

Question 19.
How is ozone estimated quantitatively ?
Answer:
When ozone reacts with an excess of KI solution buffered with a borate buffer (pH = 9.2), iodine is liberated which can be titrated against standard solution of sodium thiosulphate. This is used as a method of estimation of ozone quantitatively.

Question 20.
What happens when sulphur dioxide gas is passed through an aqueous solution of Fe(III) salt ?
Answer:
The gas is a reducing agent and reduces a Fe(III) salt to Fe(II) salt.

Question 21.
Comment on the nature of two S—O bonds formed in SO₂ molecule. Are the two bonds in the molecule equal ?
Answer:
The two S—O bonds in the molecule are equal with bond length equal to 143 pm. This means that SO₂ molecule exhibits two resonating structures as shown below.

Question 22.
How is presence of SO₂ detected ?
Ans.
Presence of SO₂ is detected by bringing a paper dipped in acidified potassium dichromate near the gas. If the paper turns green, it shows the presence of SO₂ gas.

Question 23.
Mention three areas in which H₂SO₄ plays an important role.
Answer:
(i) It is used in the manufacture of fertilizers such as (NH₄)² SO₄ , calcium superphosphate.
(ii)It is used as an electrolyte in storage batteries.
(iii)It is used in petroleum refining, detergent industry and in the manufacture of paints, pigments and dyes.

Question 24.
Write the conditions to maximise the yield of sulphuric acid by Contact process.
Answer:
Catalytic oxidation of sulphur dioxide into sulphur trioxide. Sulphur dixocide is oxidised to sulphur trioxide with air in the presence of V₂O₅ or platinised asbestos acting as the catalyst

Question 25.
Why is \({ K }_{ { a }_{ 2 } }\) < \({ K }_{ { a }_{ 1 } }\) for H₂SO₄ in water ?
Answer:
H₂SO₄ is a very strong acid in water largely because of its first ionisation to H₃O⁺ and HSO₄^– The ionisation of HSO₄^– to H₃O⁺ and SO₄^2- is very very small. That is why, \({ K }_{ { a }_{ 2 } }\) < \({ K }_{ { a }_{ 1 } }\).

Question 26.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and Cl₂.
Answer:
Fluorine is a better oxidising agent than chlorine because E°_F2/F- is higher than E°_Cl2/Cl- It is mainly due to low bond dissociation energy, high hydration energy and lower electron gain enthalpy, non-availability of d-orbitals in valence shell, that results in higher reduction potential of F₂ than chlorine.

Question 27.
Give two examples to show the anomalous behaviour of fluorine.
Answer:
The anomalous behaviour of fluorine, the first member of the halogen family as compared to the rest of the members is due to its very small size, very high electronegativity and absence of vacant d-orbitals in the valence shell. It is supported by the following points.
(i) Fluorine shows only negative oxidation state of -1 in its compounds. The other members exhibit both positive and negative oxidation states.
(ii) Fluorine forms hexafluoride with sulphur (SF₆). No other member of the family forms hexahalide with sulphur.

Question 28.
Sea is the greatest source of some halogens. Comment.
Answer:
The name halogen is a Greek Word meaning lsea salt forming’. Sea is a major source of a members of halogens particularly chlorine, bromine and iodine and they exist as the soluble salts of sodium, potassium, calcium, magnesium etc. The deposits of dried up sea water contain sodium chloride and carnallite (KClMgCl₂.6H₂O). Sea weeds contain nearly 0-5 percent of iodine. Similarly Chile saltpeter contains about 0-2% of sodium iodate (NaIO₃).

Question 29.
Give reason for the bleaching action of Cl₂.
Answer:
Bleaching by chlorine occurs in the presence of moisture. In fact, it releases nascent oxygen on reacting with HO which carries bleaching. Since the reaction cannot be reversed, the bleaching by chlorine is permanent.
Chlorine bleaches by oxidation Cl₂ + H₂O → HCl + HOCl → HCl + [O]
The nascent oxygen reacts with dye to make it colourless.

Question 30.
Name two poisonous gases which can be prepared from chlorine gas.
Answer:
Two poisonous gases are phosgene and mustard gas.

Question 31.
Why is ICl more reactive than I₂ ? (C.B.S.E. Outside Delhi 2012)
Answer:
The reactivity of ICl is due to its polar nature (I—Cl). Iodine (I₂) being non-polar is comparatively less reactive chemically.

Question 32.
Why is helium used in diving apparatus?
Answer:
A mixture of helium and oxygen does not cause pain due to very low solubility of helium in blood as compared to nitrogen.

Question 33.
Balance the equation : XeF₆ + H₂O → XeO₂F₂ +HF (H.P. Board 2013)
Answer:

Question 34.
Why has it been difficult to study the chemistry of radon ?
Answer:
Radon (Rn) is a radioactive element with very short half period of 3-82 days. Therefore, it becomes quite difficult to study the details about the chemistry of the element.

NCERT EXERCISE

Question 1.
Discuss the general characteristics of group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
Answer:
(i) Electronic configuration: All the elements in group 15 have 5 valence electrons. Their general electronic configuration is ns²np³

(ii) Oxidation states:
All these elements have 5 valence electrons and require three more electrons to complete octets. However, gaining electrons is very difficult as the nucleus will have to attract three more electrons. This can take place only with nitrogen as it is the smallest in size and the distance between the nucleus and the valance shell is relatively small. The remaining elements of this group show a formal oxidation state of -3 in their covalent compounds.

In addition to the -3 state, N and P also show -1 and.-2 oxidation states. All the elements present in this group show +3 and +5 oxidation states. However, the stability of the +5 oxidation state decreases down the group, whereas the stability of+3 oxidation state increases. This happens because of the inert pair effect.

(iii) Ionization energy and electronegativity First ionization energy decreases on moving down a group. This is because of increasing atomic sizes. As we move down a group, electronegativity decreases, owing to an increase in size.

(iv) Atomic size:
On moving down a group, the atomic size increases. This increase in the atomic size is attributed to an increase in the number of shells.

Question 2.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:
N₂ exist as a diatomic molecule containing triple bond between two N-atoms. Due to the presence —of the triple bond between p  the two N-atoms, the bond dissociation energy is large (941 .4 kJ mol^-1 ). As a result of this N₂ is inert and unreactive whereas, phosphorus exists as a tetratomic molecule, containing P – P single bond. Due to the presence of single bond, the bond dissociation energy is weaker (213 kJmol^-1 ) than N a N triple bond (941 .4 kJ mol^-1 ) and moreover due to presence of angular strain in P₄ tetrahedra. As a result of this, phosphorus is much more reactive than nitrogen.

Question 3.
Discuss the trends in chemical reactivity of group 15 elements.
Answer:
(a) Hydrides: The hybrids are covalent with pyramidal structures  and the central atom is sp³ hybrilised. The presence of the lone pair of electrons on the central atom distorts the geometry of the molecules and the bond angle less than that of a regular tetrahedron.

(b) Halides: Elements of group 15 form two types of halides viz. trihalides and pentahalides. The halides are predominantly basic (Lewis bases) in nature and have lone pair of electrons (central atom is sp3 hybridized). The pentahalides are thermally less stable than the trihalides.

(c) Oxides: All the elements of this group form two types of oxides ie., M₂O₃ and M₂O₅ and are called trioxides and pentoxides

Question 4.
Why does NH₃ form hydrogen bonding while PH₃ does not?
Answer:
The N—H bond in ammonia is quite polar on account of the electronegativity difference of N (3·0) and H (2·1). On the contrary, P—H bond in phosphine is almost non-polar because both P and H atoms have almost same electronegativity (21). Due to polarity, intermolecular hydrogen bonding is present in the molecules of ammonia but not in those of phosphine.

Question 5.
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.
Answer:
Laboratory preparation: Dinitrogen is prepared in the laboratory by heating a solution containing an equivalent amount of sodium nitrite and ammonium chloride.

Question 6.
How is ammonia manufactured industrially?
Answer:
Ammonia is prepared on a commercial scale by Haber’s Process from dinitrogen and dihydrogen by the following chemical reaction
N₂ + 3H₂ ⇌ 2NH₃; ∆_fH° = – 46.1 kJ mon^-1
Dihydrogen needed for the commercial preparation of ammonia is obtained by the electrolysis of water while dinitrogen is obtained from the liquefied air as a fractional distillation. The two gases are purified and also dried.
These are compressed to about 200 atmosphere pressure and are then led into the catalyst chamber packed with the catalyst and the promoter.

