SmartSkills Sanskriti School

SMART SKILLS

2019 - 2020

CHEMISTRY

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INDEX

S.No. CONTENTS Page no.

1. Syllabus 3

2. Question paper design 6

Assignments :
3. Unit I Solutions 18
4. Unit II Electrochemistry 22
5. Unit III Chemical Kinetics 25
6. Unit IV Surface Chemistry 30
7. Unit V General Principles and processes of Isolation of Elements 45
8. Unit VI p -Block Elements 54
9. Unit VII d -and f -Block Elements 75
10. UnitVIII Coordination Compounds 94
11. Unit IX Haloalkanes and Haloarenes 97
12. Unit X Alcohols, Phenols and Ethers 103
13. Unit XI Aldehydes, Ketones and Carboxylic Acids 107
14. Unit XII Organic Compounds containing Nitrogen 112
15. Unit XIII Biomolecules 115
16. Unit XIV Polymers 117
17. Unit XV Chemistry in Everyday Life 118
Practice Papers:
18. Practice paper for Summer Vacation 129
19. Term 1 paper (2018-19) 132
20. Term 1 paper (2017-18) 136
21. Term 2 paper (2018-19) 140
22. Term 2 paper (2017-18) 145
23. Preboard Paper (2018-19) 149
24. Preboard paper (2017-18) 155
25. CBSE sample paper 2018-19,2017-18 Exam 161

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CLASS XII (2019-20)

(THEORY)

Total Periods (For Theory -150 + For Practical -60)

Time: 3 Hours 70 Marks

Unit No. Title No. of Periods Marks

Unit I Solutions 10

Unit II Electrochemistry 12 23

Unit III Chemical Kinetics 10

Unit IV Surface Chemistry 08

Unit V General Principles and Processes of 08

Isolation of Elements

Unit VI p -Block Elements 12 19

Unit VII d -and f -Block Elements 12

Unit VIII Coordination Compounds 12

Unit IX Haloalkanes and Haloarenes 10

Unit X Alcohols, Phenols and Ethers 10

Unit XI Aldehydes, Ketones and Carboxylic Acids 10

Unit XII Organic Compounds containing Nitrogen 10 28

Unit XIII Biomolecules 12

Unit XIV Polymers 08

Unit XV Chemistry in Everyday Life 06

Total 150 70

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Unit I: Solutions 10 Periods

Types of solutions, expression of concentration of solutions of solids in liquids, solubility of gases in

liquids, solid solutions, colligative properties - relative lowering of vapour pressure, Raoult's law,
elevation of boiling point, depression of freezing point, osmotic pressure, determination of
molecular masses using colligative properties, abnormal molecular mass, Van't Hoff factor.

Unit II: Electrochemistry 12 Periods

Redox reactions, conductance in electrolytic solutions, specific and molar conductivity, variations of
conductivity with concentration, Kohlrausch's Law, electrolysis and law of electrolysis (elementary
idea), dry cell -electrolytic cells and Galvanic cells, lead accumulator, EMF of a cell, standard
electrode potential, Nernst equation and its application to chemical cells, Relation between Gibb’s
energy change and EMF of a cell, fuel cells, corrosion.

Unit III: Chemical Kinetics 10 Periods

Rate of a reaction (Average and instantaneous), factors affecting rate of reaction: concentration,
temperature, catalyst; order and molecularity of a reaction, rate law and specific rate constant,
integrated rate equations and half life (only for zero and first order reactions), concept of collision
theory (elementary idea, no mathematical treatment). Activation energy, Arrhenius equation.

Unit IV: Surface Chemistry 08 Periods

Adsorption - physisorption and chemisorption, factors affecting adsorption of gases on solids,

catalysis, homogenous and heterogenous activity and selectivity; enzyme catalysis colloidal state
distinction between true solutions, colloids and suspension; lyophilic, lyophobic multimolecular and
macromolecular colloids; properties of colloids; Tyndall effect, Brownian movement,
electrophoresis, coagulation, emulsion - types of emulsions.

Unit V: General Principles and Processes of Isolation of Elements 08 Periods

Principles and methods of extraction - concentration, oxidation, reduction - electrolytic method and
refining; occurrence and principles of extraction of aluminium, copper, zinc and iron.

Unit VI: p-Block Elements 12 Periods

Group 16 Elements: General introduction, electronic configuration, oxidation states, occurrence,

trends in physical and chemical properties, dioxygen: Preparation, Properties and uses, classification
of Oxides, Ozone, Sulphur -allotropic forms; compounds of Sulphur: Preparation Properties and
uses of Sulphur-dioxide, Sulphuric Acid: industrial process of manufacture, properties and uses;
Oxoacids of Sulphur (Structures only).

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Group 17 Elements: General introduction, electronic configuration, oxidation states, occurrence,
trends in physical and chemical properties; compounds of halogens, Preparation, properties and
uses of Chlorine and Hydrochloric acid, interhalogen compounds, Oxoacids of halogens (structures
only).

Group 18 Elements: General introduction, electronic configuration, occurrence, trends in physical

and chemical properties, uses.

Unit VII: "d" and "f" Block Elements 12 Periods

General introduction, electronic configuration, occurrence and characteristics of transition metals,

general trends in properties of the first row transition metals - metallic character, ionization
enthalpy, oxidation states, ionic radii, colour, catalytic property, magnetic properties, interstitial
compounds, alloy formation, preparation and properties of K2Cr2O7 and KMnO4.

Lanthanoids - Electronic configuration, oxidation states, chemical reactivity and lanthanoid

contraction and its consequences.

Actinoids - Electronic configuration, oxidation states and comparison with lanthanoids.

Unit VIII: Coordination Compounds 12 Periods

Coordination compounds - Introduction, ligands, coordination number, colour, magnetic properties

and shapes, IUPAC nomenclature of mononuclear coordination compounds. Bonding, Werner's
theory, VBT, and CFT; structure and stereoisomerism, importance of coordination compounds (in
qualitative inclusion, extraction of metals and biological system).

Unit IX: Haloalkanes and Haloarenes. 10 Periods

Haloalkanes: Nomenclature, nature of C -X bond, physical and chemical properties, mechanism of

substitution reactions, optical rotation.

Haloarenes: Nature of C -X bond, substitution reactions (Directive influence of halogen in

monosubstituted compounds only).

Uses and environmental effects of - dichloromethane, trichloromethane, tetrachloromethane,

iodoform, freons, DDT.

Unit X: Alcohols, Phenols and Ethers 10 Periods

Alcohols: Nomenclature, methods of preparation, physical and chemical properties (of primary
alcohols only), identification of primary, secondary and tertiary alcohols, mechanism of
dehydration, uses with special reference to methanol and ethanol.

Phenols: Nomenclature, methods of preparation, physical and chemical properties, acidic nature of
phenol, electrophilic substitution reactions, uses of phenols.

Ethers: Nomenclature, methods of preparation, physical and chemical properties, uses.

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Unit XI: Aldehydes, Ketones and Carboxylic Acids 10 Periods

Aldehydes and Ketones: Nomenclature, nature of carbonyl group, methods of preparation, physical
and chemical properties, mechanism of nucleophilic addition, reactivity of alpha hydrogen in
aldehydes: uses.

Carboxylic Acids: Nomenclature, acidic nature, methods of preparation, physical and chemical
properties; uses.

Unit XII: Organic compounds containing Nitrogen 10 Periods

Amines: Nomenclature, classification, structure, methods of preparation, physical and chemical

properties, uses, identification of primary, secondary and tertiary amines.

Cyanides and Isocyanides - will be mentioned at relevant places in text.

Diazonium salts: Preparation, chemical reactions and importance in synthetic organic chemistry.

Unit XIII: Biomolecules 12 Periods

Carbohydrates - Classification (aldoses and ketoses), monosaccahrides (glucose and fructose), D-L
configuration oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch,
cellulose,glycogen); Importance of carbohydrates.

Proteins -Elementary idea of - amino acids, peptide bond, polypeptides, proteins, structure
ofproteins - primary, secondary, tertiary structure and quaternary structures (qualitative idea
only),denaturation of proteins; enzymes. Hormones - Elementary idea excluding structure.

Vitamins - Classification and functions.

Nucleic Acids: DNA and RNA.

Unit XIV: Polymers 08 Periods

Classification - natural and synthetic, methods of polymerization (addition and condensation),

copolymerization, some important polymers: natural and synthetic like polythene, nylon polyesters,
bakelite, rubber. Biodegradable and non-biodegradable polymers.

Unit XVI Chemistry in Everyday life 06 Periods

Chemicals in medicines - analgesics, tranquilizers antiseptics, disinfectants, antimicrobials,

antifertility drugs, antibiotics, antacids, antihistamines.

Chemicals in food – preservatives, artificial sweetening agents, elementary idea of antioxidants.

Cleansing agents- soaps and detergents, cleansing action.

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CHEMISTRY (CODE-043)

QUESTION PAPER DESIGN

CLASS - XII (2019-20)
Time 3 Hours Max. Marks: 70

S. Typology of Questions Very Short Short Long Total %

No. Short Answe Answer Answer Marks Weigh-
Answ r-I –II (L.A.) tage
er (SA-I) (SA-II) (5 marks)
(VSA) (2 (3
(1 marks) marks)
mark)
1 Remembering 2 1 1 - 7 10%
(Knowledge based Simple
recall questions, to know
specific facts, terms,
concepts, principles, or
theories, Identify, define,
or recite, information)
2 Understanding- - 3 4 1 21 30%
(Comprehension -to be
familiar with meaning and
to
understandconceptually,
interpret, compare,
contrast,
explain,paraphrase
information)
3 Application - 3 4 1 21 30%
(Use abstract information
in concrete situation, to
apply knowledge to new
situations, Use given
content to interpret
asituation, provide an

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example, or solve a
problem)
4 High Order Thinking 2 - 1 1 10 14%
Skills
( Analysis &Synthesis-
Classify,compare, contrast,
or differentiate between
different pieces of
information, Organize
and/or integrate unique
pieces of information
from a variety of sources)
5 Evaluation and Multi 1 - 2 - 11 16%
Disciplinary- (Appraise,
judge, and/or justify the
value or worth of a
decision or outcome, or to
predict outcomes based on
values)
TOTAL 5x1=5 7x2=14 12x3=36 3x5=15 70(27) 100%

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QUESTION WISE BREAK UP

Type of Question(s) Mark(s) per Question Total No. of Questions Total Marks

VSA 1 5 05

SA-I 2 7 14

SA-II 3 12 36

LA 5 3 15

Total 27 70

1. Internal Choice: There is no overall choice in the paper. However, there is an internal choice in
two question of 1 mark weightage, two questions of 2 marks weightage, four questions of 3
marks weightage and all the three questions of 5 marks weightage.

2. The above template is only a sample. Suitable internal variations may be made for generating
similar templates keeping the overall weightage to different form of questions and typology of
questions same.

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PRACTICALS

Evaluation Scheme for Examination Marks

Volumetric Analysis 08

Salt Analysis 08

Content Based Experiment 06

Project work 04

Class record and viva 04

Total 30

PRACTICALS SYLLABUS 60 Periods

Micro-chemical methods are available for several of the practical experiments.

Wherever possible, such techniques should be used.

A. Surface Chemistry
(a) Preparation of one lyophilic and one lyophobic sol
Lyophilic sol - starch, egg albumin and gum
Lyophobic sol - aluminium hydroxide, ferric hydroxide, arsenous sulphide.
(b) Dialysis of sol-prepared in (a) above.
(c) Study of the role of emulsifying agents in stabilizing the emulsion of different oils.

B. Chemical Kinetics
(a) Effect of concentration and temperature on the rate of reaction between Sodium Thiosulphate
and Hydrochloric acid.
(b) Study of reaction rates of any one of the following:
(i) Reaction of Iodide ion with Hydrogen Peroxide at room temperature using different
concentration of Iodide ions.
(ii) Reaction between Potassium Iodate, (KIO3) and Sodium Sulphite: (Na2SO3) using starch
solution asindicator (clock reaction).

C. Thermochemistry
Any one of the following experiments
i) Enthalpy of dissolution of Copper Sulphate or Potassium Nitrate.
ii) Enthalpy of neutralization of strong acid (HCI) and strong base (NaOH).
iii) Determination of enthaply change during interaction (Hydrogen bond formation) between
Acetone and Chloroform.

D. Electrochemistry
Variation of cell potential in Zn/Zn2+|| Cu2+/Cu with change in concentration of electrolytes
(CuSO4 orZnSO4) at room temperature.

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E. Chromatography
i) Separation of pigments from extracts of leaves and flowers by paper chromatography and
determinationof Rf values.
ii) Separation of constituents present in an inorganic mixture containing two cations only
(constituents having large difference in Rf values to be provided).

F. Preparation of Inorganic Compounds

i) Preparation of double salt of Ferrous Ammonium Sulphate or Potash Alum.
ii) Preparation of Potassium Ferric Oxalate.

G. Preparation of Organic Compounds

Preparation of any one of the following compounds
i) Acetanilide
ii) Di -benzal Acetone
iii) p-Nitroacetanilide
iv) Aniline yellow or 2 - Naphthol Aniline dye.

H. Tests for the functional groups present in organic compounds:

Unsaturation, alcoholic, phenolic, aldehydic, ketonic, carboxylic and amino (Primary) groups.

I. Characteristic tests of carbohydrates, fats and proteins in pure samples and their detection in
givenfood stuffs.

J. Determination of concentration/ molarity of KMnO4 solution by titrating it against a standard

solution of:
i) Oxalic acid,
ii) Ferrous Ammonium Sulphate
(Students will be required to prepare standard solutions by weighing themselves).

K. Qualitative analysis
Determination of one cation and one anion in a given salt.
Cation- NH4+, Pb2+, Cu2+ As3+, Al3+, Fe3+, Mn2+, Zn2+, Co2+, Ni2+, Ca2+, Sr2+, Ba2+, Mg2+
Anions-CO32-, S2-, SO32-, SO42-, NO2-, NO3-, Cl-, Br-, I-, PO43-; CH3COO-
(Note: Insoluble salts excluded)

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PROJECT

Scientific investigations involving laboratory testing and collecting information from other
sources.

A few suggested Projects.

● Study of the presence of oxalate ions in guava fruit at different stages of ripening.
● Study of quantity of casein present in different samples of milk.
● Preparation of soybean milk and its comparison with the natural milk with respect to curd
formation, effect of temperature, etc.
● Study of the effect of Potassium Bisulphate as food preservative under various conditions
(temperature, concentration, time, etc.)
● Study of digestion of starch by salivary amylase and effect of pH and temperature on it.
● Comparative study of the rate of fermentation of following materials: wheat flour, gram
flour, potato juice, carrot juice, etc.
● Extraction of essential oils present in Saunf (aniseed), Ajwain (carum), Illaichi
(cardamom).
● Study of common food adulterants in fat, oil, butter, sugar, turmeric power, chilli powder
and pepper.

Note: Any other investigatory project, which involves about 10 periods of work, can be
chosen with the approval of the teacher.

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MONTHWISE SYLLABUS DISTRIBUTION

MONTHS: March, April, May

Unit II: Electrochemistry (Periods 14)

Redox reactions, conductance in electrolytic solutions, specific and molar conductivity

variations of conductivity with concentration, Kohlrausch's Law, electrolysis and laws of
electrolysis (elementary idea), dry cell- electrolytic cells and Galvanic cells; lead accumulator,
EMF of a cell, standard electrode potential, Nernst equation and its application to chemical
cells, fuel cells; corrosion.

Unit III: Chemical Kinetics (Periods 12)

Rate of a reaction (average and instantaneous), factors affecting rate of reaction; concentration,
temperature, catalyst; order and molecularity of a reaction; rate law and specific rate constant,
integrated rate equations and half life (only for zero and first order reactions); concept of
collision theory (elementary idea, no mathematical treatment), Activation energy, Arrhenious
equation

Unit I: Solutions (Periods 12)

Types of solutions, expression of concentration of solutions of solids in liquids, solubility of

gases in liquids, solid solutions, colligative properties - relative lowering of vapour pressure,
elevation of Boiling Point, depression of freezing point, osmotic pressure, determination of
molecular masses using colligative properties, abnormal molecular mass, Van’t Hoff factor.

Unit VI: p-Block Elements (Periods 14)

Group16 elements: General introduction, electronic configuration, oxidation states,

occurrence, trends in physical and chemical properties; dioxygen: preparation, properties and
uses; simple oxides; Ozone. Sulphur - allotropic forms; compounds of sulphur: preparation,
properties and uses of sulphur dioxide; sulphuric acid: industrial process of manufacture,
properties and uses, oxoacids of sulphur (structures only).

Group 17 elements: General introduction, electronic configuration, oxidation states, trends in

physical and chemical properties; compounds of halogens: preparation, properties and uses of
chlorine and hydrochloric acid, interhalogen compounds, oxoacids of halogens I (structures
only).

Group 18 elements: General introduction, electronic configuration. Occurence, trends in

physical and chemical properties, uses.

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Practicals:

EXPERIMENT- 1-6

Qualitative analysis
Determination of one cation and one anion in a given salt.

Cations - Pb2+ Cu2+ As3+ Al3+ Fe3+ Mn2+ Zn2+ Co2+ Ni2+ Ca2+ Sr2+ Ba2+ Mg2+ NH4+
Anions – CO32-, S2-, SO32-, SO42-, NO2-, NO3-, Cl-, Br-, I-, PO43-; CH3COO-
(Note: Insoluble salts excluded)

PROJECT WORK

MONTH : JULY

Unit IX: Haloalkanes and Haloarenes (Periods 12)

Haloalkanes: Nomenclature, nature of C-X bond, physical and chemical properties,

mechanism of substitution reactions.
Haloarenes:Nature of C-X bond, substitution reactions (directive influence of halogen for
monosubstituted compounds only). Uses and environmental effects of - dichloromethane,
trichloromethane, tetrachloromethane, iodoform, freons, DDT.