Question 7.
Illustrate how copper metal gives different products on reaction with HNO₃.
Answer:
Anhydrous nitric acid is colourless, fuming, and pungent-smelling liquid. However, it acquires a yellowish-brown colour in the presence of sunlight. HNO₃ decomposes to give NO₂ gas which dissolves and imparts its yellowish-brown colour.

On strong heating, the acid decomposes to give NO₂ and O₂

Question 8.
Give the resonating structures of NO₂ and N₂O₅.
Answer:

Question 9.
The HNH angle value is higher than those of HPH, HAsH, and HSbH angles; why?
Answer:

The above trend in the H-M-H bond angle can be explained on the basis of the electronegativity tif central atom. Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes greater repulsion between the electron pairs around nitrogen, resulting in maximum bond angle. We know that electronegativity decreases on moving down a group consequently. The repulsive interactions between the electron pairs decrease, thereby decreasing the HMH bond angle.

Question 10.
Why does R₃P = O exist but R₃N = O does not (R = alkyl group)?
Answer:
Nitrogen does not contain d-orbitals. As a result, it cannot expand its covalency beyond four and cannot form pπ – dπ multiple bonds. In contrast, P contains the d-orbitals, and can expand its covalency beyond 4 and can form pπ-dπ multiple bonds. Hence R₃P = O exist but R₃N = O does not.

Question 11.
Explain why is NH₃ basic while PH₃ is feebly basic in nature. (C.B.S.E. Outside Delhi 2008, 2009, Jharkhand Board 2009)
Answer:
NH3 is distinctly basic while BiH₃ is feebly basic.
Nitrogen has a small size due to which the lone pair of electrons are concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases, and the charge gets distributed over a large area decreasing the electron density. Hence the electron-donating capacity of group 15 elements hydrides decreases on moving down the group.

Question 12.
Nitrogen exists as a diatomic molecule (N₂) while phosphorus a tetra-atomic molecule (P₄). Why?
Answer:
Nitrogen is diatomic gaseous molecule at ordinary temperature due to its ability to form pπ – pπ multiple bonds. The molecule has one σ and two π – bonds. Phosphorus exists as discrete tetratomic tetrahedral molecules as these are not capable of forming multiple bonds due to repulsion between non-bonded electrons of the inner core.

Question 13.
Write the main difference between the properties of white and red phosphorus. (C.B.S.E. Delhi 2012)
Answer:

Question 14.
Why does nitrogen show catenation properties less than phosphorus?
Answer:
The catenation properties depend upon the strength of the element-element bond. The N-N bond strength is much weaker (due to the repulsion of lone pairs on nitrogen because of its small size) than the P-P bond strength, therefore, nitrogen shows catenation less than phosphorus.

Question 15.
Give one disproportionation reaction of phosphorus acid (H₃PO₃).
Answer:
Upon heating to about 573 K, phosphorus acid undergoes disproportionation as follows :

Question 16.
Can PCl₅ act as oxidising as well as a reducing agent? Justify.
Answer:
In general, the molecules of a substance can behave as a reducing agent if the central atom is in a position to increase its oxidation number. Similarly, they can act as an oxidising agent if the central atom is in a position to decrease its oxidation number. Now, the maximum oxidation state or oxidation number of phosphorus (P) is +5. It cannot increase the same but at the same time can decrease its oxidation number. In PCl₅, oxidation number of P is already +5. It therefore, cannot act as a reducing agent. However, it can behave as an oxidising agent in certain reactions in which its oxidation number decreases. For example,

Question 17.
Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation states and hydride formation.
Answer:
The members of the oxygen family are placed in group 16 of p-block. Their inclusion in the same group is justified on the basis of the following characteristics.
1. Electronic configuration. Members of the family have ns2p4 configuration. Their group (10 + 6) is 16.
2. Oxidation states. With exception of oxygen which exhibits -2 oxidation state in its compounds (OF₂ and H₂O₂ are exceptions), rest of the members of the family show variable oxidation states (-2, +2, +4, +6) in their compounds.
3. Hydride formation. All the members of the family form covalent hydrides (MH₂) in which the central atom is sp³ hybridised. These have angular structures and their characteristics show regular gradation.
FeS + H₂SO₄ (dil.) → FeSO₄ + H₂S
Na₂Se + H₂SO₄ (dil.) → Na₂SO₄ + H₂Se

Question 18.
Why is dioxygen a gas while sulphur is a solid? (C.B.S.E. Delhi 2013)
Answer:
The oxygen atom has a tendency to form pπ-pπ multiple bonding due to its small size and high electronegativity. As a result, oxygen exists as a diatomic molecule. These mol¬ecules are held together by weak van der Waal’s forces of attraction which can be easily overcome by collisions of the molecules at room temperature. Therefore, O₂ is the gas at room temperature.

Sulphur, on the other hand, because of its bigger size and lower electronegativity, does not form pπ-pπ multiple bonds. Instead, it prefers to form an S-S single bond and form polyatomic complex molecules having eight atoms per molecule (S₈) and have puckered ring structure. Therefore, S atoms are strongly held together and it exists as a solid.

Question 19.
Knowing the electron gain enthalpy values for O → O^– and O → O^2- as – 141 kJ mol^-1 and 702 kJ mol^-1 respectively, how can you account for the formation of a large number of oxides having O^2- species and not O^– ?
Answer:
The stability of an ionic compound depends on its lattice energy. The more the lattice energy of a compound, the more stable it will be. Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving O^2- ion is much more than the oxide involving O^– ion. Hence, we can say that the formation of O^2- is energetically more favourable than the formation of O^–

Question 20.
Which aerosols deplete ozone?
Answer:
Aerosols or chlorofluorocarbons (CFC’s) such as freon (CCl₂F₂) deplete ozone layer by supplying chlorine-free radicals (Cl) which convert ozone into oxygen.

Question 21.
Describe the manufacture of H₂SO₄ by Contact process.
Answer:
Sulphuric acid is manufactured by the Contact process which involves three steps :

1. burning of sulphur or sulphide ores in air to generate SO₂.
2. conversion of SO₂ to SO₃ by the reaction with oxygen in the presr nee of a catalyst (V₂O₅), and
3. absorption of SO₃ in H₂SO₄ to give oleum (H₂S₂O₇).

A flow diagram for the manufacture of sulphuric acid is shown in the figure. The SO₂ produced is purified by removing dust and other impurities such as arsenic compounds. The key step in the manufacture of H₂SO₄ is the catalytic oxidation of SO₂ with O₂ to give SO₃ in the presence of V₂O₅ (catalyst).

The reaction is exothermic, reversible and the forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are favourable conditions for maximum yield. But the temperature should not be very low otherwise rate of reaction will become slow.

In practice, the plant is operated at a pressure of 2 bar and a temperature of 720 K. The SO₃ gas from the catalytic converter is absorbed in concentrated H₂SO₄ to produce oleum. Dilution of oleum with water gives H₂SO₄ of the desired concentration. In the industry, two steps are carried out simultaneously to make the process a continuous one and also to reduce the cost.

Question 22.
How is SO₂ an air pollutant?
Answer:

• It combines with water vapour present in the atmosphere to form sulphuric acid. This causes acid rain. Acid rain damages soil, plants and buildings, especially those made of marble.
• Even in very low concentrations, SO₂ causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.
• It is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves. This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of SO₂.

Question 23.
Why are halogens strong oxidising agents?
Answer:
The general electronic configuration of halogens is np⁵, where n = 2-6. Thus, halogens need only one more e- to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an election. Hence, they act as strong oxidizing agents.

Question 24.
Explain why does fluorine form only one oxoacid (HOF).
Answer:
The members of the halogen family with the exception of fluorine show variable oxidation states due to the availability of rf-orbitals for the bond formation. They form a number of oxoacids such as HOX, HOXO, HOXO_2, and HOXO₃. However, fluorine which is highly electronegative and has no rf-orbitals, forms only one oxoacid (HOF) in which its oxidation state is +1.

Question 25.
Explain why in spite of nearly the same electronegativity, nitrogen is involved in hydrogen bonding while chlorine is not.
Answer:
Both chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume.

Question 26.
Write two uses of ClO₂.
Answer:
ClO₂ is a strong oxidising agent. Therefore,
(i) It acts as bleaching agent for paper pulp in paper industry and in textile industry.
(ii) It acts as germicide for disinfecting water.

Question 27.
Why are halogens coloured?
Answer:
All the halogens are coloured in nature. The colour deepens with the increase in the atomic number of the element from fluorine to iodine.

The cause of the colour is due to the absorption of energy from the visible light by the molecules for the excitation of outer non -bonded electrons to higher energy levels. The excitation energy depends upon the size of the atom. Fluorine has the smallest size and the force of attraction between the nucleus and electrons is very large.

Question 28.
Write the reactions of F₂ and Cl₂ with water.
Answer:

Question 29.
How can you prepare Cl₂ from HCl and HCl from Cl₂? Write chemical equations only.
Answer:
(i) HCl can be oxidised to chlorine with the help of a number of oxidising agents like MnO₂, KMnO₄, K₂Cr₂O₇ etc.
Mn0O₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
(ii) Cl₂ can be reduced to HCl by reacting with H₂ in the presence of sunlight. The gas on passing through water dissolves to form hydrochloric acid

Question 30.
What inspired N. Bartlett for carrying out the reaction between Xe and PtF₆?
Answer:
The X-ray study of the compound has shown it be a crystalline solid consisting of O₂⁺ and (PtF₆]^– ions. In this reaction, PtF₆ has oxidised O₂ to O₂⁺ ion. Bartlett through than PtF₆ should Xe to xe⁺ since first ionisation enthalpy of xenon (1176 kJ mol^-1) is quite close to that of O₂ (1180 kJ mol^-1).

Question 31.
What is the oxidation state of phosphorus in the following?
(a) H₃PO₃
(b) PCl₃
(c) Ca₃P₂
(d) Na₃PO₄
(e) POF₃
Answer:

Question 32.
Write balanced equations for the following :
(i) NaCl is heated with sulphuric acid in the presence of MnO₂
(iii) Chlorine gas is passed into a solution of Nal in water.
Answer:

Question 33.
How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained ?
Answer:
Xenon forms three binary fluorides, XeF₂, XeF₄ and XeF₆ by the direct reaction of elements under appropriate experimental conditions.

Question 34.
With what neutral molecule is ClO^– isoelectronic? Is that molecule a Lewis base?
Answer:
OF₂ and ClF are isoelectronic to ClO^–, out of which ClF is a Lewis base.

Question 35.
How are XeO₃ and XeOF₄ prepared?
Answer:
Preparation of XeO₃. By complete hydrolysis of XeF₆.
XeF₆ + 3H₂O → XeO₃+ 6HF.
Preparation ofXeOF₄. By partial hydrolysis of XeF₆.
XeF₆ + H₂O → XeOF₄+ 2HF.

Question 36
Arrange the following in the order of property indicated for each set:
(i) F₂, Cl₂, Br₂, I₂ – increasing bond dissociation enthalpy.
Answer:
Bond dissociation energy usually decreases on moving down a group as the atomic size increases. However, the bond dissociation energy of F₂ is lower than that of Cl₂ and Br₂. This is due to the small atomic size of fluorine, Thus, the increasing order for bond dissociation energy among halogens is as follows:
I₂ < F₂ < Br₂ < Cl₂

(ii) HF, HCl, HBr, HI – increasing acid strength.
Answer:
HF < HCl < HBr < HI
The bond dissociation energy of HX molecules where X – F, Cl, Br, I, decreases with an increase in the atomic size. Since H – I bond is the weakest, HI is the strongest acid.

(iii) NH₃, PH₃, ASH₃, SbH₃, BiH₃ – increasing base strength.
Answer:
BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃ On moving from nitrogen to bismuth, the size of the atom increases while the electron density on the atom decreases. Thus, the basic strength decreases.

Question 37.
Which one of the following does not exist?
(i) XeOF₄
(ii) NeF₂
(iii) XeF₂
(iv) XeF₆.
Answer:
NeF₂ does not exist because the element Ne(Z = 10) with 1s²2s²2p⁶ does not have vacant 2d orbitals. As such, there is no scope of any electron promotion even by a highly electronegative element.

Question 38.
Give the formula and describe the structure of a noble gas species which is isostructural with :
(i) \(IC{ l }_{ 4 }^{ – }\)
(ii) \(IB{ r }_{ 2 }^{ – }\)
(iii) \(Br{ O }_{ 3 }^{ – }\)
Answer:

(i) Structure of \(IC{ l }_{ 4 }^{ – }\). The central I atom has in all 8 electrons (7 valence electrons +1 due to negative charge). Out of these, it shares 4 electrons with four atoms of Cl and the remaining four electrons constitute two lone pairs. In all, there are six pairs. The structure of the ion must be octahedral or distorted square planar in order to minimise the forces of repulsion among the two electrons pairs. \(IC{ l }_{ 4 }^{ – }\) has (7 + 4 x 7 + 1) = 36 valence electrons and it iso-electronic and iso-structural with XeF₄ (8 + 4 x 7) which has also 36 valence electrons.

(ii) Structure of \(IB{ r }_{ 2 }^{ – }\). In \(IB{ r }_{ 2 }^{ – }\) ion, the central I atom has 8 valence electrons (7 + 1). Out of these, it shares 2 electrons with two atoms of Br and the remaining 6 electrons constitute three lone pairs. In all, there are five pairs. The structure of the ion must be trigonal bipyramidal or linear in order to minimise the force of repulsion among the three lone electrons pairs. IBrf has (7 + 2 x 7 + 1) = 22 valence electrons and is isoelectronic as well as iso-structural with noble gas species XeF₂ which has also 22 (8 + 2 x 7) electrons.

(iii) Structure of \(Br{ O }_{ 3 }^{ – }\) ion. In \(Br{ O }_{ 3 }^{ – }\) ion, the central Br atom has 8 valence electrons (7 + 1). Out of these, it shares 4 with two atoms of O forming Br=0 bonds. Out of the remaining four electrons, 2 are donated to the third O atom and account for its negative charge. The remaining 2 electrons constitute one lone pair. In order to minimise the force of repulsion, the structure of \(Br{ O }_{ 3 }^{ – }\) ion must be pyramidal. \(Br{ O }_{ 3 }^{ – }\) ion has (7 + 3 x 6 + 1) = 26 valence electrons and is isoelectronic as well as iso-structural with noble gas species Xe03 which has also 26 (8 + 3 x 6) electrons.

Question 39.
Why do noble gases have comparatively large atomic sizes?
Answer:
Noble gases do not form molecules. In the case of noble gases. the atomic radii correspond to Van der Waa1s radii. On the other hand, the atomic radii of other elements correspond to their covalent radii. By definition, Yan der Waal’s radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period.

Question 40.
List the uses of neon and argon gases.
Answer:
Uses of Neon:

• Neon lights are used for commercial advertisements.
• It consists of a long tube fitted with electrodes at both ends.
• On filling the tube with neon gas and passing electric discharge of about 1000 volt potential, a bright red light is produced.
• Different colours can be obtained by mixing neon with other gases.
• For producing blue or green light neon is mixed with mercury vapours.

Uses of Argon:

• Like helium, it is also used to create an inert atmosphere in welding aluminium and stainless steel.
• It is filled in electric bulbs along with 25% nitrogen.
• It is also used in radio valves.
• Argon alone or its mixture with neon is used in tubes for producing lights of different colours.

We hope the NCERT Solutions for Class 12 Chemistry Chapter 7 The p-Block Elements help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 7 The p-Block Elements, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Chemistry Chapter 4 provides excellent solutions for the questions asked in the textbook. The step wise solutions and diagrammatic representations make the concepts easy to understand. The subject experts have given the best explanations to the queries. The students appearing for the board exams or competitive exams can refer to these for better preparations.

CBSE, MP board, UP board, Gujarat board, etc. have the NCERT Solutions for reference in the curriculum. Chemistry is an important subject and requires conceptual analysis. The detailed explanations in the NCERT Solutions for Class 12 Chemistry Chapter 4 will help the students to score well in the examination.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 4
Chapter Name	Chemical Kinetics
Number of Questions Solved	39
Category	NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics

Class 12 Chemistry chapter 4 Chemical Kinetics is an important chapter and is often asked in the examination. Chemical kinetics helps to understand the chemical reactions. This chapter explains all about the rate of reaction and the factors determining the rate of reaction.