Unit X: Alcohols, Phenols and Ethers (Periods 12)

Alcohols: Nomenclature, methods of preparation, physical and chemical properties (of

primary alcohols only); identification of primary, secondary and tertiary alcohols; mechanism
of dehydration, uses of methanol and ethanol
Phenols: Nomenclature, methods of preparation, physical and chemical properties, acidic
nature of phenol, electrophillic substitution reactions, uses of phenols.
Ethers: Nomenclature, methods of preparation, physical and chemical properties, uses.

Unit XI: Aldehydes, Ketones and Carboxylic Acids (Periods 12)

Aldehydes and Ketones: Nomenclature, nature of carbonyl group, methods of preparation.

physical and chemical properties mechanism of nucleophilic addition.
EXPERIMENT No.: 7-8

Determination of concentration/molarity of KMnO4 solution by titrating it against a standard

solution of:
i) Oxalic acid,
ii) Ferrous ammonium sulphate
(Students will be required to prepare standard solutions by weighing themselves).

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EXPERIMENT No.: 9-10
Preparation of one lyophilic and one lyophobic sol.
(a) Lyophilic sol- starch
(b) Lyophobic sol- ferric hydroxide.

MONTH : AUGUST

Unit XI: Aldehydes, Ketones and Carboxylic Acids (Continued) (Periods 12)

Aldehydes and Ketones : Reactivity of alpha hydrogen in aldehydes, chemical reactivity;

uses.
Carboxylic Acids: Nomenclature, acidic nature, methods of preparation, physical and
chemical properties; uses.

Unit XII: Organic compounds containing Nitrogen (Period 10)

Amines: Nomenclature, classification, structure, methods of preparation, physical and

chemical properties, uses, identification of primary, secondary and tertiary amines.
Cyanides and Isocyanides - will be mentioned at relevant places in context.
Diazonium salts: Preparation, chemical reactions and importance in synthetic organic
chemistry.

Unit XIV: Polymers (Periods 8)

Classification - natural and synthetic, methods of polymerization (addition and

condensation), copolymerization. Some important polymers: natural and synthetic like
polythene, nylon, polyesters, bakelite, rubber. Biodegradable and non- biodegradable
polymers.

EXPERIMENT No.: 11, 12

Effect of concentration on the rate of reaction between sodium thiosulphate and hydrochloric
acid

Effect of temperature on the rate of reaction between sodium thiosulphate and hydrochloric
acid

EXPERIMENT No.: 13-18

Tests for the functional groups present in organic compounds:

Alcoholic, phenolic, aldehydic, ketonic, carboxylic and amino (primary) groups.

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MONTH: SEPTEMBER- OCTOBER

Unit XIII: Biomolecules (Periods 12)

Carbohydrates - Classification (aldoses and ketoses), monosaccal1rides (glucose and tfuctose),

oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch, cellulose, glycogen);
importance.
Proteins - Elementary idea of a - amino acids, peptide bond, polypeptides, proteins, structure
of amines-primary, secondary, tertiary structure and quaternary structures ( qualitative idea
only), denaturation of proteins; enzymes.
Vitamins -Classification and functions. Nucleic Acids: DNA and RNA.

Unit VIII: Coordination Compounds (Period 12)

Coordination 'Compounds - Introduction, ligands, coordination number, colour, magnetic

properties and shapes, IUPAC nomenclature of mononuclear coordination
compounds.bonding; isomerism, importance of coordination compounds (in qualitative
analysis, extraction of metals and biological systems).

Unit VII: d and f Block Elements (Period 14)

General introduction ,electronic configuration, occurrence and characteristics of transition

metals, general trends in properties of the first row transition metals - metallic character,
ionization enthalpy, oxidation states, ionic radii, colour catalytic property, magnetic
properties, interstitial compounds, alloy formation preparation and properties of K2Cr2O7 and
KMnO4.

Lanthanoids - electronic configuration, oxidation states, chemical reactivity and lanthanoid

contraction.
Actinoids - Electronic configuration, oxidation states.

EXPERIMENT No.: 19-20

Preparation of lnorganic Compounds:
i) Preparation of double salt of ferrous ammonium sulphate
ii) Preparation of double salt of potash alum.

EXPERIMENT No.: 21-23

Characteristic tests of carbohydrates, fats and proteins in pure samples and their detection in
given food stuffs.

MONTH: NOVEMBER

Unit V: General Principles and Processes of Isolation of Elements (Periods 8)

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Principles and methods of extraction - concentration, oxidation, reduction electrolytic method
and refining; occurrence and principles of extraction of aluminium, copper, zinc and iron.

Unit IV: Surface Chemistry

Adsorption - physisorption and chemisorption, factors affecting adsorption of gases on solids,
colloids distinction between true solutions,colloids and suspension; lyophilic, lyophobic
multi molecular and macromolecular colloids; properties of colloids; Tyndall effect, Brownian
movement, electrophoresis, coagulation, emulsion - types of emulsions.

Unit XV: Chemistry in Everyday life (Period 8)

Chemicals in medicines - analgesics, tranquilizers, antiseptics, disinfectants, antimicrobials,

antifertility drugs, antibiotics, antacids, antihistanlines.
Chemicals in food - preservatives, artificial sweetening agents.
Cleansing agents - soaps and detergents, cleansing action

EXPERIMENT No.: 24
Chromatography
To separate the constituents present in an inorganic mixture containing Fe3+ and Cu2+ using
paper chromatography and determination of their Rf values.

EXPERIMENT No.: 25
Preparation of Di-benzal Acetone

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Assignment

Chapter 1: Solutions

1. Calculate the freezing point of a solution containing 60 g of glucose . (Molar mass = 180
g mol-1) in 250 g of water. ( Kf of water = 1.86 K kg mol-1) (-2.48oC)

2. Give reasons for the following:

(i)Measurement of osmotic pressure method is preferred for the determination of molar
masses of macromolecules such as proteins and polymers.
(ii)Aquatic animals are more comfortable in cold water than in warm water.
(iii)Elevation of boiling point of 1M KCl solution is nearly double than that of 1 M sugar
solution.
3. A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the
freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K.
Given : (Molar mass of sucrose = 342 g mol , Molar mass of glucose = 180 g mol )
–1 –1

4. State the formula relating pressure of a gas with its mole fraction in a liquid solution in
contact with it. Name the law and mention its two applications.

5. Two liquids A and B boil at 145 C and 190 C respectively. Which of them has a higher
0 0

vapour pressure at 80 C?
0

6. (a) Why is the vapour pressure of a solution of glucose in water lower than that of
water?
(b) A 6.90 M solution of KOH in water contains 30% by mass of KOH. Calculate the
density of the KOH solution? ( molar mass of KOH = 56 g/mol)
(1.288 g/ml)
7. Define azeotropes. What type of azeotrope is formed by positive deviation from Raoult’s
law? Give an example.

8. Explain with suitable examples in each case why the molar masses of some substances
determined with the help of colligative properties are (i) higher (ii) lower than actual
values.

9. Given below is the sketch of a plant for carrying out a process.

(i) Name the process occurring in the above plant.

(ii) To which container does the net flow of solvent take place?
(iii) Name one SPM which can be used in this plant.
(iv) Give one practical use of the plant.
10. Calculate the freezing point of solution when 1.9 g of MgCl (M=95 g mol–1) was
2

dissolved in 50 g of water, assuming MgCl undergoes complete ionization. (Kf for water
2

= 1.86 K kg mol–1)

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11. a) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
b) What happens when the external pressure applied becomes more than the osmotic
pressure of solution ?

12. State Raoult’s law for solutions of volatile liquids. Taking suitable examples explain the
meaning of positive and negative deviations from Raoult’s law. What is the sign of
∆H for positive deviation?
mix

13. a) Define the term osmotic pressure. Describe how the molecular mass of a substance
can be determined by a method based on measurement of osmotic pressure.
b) Determine the osmotic pressure of a solution prepared by dissolving 0.025g of K SO 2 4

in 2L of water at 25 C, assuming that is completely dissociated.

o

(R=0.0821 L atm/K/mol, molar mass of K SO = 174g/mol)

2 4

14. 15 g of an unknown molecular material was dissolved in 450 g of water. The resulting
solution was found to freeze at -0.34 C. What is the molar mass of this material? K for
o
f

water = 1.86 K Kg mol ).

-1 (182 g mol ) -1

15. A solution is prepared by dissolving 1.25g of oil of winter green (methyl salicylate) in
99.0g of benzene has a boiling point of 80.31 C. Determine the molar mass of this
0

compound. (B.P. of pure benzene = 80.10 C and Kb for benzene = 2.53 C kg mol )
0 0 -1

(152.21 g/mol)
16. A 1.00 molal aqueous solution of trichloroacetic acid ( CCl COOH) is heated to its boiling
3

point. The solution has the boiling point of 100.18 C. Determine the van’t Hoff factor for
o

trichloroacteic acid (Kb for water = 0.512 K Kg mol ). -1 (i= 0.35)

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More Practice
Chapter 1: Solutions

1. Calculate the temperature at which a solution containing 54 g of glucose, (C H O ), in 6 12 6

250g of water will freeze (k for water = 1.86 k mol Kg).

f
-1

2. Why it is better to find molality of a solution than its molarity?

3. The Henry law constant for oxygen dissolved in water is 4.34 x 10 atm at 25 C. If the
4 0

partial pressure of oxygen in air is 0.2 atm. under ordinary atmospheric conditions.
Calculate the concentration (in moles per litre) of dissolved oxygen in water in
equilibrium with air at 25 C.
0

4. Define the following terms:

a ) Mole fraction
b) Isotonic solutions
c) Van’t Hoff factor
d) ideal solution.
e) Colligative properties
f) molality

5. What is the Van’t Hoff factor for a compound which undergoes tetramerization in an
organic solvent?

6. Benzoic acid completely dimerizes in benzene. What will be the vapour pressure of a
solution containing 61 g of benzoic acid per 500 g benzene when the vapour pressure of
pure benzene at the temperature of experiment is 66.6 torr? What would have been the
vapour pressure in the absence of dimerisation?

7. Two elements A and B form compounds having molecular formulae AB and AB . When 2 4

dissolved in 20 g of benzene, 1 g of AB lowers the freezing point by 2.3 K whereas 1g of

2

AB lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K Kg mol .
4
-1

Calculate the atomic masses of A and B.

8. Phenol associates in benzene to a certain extent to form a dimer. A solution containing

20 X 10 kg of phenol in 1 kg of benzene has its freezing point lowered by 0.69K.
-3

Calculate the fraction of phenol that has dimerised (K = 5.1 KKg mol )
f
-1

9. 100 g of a protein is dissolved in just enough water to make 10.0 ml of solution. If this
solution has an ostomic pressure of 13.3 mm Hg at 25 C, what is the molar mass of the
0

protein?

10. Calculate the amount of KCl which must be added to 1 Kg of water so that the freezing
point is depressed by 2K. (Kf for water = 1.86 /K Kg mol ) -1 (40.05 g)

11. A decimolar solution of K [Fe(CN) ] is 50% dissociated at 300K. Calculate the osmotic
4 6

pressure of the solution in atm.

12. Heptane and Octane form an ideal solution at 373 K. The vapour pressures of the pure
liquids at this temperature are 105.2 KPa and 46.8 KPa respectively. If the solution
contains 25 g of heptane and 28.5 g of octane, calculate
(i) vapour pressure exerted by heptane.

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(ii) vapour pressure exerted by solution.
(i) mole fraction of octane in the vapour phase.

13. A solution is made by dissolving 30 g of a nonvolatile solute in 90 g of water. It has a

vapour pressure 2.8 KPa at 298 K. At 298 K, vapour pressure of pure water is 3.64 KPa.
Calculate the molar mass of the solute.
14. What type of azeotrope is formed on mixing nitric acid and water?

15 An antifreeze solution is prepared from 222.6 g of ethylene glycol (C H (OH) ) and 200 g
2 4 2

of water. Calculate the molality of the solution. If the density of this solution be 1.072 g
ml , what will be the molarity of the solution?
-1

(9.1 M; 17.95 m)

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Assignment
Chapter 2: Electrochemistry

1. Write the cell reaction and calculate the e.m.f of the following cell at 298 K.
Sn(s)|Sn2+(0.004 M) ||H+ (0.020 M) | H2(g) (1 bar) | Pt (s)
(Given : EoSn2+/Sn = -0.14 V) (0.11 V)

2. For the reaction

2AgCl (s) + H2 (g) (1 atm) 2Ag(s) + 2 H+(0.1M ) + 2Cl- (0.1M)
0 o
G = -43600 J at 25 C. Calculate the emf of the cell.
(0.344 V)
3. Give reasons:
a) On the basis of E0 values, O2 gas should be liberated at anode but it is Cl2 gas which is
liberated in the electrolysis of aqueous NaCl.
b) Conductivity of CH3COOH decreases on dilution.

4. (a) Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed
through a solution of AgNO for 15 minutes.
3

(Given : Molar mass of Ag = 108 g mol 1F = 96500 C mol )

–1 –1

(b) Define fuel cell.

5. Calculate the degree of dissociation (α) of acetic acid if its molar conductivity (Λ ) is 39.05 m

S cm mol . Given λ (H ) = 349.6 S cm mol and λ (CH COO ) = 40.9 S cm mol

2 –1 o + 2 –1 o 3 – 2 –1

6. i. Define molar conductivity of a solution and write their units and the relation
between the two. How does molar conductivity changes with change in concentration
of solution for weak and strong electrolyte.
ii. Define limiting molar conductivity. Why is the conductivity of an electrolyte
solution decrease with the decrease in concentration?

7. Define fuel cells? Give electrode reactions of H -O fuel cell. Name any other fuel which
2 2

can be used instead of H . Write its two advantages.

2

8. From the given cells :

Lead storage cell, Mercury cell, Fuel cell and Dry cell
Answer the following :
(i) Which cell is used in hearing aids ?
(ii) Which cell was used in Apollo Space Programme ?
(iii)Which cell is used in automobiles and inverters ?
(iv) Which cell does not have long life ?

9. The resistance of a conductivity cell containing 0.001 M KCl solution is 1500 Ω at 298K.
What is the cell constant, if the conductivity of 0.001 M KCl solution at 298K is 0.146X10 -3

Scm ? -1 (0.219 cm ) -1

10. Account for the following :

1. Alkaline medium inhibits the rusting of iron.
2. Iron does not rust even if the zinc coating is broken in a galvanized iron pipe.
11. Three iron sheets have been coated separately with three metals ( A, B and C) whose
standard electrode potentials are given below:
Metal A B C Iron
E values
o -0.46 V -0.66 V -0.20 V - 0.44 V
Identify in which case rusting will take place faster when coating is damaged.

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12. Write the reactions occurring during the electrolysis of ;
a) Solution of dil. Sulphuric acid using platinum electrodes.
b) Aqueous Silver Nitrate solution using Silver electrodes.
c) Aqueous Sodium Chloride solution.
13.
Cu 2+ + 2e - Cu E = +0.34 V
o

Ag + + e - Ag E = +0.80 V
o

a. Construct a galvanic cell using the above data.

b. For what concentration of Ag ions will the emf of the cell be zero at 25 C, if
+ o

the concentration of Cu is 0.01 M ? [ log 3.919 = 0.593] (calc. not reqd.)

2+

14. (i) State Kohlaursch law of independent migration of ions. Write an expression for the
molar conductivity of acetic acid at infinite dilution according to Kohlaursch law.
0 0
(ii) Calculate Λ for acetic acid. ( Given that Λ m HCl = 426 Scm mol Λ m NaCl = 126
0
m
2 -1

Scm mol 2 Λ0m CH COONa = 91 Scm mol )

-1
3
2 -1 (391 Scm mol ) 2 -1

15. What type of battery is lead storage battery? Write the anode and the cathode reactions
and overall reaction occurring in a lead storage battery when current is drawn from it.
16. Following reactions occur at the cathode during the electrolysis of aqueous silver chloride
solution:
Ag + e
+ - Ag E = +0.80 V o

H + e
+ - ½H E = +0.00 V
2
o

On the basis of standard reduction potential values, which reaction is feasible at the
cathode and why?

Hands –on / IT Enabled work:

1) working of Cu-Zn daneill cell
https://www.youtube.com/watch?v=LahawEMMvvY
2) batteries:
https://drive.google.com/open?id=0B7-07wgSVzqTek0yQ2tOWG5MZW8
3) Demonstration of electrochemical cell.

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More Practice
Chapter 2: Electrochemistry

1. Conductivity of 0.00241 M acetic acid is 7.896 X 10 Scm . Calculate its molar -5 -1

conductivity. If Λo for acetic acid is 390.5 Scm mol , what is its dissociation constant? 2 -1

2. Explain with examples the terms weak and strong electrolytes? How can these be
distinguished? With the help of a diagram explain the difference in the variation of
molar conductivity with concentration for strong and weak electrolytes

3. Calculate the emf of the cell Mg|Mg (0.1 M)||Cu (0.0001 M)|Cu at 298 K. Given 2+ 2+

E o = -2. 37 V and E
Mg2+/Mg = + 0. 34 V. o
Cu2+/Cu

4. A voltaic cell is set up at 25 C with the following half –cells; o

Al| Al (0.001 M) and Ni| Ni (0.50 M)

3+ 2+

Calculate the cell voltage [E = -0.25 V, E = -1.66 V] 0

Ni2+|Ni
0
Al3+|Al (1.45 V)
5. The following chemical reaction is occuring in an electrochemical cell
Mg(s) + 2Ag (0.0001 M) +Mg ( 0.10 M) + 2Ag(s) 2+

The E values are Mg /Mg = -2.36 V and Ag / Ag = 0.80V

o +2 +

For this cell calculate / write

(a) The carriers of current within this cell.
(b) E value for the electrode 2 Ag /2 Ag .
o +

(c) Standard cell potential E . o

cell

(d) Cell potential E cell

(e) How will the value of E change if the concentration of Ag (aq.) is increased? cell
+

(f) Symbolic representation of the above cell.