NCERT IN-TEXT QUESTIONS

Question 1.
For a reaction R → P, the concentration of a reactant changes from 0·03 M to 0·02 M in 25 minutes. Calculate the average rate of the reaction using the units of seconds.
Answer:
For a reaction, R → P

Question 2.
In a reaction, 2A → Products, the concentration of A decreases from 0·5 mol L^-1 to 0·4 mol L^-1 in 10 minute. Calculate the rate during this interval.
Answer:
For the reaction: 2A → Products

Question 4.
Fora reaction,A+B —> Product; the rate law is given by, r =k [ A]^1/2 [B]². What is the order of the reaction?
Answer:
Order of reaction. = 1/2+ 2 = 2^1/2 or 2.5

Question 4.
The conversion of the molecules X to Y follows second order kinetics. If concentration of X is increased to three times, how will it affect the rate of formation of Y ?
Answer:
For the reaction X → Y
Reaction rate (r) = k[X]²
If the concentration be increased to three times, then
Reaction rate (r’) = k [3X]²
\(\frac { { r }^{ ‘ } }{ r } =\frac { k\left[ 3X \right] ^{ 2 } }{ k\left[ X \right] ^{ 2 } } =9\)

Question 5.
A first order reaction has rate constant of 1·15 x 10^-3 s^-1. How long will 5 g of this reactant take to reduce to 3g?
Answer:
For the first order reaction,

Question 6.
Time required to decompose SO₂Cl₂ to half of its initial concentration is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Answer:
For the first order reaction ;
Rate constant (k) = \(\frac { 0.693 }{ { t }_{ 1/2 } } =\frac { 0.693 }{ \left( 60min \right) } \)
= \(\frac { 0.693 }{ \left( 60\times 60s \right) } =1.925\times { 10 }^{ -4 }{ s }^{ -1 }\)

Question 7.
What will be the effect of temperature on rate constant?
Answer:
With the rise in temperature by 10°, the rate constant of a reaction is nearly doubled. The dependence of rate constant on temperature is given the Arrhenius equation, k = A e^-Ea/RT where A is the Arrhenius constant and Ea is activation energy of the reaction.

Question 8.
In general, it is observed that the rate of a chemical reaction doubles with every 10° rise in temperature. If the generalisation holds for a reaction in the temperature range 295 K to 305 K, what would be the value of activation energy for the reaction? (C.B.S.E. Delhi2005 Supp., Pb. Board2007)
Answer:
According to Arrhenius equation,

Question 9.
The activation energy for the reaction, 2HI(g) → H₂(g) + I₂(g) is 208·5 kJ mol^-1. Calculate fraction of molecules of reactants having energy equal to or greater than activation energy.
Answer:
The fraction of the molecules (x) having energy equal to or more than activation energy may be calculated as follows :

NCERT Exercise

Question 1.
The rate expression for the following reactions determine the order of reaction and the dimensions of the rate constant.

Answer:
(a) 2
(b) 2
(c) 3/2
(d) 1

Question 2.
For the reaction, 2A + B → A₂ B, the rate = k [AJ[B]2 with k = 2.0 x 10^-6 mol^-2 L² s^-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1, [B] = 0.2 mol L^-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1.
Answer:
Initial rate of reaction = k [A] [B]²
= (20 x 10^-6 mol^-2 s^-1) (0.1 mol L^-1) (0.2 mol L^-1)² = 8 x 10-9molL^-1 s^-1.
When [A] is reduced from 010 mol L^-1 to 0.06 molL^-1, i.e., 0.04 mol L^-1 of A has reacted, the concentration of B reacted, is = 1/2 x 0.04 mol L^-1 = 0.02 mol L^-1
Concentration of B, remained after reaction with A = 0.2 – 0-02=0.18 mol L^-1
Now, rate=(20 x 10^-6 mol^-2 L² s^-1) (0.06 mol L^-1) (0.18 molL^-1)²
= 3-89 x 10^-9mol L^-1 s^-1

Question 3.
The rate of decomposition of NH₃ on the platinum surface is zero order. What is the rate of production of N₂ and H₂ if k = 2·5 x 10^-4 Ms^-1? (C.B.S.E. Delhi 2008)
Answer:

Question 4.
The decomposition of dimethyl ether leads to the formation of CH₄, H_2, and CO, and the reaction rate is given by the expression:
rate = k [CH₃OCH₃]^3/2
The rate of reaction is followed by increase in pressure in a close vessel and the rate can also be expressed in terms of partial pressure of dimethyl ether :
rate = k [pCH₃OCH₃]^3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constant ?
Answer:

Question 5.
Mention the factors that affect the rate of a chemical reaction.
Answer:

• Concentration of reactants
• Temperature
• Nature of reactants and products
• Exposure to light (Radiation)
• Presence of catalysts.

Question 6.
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
(i) doubled
(ii) reduced to 1/2? (C.B.S.E. Outside Delhi 2008, 2009)
Answer:
Let the reaction be; A → Products
Reaction rate (r) = k [A]² (for second order reaction)
(i) When concentration is doubled, the rate of reaction may be expressed as :
Reaction rate (r’) = k [2A]²

reaction rate becomes four times.
(ii) When concentration is reduced to half, the rate of reaction may be expressed as :

reaction rate will be reduced to 1/4.

Question 7.
What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?
Answer:
Increasing the temperature on decreasing the activation energy will result in an increase in the rate of reaction and an exponential increase in the rate constant. On increasing the temperature the fraction of molecules which collide with energy greater than Ea increases and hence the rate constant (exponentially)
K = A ^-ea/RT, quantitative representation of temperature effect on rate constant.

Question 8.
In pseudo-first-order hydrolysis of ester in water, the following results were obtained.

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Answer:

Question 9.
A reaction is first order in A and second order in B
(i) Write differential rate equation.
(ii) How is rate affected when the concentration of B is tripled?
(iii) How is rate affected when the concentration of both A and B are doubled? (C.B.S.E. Outside Delhi 2010, 2013)
Answer:

Question 10.
In a reaction between A and B, the initial rate of reaction was measured for different initial concentration of A and B as given ahead :

What is the order of reaction with respect to A and B?
Answer:

Question 11.
The following data were obtained at 300 K for the reaction 2A + B → C + D:

Calculate the rate of formation of D when [A] = 0·5 mol L^-1 and [B] = 0·2 mol L^-1.
Answer:

Question 12.
The reaction between A and B is first order with respect to A and zero-order with respect to B. Fill in the blanks in the following table:

Answer:
The rate equation for the reaction is: r = k [A]¹ [B]⁰
(i) Comparing experiments I and II,

Thus, the concentration of A in experiment II is 0·2 M
(ii) Comparing experiments II and in.
When the concentration of A is made double, the reaction rate will also become twice.
∴ Rate of reaction in experiment III is 8·0 x 10^-2
(iii) Comparing experiments I and IV.
Since the reaction rates are the same in both the experiments, the molar concentration of A in experiment IV must be the same as in experiment I i. e., it must be 0·1 M.

Question 13.
Calculate the half-life of the first-order reaction from their rate constants given as
(a) 200 s^-1
(b) 2 min^-1
(c) 4 year^-1.
Answer:

Question 14.
The half-life for the radioactive decay of ¹⁴C is 5730 Y. An archaeological artifact contained wood had only 80% of the 14 C found in a living tree. Estimate the age of the sample. (C.B.S.E. Delhi 2008)
Answer:

Question 15.
The experimental data for decomposition of N₂O₂ [2N₂O₅ → 4NO₂ + O₂] in gas phase at 318 K are given below :

(a) Plot [N₂O₅] against t
(b) Find the half-life period for the reaction
(c) Draw a graph between log [N₂O₅] and t
(d) What is rate law?
(e) Calculate the rate constant
(f) Calculate the half-life period from k and compare it with (b).
Answer:
The available data is:

(b) Initial cone. of N₂O₅ = 1·63 x 10^-2M. Half of initial cone. = 1/2 x (1·63 x 10^-2 M) = 0·815 x 10^-2 M Time corresponding to half of inital concentration (t/2) from the plot (a) = 1400 (s) approximately
(c) The graph of log [N₂O₅] Vs. time has been plotted.
(d) Since the graph between log [N₂O₅] and time is a straight line the reaction is of the first order
The rate equation : rate (r) = k[N₂O₅]

Question 16.
The rate constant for the first-order reaction is 60 s^-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value? (C.B.S.E. Delhi 2013)
Answer:
For the first-order reaction

Question 17.
During a nuclear explosion, one of the products is ⁹⁰Sr with a half period of 28·1 Y. If 1 pg of ⁹⁰Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically?
Answer:

Question 18.
For a first-order reaction, show that the time required for 99% completion is twice the time required for the completion of 90% of the reaction.
Answer:

Question 19.
A first-order reaction takes 40 minutes for 30% decomposition. Calculate its half-life period. (C. B. S. E. Outside Delhi 2013)
Answer:

Question 20.
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data is obtained.

Calculate the rate constant.
Answer:
The decomposition reaction is of gaseous nature and the expression of the rate equation for the reaction is :

Question 21.
The following data were obtained during the first-order thermal decomposition of SO₂Cl₂ at a constant volume.
SO₂Cl₂ (g) → SO₂ (g) + Cl₂(g)

Calculate the rate of the reaction when the total pressure is 0·65 atm. (C.B.S.E. Sample Paper 2011)
Answer:

Question 22.
The rate constant for the decomposition of N₂O₅ at various temperatures is given below :

Draw a graph between ln k and 1/T and calculate the values of A and E_a. Predict the rate constant at 30° and 50°C.
Answer:
To draw the plot of log K versus 1/T, we can re-write the given data as follows :

Question 23.
The rate constant for the decomposition of a hydrocarbon is 2.418 × 10^-5 s^-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of the pre-exponential factor.
Answer:

Question 24.
Consider a certain reaction A → Products with k = 2·0 x 10^-2 s^-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1·0 mol L^-1.
Answer:
For the first-order reaction :

Question 25.
Sucrose decomposes in acid solution into glucose and fructose according to the first-order rate law with t_1/2 = 3·0 hrs. What fraction of the sample of sucrose remains after 8 hours? (C.B.S.E. Sample Paper 2011)
Answer:

Question 26.
The decomposition of a hydrocarbon follows the equation :
k = (4·5 x 10¹¹ s^-1)e^-28000k/T.
Calculate the energy of activation (E_a).
Answer:

Question 27.
The rate constant for the first-order decomposition of H₂O₂ is given by the following equation:
log k = 14·34 – 1·25 x 10⁴K/T.
Calculate the Ea for the reaction. At what temperature will the half-life period be 256 minutes?
Answer:

Question 28.
The decomposition of A into the product has a value of k as 4·5 x 10³ s^-1 at 10°C and energy of activation 60 kJ mol^-1. At what temperature would k be 1·5 x 10⁴ s^-1? (C.B.S.E. Sample Paper 2011)
Answer:
According to Arrhenius equation,

Question 29.
The time required for 10% completion of a first-order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 x 10¹⁰s^-1, calculate k at 318 K and E_a.
Answer:

Question 30.
The rate of a particular reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction (R = 8·314 JK^-1 mol^-1). (C.B.S.E. Outside Delhi 2013)
Answer:
According to Arrhenius equation,

We hope the NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics, drop a comment below and we will get back to you at the earliest.

NCERT Class 12 Chemistry Solutions for Chapter 8  d-and f-Block Elements provides solutions to the questions provided in the textbook. These solutions are provided by the subject experts and are helpful for the students appearing for boards or competitive exams. The students can refer to these and enhance their conceptual knowledge.

NCERT Solutions are provided in the CBSE, MP, UP, and Gujarat boards. The students appearing for these boards can practice through the NCERT Solutions to score well in the examination.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 8
Chapter Name	d-and f-Block Elements
Number of Questions Solved	48
Category	NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements

The chapter d and f block elements is very important from the examination perspective. It explains the elements in the groups 3-12. A detailed understanding of this chapter will help the students to differentiate between the characteristics of d and f block elements. It also explains a comparative account of lanthanoids and actinoids.

The NCERT Solutions for Class 12 Chemistry Chapter 8 provides in-depth details of d and f blocks elements. The students are advised to refer to these solutions for better results.

NCERT IN-TEXT QUESTIONS

Question 1.
Silver has completely filled d-orbitals (4d10) in its ground state. How can you say that it is a transition metal ?
Answer:
Silver (Z = 47) belongs to group 11 of (7-block (Cu, Ag, Au) and its outer electronic configuration is 4d¹⁰5s¹. It shows + 1 oxidation state (4d10 configuration) in silver halides (e.g. AgCl). However, it can also exhibit + 2 oxidation state (4d⁹ configuration) in compounds like AgF₂ and AgO. Due to the presence of half filled d-orbital, silver is a transition metal.

Question 2.
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol^-1. Why?
Answer:
In 3d series from Sc to Zn, all elements have one or more unpaired e^-1 s except Zn which has no unpaired electron as its outer EC is 3d¹⁰4s². Hence, the intermetallic bonding is weakest in zinc. Therefore, enthalpy of atomisation is lowest.

Question 3.
Which of the 3d series of transitional metals exhibits largest oxidation states and why?
Answer:
Mn (Z = 25) with electronic configuration [Ar]3d⁵4s² shows maximum oxidation state (+ 7) in its compounds since it has the maximum number of unpaired five i.e., seven. It shows largest variable oxidation state from + 2 to + 7 ( + 2, + 3, + 4, + 5, + 6, + 7) in its compounds.

Question 4.
The E°(M²⁺/M) value for copper is positive (+ 0·34 V). What is possibly the reason for this ? (C.B.S.E. Outside Delhi 2012, Sample Paper 2012)
Answer:
E°(M²⁺/M) for any metal is based upon three factors which have been discussed in the text part.
M(s) + ∆_aH → M(g) ; (∆_aH = Enthalpy of atomisation)
M(g) + ∆_fH → M²⁺(g) ; (∆_fH = Ionisation enthalpy)
M²⁺(g) + aq → M²⁺(aq); (∆_hydH = Hydration enthalpy)
Copper has very high enthalpy of atomisation (energy required) and low enthalpy of hydration (energy released). In nut shell, the ∆_fH i.e. ionisation enthalpy needed is not compensated by the energy released. Therefore E°(Cu²⁺/Cu) is positive.

Question 5.
How would you account for the irregular variation in ionisation enthalpies (first and second) in first series of transition elements ?
Answer:
The ionisation enthalpies of the transition metals are higher than those of s-block elements and less than the elements of p-block. Thus, these are less electropositive than the elements of s-block and at the same time more electropositive than the elements belonging to p-block present in the same period. In a transition series, the ionisation enthalpies increase from left to right. However, the gaps in the values of the two successive elements are not regular.

Question 6.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer:
Oxygen and fluorine have small size and high electronegativity. Hence, they can oxidise the metal to the highest O.S.

Question 7.
Which is a stronger reducing agent Cr²⁺ or Fe²⁺ and why ? (SamplePaper 2011, C.B.S.E. Outside Delhi 2010, 2014)
Answer:
Cr²⁺ is a stronger reducing agent than Fe²⁺. This is quite evident from the E° values ;
E°Cr³⁺/Cr²⁺ = – 0-41 V and E°Fe³⁺/Fe²⁺ = 0-77 V.
Reason : d⁴ → d³ occurs when Cr²⁺ changes to Cr³⁺ ion while d⁶ → d⁵ takes place when Fe²⁺ gets converted to Fe³⁺ ion.

Now, d⁴ → d³ transition is easier as compared to d⁶ → d⁵ transition because in the second case, an electron is removed from a paired orbital which is rather difficult. Therefore, Cr²⁺ is a stronger reducing agent than Fe²⁺.

Question 8.
Calculate the spin magnetic moment of M²⁺(aq) ion (Z = 27).
Answer:
Electronic configuration of element M(Z = 27) : [Ar] 3d⁷4s²
Electronic configuration M²⁺ (aq) ion : 3d⁷ or
Magnetic moment of M²⁺ (aq) ion with n = 3 ; \(\mu =\sqrt { n\left( n+2 \right) } \)
\(=\sqrt { 3\left( 3+2 \right) } =\sqrt { 15 } =3\cdot 87BM.\)

Question 9.
Explain why Cu⁺ ion is not stable in an aqueous solution. (C.B.S.E. Delhi 2011)
Answer:
In aqueous solution Cu⁺ (aq) undergoes disproportionation to form Cu²⁺ (aq) ion and Cu.
2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)
The higher stability of Cu²⁺ (aq) in an aqueous solution may be attributed to its greater negative ∆_hydH^⊝ than that of Cu⁺ (aq). It compensates for the second ionisation enthalpy of Cu involved in the formation of Cl²⁺ ion. Thus, Cu⁺ (aq) ion changes to Cu²⁺ (aq) ion which is more stable.

Question 10.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Answer:
This is due to poor shielding by 5f-electrons in the actinoids than that by 4f e^-1s in lanthanoids.