(g) Will the above cell reaction be spontaneous?

6. In the button cell , widely used in watches , the following reaction takes place
Zn Ag O + H O
(s)
+
2 Zn + 2 Ag + 2OH
(s) 2 (l)
2+
(aq) (s)
-

Determine E and ΔG for the reaction. 0 0

( given: E 0.80 V , E = -0.76 V)

0
Ag+ / Ag = (E=1.56 V, ΔG = -301.08 KJ mol )
0
Zn2+/Zn
o -1

7. Three electrolytic cells A, B and C containing solutions of zinc sulphate, silver nitrate
and copper sulphate, respectively are connected in series. A steady current of 1.5
ampere was passed through them until 1.45 g of silver were deposited at the cathode of
cell B. How long did the current flow? What mass of copper and what mass of zinc
were deposited in the concerned cells? ( Atomic masses of Ag = 108, Zn = 65.4, Cu =
63.5)
8. Calculate the emf of the following cell at 298K:
Fe(s)|Fe (0.001M)||H (1M)|H (g)(1bar), Pt(s)
2+ +
2

Given E =0.44V o
cell

9. How many coulombs of electric charge must be passed through a solution of silver
nitrate to coat a silver sheet of area 100 cm on both the sides with a 0.005 mm thick layer. 2

Density of silver is 10.5 g/cm . Relative atomic mass of silver is 108. ( 938.2 C) 3

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Assignment
Chapter 3: Chemical Kinetics

1. For the reaction:

2N2O5 (g)  4NO2 (g) + O2(g)
The rate of formation of NO2 (g) is 2.8 X 10-3 M s-1. Calculate the rate of disappearance of
N2O5 (g). (1.4 X 10-3 M/s)

2. A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K.
Calculate the activation energy of the reaction. (Given: log 2 = 0.3010, log 4 =0.6021, R =
8.314 JK-1mol-1) (27.66 kJ/mol)

3. For a reaction R ––> P, half-life (t1/2) is observed to be independent of the initial concentration
of reactants. What is the order of reaction ?

4. For a reaction :
2NH3(g) Pt N2(g) + 3H2(g)
Rate = k
(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k.

5. Explain the following terms:

(i) Rate constant (k)
(ii) Half life period of a reaction(t1/2)
(iii) Order of the reaction
(iv) pseudo first order reaction.

6. The rate of a particular reaction triples when temperature changes from 500C to 1000C.
Calculate the activation energy of the reaction.
[log 3 = 0.4771, R = 8.314 JK-1 mol-1] ( 22.01 kJ/mol)

7. A reaction, reactant product is represented by the graph below. predict

y

Ln [R]

Time (s) x

(i)The order of the reaction in this case.

(ii)What does the slope of the graph represent?
(iii)What are the units of rate constant k?
(iv)Give the relationship between k and t1/2 (half life period).
(v) Draw the plot of log [R]o/[R] vs. time (s)
8. For the reaction
2 NO(g) + Cl 2(g) 2NOCl (g)
The followingdata were collected . All the measurements were taken at 263 K:

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Experiment No. Initial(NO) (M) Initial(Cl2) (M) Intial rate of disappearance
of Cl2 (M/min)
1 0.15 0.15 0.60
2 0.15 0.30 1.20
3 0.30 0.15 2.40
4 0.25 0.25 ?

Write the expression for the rate law.

(a) Calculate the value of rate constant and specify in units.
(b) What is initial rate of disappearance of Cl2 in exp.4 ?
[(a) k= 177,7 M-2 min-1; (b) Rate = 2.8 M min-1]

9. For a certain chemical reaction:

2A + 2 B 2C + D
The experimentally obtained information is tabulated below.
Experiment [A]o [B]o Initial rate of reaction
1. 0.30 0.30 0.096
2. 0.60 0.30 0.384
3. 0.30 0.60 0.192
4. 0.60 0.60 0.768
For this reaction
(i) derive the order of reaction w.r.t both the reactants A and B.
(ii) write the rate law.
(iii) calculate the value of rate constant k. (k = 3.5 l2mol-2 s-1)
(iv) write the expression for the rate of reaction in terms of A and C.

10. For an elementary reaction

2A + B 3C
the rate of appearance of C at time ‘t’ is 1.3 x 10-4 mol l-1s-1
Calculate at this time
i) rate of the reaction.
ii) Rate of disappearance of A.
{ (i) 4.33 x 10-5 mo/l/s, (ii) 8.66 x 10-5 mol/l/s) }
11. The following data were obtained during the first order thermal decomposition of SO2Cl2 at
a constant volume:
SO2Cl2 (g) SO2 (g) + Cl2(g)

Experiment Time/s-1 Total pressure/atm

1 0 0.4
2 100 0.7
Calculate the rate constant. (Given log 4 = 0.6021, log 2 = 0.3010)
12. For a reaction, A+ B → P, the reaction is of first order in reactant A and second order in
reactant B.
(i) How is the rate of this reaction affected when the concentration of B doubled.
(ii) What is the overall order of rection if A is present in large excess.

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13. The rate constant for the first order decomposition of H2O2 is given by the following
equation:
Log k = 14.2 - 1.0 x 104 K
T
Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes.
(Given: R = 8.314 J K-1mol-1)

14. The rate constant for a reaction of zero order in A is 0.0030 mol L-1 s-1. How long will it take
for the initial concentration of A to fall from 0.10 M to 0.075 M.
(t= 8.33 sec)
15. The half life for decay of radioactive 14C is 5730 years. An archaeological arttefact containing
wood has only 80% of the 14C activity as found in living trees. Calculate the age of the
artefact.

16. For hydrolysis of methyl acetate, the following data were obtained

Time/s-1 0 30 60
[CH3COOCH3]/mol L-1 0.60 0.30 0.15

(i) Show that it follows pseudo first order reaction, as

concentration of water remains constant.

(ii) Calculate the average rate between the time interval 30 to 60 seconds.
[log2=0.3010, log4=0.6020]

Hands-on/ IT Enabled work:

1) Collision theory https://www.youtube.com/watch?v=wbGgIfHsx-I

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More Practice

Chapter 3: Chemical Kinetics

1. Following data are obtained for the reaction :

N2O5 2NO2 + ½O2

t/s 0 300 600

[N2O5]/mol 1.6 × 10–2 0.8 × 10–2 0.4 × 10–2
L–1
(a) Show that it follows first order reaction.

(b) Calculate the half-life.

( c) calculate its concentration after 2 min, if rate costant is 5 x 10-4 s-1

(Given log 2 = 0.3010 log 4 = 0.6021)

2. The decomposition of NH3 on platinum surface, 2NH3 (g) Pt N2 (g) + 3H2(g) is a

zero order reaction with k = 2.5 x 10 MS . What are the rates of production of N2 and
-4 -1.

H2 ? ( 2.5 x 10-4 Ms-1, 7.5 x 10-4 Ms-1)

3. A first order reaction takes 69.3 minutes for 50% completion. Set up an equation for the
determining the time needed for 80% completion of this reaction.
(Calculation of result is not required).

4. For the reaction :

Cl2 (g) + 2 NO (g) 2NOCl (g)
The rate law is expressed as rate = k [Cl2][NO]2
What is the overall order of the reaction?

5. Answer the following questions on the basis of the curve for a first order reaction
A P

Log [R]o /[R]

Time
a) What is the relation between slope of this line and rate constant?
b) Calculate the rate constant of the above reaction if the slope is 2 x 10-4 S-1

6. A first order decomposition reaction takes 40 minutes for 30% decomposition. Calculate its
t1/2 value.

7. For the reaction A B , the rate of reaction becomes 27 times when the concentration
of A is increased three times. What is the order of the reaction?

8. At elevated temperatures, HI decomposes according to the chemical equation;

2HI (g) H 2(g) + I2(g) at 443oC. The rate of the reaction increases with
concentration of HI, as shown in the following table:

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HI (mol/ l) 0.005 0.01 0.02
Rate (mol/l/s) 7.5 X 10 -4 3.0 X 10-3 1.2 X 10-2
(a) Determine (i) order of this reaction and
(ii) write the rate expression
(b) Calculate the rate constant and give its units.

9. A first order reaction has a rate constant of 0.0051 min-1. If we begin with 0.10 M
concentration of the rectant, what concentration of reactant will remain in solution after 3
hours?

10. Consider the reaction A k P. The change in concentration of A with time is shown
in the following plot.

time, t (s) x

(i) Predict the order of the reaction.

(ii) Derive the expression for the time required for the completion of the reaction.

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Notes :Chapter 4: Surface Chemistry

 Adsorption:
(i) The accumulation of molecular species at the surface rather than in the bulk of a solid or
liquid is termed as adsorption.
(ii) It is a surface phenomenon.
(iii) The concentration of adsorbate increases only at the surface of the adsorbent.
 Adsorbate: It is the substance which is being adsorbed on the surface of another substance.
 Adsorbent: It is the substance present in bulk, on the surface of which adsorption is taking
place.
 Desorption: It is the process of removing an adsorbed substance from a surface on which it is
adsorbed.
 Absorption:
(i) It is the phenomenon in which a substance is uniformly distributed throughout the bulk of
the solid.
(ii) It is a bulk phenomenon.
(iii) The concentration is uniform throughout the bulk of solid.
 Sorption: When adsorption and absorption take place simultaneously, it is called sorption.
 Enthalpy or heat of adsorption: Since, adsorption occurs with release in energy, i.e., it is
exothermic in nature. The enthalpy change for the adsorption of one mole of an adsorbate on
the surface of adsorbent is called enthalpy or heat of adsorption.

Mechanism of Adsorption
Inside the Adsorbent (in bulk) the force acting between the particles are mutually balanced but on the
surface, the particles are not surrounded by atoms or molecules of their kind on all sides and hence
they posses attraction force so particle stick on the surface of the Adsorbent.

The extent of adsorption increases with increase in surface area per unit mass of the adsorbent at a
given temperature and pressure.

Heat of adsorption: - With increase in heat Adsorption process decreases.

Adsorption equilibrium: - As the molecules of the adsorb ate are held on the surface of the solid
adsorbent.

Entropy decreases, i.e. S is negative

For the process of adsorption to occur, G must be negative which is possible only when, S keeps
on decreasing and T S keeps on increasing till ultimately H becomes equal.

To T S so that G = 0, this state is called adsorption equilibrium.

 Types of adsorption: There are different types of adsorption namely,

1. Physical adsorption
2. Chemical adsorption

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 Physical adsorption
(i) If the adsorbate is held on a surface of adsorbent by weak van der Waals’ forces, the
adsorption is called physical adsorption or physisorption.
(ii) It is non-specific.
(iii) It is reversible.
(iv) The amount of gas depends upon nature of gas, i.e., easily liquefiable gases like NH3,
CO2, gas adsorbed to greater extent than H2 and He. Higher the critical temperature of gas,
more will be the extent of adsorption.
(v) The extent of adsorption increases with increase in surface area, e.g. porous and finely
divided metals are good adsorbents.
(vi) There are weak van der Waals’ forces of attraction between adsorbate and adsorbent.
(vii) It has low enthalpy of adsorption (20 – 40 kJ mol-1).
(viii) Low temperature is favourable.
(ix) No appreciable activation energy is needed.
(x) It forms multimolecular layers.
 Chemical adsorption or chemisorption:

(i) If the forces holding the adsorbate are as strong as in chemical bonds, the adsorption process is
known as chemical adsorption of chemisorption.
(ii) It is highly specific.
(iii) It is irreversible.
(iv) The amount of gas adsorbed is not related to critical temperature of the gas.
(v) It also increases with increase in surface area.
(vi) There is strong force of attraction similar to chemical bond.
(vii) It has enthalpy heat of adsorption (180 – 240 kJ mol-1).
(viii) High temperature is favourable.
(ix) High activation energy is sometimes needed.
(x) It forms unimolecular layers.

 Factors affecting adsorption of gases on solids:

a. Nature of adsorbate: Physical adsorption is non-specific in nature and therefore every gas gets
adsorbed on the surface of any solid to a lesser or greater extent. However, easily liquefiable gases
like NH3,HCl, CO2, etc. which have higher critical temperatures are absorbed to greater extent
whereas H2, O2, N2 etc. are adsorbed to lesser extent. The chemical adsorption being highly specific,
therefore, a gas gets adsorbed on specific solid only if it enters into chemical combination with it.

b. Nature of adsorbent: Activated carbon, metal oxides like aluminum oxide, silica gel and clay are
commonly used adsorbents. They have their specific adsorption properties depending upon pores.

c. Specific area of the adsorbent: The greater the specific area, more will be the extent of
adsorption. That is why porous or finely divided forms of adsorbents adsorb larger quantities of
adsorbate. The pores should be large enough to allow the gas molecules to enter.

d. Pressure of the gas: Physical adsorption increases with increase in pressure.

 Adsorption isotherm:
 The variation in the amount of gas adsorbed by the adsorbent with pressure at constant
temperature can be expressed by means of a curve is termed as adsorption isotherm.

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 Freundlich Adsorption isotherm: The relationship between and pressure of the gas at constant
temperature is called adsorption isotherm and is given by

Where x- mass of the gas adsorbed on mass m of the adsorbent and the gas at a particular temperature
k and n depends upon the nature of gas

 The adsorption first increases with increase in pressure at low pressure but becomes
independent of pressure at high pressure.

Taking logarithm on both sides, we get,

 If we plot a graph between log x/m and log P, we get a straight line.

The slope of the line isand intercept will be equal to log k.

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Applications of Adsorption
1) Production of high vacuum
2) Gas masks
3) Control of humidity
4) Removal of coloring matter from solution
5) Separation of inert gases
6) Froth floatation process
7) Chromatographic analysis

 Catalyst: These are substances which alter the rate of a chemical reaction and themselves
remain chemically and quantitatively unchanged after the reactionand the phenomenon is
known as catalysis.

 Promoters: These are the substances which increase the activity of catalyst. Example – Mo is
promoter whereas Fe is catalyst in Haber’s Process.

 Catalytic poisons (Inhibitors): These are the substances which decrease the activity of catalyst.
Example -Arsenic acts as catalytic poison in the manufacture of sulphuric acid by ‘contact
process.’

 Types of catalysis:
 There are two types of catalysis namely,
1. Homogeneous catalysis: When the catalyst and the reactants are in the same phase, this kind of
catalytic process is known as homogeneous catalysis.

2.Heterogeneous catalysis: When the catalyst and the reactants are in different phases, the
catalytic process is said to be heterogeneous catalysis.

3. Activity of catalyst:

It is the ability of a catalyst to increase the rate of a chemical reaction.Catalyst has an ability
to increase the rate of reaction. This ability of catalyst is known as the activity of catalyst. It
depends upon adsorption of reactants on the surface of catalyst. Chemisorption is the main

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factor governing the activity of catalysts. The bond formed during adsorption between the
catalytic surface and the reactants must not be too strong or too weak.

It must be strong enough to make the catalyst active whereas, not so strong that the reactant
molecules get immobilized on the catalytic surface leaving no further space for the new
reactants to get adsorbed. Generally for the hydrogenation reaction, from Group 5 to Group 11
metals, the catalytic activity increases. The catalytic activity is found to be highest for group
7-9 elements of the periodic table.

Pt
H2(g) + O2(g) 2H2O(l)

4. Selectivity of Catalyst

Catalysts are highly specific compounds. They have an ability to direct the reaction to yield a
particular product. The reaction with same reactants but different catalyst may yield different
products. This is termed as the selectivity of catalyst. Catalysts are highly selective in nature.
They can accelerate a particular reaction while inhibit another reaction. Hence, we can say a
particular catalyst can catalyse one particular reaction only. It may fail to catalyse another
reaction of the same type. For example: reaction of hydrogen and carbon monoxide yields
methane when nickel is used as catalyst, methanol when a mixture of zinc oxide and
chromium oxide is used as catalyst and methanal when only copper is used as catalyst.

For example: CO and H2 react to form different products in presence of different catalysts as
follows:
Ni
(i) CO(g) + 3H2(g) −→ CH4(g) + H2O(g)
Cu/ZnO Cr2O3
(ii) CO(g) + 2H2(g) − −→ CH3OH(g)
Cu
(iii)CO(g) + H2(g) −→ HCHO(g)

 Shape – selective catalysis: It is the catalysis which depends upon the pore structure of the
catalyst and molecular size of reactant and product molecules. Example – Zeolites are shape –
selective catalysts due to their honey- comb structure. They are alumino silicates with Al-O-Si
network. Example- ZSM-5 is a shape selective catalyst which catalysis dehydration of
alcohols to for gasoline.
 Enzymes: These are complex nitrogenous organic compounds which are produced by living
plants and animals. They are actually protein molecules of high molecular mass. They are
biochemical

catalysts

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 Steps of enzyme catalysis:

(i) Binding of enzyme to substrate to form an activated complex.

(ii) Decomposition of the activated complex to form product.

 Characteristics of enzyme catalysis:

(i) They are highly efficient. One molecule of an enzyme can transform 106 molecules of reactants
per minute.
(ii) They are highly specific in nature. Example – Urease catalysis hydrolysis of urea only.
(iii) They are active at optimum temperature (298 – 310 K). The rate of enzyme catalysed reaction
becomes maximum at a definite temperature called the optimum temperature.
(iv) They are highly active at a specific pH called optimum pH.
(v) Enzymatic activity can be increased in presence of coenzymes which can be called as
promoters.
(vi) Activators are generally metal ions Na+, Co2+ and Cu2+ etc. They weakly bind to enzyme and
increase its activity.
(vii) Influence of inhibitors (poison): Enzymes can also be inhibited or poisoned by the presence of
certain substances.