NCERT EXERCISE

Question 1.
Write down the electronic configuration of :
(a) Cr³⁺
(b) Cu⁺
(c) Co²⁺
(d) Mn²⁺
(e) Pm³⁺
(f) Ce⁴⁺
(g) Lu²⁺
(h) Th⁴⁺
Answer:

Question 2.
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their+3 state?
Answer:
A half-filled d orbital or a completely filled d orbital is more stable than any other state. Mn²⁺ already has a half-filled stable state hence would not undergo oxidation to form Mn⁺³. On the other hand, Fe²⁺ on oxidation to Fe³⁺ will have half-filled d orbitals which are more stable.

Question 2.
Why are Mn²⁺ compounds more stable than Fe²⁺ compounds towards oxidation to their +3 state?
Answer:
Electronic configuration of Mn²⁺ is 3d⁵ while that of Fe²⁺ is 3d6. This shows that the Fe²⁺ ion has an urge to change to Fe³⁺ ion by losing an electron whereas the Mn²⁺ ion has no such tendency. Thus, the + 2 oxidation state of Mn is more stable as compared to the + 2 oxidation state of Fe.

Question 3.
Explain briefly how + 2 oxidation state becomes more and more stable in the first half of the first-row transition elements with increasing atomic number.
Answer:
In all the elements listed, with the removal of valence 45 electrons (+2 oxidation state), the 3d-orbitals get gradually occupied. Since the number of empty d-orbitals decreases or the number of unpaired electrons in 3d orbitals increases, the stability of the cations (M²⁺) increases from Sc²⁺to Mn²⁺.

Question 4.
To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with an example.
Answer:
In the transition series, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, the electronic configuration of Fe(Z = 26) is [Ar] 3d⁶4s². It shows various oxidation states but Fe (III) is most stable because it has the configuration [Ar]3d⁵.

Question 5.
What must be the stable oxidation state of the transition elements with the following electronic configuration in the ground states of their atoms: 3d³, 3d⁵, 3d⁸, 3d⁴?
Answer:
The maximum oxidation states of reasonable stability in the transition metals of 3d series correspond to the sum of s and d-electrons upto Mn. However, after Mn there is an abrupt decrease in oxidation states. In the light of this, most stable oxidation states of the elements are:
3d³ : 3d³s² (+ 5);3 : 3d⁵4s¹ (+ 6) and 3d⁵4s² (+ 7)
3d⁸ : 3d⁸4s² (+ 2); 3d⁴ : 3d⁴4s² or 3d⁵4s¹ (+6)

Question 6.
Name the oxometal anions in the first transition series of transition metals in which the metal exhibits an oxidation state equal to its group number.
Answer:
Vanadate: VO₃^–
chromate: CrO₄^2-
permanganate: MnO₄^–

Question 7.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction? (C.B.S.E. Delhi 2013)
Answer:
One common property associated with the elements in the periodic table is the variation in their atomic and ionic radii down the group and along a period. In general, these increase down the group due to the increase in the number of shells and decrease along a period considerably because of the increase in the magnitude of the effective nuclear charge.
Consequences of Lanthanoid Contraction
(a) Separation of Lanthanoids: Separation of lanthanoids is possible only due to lanthanoid contraction. All the lanthanoids have quite similar properties and due to this reason, they are difficult to separate. However, because of lanthanoid contraction, their properties (such as ionic size, ability to form complexes etc.,) vary slightly.

(b) Variation in basic strength of hydroxides: The basic strength of oxides and hydroxides decreases from La(OH)₃ to Lu(OH)₂. Due to lanthanoid contraction, size of M³⁺ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. The acidic strength which involves the cleavage of the O—H bond follows the reverse trend i.e. it increases along the series.

(c) Similarly in the atomic sizes of the elements of the second and third transition series present in the same group: We know that the atomic sizes of the elements generally increase appreciably down a group. Similar trend is also expected in the elements present in the different groups of d-block.

(d) Variation in standard reduction potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.
M³⁺ (aq) + 3e^– → M (aq)

(e) Variation in physical properties like melting point, boiling point, hardness etc: Various physical properties like m.pt., b.pt., hardness etc., increase with the increase in atomic number. This is because the attraction forces between the atoms increase as the size decreases.

Question 8.
What are the characteristics of transition elements and why are they called transition elements? Which of the d- block elements may not be regarded as the transition elements?
Answer:
In the transition elements, d-orbitais are successively filled. The general electronic configuration of transition elements is (n – 1) d^1-10 ns^1-2. There are three transition series. The first transition series involves the filling of 3d-orbitais. It starts from scandium (Z = 21) and goes upto zinc (Z = 30).

The second transition series involves the filling of 4 d-orbitais and includes 10 elements from yttrium (Z = 39) to cadmium (Z = 48). The third transition series invokes the filling of 5d-orbitals. The first element of this series lanthanum (Z = 57). It is followed by fourteen elements called lanthanides which involve the filling of 4f-orbitais. The net nine elements from hafnium (Z = 72) to mercury (Z = 80) belong to the third transition series.

There is an incomplete fourth transition series. it involves the filling of 6d- subshell starting from actinium (Z = 89) followed by elements with atomic number 104 onwards.

Question 9.
In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
Answer:
Transition elements contain partially filled d-orbitals whereas non-transition elements have no d-orbitals or have completely filled d -orbitals.

Question 10.
What are the different oxidation states exhibited by lanthanoids?
Answer:
+3 is the common oxidation state of the lanthanoids.
In addition to +3, oxidation states +2 and +4 are also exhibited by some of the lanthanoids.

Question 11.
Explain giving the reason:
(a) Transition metals and many of their compounds show paramagnetic behaviour. (H.P. Board 2014)
(b) The enthalpies of atomisation of transition metals are high. (Jharkhand Board 2010, C.B.S.E. Outside Delhi 2008, 2012, H.P. Board 2014)
(c) The transition metals generally form coloured compounds. (C.B.S.E. 2010. 2012)
(d) The transition metals and their compounds act as good catalysts. (C.B.S.E.Outside Delhi 2010) (C.B.S.E. Delhi 2008, Sample Paper 2010, H.P. Board 2017)
Answer:
(i) Paramagnetic arises due to the presence of unpaired electrons, each such electron has a magnetic moment associated with it due to its spin angular momentum. Transition metals have in its ground state or ionized state has a number of unpaired d-electrons which gives them a paramagnetic behaviour.

(ii) Transition metals have very high interatomic metallic interaction due to the involvement of greater number of electrons from (n – l)d in addition to the ns electrons. The greater the number of valence electrons, the stronger is the resultant bonding due to the greater overlapping of half-filled orbitals. Hence, more amount of energy is required to break these metallic bonds. Thus enthalpy of atomisation of transition metal is very high.

(iii) Colour of transition metal – compounds is due to the excitation of an electron from a lower energy d-orbital to a higher energy d orbital. The energy of excitation corresponds to the frequency of light absorbed and the colour observed corresponds to the complementary colour of the light absorbed (whose frequency lies on the visible region). The frequency of the light absorbed depends on the nature of ligand. Transition metals form coloured compounds due to the presence of vacant d-orbitals for the d-d transition of e’ which causes the colour.

(iv) The catalytic activity of transition metals is ascribed to their ability to adopt multiple oxidation states and to form complexes. Catalyst at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst this has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules (activation energy is lowered. Transition metals have d and s orbitals to from these bonds.

Question 12.
What are interstitial compounds? Why are such compounds well known for transition metals?
Answer:
Interstitial compounds are the compounds formed as a result of the trapping of atoms of small elements like H, N, C, B etc. in the crystal lattices of certain metals. These are non-stoichiometric in nature and are neither ionic nor covalent. In fact, no proper bonds exist in the atoms of metals and non-metals involved in these compounds. Transition metals have a tendency to form such compounds. A few examples are: TiC,
M_n4N, Fe₃H, VH_0·56, V_se0·98 and Fe_0·94O etc.

1. These are generally non-stoichiometric in nature. Therefore, they cannot be represented by a definite structure or formula.
2. The compounds are neither covalent nor ionic and they don’t represent the normal oxidation states of the metals.
3. Since the strengths of the metallic bonds in these compounds increase due to greater electronic interactions, they show high melting points and high metallic conductivity. However, these compounds are chemically inert.
4. The conductivity of the metals remains unaffected in the corresponding interstitial compounds.

Question 13.
How is the variability in oxidation states of transition metals different from that of non-transition metals? Illustrate with examples.
Answer:
The variability of oxidation states, characteristics of transition elements arises out of incomplete filling of d orbitals in such a way that their oxidation states differ from each other by unity This is in contrast with the variability of oxidation states of non-transition elements where oxidation states normally differ by a unit of two.
eg: Vanadium: V⁺², V⁺³, V⁺⁴, V⁺⁵
Chromium : Cr⁺², Cr⁺³, Cr⁺⁴, Cr⁺⁵, C⁺⁶
Nitrogen: +5, +3, +1, -1, -3.

Question 14.
Describe the preparation of potassium dichromate from chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Answer:
Preparation from chromite:
Potassium dichromate is generally prepared from chromite ore (FeCr₂O₄). It is in fact, a mixed oxide Fe0.Cr₂O₃ of iron and chrome also called ferrochrome or chrome iron.
(i) Conversion of chromite ore into sodium chromate: Chromite ore is fused with sodium hydroxide or sodium carbonate in the presence of air.

(ii) Conversion of sodium chromate into sodium dichromate : The fused mass obtained above is extracted with water. Sodium chromate which is soluble in water goes into the solution leaving behind the insoluble ferric oxide (Fe₂O₃). The yellow solution of sodium chromate obtained above is treated with concentrated H₂SO₄ to form sodium dichromate which has an orange colour.

(iii) Conversion of sodium dichromate into potassium dichromate: Sodium dichromate is more soluble and less stable than potassium dichromate.

Effect of increasing pH : The solution of potassium dichromate (K₂Cr₂O₇) in water is orange in colour. On increasing the pH i.e. on adding the base, the potassium dichromate changes to potassium chromate (K₂CrO₄) which is yellow in colour. Thus, on increasing the pH, the colour of the solution changes from orange to yellow.

Question 15.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
(a) iodide,
(b) iron (II) solution
(c) H₂S.
Answer:

Question 16.
Describe the preparation of potassium permanganate. How does acidified permanganate solution react with:
(a) iron (II) solution
(b) SO₂
(c) oxalic acid?
Write the ionic equations for the reactions.
Answer:
Potassium permanganate is prepared on a large scale from the mineral pyrolusite, MnO₂.
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O

Question 17.
For M²⁺/M and M³⁺/M²⁺ systems, the E° values of some metals are given :

Use this data to comment upon :
(a) The stability of Fe³⁺ in acid solution as compared to that of Cr³⁺ or Mn³⁺.
(b) The ease with which iron can be oxidised as compared to the similar process for either chromium or manganese metal.
Answer:
As \({ E }_{ { Cr }^{ 3+ }/{ Cr }^{ 2+ } }^{ \circ }\) is negative (-0·4 V), this means that Cr³⁺ ions in solution cannot be reduced to Cr²⁺ ions or Cr³⁺ ions are very stable. As farther comparison of E° values shows that Mn³⁺ ions can be reduced to Mn²⁺ ion more readily than Fe³⁺ ions. Thus, in the light of this, the order of relative stabilities of different ions is Mn³⁺ < Fe³⁺ < Cr³⁺.
(b) From the E° values, the order of oxidation of the metal to the divalent cation is Mn > Cr > Fe.

Question 18.
Predict which of the following will be coloured in an aqueous solution? Ti³⁺, V³⁺,Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and Co²⁺ Give reasons for each.
Answer:
Among the above mentioned, ions Ti³⁺, V³⁺ Mn²⁺, Fe³⁺, CO²⁺ will be coloured in its aqueous solution due to the ability of e’ to jump from a lower energy d orbital to a higher energy d orbital. In case of the ions Cu⁺, Sc³⁺ this d-d transition cannot take place either due to the absence of any e- in 3d orbital or due to complete filling of d orbital.

Question 19.
Compare the stability of the +2 oxidation state for the elements of the first transition series.
Answer:
The common oxidation state of 3d series elements is + 2 which arises due to the participation of only 4s electrons. The tendency to show the highest oxidation state increases from Sc to Mn then decreases due to the pairing of electrons in 3d subshell. Thus in the series Sc(II) does not exist, Ti(II) is less stable than Ti(IV). At the other end of the series, the oxidation state of Zn is +2 only.

Question 20.
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
(i) electronic configuration
(ii) atomic and ionic sizes and
(iii) oxidation state
(iv) chemical reactivity.
Answer:
(i) Electronic configuration:
The general electronic configuration for lanthanoids is [Xe] ^0-14 5d ^0-1 6s² and that of actinoids is [Rn] 5f ^0-14 6d^0-1 7s². Unlike 4f orbitals, 5 f orbitals are not deeply buried and participate in bonding to a greater extent.

(ii) Oxidation states:
The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of +2 and +4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.

(iii) Atomic and ionic size Similar to lanthanoids, actinoids also exhibit actinoid contraction overall decrease in atomic and ionic radii. The contraction is greater due to the poor shielding effect of 5f orbitals. Hence there is an unexpected in the atomic and ionic sizes of actinoids.

(iv) Chemical activity
In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In the case of acids, they are slightly affected by nitric acid (because of the -formation of a protective oxide layer).

Question 21.
How would you account for the following :
(a) Of the d⁴ species, Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidising in nature.
(b) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents, it is easily oxidised.
(c) d¹ configuration is very unstable in ions.
Answer:
(a) E° value of Cr³⁺/Cr²⁺ is negative (-0·41 V) while that of Mn³⁺/Mn²⁺ is positive (+ 1·57 V). This means Cr²⁺ ions can lose electrons to form Cr³⁺ ions and act as a reducing agent while Mn³⁺ ions can accept electrons and can act as oxidising agent.
(b) Cobalt (II) is stable in aqueous solution but in the presence of complexing agent, it undergoes change in oxidation state from +2 to +3 and is easily oxidised.
(c) The ion with d¹ configuration is expected to be extremely unstable and has a great urge to acquire d° configuration (very stable) by losing the only electron present in the d-subshell.

Question 22.
What is meant by ‘disproportionation’ ? Give two examples of disproportionation reaction in aqueous solution.
Answer:
In a disproportionation reaction, an element undergoes an increase as well as decrease in its oxidation state forming
two different compounds. In other words, we can say that it can act both as reducing agent as well as oxidising agent.

Question 23.
Which metal in the first series of transition metals exhibits + 1 oxidation state most frequently and why?
Answer:
Copper, because with +1 oxidation state an extra stable configuration, 3d¹⁰ results.

Question 24.
Calculate the number of unpaired electrons in the following gaseous ions :
Mn³⁺, Cr³⁺, V³⁺.
Answer:

Question 25.
Give examples and suggest reasons for the following features of the transition metal chemistry:
(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(ii) A transition metal exhibits the highest oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.
Answer:
(i) In case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atoms are not involved in bonding. As a result, it can donate electrons and behave as a base.

On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding. As a result, it can accept electrons and behave as an acid. For example, Mn^IIO is basic and Mn^VIIO is acidic.

(ii) Oxygen and fluorine act as strong oxidising agents because of their high electronegativities and small sizes. Hence, they bring out the highest oxidation states from the transition metals. In other words, a transition metal exhibits oxidation states in oxides and fluorides. For example, in OsF₆ and V₂O₅, the oxidation states of Os and V are + 6 and respectively.

(iii) Oxygen is a strong oxidising agent due to its high electronegativity and small size so, oxoanions of metal have the highest oxidation state. For example, in MnO₄^–, the oxidation state of Mn is +7.

Question 26.
Give the steps in the preparation of (C.B.S.E. Delhi 2009 comptt.)
(a) K₂Cr₂O₇ from chromite ore
(b) KMnO₄ from pyrolusite ore.
Answer:
(a)

(b)

Question 27.
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
Answer:
An alloy is a homogeneous mixture of two or more metals or metals and non-metals. An important alloy that contains lanthanoid metal is mischmetal which contains 50% Cerium and 25 % Lanthanum, with small amounts of Nd (Neodymium) and Pr (Praseody-mium). It is used in Mg-based alloy to produce bullets, shells, and lighter flints.

Question 28.
What are inner transition elements? Decide which of the following atomic numbers belong to inner transition elements :
29, 59, 74, 95, 102, 104.
Answer:
The inner transition elements also called/-block elements include the series of lanthanoids (Z = 58 to 71) and actinoids (Z = 90 to 103). This means that the elements with atomic numbers 59, 95, and 102 belong to inner transition elements.

Question 29.
The chemistry of actinoid elements is not so smooth as that of lanthanoids. Justify this statement by giving some examples of the oxidation states of these elements.
Answer:
Lanthanoids show a limited number of oxidation states, such as +2, +3 and +4 (+3 is the principal oxidation state). This is because of large energy gap between 5d and 4f- subshells. On the other hand, actinoids also show a principal oxidation state of +3 but show a number of other oxidation state also. For example, Uranium (Z = 92) exhibits oxidation states of +3, +4, +5, +6 and Neptunium (Z = 94) shows oxidation states of +3, +4, +5, +6 and +7. This is because of small energy difference between 5f and 6d orbitals.