 True solution:
(i) It is homogeneous.
(ii) The diameter of the particles is less than 1 nm.
(iii)It passes through filter paper.
(iv) Its particles cannot be seen under a microscope.

 Colloids:
(i) It appears to be homogeneous but is actually heterogeneous.
(ii) The diameter of the particles is 1 nm to 1000 nm.
(iii) It passes through ordinary filter paper but not through ultra-filters.
(iv) Its particles can be seen by a powerful microscope due to scattering of light.

 Suspension:
(i) It is heterogeneous.
(ii) The diameter of the particles are larger than 1000 nm.
(iii) It does not pass through filter paper.
(iv) Its particles can be seen even with naked eye.

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 Dispersed phase: It is the substance which is dispersed as very fine particles.

 Dispersion medium: It is the substance present in larger quantity.

 Classification of colloids on the basis of the physical state of dispersed phase and dispersion
medium:

Name Dispersed phase Dispersed medium Examples

Solid sol solid Solid Coloured gem stones
Sol Solid Liquid Paints
Aerosol Solid Gas Smoke, dust
Gel Liquid Solid Cheese, jellies
Emulsion Liquid Liquid Hair cream, milk
Aerosol Liquid Gas Mist, fog, cloud
Solid sol Gas Solid Foam rubber, pumice stone
Foam Gas Liquid Whipped cream

 Classification of colloids on the basis of nature of interaction between dispersed phase and
dispersion medium, the colloids are classified into two types namely,

1. Lyophobic sols
2. Lyophilic sols

 Lyophobic sols:

(i) These colloids are liquid hating.

(ii) In these colloids the particles of dispersed phase have no affinity for the dispersion medium.
(iii) They are not stable.
(iv) They can be prepared by mixing substances directly.
(v) They need stabilizing agents for their preservation.
(vi) They are irreversible sols.

 Lyophilic sols:

(i) These colloids are liquid loving.

(ii) In these colloids, the particles of dispersed phase have great affinity for the dispersion medium.
(iii) They are stable.
(iv) They cannot be prepared by mixing substances directly. They are prepared only by special
methods.
(v) They do not need stabilizing agents for their preservation.
(vi) They are reversible sols.

 Classification of colloids on the basis of types of particles of the dispersed phase:

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There are three types of colloids based on the type of dispersed phase, namely,

1. Multimolecular colloids: The colloids in which the colloidal particles consist of aggregates of
atoms or small molecules. The diameter of the colloidal particle formed is less than 1 nm.
2. Macromolecular colloids: These are the colloids in which the dispersed particles are
themselves large molecules (usually polymers). Since these molecules have dimensions
comparable to those of colloids particles, their dispersions are called macromolecular colloids,
e.g., proteins, starch and cellulose form macromolecular colloids.
3. Associated colloids (Micelles): Those colloids which behave as normal, strong electrolytes at
low concentrations, but show colloidal properties at higherconcentrations due to the formation
of aggregated particles of colloidal dimensions. Such substances are also referred to as
associated colloids.

 Kraft Temperature (Tk):Micelles are formed only above a certain temperature called Kraft
temperature.

 Critical Micelle Concentration (CMC): Micelles are formed only above a particular
concentration called critical micelle concentration.

 Soaps: These are are sodium or potassium salts of higher fatty acids e.g., sodium stearate
CH3(CH2)16COO-Na+
 Cleansing action of soap:

When soap is dissolved in water, its hydrophobic ends attach themselves to dirt and remove it
from the cloth.First, the molecules of soap arrange themselves in micelle formation and trap
the dirt at the centre of the cluster. These micelles remain suspended in water like particles in
a colloidal solution.

Micelles

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Methods of preparation of colloids

Chemical Methods:

Bredig's method:

An electric arc is struck between two metallic electrodes immersed in

dispersion medium. The arc produced vapourises the metal which onfurther
condensation produces particles of colloidal size.

Peptization:

Process of converting a freshly prepared precipitate into colloidal sol by shaking it with electrolyte in
dispersion medium is called as peptization. The electrolyte used for this purpose is called peptizing
agent.

 Purification of colloids:

1. Dialysis: It is a process of removing a dissolved substance from a colloidal solution by means

of diffusion through a suitable membrane.
2. Electro dialysis. The process of dialysis is quite slow. It can be made faster by applying an
electric field if the dissolved substance in the impure colloidal solution is only an electrolyte.
3. Ultrafiltration: It is the process of separating the colloidal particles from the solvent and
soluble solutes present in the colloidal solution by specially prepared filters, which are
permeable to all substances except the colloidal particles. Ultra filter paper is made by dipping
filter paper in 4% solution of nitro cellulose in alcohol and ether. This filter paper is hardened
by dipping in formaldehyde.
4. Ultracentrifugation: In this process, the colloidal solution is taken in a tube which is placed in
ultracentrifuge. On rotating the tube at very high speed, the colloidal particles settle down at
the bottom of the tube and the impurities remain in solution. The settled particles are mixed
with dispersion medium to regenerate the sol.

 Properties of colloids:
1. Colour: The colour of colloidal solution depends upon the wavelength of light
scattered by the colloidal particles which in turn depends upon the nature and size of
particles. The colour also depends upon the manner in which light is received by the
observer. Example- Finest gold sol is red in colour and as the size of the particle
keeps increasing its colour changes to blue, then purple and finally gold.
2. Brownian movement: Colloidal particles move in zig – zag path. This type of motion
is due to colliding molecules of dispersion medium constantly with colloidal particles.
3. Colligative properties: The values of colligative properties (osmotic pressure,
lowering in vapour pressure, depression in freezing point and elevation in boiling

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point) are of small order as compared to values shown by true solutions at the same
concentrations.
4. Tyndall effect: The scattering of a beam of light by colloidal particles is called
Tyndall effect. The bright cone of light is called the Tyndall cone.
5. Charge on colloidal particles: Colloidal particles always carry an electric charge. The
nature of this charge is the same on all the particles in a given colloidal solution and
may be either positive or negative.

Positively charged colloidal particles

(i) These include hydrated metallic oxides such as Fe2O3.H2O, Cr2O3.H2O, Al2O3.H2O
(ii) Basic dye stuff like malachite green, methylene blue sols.
(iii) Example – Haemoglobin (blood).

Negatively charged colloidal particles:

(i) Metallic sulphides like As2S3, Sb2S3 sols.

(ii) Acid dye stuff like eosin, methyl orange, Congo red sols.
(iii) Examples – Starch sol, gum, gelatin, clay, charcoal, egg albumin, etc.

6. Helmholtz electrical double layer: When the colloidal particles acquire negative or
positive charge by selective adsorption of one of the ions, it attracts counter ions from
the medium forming a second layer. The combination of these two layers of opposite
charges around colloidal particles is called Helmholtz electrical double layer.
7. Electrokinetic potential or zeta potential: The potential difference between the fixed
layer and the diffused layer of opposite charges is called electrokinetic potential or
zeta potential.
8. Electrophoresis: The movement of colloidal particles under an applied electric
potential is called electrophoresis.

9. Coagulation- It is process of settling of colloidal particles. Also called

precipitation of sol

Coagulation of Lyophobic Sols-

Coagulation of lyophobic sols can be done by the following methods:

 By electrophoresis - The colloidal particles move towards oppositely changed electrodes get
discharged and precipitate.
 By mixing two oppositely charged sols - Oppositely charged sols when mixed together in
almost equal proportion, neutralise their charges and get partially or completely precipitated.
 By Boiling- When a sol is boiled the adsorbed layer is disturbed due to increased number of
collisions with the molecules of the dispersion medium. This reduces the charge on the
particles and they ultimately settle down in the form of a precipitate.
 By Persistent dialysis - On prolonged dialysis, traces of the electrolyte present in the sol are
removed almost completely. Colloids become unstable and coagulate.

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 By addition of electrolyte - When excess of electrolyte is added, colloidal particles precipitate
as colloids interact with ions carrying charge opposite to that present on themselves. This
causes neutralisation leading to their coagulation.

Example- A negatively charged ion when added to a positively charged sol causes coagulation. The
negatively charged ion is called coagulating ion/flocculative ion as it neutralises the colloid to cause
coagulation.

Coagulation of Lyophillic Sols-

Lyophilic sols are stable because of charge and solvation of colloidal particles. So we remove these
two factors to coagulate them. This is done by

 Addition of an electrolyte
 Addition of a suitable solvent

Protection of colloids -

 Lyophilic sols are more stable than lyophobic sols

 Lyophilic colloids have a unique ability to protect lyophobic colloids from electrolytes
 When a lyophilic sol is added to lyophobic sol, the lyophilic particles (colloids) form a layer
around the particles of lyophobic sol
 Lyophilic colloids are also called protective colloids

Hardy-Schulze Rule-

 The greater the valency of the flocculating ion added, the greater is its precipitation.
 For negative sols, when positive ions are added

Al3+> Ba2+>Na+ is the order in terms of flocculating power

 For positive sols, when negative ions are added

[Fe(CN)6] 4-> PO43->SO42->Cl- is the order in terms of flocculating power

Coagulation Value/ Flocculation value: The number of millimoles of an electrolyte required to

bring about the coagulation of one litre of a colloidal solution is called its flocculation value.

Emulsions: Emulsions are colloidal solutions where the dispersed phase and dispersion medium,
both , are in liquid state.

 Types of emulsions:
1. Water dispersed in oil (W/O): When water is the dispersed phase and oil is the
dispersion medium. E.g. butter
2. Oil dispersed in water (O/W): When oil is the dispersed phase and water is the
dispersion medium. E.g. milk
 Emulsification: It is the process of stabilizing an emulsion by means of an emulsifier.

 Emulsifying agent or emulsifier: These are the substances which are added to stabilize the
emulsions. They form an interfacial layer between the dispersed phase and dispersion

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medium. Examples – soaps, gum for O/W type emulsion and lamp black and long chain
alcohols act as emulsifying agents for W/O type

 Demulsification: It is the process of breaking an emulsion into its constituent liquids by

freezing, boiling, centrifugation or some chemical methods.

Certain reasons related to colloids around us:

 Sky appears blue due to scattering of light by air molecules, water droplets, and other
colloidal particles in the sky. Blue colour scatters the most.
 Deltas are formed when river carrying silt, clay colloidal particles meets sea, as there a
lot many salts dissolved in sea water, they neutralize the charge on colloidal particles
leading to the coagulation of clay and silt forming deltas.
 Albimunoids in blood are negatively charged colloids. If there is bleeding happening
from a cut in the body, rubbing with alum (phikari), FeCl3 salt leads to coagulation of
blood due to neutralization of charged albimunoid colloidal particles.
 Artificial rain can be produced by spraying oppositely charged sol on the clouds which
are colloids leading to their precipitation.

Applications of Colloids

1. Electro precipitation of smoke – The smoke is led through a chamber containing plates having
a charged opposite to that carried by smoke particles. The particles on coming in contact with
these plates lose their charge and get precipitated. The particles settle down on the floor of the
chamber. The precipitator is called Cottrell precipitator.

2. Purification drinking water – Alum is added to impure water to coagulate the suspended
impurities and make water fit for drinking.

3. Medicines – Most of the medicines are colloidal in nature. Colloidal medicines are more
effective because they have a larger surface area and are more easily absorbed by the body.
Eg- Argyrol is a silver sol used as an eye lotion, milk of magnesia is used to cure stomach
disorders, Antimony sol is used to cure Kalazaar, Gold sol is used in intramuscular injections.

4. Tanning – Animal hides are colloidal in nature. When a hide that has positively charged
particles is soaked in tannin/chromium salts, which contains negatively charged particles ,

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mutual coagulation takes place. This results in the hardening of leather. This process is termed
as tanning.

5. Cleansing action of soaps- already explained

6. Photographic plates and films – Photographic plates and films are prepared by coating an
emulsion of the light sensitive silver bromide in gelatin over glass plates or celluloid films.

7. Rubber industry- Latex is a colloidal solution of rubber particles which are negatively
charged. Rubber is obtained by coagulation of latex.

8. Industrial products- Paints inks, synthetic plastics, rubber, cement, graphite lubricants are all
colloids.

Hands-on/ IT Enabled work:

Whole chapter will be done with the help of power point presentation.

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Assignment

Chapter 4: Surface Chemistry

1. CO(g) and H2(g) react to give different products in the presence of different catalysts.
Which ability of the catalyst is shown by these reactions?

2. What happens when

(i) a freshly prepared precipitate of Fe(OH)3 is shaken with a small amount of FeCl3
solution?
(ii) persistent dialysis of a colloidal solution is carried out?
(iii)size of dispersed phase changes in gold sol.
3. (a) Define adsorption. Write any two features which distinguish physisorption and
chemisorption. Which has higher enthalpy of adsorption?
(b) Write one similarity between Physisorption and Chemisorption.
(c) List four applications of adsorption.

4. Write one difference in each of the following :

(i) Lyophobic sol and Lyophilic sol
(ii) Solution and Colloid

5. (a) How can a colloidal solution and true solution of the same colour be distinguished
from each other?
(b) Why is ferric chloride preferred over potassium chloride in case of a cut leading to
bleeding?
(c)Name the physical states of dispersed phase and dispersion medium of froth?

6. How does an increase in temperature affect both physical and chemical adsorption?

7. What causes Brownian movement in a colloidal solution?

8. Explain the following observations:

a) Lyophilic colloid is more stable than lyophobic colloid.
b) Coagulation takes place when sodium chloride solution is added to a colloidal solution
of ferric hydroxide.
c) Sky appears blue in colour.
d) Adsorption of a gas on the surface of solid is generally accompanied by a decrease in
entrop, still it is a spontaneous process.
e) Enzyme catalysts are highly specific in their action.
f) The enthalpy in case of chemisorption is usually higher than that of physisorption.
g) Cottrell’s smoke precipitator is fitted at the mouth of the chimney used in factories.
h) Physical adsorption is multilayered, while chemisorption is monolayered.

9. (i) Differentiate between adsorbtion and absorption.

(ii) Out of MgCl2 and AlCl3, which one is more effective in causing coagulation of
negatively charged sol and why?
(iii) Out of sulphur sol and proteins, which one forms multimolecular colloids ?
10. A colloidal solution of AgI is prepared by two different methods shown below:-
(A) (B)

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(a) What is the charge of AgI colloidal particles in the two test tubes (A) and (B)?
(b) Give reasons for the origin of charge.

11. What are the two classes of emulsion? Give one example of each class. State one activity
to test the type of emulsion.

12. Describe the following giving an example each:

a) Mechanism of heterogeneous catalysis. Give a point of difference between
Homogeneous catalysis and Heterogeneous catalysis.
b) Hardy Schulze Rule
c)Emulsification

13. Consider the adsorption isotherms given below and interpret the variation in the extent of
adsorption ( x/m) when:

(a) (i) temperature increases at constant pressure.

(ii) pressure increases at constant temperature.
(b) Name the catalyst and the promoter used in Haber’s process for manufacture of
ammonia.

14. In reference to Freundlich adsorption isotherm write the expression for adsorption of gases
on solids in the form of an equation.
Based on type of particles of dispersed phase, give one example each of associated colloid
and multimolecular colloid.

15. Give reason for the following observations;

(i) Leather gets hardened after tanning.
(ii) It is necessary to remove CO when ammonia is prepared by Haber’s process.

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Notes- Chapter -5
PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

Minerals: These are naturally occurring chemical substances which are obtained from earth’s crust
by mining. In minerals metals are present in either native state or combined state.
Ores: The mineral from which metal can be economically and conveniently extracted is called ore.
Metallurgy: The entire scientific and technological process used for isolation of the metal from the
ores is known as metallurgy. The major steps for extraction and isolation of metals are:--
 Concentration of the ore
 Isolation of metal from the concentrated ore
 Purification of metal

I) CONCENTRATION OF ORES
Removal of unwanted materials (ie; clay, sand etc.) from ores is known as concentration. These
earthly or undesirable impurities are called GANGUE. Some important procedures are:-

1) Hydraulic washing: This is based on the difference in gravities of the ore and the gangue
particles. In this an upward stream of running water is used to wash the powdered on. The
lighter gangue particles are washed away and heavier ores are left behind.

2) Magnetic Separation: This is base on the differences in the magnetic properties of the ore
components. One of the two, ore or gangue is capable of being attracted by magnetic field.

3) Froth Floatation process: This method is used to remove gangue from sulphide ores. A
suspension of powdered ore is made with water. To it collectors (E.g- Pine oil) and Froth
stabilizers (e.g., cresol, aniline) are added. Collectors enhance wettability of pine oil and
froth stabilizers froth. Mineral particles wetted by oil are carried away with froth and
gangue particles move into water.

Two sulphide ores can be separated using depressants. For eg. NaCN is used to separate
ZnS and PbS present in a ore.

4) Chemical Methods(Leaching):
a) Leaching of Alumina from Bauxite: Bauxite contains SiO2, iron oxides and titanium oxide
as impurities. Powdered ore is treated with concentrated solution of NaOH at 478-523 K
and 35-36 bar pressure.
Al2O3(s) + 2NaOH(aq.) + 3H2O(l) 2Na[Al(OH)4] (aq.)

The aluminate solution is neutralized by passing CO2 gas and hydrated Al2O3 separated out.
2Na[Al(OH)4](aq.) + CO2(g) Al2O3.xH2O(s) + 2NaHCO3g)

The sodium silicate remains in the solution and hydrated alumina is filtered, dried and
heated to give pure Al2O3.
Al2O3.xH2O (S) Al2O3 (s) + x H2O (g)
b) Other example: In metallurgy of silver and gold, the respective metal is leached with
dilute solution of NaCN or KCN in presence of air (Or O2)

4M(S) + 8 CN-(aq.) + 2H2O(aq.) + O2(g) 4[M(CN)2]-(aq.) + 4 OH-(aq.)

2[M(CN)2]- (aq.) + Zn(s) [Zn(CN)4]2- + 2 M(s) (M=Ag or Au)

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II) CALCINATION/ROASTING: The concentrated ore is then converted to its oxide form by
heating .
If the concentrated ore is heting strongly in the absence of air to get rid of impurities, and the
ore changes to oxide form, it is called calcination. It is generally observed for carbonate ore.
CaCO3 CaO + CO2
If the concentrated ore is heated in the presence of oxygen, ore gets converted to oxide form
and small molecules like SO2 are released. It generally takes place for sulphide ore. ZnS+
O2 ZnO + SO2

III) REDUCTION

Reduction of oxide to metal:

This involves heating of the oxide with a reducing agent like C or CO or even another metal
based on reactivity of the metal to be extracted.
MxOy + yC xM + yCO

Extraction of iron from its oxides

FeO(S) + C(s) Fe(S/l) + CO(g)
(Discussed later under extraction of Iron)

Extraction of copper from cuprous oxide:

The sulphide ores are roasted/smelted to give oxides (Most sulphide ores contain iron)
2Cu2S + 3O2 2Cu2O + 2SO2
The oxide is then reduced to metallic copper using coke.
Cu2O + C 2Cu + CO
In actual process, the ore is heated in a reverberatory furnace after mixing with silica. The
iron oxide slags off as iron silicate and copper is produced in the form of copper matte.
This contains Cu2S and FeS.
FeO+ SiO2 FeSiO3 (slag)

Copper matte is then charged into silica lined convertor. Some silica is also added and hot
air blast is blown to convert the remaining FeS, FeO and Cu2S/Cu2O to metallic copper.
The reactions taking place are:
2FeS + 3O2 2FeO + 2SO2
FeO + SiO2 FeSiO3
2Cu2S + 3O2 2Cu2O + 2SO2
2Cu2O + Cu2S 6Cu + SO2
The solidified copper obtained has blistered appearance due to evaluation of SO2& so is
called blister copper.

Extraction of zinc from zinc oxide:

The reduction of zinc oxide is done using coke. For the purpose of heating, the oxide is made
into brickettes with coke and clay.
ZnO + C coke, 673K Zn + CO
The metal is distilled off and collected by rapid chilling.

Thermodynamic principles of metallurgy:

Gibb’s energy is described by the equation ΔG = ΔH –TΔS.
For any reaction, this change can be described by ΔG = -RTlnK

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Negative ΔG implies positive K and this can happen when reaction proceeds towards
products.
(i) If ΔS is positive, on increasing T the value of TΔS would increase (ΔH < TΔS) and
ΔH<TΔS) and ΔG becomes negative.
(ii) If reactants and products of two reactions are put together in a system and net ΔG is
negative, overall reaction would occur. So the process involves coupling of two
reactions and getting sum of their ΔG as negative.
For eg: In reduction of FeO
FeO (s) Fe(s) + ½ O2 (g) [ ΔG(FeO, Fe)]
C (s) + ½ O2 (g) CO (g) [ ΔG (C, CO)]
We couple the two so that net Gibb’s energy change becomes
ΔG(C,CO) + ΔG (FeO, Fe) = ΔG
The resultant will have ΔG as negative.
According to Ellingham diagram, the ΔGo Vs. T plot representing the reaction goes upward
and that representing C CO goes downward. At temp.above 1073 K the (C,CO) line
come below the Fe, FeO line [ ΔG(C, CO) < ΔG( Fe, FeO)]. So coke will be reducing FeO
and is itself oxidized to CO.

IV) REFINING

A metal extracted by any method is usually contaminated with some impurity. Some
methods are
a) Distillation: The impure metal is evaporated to obtain pure metal as distillate. Used
for low boiling metals like Zn and mercury.
b) Liquation: The low melting metal, like tin is separated from high melting impurities by
this method. The metal is made to flow on sloping surface.
c) Electrolytic refining: In this method impure metal is made to act as anode. A strip of
same metal in pure form is made cathode. The electrolyte used contains soluble salt of
the same metal. The more basic metals remain in the solution and less basic form anode
mud.
For e.g. Copper is refined by electrolytic method. Anode is impure copper and pure
copper strips are taken as cathode and electrolyte is acidified solution of copper sulphate.
Anode: Cu Cu2+ + 2e-
Cathode: Cu2+ + 2e- Cu
d) Zone refining: It is based on the fact that melting point of a substance is lowered by the
presence of impurities. Consequently when an impure metal in molten state is cooled,
crystals of pure metal are solidified first and impurities remain behind in the molten
metal which crystallizes later. The semiconductors –silicon and germanium are purified
by this method.
e) Vapour Phase refining: This method is used for preparing ultrapure metals by forming
vapours of the compound of metal and later decomposition to get pure metal.
Eg- 1) Mond process for refining Ni
Ni + 4CO 330-350 K Ni(CO)4
volatile complex- nickel tetracarbonyl
At higher temperature,
Ni(CO)4 450-470 K Ni + 4CO

2) Van Arkel method for refining of Zr or Ti : This method is useful for

removing all oxygen and nitrogen present as impurity. Crude metal is heated in
evacuated vessel with iodine.

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Zr + 2I2 ZrI4
Metal iodide is then heated on a tungsten filament to about 1800 K.
ZrI4 ∆ Zr + 2I2

f) Chromatographic methods: This method is based on the principle that different

components are differently adsorbed on adsorbent. The mixture is put in a suitable
solvent and applied on top of the column. The adsorbent is packed in a glass column.
The adsorbed components are removed (eluted) out using suitable solvent (eluent). The
weakly adsorbed component is eluted first followed by the more strongly adsorbed and
so on. This is called column chromatography. This method is used for purification of
elements available in minute quantities and impurities not very different in chemical
properties.

EXTRACTION OF ALUMINIUM:

Aluminium is extracted from bauxite ore, Al2O3.2H20. It involves two steps:

1. Concentration (by Baeyer’s Process) : Purification of Bauxite ore is done by Baeyer’s process.
This Chemical method is called leaching.
(Explained under methods of concentration)
2. Reduction { byElectrolysis of fused alumina (Hall-Heroult process)}: The purified alumina is
dissolved in molten cryolite and is electrolyse in an iron tank lined inside with carbon. The
molten cryolite decreases the melting point to about 1173 K and also increases conductivity. The
anode consists of a number of carbon rods which dip in fused electrolyte. The electrolyte is
covered with a layer of powdered coke.

2Al2O3 + 3C 4Al + 3CO2

Reactions at electrodes are:

At Cathode: Al3+ (melt) + 3e- Al(l)
At Anode: C(s) + O2- (melt) CO(g) +2e-
C(s) + 2O2- (melt) CO2 (g) + 4e-
Therefore, aluminium is liberated at the cathode and gets collected at the bottom of the tank from
where it is removed. The oxygen evolved combines with carbon of anode to form CO or CO2 and
escapes out. Because of reaction at carbon anodes, these need to be replaced periodically.
3. Refininig of Aluminium:
The aluminium metal obtained above is 99% pure, which is further purified by Hoop’s electrolytic
method. The process is carried out in an iron tank lined with carbon. It has 3 layers of molten liquid
having different densities.
(i) The top layer consists of pure Al having carbon electrodes dipping in it. The carbon
electrodes act as cathode.
(ii) The middle layer has fluorides of sodium, barium and Aluminium in molten state. This acts
as an electrolyte.
(iii) The bottom layer consists of impure Al along with the carbon lining acts as anode.

On passing electric current aluminium ions from the middle layer are discharged at cathode as pure
Al. The pure Al is removed from the tapping hole. An equivalent amount of Al from bottom layer
moves into the middle layer leaving behind impurities.

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EXTRACTION OF IRON
Iron is the second most abundant element occurring in earth’s crust. The common ores are:
Haemite Fe2O3
Magnetite Fe3O4
Limonite Fe2O3.3H2O
Iron Pyrites FeS2
Siderite FeCO3
Cast Iron is usually extracted from its oxide ore (haematite). It involves the following steps---
1) Concentration: The ore is first crushed and crushed ore is concentrated by gravity
separation; ie hydraulic washing
2) Calcination: The concentrated ore is calcined, ie, heated strongly in limited supply of air in a
reverberatoryfurnance. The following changes take place:
i) Moisture is removed
ii) Impurities of S, P4 and As are converted to their gaseous oxides; sO2 , As2O3 and
P4O10 which are volatile and escape out.
3) Reduction (by Smelting) : The calcined ore is reduced with carbon, ie smelted in a blast
furnance. It is a tall cylindrical furnance made of steel llined with fire bricks. It is narrow at
the top and has cup and cone arrangement for the introduction of charge and outlet for waste
gases. At the base of furnace, it is provided with ---
i) Tuyeres arrangement for introduction of hot air
ii) A tapping hole for withdrawing molten iron and
iii) An outlet through which slag is flown out.

The calcined ore ( 8 parts ) is mixed with coke (4 parts) and limestone (1part) is introduced from top.
At the same time a blast of hot air preheated at 1000 K is blown upwards with the help of tuyers
arrangement. The added coke acts as a reducing agent and lime serves as flux. The burning of coke
to carbon monoxide supplies most of the heat required for working temperature of furnace and give
temp. upto 2200 K at the bottom of furnace. As the gases move up, they meet the descending
charge and temp. falls. At the bottom reducing agent is carbon but at the top the reducing agent is
CO.

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The reactions occurring are:---
1) Combustion zone: At the base coke burns to produce CO2 which starts rising upward. The
reaction is exothermic and heat produced raises the temperature to about 2200 K.
C + O2 CO2 ∆H = -393.4 KJ
2) Fusion zone: As CO2 rises upward, it comes in contact with coke and gets reduced to CO.
CO2 + C 2CO ∆H = +63.2 KJ
The reaction is endothermic , the temperature is lowered to 1570 K. The iron produced in the upper
region melts. Any Fe2O3 if present is reduced by hot coke to iron.
Fe2O3 + 3C 2Fe + 3CO + heat
3) Slag formation zone; In the middle temperature is about 1270K. In this region limestone
decomposes.
CaCO3 CaO + CO2

The lime acts as flux and combines with silica (present as an impurity) to produce slag.
CaO + SiO2 CaSiO3
The molten slag forms a separate layer above molten iron.
4) Reduction Zone: The temperature near the top of furnace is 875 K. The oxide is reduced by
carbon monoxide to iron.
FeO + CO Fe + CO2
3Fe2O3+ CO 2Fe3O4 + CO2
Fe3O4 + 4CO 3Fe + 4CO2
Fe2O3 + CO 2FeO + CO2
The spongy iron produced moves down slowly and melts in fusion zone.
At lower hotter part, reaction is
FeO + C Fe + CO.
It dissolves some carbon, silicon, phosphorus and manganese and forms a layer at the
bottom. The iron obtained is called Pig iron.
Cast iron is different from Pig Iron and is made by melting pig iron with scrap iron and coke
using hot air blast. It has slightly less carbon content (about 3%) . It is extremely hard but
brittle.
Wrought iron is purest form of iron and is prepared from cast iron by oxidizing impurities in
reverberatory furnace lined with haematite. Haematite oxidizes carbon.
Fe2O3 + 3C 2Fe + 3CO
Limestone is added as flux and S, Si and P are oxidized and passed into slag. The metal is
then removed.

EXTRACTION OF COPPER:
Copper is mainly extracted from copper pyrites (CuFeS2). The various steps are:
1) Crushing and concentration: The ore is crushed in haw crushers and is finally powdered. It is
concentrated by froth floatation process.
2) Roasting: The concentrated ore is roasted ,i.e, heated strongly in the presence of excess of air in
a reverberatory furnace.
a) Moisture is removed from ore and it becomes dry.
b) The impurities of S, P4, As and Sb are removed as their volatile oxides.
S + O2 SO2
P4 + 5O2 2P2O5
4As + 3O2 2As2O3
4Sb + 3O2 2Sb2O3
c) Copper pyrites is converted to ferrous sulphide (FeS) and cuprous sulphide (Cu2S)
2CuFeS2 + O2 Cu2S + FeS + SO2
2FeS + 3O2 2FeO + 2SO2

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2Cu2S + 3O2 2Cu2O + 2SO2
3) Smelting: The roasted ore is mixed with powdered coke and sand and is strongly heated in a
blast furnace . The blast furnace is made of steel and is lined with fire bricks. A blast of hot air is
introduced at the lower part of the furnace and changes occurring are:
a) Ferrous sulphide is oxidized to ferrous oxide which combines with silica to form slag.
2FeS + 3O2 2FeO + 2SO2
FeO+ SiO2 FeSiO3 (slag)
The slag being lighter forms the upper layer and is removed from time to time.
b) During roasting if any oxide of copper is formed, it combines with FeS and is changed back
into its sulphide
2Cu2S + 3O2 2Cu2O + 2SO2
Cu2O + FeS Cu2S + FeO (This changes to slag by combining with SiO2)
As a result two separate layers are formed at the bottom of furnace. Upper layer is slag
which is removed as waste. The lower layer of molten mass contains mostly cuprous
sulphide and some traces of ferrous sulphide. It is called matte and is taken out from
tapping hole at bottom.

4) Bessemerisation: The molten matte from Blast-furnace is transferred to Bessemer converter. The
vessel is silica lined from inside. A blast of hot air is mixed with sand is blown into molten
matte. During this process ….
a) Traces of ferrous sulphide present in matte are oxidized to FeO which combines with silica to
form slag.
2 FeS + 3O2 2FeO + 2SO2
FeO+ SiO2 FeSiO3 (slag)
b) Copper sulphide is oxidized to cuprous oxide which further reacts with remaining copper
sulphide to form copper and sulphur dioxide.
2Cu2S + 3O2 2Cu2O + 2SO2
2Cu2O + Cu2S 6Cu + SO2
After the reaction has been completed, the converter is tilted and molten copper is put in
moulds. The copper thus obtained is 99% pure and is known as blister copper. The name
blister is given because as metal solidifies, the dissolved SO2 escapes out producing blisters
on metal surface.

5) Refining: Blister copper is purified by:

a) Poling: Heating strongly in a reverberatory furnace in the presence of excess of air.
Impurities are either converted to oxides or converted to slag. Some copper also changes to
cuprous oxide. This is reduced back to copper by stirring the molten metal with green poles
of wood. This gives 99.5 % pure Cu, which is then purified by electrolytic refining.
b) Electrolytic refining: A thin sheet of metal is made cathode and block of crude metal is made
as anode. Both the electrodes are placed in an acidified CuSO4 solution when electric current
is passed through the solution, impure Cu from anode goes into the solution and pure Cu
from the solution gets deposited on the cathode.
At anode: Cu Cu2+ +2e-
At Cathode Cu + 2e2+ - Cu
The impurities of Zn, Ni, Fe etc. gets collected below as anode mud.

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EXTRACTION OF ZINC

Principal ore of Zinc is Zinc Blende. Extraction is carried out in the following steps:-
1) Concentration: Ore is concentrated by froth floatation process.
2) Roasting: Concentrated ore is roasted in excess of air at about 1200 K andZnS is converted to
ZnO.
2ZnS + 3O2 2ZnO + 2SO2
3) Reduction: ZnO is reduced by heating with crushed coke at 673 K. For the purpose of heating,
the oxide is made into brickettes with coke and clay.
ZnO + C coke, 673K Zn + CO
The metal is distilled off and collected by rapid chilling.
4) Refining: Impure metal is refined by electrolysis. In this process, impure zinc is made the anode
and a plate of pure zinc is made the cathode. The electrolyte is zinc sulphate with little dil.
H2SO4. On passing current zinc is deposited at cathode while equal amount of zinc from anode
goes into electrolyte. Thus, pure zinc is obtained on cathode.

Some extractions based on oxidation:

1) EXTRACTION OF CHLORINE FROM BRINE
Chlorine is abundant in sea water and hence is obtained from electrolysis of brine solution.
2Cl- (aq.) + 2H2O (l) 2OH- (aq.) + H2(g) + Cl2(g)
ΔG = +422 KJ and using
o ΔG =-nFEoCell , Eo=-2.2 V .
o

So it requires emf greater than 2.2 V. But electrolysis requires an excess potential to overcome
some hindering reaction. Thus, Cl2 is obtained by electrolysis giving out H2 and aqueous NaOH
as by products.

2) EXTRACTION OF GOLD AND SILVER:

Extraction of Gold involves leaching the metal with CN- (OXIDATION)
4Au(s) + 8CN-(aq.) + 2 H2O (aq.) + O2 (g) 4[Au(CN)2]-(aq.) + 4OH- (aq.)[oxidation,
NaCN here acts as a leaching agent or complexing agent.
2[Au(CN)2]-(aq.) + Zn (s) 2Au(s) + [ Zn(CN)4]2- (aq.) [ Reduction , Zn acts as a
reducing agent]

Similarly for Silver

4Ag(s) + 8CN-(aq.) + 2 H2O (aq.) + O2 (g) 4[Ag(CN)2]-(aq.) + 4OH- (aq.)
2[Ag(CN)2] (aq.) + Zn (s)
- 2Ag(s) + [ Zn(CN)4]2- (aq.)

Hands-on/ IT Enabled work

Whole chapter is taught with the help of a presentation with animations.

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Assignment
Chapter 5: General Principles and Processes of Isolation of Elements

1. Write the chemical reactions involved in the process of extraction of Gold. Explain the role of
dilute NaCN and Zn in this process.