Question 30.
Which is the last element in the series of actinoids? Write the electronic configuration of the element. Comment upon its possible oxidation state.
Answer:
Lawrencium (Lr = 103); [Rn] 5f¹⁴6d¹7s² oxidation state = +3.

Question 31.
Use Hund’s rule to derive the electronic configuration of Ce3+ ion and calculate its magnetic moment on the basis of the spin-spin formula.
Answer:
Cerium electronic configuration = [Xe]4f¹ 5d¹ 16s²
Ce⁺³ ion = [Xe]4f¹
i.e., one unpaired electron is present
Magnetic moment, µ = \(\sqrt{n(n+2)}\) = \(\sqrt { 3 }\) = 1.73BM.

Question 32.
Name the members of the lanthanoid series which exhibit+4oxidatk>nstatesandthosewhichexhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.
Answer:
+ 4 oxidation state in Ce (Z = 58), Pr (Z = 59), Tb (Z = 65).
+ 2 oxidation state in Nd (Z = 60), Sm(Z=62), Eu (Z = 63), Tm (Z=69), Yb (Z = 70).
+ 2 oxidation state is exhibited when the lanthanoid has the configuration 5cf 6s2 so that two electrons are-easily lost.
+ 4 oxidation state is exhibited by the elements which after losing four electrons acquire configuration 4f° or 4f¹

Question 33.
Compare the chemistry of the actinoids with that of lanthanoids with reference to:
(i) electronic configuration
(ii) oxidation states and
(iii) chemical reactivity.
Answer:
(i) Electronic configuration:
The general electronic configuration for lanthanoids is [Xe] ^0-14 5d ^0-1 6s² and that of actinoids is [Rn] 5f ^0-14 6d^0-1 7s². Unlike 4f orbitals, 5 f orbitals are not deeply buried and participate in bonding to a greater extent.

(ii) Oxidation states:
The principal oxidation states of lanthanoids is (+3). However, sometimes we also encounter oxidation states of +2 and +4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in the +3 state than in the +4 state.

(iii) Atomic and ionic size Similar to lanthanoids, actinoids also exhibit actinoid contraction overall decrease in atomic and ionic radii. The contraction is greater due to the poor shielding effect of 5f orbitals. Hence there is an unexpected in the atomic and ionic sizes of actinoids.

(iv) Chemical reactivity
In the lanthanoid series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In the case of acids, they are slightly affected by nitric acid (because of the -formation of a protective oxide layer).

Question 34.
Write the electronic configuration of the elements with atomic numbers 61, 91, 101, 109.
Answer:
Promethium or Pm (Z = 61) [Xe]⁵⁴4f⁵5d⁰6s²
Protactinium or Pa (Z = 91) [Rn] 4f²6d¹7s²
Mendelevium or Md (Z = 101) [Rn] 5f¹6d⁰7s²
Meitnerium or Mt (Z = 109) [Rn] 5f¹⁴6d⁷7s²

Question 35.
Compare the general characteristics of the first transition series of transition metals with those of the second and third transition series metals in the respective vertical columns. Give special emphasis on the following points :
(i) electronic configuration
(ii) oxidation states
(iii) ionisation enthalpies
(iv) atomic sizes.
Answer:
(i) Electronic configuration. There are some exceptions in the electronic configurations in all the three series.

(ii) Oxidation state. The elements belonging to the different series but present in the same group have similar electronic configurations and therefore, exhibit almost same variable oxidation states. In general, these are maximum in the middle of the series while minimum towards the end. Transition elements show variable oxidation states due to the participation of ns and (n – 1) d electrons in bonding because the energies of ns and (n – 1) d-subshells are quite close. The stability of a particular oxidation state depends upon the nature of the element with which the transition metal forms the compound.

(iii) Ionisation enthalpies. In general, the ionisation enthalpies in all three transition series increase from left to the right. However, the gaps in the two successive elements in a particular series are small and are also not regular. The first three ionisation enthalpies of the elements present in the first transition series are given in the text part. The ∆_iH₁ [ values of the elements belonging to the 5d series and higher as compared to those belonging to 3d and Ad series in the same group because of poor shielding by intervening 4f electrons present.

(iv) Atomic size. In all three transition series, the atomic, as well as ionic radii of the elements, increase from left to right. The values for 3d series are given in the text part. However, the increase in their values is not as much as expected since the shielding by (n – 1 )d electrons is not as much as expected. In a particular group, the atomic radius of the elements belonging to Ad series is more than the elements in the 3d series. However, the gaps in the elements belonging to Ad and 5d series are negligible on account of lanthanoid contraction which the elements of the 5d experience.

Question 36.
Write down the number of 3d electrons in each of the following ions :
Ti²⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Fe³⁺, Co²⁺, Ni²⁺ and Cu²⁺.
Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).
Answer:
The number of 3d electrons in the ions are :

For the explanation of the involvement of 3d orbitals in the hydrated ions (octahedral in nature) consult the next unit on coordination compounds.

Question 37.
Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
Answer:
The properties of the elements of the first transition series differ from those of the heavier transition elements in many ways.
(i) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements of the 2^nd and 3_rd transition series). However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series this is due to lanthanoid configuration.

(ii) +2 and +3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements.

(iii) The enthalpies of atomization of the elements in the first transition series are lower than those of the corresponding elements in the second and third transition series.

(iv) The melting and boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of strong metallic bonding (M – M bonding)

(v) The dements of the first transition series from low-spin or high-spin complexes depending upon the strength of the ligand field. However, the heavier transition elements form only low-spin complexes, irrespective of the strength of the ligand field.

Question 38.
What can be inferred from the magnetic moment values of the following complex species?

Answer:
The magnetic moment of a compound is given by the relation (µ) = \(\sqrt { n\left( n+2 \right) } \) B.M, where n is the number of unpaired electrons.
For one unpaired electron (n = 1) ; µ = \(\sqrt { 1\left( 1+2 \right) } =\sqrt { 3 } =1\cdot 73\quad B.M.\)
For two unpaired electrons (n – 2) ; µ = \(\sqrt { 2\left( 2+2 \right) } =\sqrt { 8 } =2\cdot 83\quad B.M.\)
For three unpaired electrons (n = 3); µ = \(\sqrt { 3\left( 3+2 \right) } =\sqrt { 15 } =3\cdot 87\quad B.M.\)
For four unpaired electrons (n = 4) ; µ = \(\sqrt { 4\left( 4+2 \right) } =\sqrt { 24 } =4\cdot 9\quad B.M.\)
For five unpaired electrons (n = 5) ; µ = \(\sqrt { 5\left( 5+2 \right) } =\sqrt { 35 } =5\cdot 92\quad B.M.\)
*In the light of the above value, let us gather the desired information about the complex species that are mentioned
(i) K₄[Mn(CN)₆]
Oxidation state of Mn : [Mn(CN)₆]^4- , x + 6(-l) = -4 or x = -4 + 6 = + 2
The magnetic value of 1·73 B.M. indicates the presence of one unpaired electron in the complex. When six, CN^– ions (or ligands) approach Mn²⁺ ion, electrons in 3d orbitals pair up to make available six vacant orbitals involving d²sp³ hybridisation.

The complex is octahedral and is paramagnetic due to one unpaired electron.
(ii) [Fe(H₂O)₆]²⁺
Oxidation state of Fe : [Fe(H₂O)₆]²⁺ ; r + 6 (0) = +2
The magnetic moment value of 5·3 B.M. indicates that there are four unpaired electrons in the complex. This means that the electrons in Fe²⁺ ion do not pair up when six H20 molecules (or ligands) approach it. Since the desired number of vacant orbitals (six) are available, die complex formed is sp³d² hybridised.

The complex is octahedral and is paramagnetic due to four unpaired electrons. It is also called outer orbital complex because 4d (n = 4) orbitals are involved.
(iii) K₂[MnCl₄]
Oxidation state of Mn : [MnCl₄]^2-, x + 4(-1) = -2 or x = -2 + 4= + 2
The magnetic moment value of 5·9 B.M. indicates that there are five unpaired electrons in the complex. This means that all the five 3d orbitals in Mn²⁺ ion are involved in the bond formation. The complex is sp³ hybridized in which one vacant 4s and three vacant 4p orbitals participate.

The complex is therefore, tetrahedral in nature.

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