2. (a) Write the principle of method used for the refining of germanium.
(b) Out of PbS and PbCO3 (ores of lead), which one is concentrated by froth floatation process
preferably?
(c) What is the significance of leaching in the extraction of aluminium?

3. 2What do you understand by the following terms?

(a) Roasting (b) Flux (c) Calcination (d) Smelting (e) Slag

4. a) In the extraction of Al, impure Al2O3 is dissolved in conc. NaOH to form

sodium aluminate and leaving impurities behind. What is the name of this
process? also write the reactions involved
b) What is the role of coke in the extraction of iron from its oxides?

5. Write
5 the reactions occurring in the different zones of blast furnance during extraction of iron
from concentrated Haemetite ore.
6. 6Explain Hall Heroult process of reduction of aluminium oxide. What is the role of graphite and
cryolite in electrometallurgy of aluminium?

7. 7(a) Why is the froth floatation method selected for the concentration of Sulphide ores? Write
reactions taking place in the extractions of zinc from zinc blende.
(b) An ore sample of galena (PbS) is contaminated with zinc blende (ZnS). Name one chemical
compound which can be used to concentrate galena selectively by froth floatation method.
What are such substances called?

8. (a)
8 What is the role of silica in the extraction of copper?
(b) Explain electrorefining of copper. Name the common metals present as anode mud in
electrorefining of copper.

9. 1Account for the following facts :

0
a) The reduction of a metal oxide is easier if the metal formed is in liquid state at the
temperature of reduction.
b) The reduction of Cr2O3 with Al is thermodynamically feasible, yet it does not occur at
room temperature.
c) Pine oil is used in froth floatation method

10. 1a) Indicate the principle behind the method used for refining of zinc.
1 Out of C and CO, which is a better reducing agent at the lower temperature range in the
b)
blast furnace to extract iron from the oxide ore?
c) Which form of iron is the purest form of iron?

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p Block Elements

General Configuration: ns2 np1-6

Maximum oxidation state = Group 10

Across a period: Covalent radii and metallic character decreases, but electro negativity, electron
affinity, oxidizing power and ionization energy increases.

Down the group: Covalent radii and metallic character increases, but electro negativity, electron
affinity, oxidizing power and ionization energy decreases.

Inert pair effect: While going down the group, the ns2 electrons become more and more reluctant to
participate in bond formation. This is because down the group bond energy decreases and so the
energy required to un-pair ns2 electrons is not compensated by the energy released in forming two
additional bonds.

Group 16

Elements in group 16 are Oxygen (O), Sulphur (S), Selenium (Se), Tellurium (Te) and Polonium (Po).
Also called Chalcogens.

Occurrence:
1. Oxygen is the most abundant element.
2. Sulphur exists as gypsum CaSO4.2H2O, Epsom salt MgSO4.7H2O or galena PbS, Zinc blende ZnS
etc.
3. Selenium and Tellurium as Selenides and Tellurides in sulphur ores.
4. Polonium exists as decay product of thorium and uranium minerals.

Electronic Configuration:
General electronic config – ns2np4

Atomic and Ionic Radii:

Due to increase in number of shells, atomic and ionic radii increase from top to bottom in the group.

Ionization Enthalpy:
1. Decreases down the group due to increase in size.
2. Group 16 elements have lower ionization enthalpy than corresponding elements of group 15 due
to stability of half filled p-orbital electronic configuration in group 15.

Electron Gain Enthalpy:

1. Oxygen has less negative electron gain enthalpy due to its compact nature.
2. Sulphur onwards the value again becomes negative upto Po.

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Electronegativity:
Oxygen is the second most electronegative element, next to fluorine. Within a group electro
negativity decreases with increase in atomic number.

Metallic Character:
Increase from oxygen to polonium.

Physical Properties:
O,S - Non-Metals 1. All of them exhibit Allotropy.
Se, Te - Metalloids 2. M.P and B.P increases with increase in atomic number but there
Po - Metal is large difference in m.p of O and S due to its atomicity – O2 and
S8.

Chemical Properties:
1. Stability of -2 oxidation state decreases down the group.
2. Oxygen shows only negative oxidation state of -2 and in case of OF2, oxidation state of +2.
3. Other elements of the group easily show +2, +4, and +6 oxidation state.
4. Stability of +4 oxidation state increase down the group due to inert pair effect.
Reactivity with Hydrogen:
All the elements form hydrides of the formula H2E (E = S, Se, Te, Po).

H2O 1. Acidic nature increases from H2O to H2Te due to decrease in bond dissociation
H2S enthalpy.
H2Se 2. Thermal Stability decreases from H2O to H2Po.
H2Te 3. All hydrides except water posses reducing property and this characteristic
H2Po increases from H2S to H2Te.

Reactivity with Oxygen:

All these elements form oxides of the formula EO2 and EO3 where E = S, Se, Te or Po.
1. SO2 is gas but SeO2 is solid.
2. Reducing property decreases from SO2 and TeO2. SO2 is reducing and TeO2 is an oxidising agent.
3. Also form EO3 type oxides. Both types are acidic.

Reactivity towards halogens:

1. Form halides of the formula EX6, EX4 and EX2.
2. Stability of halides decrease in the order F- > Cl- > Br- > I-.
3. Among hexahalides, hexafluorides are only stable. They (hexafluorides) are gaseous octahedral
in nature. Most stable is SF6 due to steric reasons.
4. Amongst tetrafluorides, SF4 is gas, SeF4 is liquid and TeF4 a solid. They have Sp3d hybridization
and have trigonal bipyramidal structures and are regarded as see-saw geometry.
5. All elements except Se form dichlorides and dibromides. Dihalides have Sp3 hybridisation.
6. Monohalides are dimeric in nature. E.g – S2Fe, S2Cl2, S2Br2, Se2Cl2 and Se2Br2. Dimeric halides
undergo disproportionation as:- 2Se2Cl2 → SeCl4 + 3Se

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DIOXYGEN

Methods of Preparation:
1. By heating chlorates, nitrates and permanganates.

2. Thermal decomposition of oxides.

3. 2H2O2→ 2H2O + O2 Catalyst used – finely divided metals and manganese dioxide.
4. On large scale, prepared by electrolysis of water.
5. Industrially, from air → first carbon dioxide and water vapour are removed and remaining gases
liquefied are fractionally distilled.

Properties of DIOXYGEN:

1. Colourless and odourless gas.

2. Directly reacts with nearly all metals and non metals except Au, Pt and some noble gases. Its
combination with other elements is exothermic.

3. Some reactions with metals and non-metals are :

2Ca + O2 2CaO
4Al + 2O2 2Al2O3
P4 + 5O2 P4O10
C + O2 CO2
2ZnS + 3O2 2ZnO + 2SO2
CH4 + 2O2 CO2 + 2H2O
4. Some compounds are catalytically oxidized.
2SO2 + O2 V2O5 2SO3
4HCl + O2 CuCl2 2H2O

Uses:
1. Importance in normal respiration and combustion.
2. Oxyacetylene welding.
3. Manufacture of steel.
4. Oxygen cylinders are used in hospitals, high altitude flying and in mountaineering.
5. Combustion of fuels, e.g, hydrazine in liquid oxygen provides thrust in rockets.

OXIDES
A binary compound of oxygen with another element is called oxide.

Oxides can be simple (MgO, Al2O3) or mixed (Pb3O4, Fe3O4). Simple oxides can be acidic, basic or
amphoteric oxides.

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An oxide which combines with water to give an acid is termed as acidic oxide. (e.g SO2, CO2, N2O5)
SO2 + H2O H2SO3 (Non metal oxides are acidic)
The oxide which gives a base with water is called basic oxides. Metal oxides are basic (e.g- Na2O,
CaO, BaO)
CaO + H2O Ca(OH)2
Some metal oxides exhibit dual behavior. They show characteristics of both acidic as well as basic
oxides. They are known as amphoteric oxides. E.g Al2O3

Al2O3 + 6HCl + 9H2O 2[Al(H2O)6]3+ + 6Cl-

Al2O3 + 6NaOH + 3H2O 2Na3[Al(OH)6]

Some oxides are neither acidic nor basic. E.g CO, NO and N2O

OZONE

Ozone is an allotrope of oxygen. It is too reactive to remain at sea level. At a height of 20km above
sea level it is formed from atmospheric oxygen in the presence of sunlight. The ozone layer protects
earth’s surface from excessive concentration of UV radiation.

Preparation:
When silent electric discharge is passed through dry oxygen, ozonised oxygen (10%) is produced.
3O2 2O3 (∆H=142KJMol-1)

Properties:

1. Pale blue gas, dark blue liquid and violet black solid.
2. In small concentrations, it is harmless and if concentration rises above 100ppm breathing
becomes uncomfortable.
3. Ozone is thermodynamically unstable. Its decomposition to oxygen results in liberation of heat
and increase in entropy. The two effects results in negative Gibb’s energy change for conversion
into oxygen. High concentration of ozone can be dangerously explosive.
4. Acts as good oxidising agent because [O3 → O2 + O] due to liberation of nascent oxygen.
PbS + 4O3 PbSO4 + 4O2
2I- + H2O + O3 2OH- + I2 + O2
This reaction can be used for estimating O3 gas. I2 liberated titrated against standard Na2S2O3
solution helps in estimation.
5. Nitrogen oxides emitted from exhaust system of supersonic aeroplanes might be slowly
depleting ozone layer in upper atmosphere.
NO + O3 NO2 + O2
6. Ozone layer is also depleted by refrigerants and aerosol sprays.
7.

Bond length is 128pm and bond angle is 117o.

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Uses:

1. Used as germicide, disinfectant and sterilizing water.

2. Used for bleaching oils, ivory, flour and starch.
3. Acts as oxidising agent in manufacture of KMnO4.

SULPHUR

Allotropic forms of Sulphur:

1. Yellow rhombic sulphur (α-sulphur)

2. Monoclinic sulphur (β-sulphur)

Rhombic is stable at room temperature which transforms to monoclinic when heated above 369k.
Rhombic sulphur is yellow in colour, insoluble in water and dissolves to some extent in benzene,
alcohol and ether. It is readily soluble in CS2.

Monoclinic sulphur is stable above 369K and α-sulphur below 369K, called transition temperature,
both forms are stable. Both exist as S8 molecules. At high temperature (~1000k), S2 is dominant and it
is para magnetic.

SULPHUR DIOXIDE

Preparation:
1. S + O2 SO2 Suplhur burnt in air or oxygen gives SO2 along with a little (6-8%) SO3.
2. Suplhites with dilute sulphuric acid.
SO32-(aq) + 2H+ (aq) H2O(l) + SO2(g)
3. By product of roasting sulphide ores.
4FeS2 + 11O2(g) 2Fe2O3(s) + 8SO2(g)

Properties:
1. Colourless gas with pungent smell.
2. Highly soluble in water.
3. Liquifies at room temp. under a pressure of 2atm and boils at 263K.
4. SO2(g) + H2O(l) H2SO3(aq) (sulphurous acid)
5. NaOH + SO2 Na2SO3 (sodium sulphite) + H2O
Na2SO3 + H2O + SO2 2NaHSO3 (sodium hydrogen sulphite)
6. Reacts with chlorine in the presence of charcoal as catalyst to give sulphuryl chloride.

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SO2(g) + Cl2(g) SO2Cl2(l)
7. Oxidised to sulphur trioxide in the presence of V2O5
2SO2(g) + O2(g) V2O5 2SO3(g)
8. Moist SO2 behaves as reducing agent. Converts Iron(III) ions to Iron(II) ions and
2Fe3+ + SO2 + 2H2O 2Fe2+ + SO42- + 4H+
decolourizes acidified KMnO4 solution
5SO2 + 2MnO4- + 2H2O 5SO42- + 4H+ + 2Mn2+
9. SO2 is angular.

Uses of SO2:
1. In refining petroleum.
2. In bleaching wool and silk.
3. As an anti-colour, disinfectant and preservative.
4. Sulphuric acid, NaHSO3 and Ca(HSO3)2 are manufactured from sulpur dioxide.
5. Liquid SO2 is used as a solvent to dissolve a number of organic and inorganic chemicals.

OXOACIDS OF SULPHUR
Forms oxoacids of the formula H2SO3, H2S2O3, H2S2O4, H2S2O5, H2SxO6 (x=2 to 5), H2SO4, H2S2O7,
H2SO5, H2S2O8.

SULPHURIC ACID
Manufacture:
By contact process. The steps are:
1. Burning of sulphur or sulphide ores in air to generate SO2.
S + O2 SO2
4FeS2 (Iron pyrites) + 11O2 2Fe2O3 + 8SO2
2. Conversion of SO2 to SO3 by the reaction with oxygen in the presence of catalyst V2O5.
2SO2(g) + O2(g) V2O5 2SO3(g) ∆Ho=-196.6KJmol-1
This is the key reaction in the process. High yield of SO3 will lead to more production of H2SO4.
3. Absorption of SO3 in 98% H2SO4 to give Oleum (H2S2O7).
SO3 + H2SO4 H2S2O7
4. Dilution of oleum with water to get desired concentration of sulphuric acid.
H2S2O7 + H2O 2 H2SO4 (Sulphuric acid prepared is 96-98% pure)

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Conditions favouring maximum yield of sulphur trioxide:

Key step in the manufacture of sulphuric acid is catalytic oxidation of SO2 with O2 to give SO3.
2SO2(g) + O2(g) V2O5 2SO3(g) ∆Ho=-196.6KJmol-1
The reaction is exothermic and reversible.
a) Low temperature: Favours oxidation of SO2 as reaction is exothermic (according to Le-
Chateliar’s principle). But it is essential to have a minimum temperature of 720K, to give
maximum yield.
b) High Pressure: Favours oxidation as volume of gaseous products are less. A pressure of 2 to 3
bar is sufficient. Very high may cause corrosion of the vessel.
c) Use of catalyst: V2O5 increases the speed of reaction.

Description of the Plant (Manufacture of H2SO4 by Contact Process):

1. Sulphur burners: Sulphur or Iron pyrites are burnt.

2. Purification unit: The gaseous mixture coming out of sulphur burner is generally impure.
Purified as:
I. Dust Chambers: Steam is introduced to remove dust.
II. Coolers: The gases are cooled to about 373K by passing them through cooling pipes.
III. Scrubber: Gases are passed into washing tower which dissolves mist and other soluble
impurities.
IV. Drying Tower: A spray of conc. H2SO4 used for drying gases.
V. Arsenic purifier: This chamber contains shelves with gelatinous ferric hydroxide. The
impurities of arsenic oxide are absorbed by ferric hydroxide.
3. Testing Box: The purifies gases are tested by passing a strong beam of light. Impurities present
will scatter the light.
4. Contact Chamber or Converter: Pure gases are then heated to about 723-823K in a pre-heater
and then introduced into contact chamber. It is a cylindrical chamber fitted with iron pipes
packed with catalyst V2O5. In this SO2 is oxidized to SO3.
As the forward reaction is exothermic, the pre-heating is stopped once the oxidation has started.

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5. Absorption tower: It is a cylindrical tower packed with acid proof flint. SO3 escaping out from
converter is led from the bottom of the tower and conc. H2SO4 (98%) is sprayed from top. SO3
gets absorbed by H2SO4 to form oleum or fuming sulphuric acid.
H2SO4 + SO3 H2S2O7 (oleum)
Oleum is then diluted with calculated amount of water to get acid of desired concentration.
H2S2O7 + H2O 2H2SO4

PROPERTIES OF H2SO4:
1. Colourless, dense and oily liquid.
2. Conc. H2SO4 dissolves in water with evolution of large quantity of heat. Hence conc. H2SO4 must
be added slowly in water with constant stirring.
3. Chemical reactions of H2SO4 are as a result of:
a) Volatility
b) Strong acidic character
c) Strong affinity of water
d) Ability to act as oxidising agent
4. In aqueous solution, sulphuric acid ionizes as:
H2SO4(aq) + H2O(l) H3O+(aq) + HSO4-(aq) Ka1=very large (>10)
H2SO4-(aq) + H2O(l) H3O+(aq) + SO42-(aq) Ka2=1.2x10-2
Larger Ka1 means – easily dissociated into H+ and HSO4- and is a stronger acid.
5. Forms two types of salt:
Normal sulphates (e.g. sodium sulphate, copper sulphate)
Acid sulphates (e.g. hydrogen sulphate)
6. Can be used to form more volatile acids from their corresponding salts.
2MX + H2SO4 2HX + M2SO4 (X=F,Cl, NO3)
7. Acts as dehydrating agent. Can dry gases and also removes water from organic compounds.
C12H22O11 H2SO4 12C + 11H2O
8. Hot conc. H2SO4 is a strong oxidising agent. Both metals and non-metals are oxidized and itself
is reduced to SO2.
Cu + 2H2SO4 (conc.) CuSO4 + SO2 + 2H2O
3S + 2H2SO4 (conc.) 3SO2 + 2H2O
C + 2H2SO4 (conc.) CO2 + 2SO2 + 2H2O

USES:
1. Important industrial chemical.
2. Used in manufacture of fertilizers (e.g. ammonium sulphate, superphosphate)
3. Petroleum refining
4. Paints, pigments and dyestuff
5. Detergent industry
6. Metallurgical applications
7. storage batteries
8. Laboratory reagent
9. In the manufacture of nitrocellulose products

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GROUP 17

F, Cl, Br, I, At (radioactive) are called halogens. Most reactive non-metals and most electronegative.

F2: Pale yellow gas

Cl2: Greenish yellow gas
Br2: Reddish brown liquid, sparingly soluble in H2O
I2: Lustrous, grayish black solid, sublimes on heating sparingly soluble in H2O, soluble in org.
solvents.

Bond Energy:

F2< Cl2> Br2> I2

Anomalous behavior of F2 is due to the fact that the lone pair of electrons are very close which lead
to repulsions.

Electron Affinity:

F < Cl > Br > I

This is because F being small in size repels the incoming electron.
Occurs as:
F – Fluorides
Cl – NaCl in oceans
Br – Bromides
I – Iodides and iodates (IO3-)

Oxidizing power F2> Cl2> Br2> I2

Radii F < Cl < Br < I
I.E F > Cl > Br > I
E.N F > Cl > Br > I

Preparation of Chlorine:

Cl2:
*Electrolysis of natural brine (NaCl)
* Deacon’s process: Oxidation of HCl(g) in presence of CuCl2 at 723K
4HCl(g) + O2 CuCl2 2Cl2 + 2H2O
By oxidation of HCl by MnO2 or KMnO4 (Lab Method)
MnO2 + 4 HCl MnCl2 + Cl2 +2H2O
2KMnO4 + 16HCl 2KCl + 2MnCl2 + 8H2O + Cl2
By oxidation of NaCl
2NaCl + 2H2SO4 + MnO2 Na2SO4 + MnSO4 + 2H2O + Cl2

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Trends in Group 17:

F is the most electronegative element i.e. has good acceptance of an electron and is
F2
therefore the strongest oxidizing agent.
Cl2
Br2
Oxidizing character decreases down the group. Thus one halogen oxidises halide
I2
ions of higher atomic number halogens.

Oxidizing Action:
F2 > Cl2> H2SO4> Br2> I2

F2 + 2X- 2F- + X2 (X=Cl, Br, I)

Cl2 + 2X - 2Cl + X2 (X=Br, I)
-

Br2 + 2I - 2Br- + I2
Similar reactions as above cannot be used for the preparation of HBr and HI as H2SO4 is a stronger
oxidizing agent and will oxidize the HBr, HI formed to BR2, I2 respectively.
NaBr + H2SO4 HBr Br2
NaI + H2SO4 HI I2
The decreasing oxidizing ability of the halogens in aqueous solution is evident from their standard
electrode potentials which is dependent on the dissociation enthalpy of X2(g). electron gain enthalpy
of X(g) and hydration enthalpy to form X-(aq).
The relative oxidizing power of halogens can be further illustrated by their reaction with water:
2F2(g) + 2H2O(l) 4H+(aq) + 4F- + O2(g)
X2(g) + H2O(l) HX(aq) + HOX(aq) ( Where X= Cl or Br)
I can be oxidized with O2 in acidic medium:
-

4I-(aq) + 4H+(aq) + O2(g) 2I2(s) + 2H2O(l)

Reverse reaction is true for F.
HX are colourless, irritating gases.
HF has a higher b.p due to H-bonding.

Reactivity of halogens with metals or non-metals decreases down the group. F2> Cl2> Br2> I2
e.g. Mg + Br2 MgBr2
Xe + 3F2 XeF6
Ionic character of M-X bond decreases down the group. M-F > M-Cl > M-Br > M-I
Low O.S of M MCl2
High O.S of M MCl4 more covalent than MCl2

Acidic Character:
HF << HCl < HBr < HI
Low acidic character of HF is due to strong H-bonding and higher bond dissociation enthalpy. HF is
corrosive and attacks glass.

NF3 is an exothermic compound but BCl3 is endothermic because bond energy of F2 is lower than Cl2
and N-F bond is smaller and stronger than N-Cl bond.

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Stability:
Decreases down the group due to decreased bond dissociation enthalpy.
HF > HCl > HBr > I

Reducing Nature:
HF < HCl < HBr < HI

Oxides:
Halogens form many oxides with oxygen. Fluorine forms only OF2 and O2F2 – oxygen fluorides, they
are good fluorinating agents.
Pu + 3O2F2 PuF6 + 3O2 (removed as PuF6)
O2F2 oxidises Pu to PuF6 and the reaction is used in removing Pu as PuF6 from spent nuclear fuel.

Chlorine, Bromide and iodine form oxides where o.s ranges from +1 to +7. The higher oxides of
halogens tend to be more stable than the lower ones.

Chlorine oxides, Cl2O, ClO2, Cl2O6 and Cl2O7 are highly reactive oxidizing agents. ClO2 is used as
bleaching agent for paper pulp and textiles and in water treatment.

Bromine oxides Br2O, BrO2, BrO3 are least stable halogen oxides. They are powerful oxidizing
agents.

The iodine oxides, I2O4, I2O5, I2O7 are insoluble solids and decompose on heating. I2O5 is a very good
oxidizing agent and is used in the estimation of CO.

Reactivity towards Metals:

Halogens react with metals to form metal halides. Mg(s) + Br2(l) MgBr2(s)
Ionic character of the halides decreases in the order: MF > MCl > MBr > MI where M is a
monovalent metal.

If metal exhibits more than one O.S, higher O.S halides are more covalent than low O.S halides

Oxoacids of Halogens:
Halic (I) acid HOF HOCl HOBr HOI
(Hypohalous (Hypofluorous (Hypochlorous (Hypobromous (Hypoiodous
acid) acid) acid) acid) acid)
Halic (III) acid – HOCIO – –
(Halous acid) – (chlorous acid) – –
Halic (V) acid – HOCIO2 HOBrO2 HOIO2
(Halic acid) – (chloric acid) (bromic acid) (iodic acid)
Halic (VII) acid – HOCIO3 HOBrO3 HOIO3
(Perhalic acid) – (perchloric acid) (perbromic acid) (periodic acid)

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Inter Halogen Compounds:

When two different halogens react with each other, interhalogen compounds are formed. E.g. XX′,
XX’3, XX’5 and XX’7 where X is halogen of larger size and X′ of smaller size and X is more
electropositive than X′.

Preparation:
They are prepared by the direct combination or by the action of halogen on lower interhalogen
compounds. For e.g.,

Properties:
XX′ ClF, BrF, IF (unstable), BrCl, ICl, IBr
ClF3 (Bent T-Shaped,, because it is sp3d hybridized)
XX’ BrF3 (Bent T-Shaped,, because it is sp3d hybridized)
3 IF3 (Bent T-Shaped,, because it is sp3d hybridized)
ClF3 (Bent T-Shaped,, because it is sp3d hybridized)

IF5 (Square pyramid)

XX’ BrF5 (Square pyramid)
5 ClF5 (Square pyramid)

XX’ IF7 (Pentagonal bipyramidal)

7

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1. They are all covalent and diamagnetic in nature. They are volatile solid or liquid except ClF
which is a gas.
2. They are more reactive than halogens because X-X’ bond in inter halogens is weaker than X-X
bond in halogens except F-F bond.
3. All these undergo hydrolysis giving halide ion.
XX' + H2O → HX' + HOX the bigger one forms hypohalous acid.

Uses:
The compounds are used as non aqueous solvents. They are useful fluorinating agents. ClF3 and
BrF3 are used for the production of UF6 in the enrichment of 235U.
U(s) + 3ClF3(l) → UF6(g) + 3ClF(g)

Properties of Chlorine:
1. Greenish yellow gas with pungent and suffocating odour.
2. Reacts with metals and non-metals to form chlorides.
2Al + 3Cl2 2AlCl3
2Na + Cl2 2NaCl
2Fe + 3Cl2 2FeCl
P4 + 6Cl2 4PCl3
S8 + 4Cl2 4S2Cl2

3. Can react with hydrogen to form HCl.

H2 + Cl2 2HCl
H2S + Cl2 2HCl + S
C10H16 + 8Cl2 16HCl + 10C
4. With excess of NH3, chlorine gives nitrogen and ammonium chloride while with excess of
chlorine, nitrogen trichloride is formed.
8NH3 + 3Cl2 → 6NH4Cl + N2; NH3 + 3Cl2 → NCl3 + 3HCl
(excess) (excess)
5. With cold and dilute alkalies chlorine produces a mixture of chloride and hypochlorite but with
hot and concentrated alkalies it gives chloride and chlorate.
2NaOH + Cl2 → NaCl + NaOCl + H2O
(cold and dilute)
NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O
(hot and conc.)
6. With dry slaked lime it gives bleaching powder.
2Ca(OH)2 + 2Cl2 → Ca(OCl)2 + CaCl2 + 2H2O
The composition of bleaching powder is Ca(OCl)2.CaCl2.Ca(OH)2.2H2O.
7. Chlorine reacts with hydrocarbons and gives substitution products with saturated hydrocarbons
and addition products with unsaturated hydrocarbons. For example,
CH4 + Cl2 UV CH3Cl + HCl
C2H4 + Cl2 Room temp. C2H4Cl2
8. Chlorine water on standing loses its yellow colour due to the formation of HCl and HOCl.

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9. It oxidises ferrous to ferric, sulphite to sulphate, sulphur dioxide to sulphuric acid and iodine to
iodic acid.
2FeSO4 + H2SO4 + Cl2 → Fe2(SO4)3 + 2HCl
Na2SO3 + Cl2 + H2O → Na2SO4 + 2HCl
SO2 + 2H2O + Cl2 → H2SO4 + 2HCl
I2 + 6H2O + 5Cl2 → 2HIO3 + 10HCl
10. It is a powerful bleaching agent; bleaching action is due to oxidation.
Cl2 + H2O → 2HCl + O
Coloured substance + O → Colourless substance

Uses:
1. Used for bleaching woodpulp, bleaching cotton and textiles.
2. In extraction of gold and platinum.
3. In manufacture of dyes, drugs and organic compounds.
4. In sterilizing drinking water.
5. Preparation of poisonous gases such as phosgene (COCl2), tear gas and mustard gas.

HYDROGEN CHLORIDE:
Preparation:

Prepared by heating sodium chloride with concentrated sulphuric acid.

NaCl + H2SO4 420K NaHSO4 + HCl
NaHSO4 + NaCl 823K Na2SO4 + HCl
HCl gas can be dried by passing through concentrated sulphuric acid.

Properties:
1. Colourless and pungent smelling gas. Easily liquefied.
2. Extremely soluble. HCl(g) + H2O(l) H3O+(aq) + Cl-(aq)
Its aqueous solution is called hydrochloric acid.
3. Reacts with NH3 and gives white fumes. NH3 + HCl NH4Cl
4. 3 parts of conc. HCl and 1 part of con. HNO3 forms aqua regia.
Au + 4H+ + NO3- + 4Cl- AuCl4- + NO + 2H2O
3Pt + 16H+ + 4NO3- + 18Cl- 3PtCl62- + 4NO + 8H2O
5. It decomposes salts of weaker acids, e.g., carbonates, hydrogencarbonates, sulphites, etc.
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
NaHCO3 + HCl → NaCl + H2O + CO2
Na2SO3 + 2HCl → 2NaCl + H2O + SO2

Uses:
1. In manufacture of chlorine, NH4Cl and glucose.
2. In medicine and as laboratory reagent.

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Uses of halogens and its compounds:

● Manufacture of UF6 for nuclear power generation and SF6 for dielectrics.
F2

● Chemicals obtained are chlorofluorocarbons like Freon used as a refrigerant

and polytetra-fluorethylene called Teflon used in nonstick pans.
HF ● Also used in glass industry as an etching agent and in the manufacture of
fluoride salts.

● Used for fluoridation of water, (1ppm of fluoride in drinking ware prevents

NaF tooth decay)

● Used in fluoride toothpastes.

SnF2

● Used for bleaching paper, pulp, textiles.

● Used as disinfectant for sterilizing drinking water.
Cl2 ● Used in production of organic compounds- PVC, chlorinated hydrocarbons,
pharmaceuticals.
● Used in production of inorganic compounds. E.g. HCL, PCl3, NaOCl
C2H2Br2 ● Used as gasoline additive.
Br2 ● Used to make AgBr for photography.
I2 ● Used for preparation of iodoform and KI
NaI,
NaIO3 Added to table salt and is called iodised salt. (Insufficient iodine in the diet leads
or to Goitre.
KI, KIO3

GROUP 18

Known as Noble gases as their valance shell orbitals are completely filled and react with a few
elements under certain conditions.
He 1s2 ● Atmospheric abundance of the noble gases in dry air ~1% by volume
Ne 2s 2p
2 6 of which Ar is major constituent.
Ar 3s 3p ● Helium or Neon is also found in radioactive minerals, e.g. pitchblende,
2 6

Kr 4s24p6 monazite, cleveite.

Xe 5s 5p
2 6 ● Natural gas is commercial source of helium.
Rn 6s26p6 ● Radon is obtained as decay product of 226Ra.
226Ra 222Rn + 4He
88 86 2

Electronic Configuration:
General electronic configuration is ns2np6 except He (1s2). Due to fully filled configuration noble
gases are inactive in nature

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IonizationEnthalpy:
Due to stable electronic configuration, they have high I.E. Decreases down the group due to increase
in atomic size.

Atomic Radii:
Increases down the group due to increase in number of shells.

Electron Gain Enthalpy:

Noble gases have stable electronic configuration, so no tendency to accept electron and hence have
large positive values of ∆egH.

Physical Properties:
1. Monoatomic, colourless, odourless and tasteless.
2. Sparingly soluble in water.
3. Low m.p and b.p due to weak dispersion forces.
4. Can be liquefied at low temperature due to weak forces.

Chemical Properties:
Chemically inert due to:
1. Completely filled electronic config. ns2np6 in their valance shell.
2. High ionization enthalpy.
3. More positive electron gain enthalpy.

Uses of Noble Gases:

Helium:
● Non inflammable and light gas. Used in filling balloons for meteorological observations.
● Used in gas cooled nuclear reactors.
● Liquid He (b.p. 4.2K) is used as cryogenic agent for carrying out various experiments at low
temperatures.
● Used to produce and sustain powerful superconducting magnets which form an essential part of
modern NMR spectrometers and Magnetic Resonance Imaging (MRI) systems.

Neon:
● Used in discharge tubes.
● Used in fluorescent bulbs for advertisement display purpose.

Argon:
● Used to provide an inert atmosphere in high temperature metallurgical processes.
● Used for filling electric bulbs.
● Used in lab for handling substances that are air sensitive.
Kr, Xe:
● Used in light bulbs used for special purposes.

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Energy of Compounds of Noble Gases:

In 1962, Neil Bartlett prepared a red compound O2+PtF6–. He realised that the first I.E of O2=1175
kJmol–1 was almost identical to I.E of Xe=1170 k Jmol–1. He prepared a similar red colour compound
by mixing Xe and PtF6.
Xe + PtF6 Xe+[PtF6]-

After this, a number of compounds of Xe with electronegative elements like fluorine and oxygen
have been synthesised.

Xenon-fluorine compounds

Xenon forms three binary fluorides, XeF2, XeF4 and XeF6 by the direct reaction of elements.

Xe (g) + F2 (g) 673k, 1bar XeF2 (s)

(Xe in excess)
Xe (g) + 2F2 (g) 873k, 7bar XeF4 (s)
(1:5 ratio)
Xe (g) + 3F2 (g) 573k, 60-70bar XeF6 (s)
(1:20 ratio)
XeF6 is also prepared as XeF4 + O2F2 XeF6 + O2 at 143K.

Structure:

Properties:
1. XeF2, XeF4 and XeF6 are colourless crystalline solids.
2. Sublime readily at 298 K.
3. They are powerful fluorinating agents. They react with fluoride ion acceptors to form cationic
species and fluoride ion donors to form fluoroanions.
XeF2 + PF5 → [XeF]+ [PF6]– ; XeF4 + SbF5 → [XeF3]+ [SbF6]–
XeF6 + MF → M+ [XeF7]– (M = Na, K, Rb or Cs)
4. They are readily hydrolysed even by traces of water.
2XeF2 (s) + 2H2O(l) → 2Xe (g) + 4HF(aq) + O2(g)

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Xenon-oxygen compounds:
Hydrolysis of XeF4 and XeF6 with water gives XeO3.
6XeF4 + 12H2O → 4Xe + 2XeO3 + 24HF + 3O2
XeF6 + 3H2O → XeO3 + 6HF
Partial hydrolysis of XeF6 gives oxyfluorides
XeF6 + H2O → XeOF4 + 2HF
XeF6 + 2H2O → XeO2F2 + 4HF
Hydrolysis of XeF6 is not a redox reaction as there is no change in oxidation state.

Structure:

Hands –on / IT Enabled work:

Whole chapter will be done with the help of presentation.

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Assignment

Chapter 6: p -Block Elements

1. (a) Give reasons:

(viii) When Cl2 reacts with excess of F2, ClF3 is formed and not FCl3.
(ix) Dioxygen is a gas while sulphur is a solid at room temperature.
(b) Draw the structures of the following:
(i) XeF4
(ii) HClO3
(iii) H2S2O7
(iv) XeF6

2. (a) Arrange the following in the decreasing order of their reducing character:
HF, HCl, HBr, HI
(b) Complete the following reaction:
(i) XeF4 + SbF5
(ii)NH3 + 3Cl2(excess) –––––>
(iii) XeF6 + 2H2O –––––>
3. Give reasons:
(i) Thermal stability decreases from H2O to H2Te.
(ii) Fluoride ion has higher hydration enthalpy than chloride ion.

4. Assign appropriate reason for each of the following observations:

a) Sulphur in vapour state exhibits some paramagnetic behaviour.
b) H2O is a liquid but H2S is a gas.
c) Hydrogen bonding in Hydrogen fluoride is much stronger than that in water, yet
water has much higher boiling point.
d) The majority of known noble gas compounds are those of Xenon.
e) Halogens are strong oxidants.
f) The value of electron gain enthalpy with negative sign for sulphur is higher than that
for oxygen.
g) ClF3 molecule has a T-shaped structure and not a trigonal planar one.
h) O2 and F2 both stabilize higher oxidation states of metals but O2 exceeds F2 in doing
so.
i) Structures of xenon fluorides cannot be explained by Valence Bond Approach.
j) SF6 is kinetically inert.
k) All the bonds in SF4 are not equivalent..
l) ICl is more reactive than I2.
m) Despite lower value of its electron gain enthalpy with negative sign, F2 is a stronger
oxidizing agent than Cl2.
n) Bond dissociation enthalpy of F2 is lower than that of Cl2.
o) Ozone is thermodynamically unstable.
p) Fluorine forms only one oxoacid HOF.

5. Write the reaction of preparation of XeF4, XeF6, XeO3

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6. Draw the structures of the following:
a) XeF2 b). XeF4 c) XeOF4 d) BrF3 e)XeO3 f) XeOF4 g) SF4 h) H2SO5 i) ClF3
j) H2S2O7 k) H2S2O8 l) BrF5

7. Complete the equations:

a. XeF2 + PF5 →
b. Cl2 + NaOH ( cold and dil.) →
c. F2 (g) + H2O (l) →
d. I-(aq.) + H2O(l) + O3(g) →
e. XeF4 + O2F2 →
f. NaOH (hot & conc) + Cl2→
g. Xe(g) (excess)+ F2(g) 673K, 1bar

8. (i)Compare the oxidizing action of F2 and Cl2 by considering parameters such as bond
dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
(ii)Write the conditions to maximize the yield of H2SO4 by contact process.
(iii) How is SO2 an air pollutant?

9. Predict the shape and the asked angle(90o or more or less) in each of the following cases:
(a) SO32- and the angle O-S-O.
(b) XeF2 and the angle F-Xe-F.

10. Which noble gas is used for filling balloons for meteorological observations?

11. How does supersonic jet aeroplanes responsible for depletion of ozone layers?

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More Practice

Chapter 6: p -Block Elements

1. Assign appropriate reason for each of the followaing observations:

1) Hydrogen fluoride is a much weaker acid than HCl in aqueous solution.
2) SCl6 is not known where as SF6 is known
3) Halogens are coloured .
4) Dioxygen is a gas but sulphur is a solid.
5) Fluorine forms only one oxoacid, HOF .
6) Sulphur has a greater tendency for catenation than oxygen.
7) The lower oxidation statebecomes more stable with increasing atomic number
in group 18.
8) Fluorine exhibits only -1 oxidation state whereas halogens exhibit higher
positive oxidation states also.
9) On addition of ozone gas to KI solution, violet vapours are obtained.
10) Chlorine act as both oxidizing and bleaching properties.
11) Chlorine loses its yellow colour on standing.
12) Bond dissociation energy of F2 is less than that of Cl2.
13) SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed.
14) XeF2 has a straight linear structure and not a bent angular structure.
15) Fluorine does not exhibit any positive oxidation state.
16) Oxygen shows catenation property less than Sulphur.
17) Helium is used in diving apparatus.

2. Describe the steps involved in the contact process for the manufacture of H2SO4.

3. Give Chemical reaction in support of each of the following statements:

a) Bleaching of flowers by Cl2 is permanent while by SO2 is temporary.
b) Fluorine is a stronger oxidizing agent than chlorine.
d) Chlorine reacts with a cold and dilute solution of sodium hydroxide.
e) Orthophosohorous acid is heated.
f) PtF6 and Xenon are mixed together.

4. Arrange the following in the decreasing order of the property indicated:

a. H2O , H2S, H2Se , H2Te boiling point

b. MF, MBr, MCl, MI Ionic character of the bond
d. HF, HCl, HBr, HI Acid strength
e. H2S, H2O, H2Se, H2Te Thermal stability
f. HClO4, HClO3, HClO2, HClO Oxidizing power
g. Cl2 , Br2 , F2 , I2 Oxidizing power
h. HClO4, HIO4, HBrO4 Oxidising ability

5. With what neutral molecule ClO- is isoelectronic? Also give the formula of a
noble gas species isostructrual with ICl-4 and IBr --.

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d-and f-block elements

d-block of the periodic table contains elements of the groups 3-12 in which the orbitals are
progressively filled in each of the four long periods. The elements constituting the f- block are those
in which the 4f and 5f are progressively in the later two long periods; these elements are formal
members of group 3 from which they have been taken outto form separate f-block of the periodic
table.

The transition metals are those elements which have incompletely filled d-subshells in there ground
stateor in any one of their oxidation state.

Cu,Ag and Au are transition metals because in their commonly occuring states, they have partly
filled d-sub shells. Zn, Cd and Hg of group 12 do not have partly filled d-sub shell in their
elementary state or commonly occuring oxidation state, and hence, are not considered as transition
elements. However, being the end members their chemistry is studied with transition elements.

Electronic Configuration- ( n-1)d 1-10 ns 1-2

Transition series- The d-block elements are called transition elements as they represent change in
properties from most electropositive s-block elements to least electropositive (or most
electropositive) p-block elements.
Transition elements consist to of 4 rows: These series are called transition series.

First transition series- 3d series

Sc Ti V Cr Mn Fe Co Ni Cu Zn

21 22 23 24 25 26 27 28 29 30
3d14s 3d 24s2 3d34s2 3d54s1 3d54s2 3d64s2 3d74s2 3d84s2 3d104s1 3d104s
2 2

Second Transition series – 4d series

Y Zr Nb Mo Tc Ru Rh Pd Ag Cd

39 40 41 42 43 44 45 46 47 48
4d15s2 4d25s2 4d45s2 4d5s1 4d65s1 4d75s1 4d85s1 4d105s 4d105s 4d105s
0 1 2

Third transition Series – 5d series

La Hf Ta W Re Os Ir Pt Au Hg
57 72 73 74 75 76 777 78 79 80

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Fourth transition Series- 6d Series

Ac Rf Db Sg Bh Hs Mt Ds Rg Uub
89 104 105 106 107 108 109 110 111 112

GENERAL POPERIES OF TRANSITION ELEMENTS

Atomic radii: The atomic radii is intermediate between those of s-and p- block elements. The

Following trends are observed:

a) The atomic radii of elements of a particular series decrease with increase in atomic number but
this decrease in atomic radii become small after midway.
Reason- The atomic radii decreases in the beginning because of increase in atomic no., the nuclear
charge goes on increasing progressively, but the electrons enter the penultimate shell and the added
d-electrons screen the outermost s-electrons. The shielding effect is small so that net electrostatic
attraction between the nuclear charge and outermost electrons increases. Consequently atomic
radius decreases. As the no. of d-electrons increase the screening effect increases. This neutralizes
the effect of increased nuclear charge due increase in atomic number and consequently atomic
radius remains almost unchanged after chromium.

b) At the end of each period, there is slight increase in atomic number

Reason- This is because of increased electron-electron repulsions between the added electrons in the
same orbital which exceeds the attractive forces due to increased nuclear charge. Therefore, electron
cloud expands and size increase.

c) The atomic radii increase while going down the group.

Reason – The atomic radii of second transition series is larger than that of first transition series
because of increase in no. of outermost shell.
The third transition series have nearly the same radii as metals of second transition series. This is
because of lanthanide contraction. This is associated with the interventions of 4f- orbitals which are
filled before 5d-series. 4f- orbital have poor screening effect, This results in regular decrease in
atomic radii which compensates the expected increase in atomic size with increase in atomic no.

Ionic Radii- The ionic radii follows the same trend as atomic radii. Since metals exhibit different
oxidation states, radii of ions also differ. The ionic radii decrease with increase in nuclear charge.

Metallic character- All transition elements are metals. They have high density, hardness, high m.p.&
b.p. & high tensile strengths, ductility, high, thermal and electrical conductivity and lustre.
Reason – The metallic character is due to their relatively low ionization enthalpies and number of
vacant orbitals in the outermost shell. The hardness of there metals suggests the presence of strong
bonding due to overlap of unpaired electrons between different metal atoms. Therefore, they exhibit
high enthalpy of atomization. Enthalpy of atomization is maximum in the middle indicates one
unpaired electron is particularly favorable for strong atomic interaction. Hence, max enthalpy of
atomization.

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Density- All metals have high density. Within a period, the densities vary inversely with atomic
radii, ie., densities increase along a period. Also densities increase upon descending down the
group.
Reason – The atomic volumes of transition elements are low because electrons are added in (n-1)d
sub shell and not in ns subshell . Therefore, increased nuclear charge is partly screened by the d-
electrons and outer electrons are strongly attracted by the nucleus. Moveover, electrons are added in
inner orbital. Conequently, densities of transition metals are high.

Melting and boiling points

The transition metals have high m.p. & b.p. The m.p. and b.p. of metals rise to a maximum value and
then decrease with increase in atomic number. However, Mn and Tc have abnormally low m.p. &
b.p.

Reason – The high m.p. &b.p. is due to strong metallic bonds between the atoms of these elements.
The metallic bond is formed due to interaction of electrons in the outermost orbital. The strength of
bonding depends on the number of outer most electrons. Greater is the number of valance electrons,
stronger is the metallic bonding and consequently, m.p. is high.

Therefore metallic strength increase up to the middle till d5 confignation and then decreases with the
decrease in availability of unpaired d- electrons(from Fe onwards) . Therefore, m.p. decrease after
the middle because of increase an pairing of electrons.

The dip in the curve in Mn and Tc is due to the fact that Mn has stable electronic configuration (3d5
452). As a result 3d electrons are more tightly held by Mn atomic nucleus and this reduces
delocalization of electrons resulting in weaker metallic bonding.

Ionization Enthalpies- The following trend is observed in the ionization enthalpies of d- block
elements.The ionization enthalpies of d-block elements are lower than those of p-block elements.
The I.E. increase along a series.
Reason - The increase in ionization enthalpy is due to the effect of increasing nuclear charge which
would tend to attract outer electron with greater force. Consequently, ionization enthalpy is
expected to increase. But the addition of electrons take place in last but one d-sub shell and this
increases the screening effect. With the increase in electrons in d-sub shell the outer electrons are

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shielded more & more. Thus, effect of increased nuclear charge is opposed by additional screening
effect of the nucleus and ionization enthalpy increases but slowly.
The irregular trend in first I.E. of first transition series elements is due to the fact that removal of one
electron alters the relatives energies of 4s and 3d orbitals. Therefore, there is reorganization energy
which gains in exchange energy due to increase in no. of electrons in dn configuration and from
transference of s-electrons in d- orbital.

Some exceptions observed in ionization enthalpies are:-

a) Cr and Cu have high I.E. This is attributed to their half filled (d5) and completely (d10)
electronic configuration .
b) The value of second I.E. for zinc is low because ionization involves removal of an electron
resulting in stable 3d10 configuration.
c) The trend in third I.E. shows high value for Mn+2 and Zn+2 because of stable 3d5 and 3d10
electronic configuration.
Similarly, I.E3 for Fe <I.E3 for Mn because of stable 3d5 in Fe. In general, third I.E. values are very
high because of filled 4f-orbitals which have poor shielding effect.

Oxidation States-

a) Transition metals exhibit a larger number of oxidation states in their compound.

Reason - This is because of participation of inner (n -1)d electrons in addition to outer ns-
electrons because the energies of ns and ( n -1)d electrons are almost equal.
The elements which gave the greatest number of oxidations states occur in or near the middle of the
series. Eg.Mn ; oxidation states from + 2 & +7

The lesser number of oxidation states at the extreme ends is either due to too few electrons to lose or
share or too many d electrons, hence fewer orbital are available in to available to share electrons
with others, thus higher valence cannot be attained.
eg. Cu can have oxidation state of +1& +2
Zn can have oxidation state of +2 only

Oxidation state of first row transition elements

Sc Ti V Cr Mn Fe Co Ni Cu Zn

21 22 23 24 25 26 27 28 29 30
3d14s2 3d 24s2 3d34s2 3d54s1 3d54s2 3d64s2 3d74s2 3d84s2 3d104s 3d104s
1 2

+3 +2 +2 +2 +2 +2 +2 +2 +1 +2
+3 +3 +3 +3 +3 +3 +3 +2
+4 +4 +4 +4 +4 +4 +4
+5 +5 +5
+6 +6 +6
+7

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b) Except Sc, the most common oxidation state of first rose correlation elements is +2 which arises
due to loss of 4s- electrons. This means Sc, 3d–orbitals are more stable and lower in energy than
4s- orbital. As a result electrons are first removed from 4s- orbital.
c) Zinc is +2 in which no d-electrons are involved.
d) Within a group, maximum oxidation state increases with atomic number. For e.g. in group 8 , Fe
shows +2 and +3 but Rutheniun and Osmium form compounds in +4 , +6 and + 8 oxidation state
.
e) In + 2 and + 3 oxidation state, bonds formed are ionic and in the compounds of higher oxidation
state bonds are covalent.
Transition elements show low oxidation states in some compounds or complexes having ligands
such as CO, for e.g. in Ni (CO)4 , Ni has zero oxidation state.

Standard electrode potential

The magnitude of ionization enthalpy gains the amount energy required to remove electrons from
particular oxidation state of metal in compounds. Smaller the I.E. the metal, the stable is its
compound.
Ni → Ni+2 + 2e- I.E = 2.49 X 103 kJ/mol
Pt → Pt+2 + 2e- I.E = 2.66 X 103 kJ/mol

Ni(II) compounds are thremodynamically more stable thaPt(II) compounds.

Ni → Ni+4 + 4e- I.E = 8.80 X103 kJ/mol

Pt → Pt+4 + 4e- I.E = 6.70 X 103 kJ/mol

Pt(IV) compounds are relatively more stable than Ni(IV) compounds. Stability of the compounds
depend on electrode.
In addition to ionization enthalpy, ∆Hsub , ∆Hhyd energy, explain the stability of a particular oxidation
state.

1) M(s) M(g) ∆subH0

2) M(g) M+(g) + e- 1. E ∆H= ∆sub H + I. E + ∆hyd H
3) M (g)+nH2O
+ M (aq) ∆hyd H
+

Smaller the value of total energy charge for a particular oxidation state in aqueous solution, greater
will be the stability of that oxidation state. The electrode potential is a measure of total energy
charge.

The lower the electrode potential, ie, more negative the standard reduction potential of the electrode,
more stable is the oxidation state of the transition metal in aqueous solution.
More negative values of E0 for Mn and Zn are due to the stability of half filled (3d5 ) in Mn+2 and
completely filled (3d10 ) configuration in Zn+2 .

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Trend in M3+ / M2+ standard electrode potential

Except copper and Zinc, all other elements of first transition series show +3 oxidation states to from
M+3 in aqueous solution.
a) Low value of Sc reflects stability of Sc+3 which has a noble gas configuration.
b) High value for Mn shows that Mn+2 (d5) is particularly stable.
c) Low value for Fe, shows extra stability for Fe+3 configuration.

Trends in stability of higher oxidation states-

The highest oxidation state is generally shown among halides and oxides.
a) Transition metals react with halogens at high temp as they have high activation energies.
High temp is required to start the reaction but heat of reaction is sufficient to continue. The
reaction.
Order of reactivity: F2> Cl2> Br2> I2
b) In general elements of I transition series react in low oxidation state.
c) Since fluorine is the most electronegative element, the transition elements show high
oxidations states with fluorine.
d) The highest oxidation states are found in TiX4, VF5, CrF6.
e) The +7 oxidation states are not shown by simple halides.
f) V(V)is shown by VFs only . Other halides undergoing hydrolysis form oxo halides OX3.
g) Fluorides are unstable in their low oxidation state. Eg –V forms VX2 (X = C1,Br or I) Cu can
form CuX (X=Cl ,I) Cu (II) halides are known except the iodide.
h) The ability of oxygen to stabilize the highest oxidation state is exhibited in their oxides. The
highest oxidation states in member of group number. Eg-Sc in Sc2O3 is +3 and is a member of
group 3. Mn in group7 has +7 oxidation state in Mn2O7.Mangnese forms highest oxidation
state fluorides as MnF4 whereas the highest oxide is Mn2O7.Tthis is due to tendency of
oxygen to form multiple bonds. In the covalent oxide Mn2O7 ,each Mn is tetrahedrally
surrounded by oxygen atoms and has Mn-O-Mn bridge. Tetrahedral MnO4-2 ions are also
known for V(V),Cr(VI), Mn(VI) and Mn(VII).

Formation of colored ions-

Most of the compo undo of transition metals are colored in solid or solution form.
Reason- The colour is due to the presence of incomplete (n-1)d sub shell. Under the influence of
approaching ions towards central metal ion, the d-ordinals of central metal split into different
energy levels. This phenomenon is called crystal field splitting. For e.g. When six ions or molecules
approach the metal ion (octahedral field) , the d-orbitals split into two sets:- One set consisting of
two d-orbital of higher energy (dx2-y2, dz2) end other set consisting of d-orbitals (dxy, dyz & dxz ) of
lower energy. The electrons are easily promoted from one to another energy level in the same d-sub
shell. There are called d-d transition. The amount of energy required to excite some of the electrons
to higher energy states within the same d-sub shell corresponds to energy of certain colours of
visible light. Therefore, when white light falls on the compounds, some part of its energy
corresponding to certain colour is absorbed and the electron gets raised from lower energy to higher
energy & the excess colour is transmited. The observed colour is complementary of colour absorbed.
Eg-Ti+3 (d1) is purple.

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Magnetic Properties-
Paramagnetism arises from the presence of unpaired electrons. Diamagnetic substances are repelled
by the applied magnetic field while the paramagnetic substances are attracted whereas the
ferromagnetic substances are attracted the most.
Each unpaired electron has a magnetic moment associated with its spin angular momentum and
orbital angular momentum. For the compounds of the 1st transition series, the contribution of orbital
angular momentum is effectively less and hence is of no significance. For these, the magnetic
moment is determined by the number of unpaired electrons and can be calculated by ‘spin only’

µ = √ n(n+2) BM
N= no. of unpaired e-
µ = Magnetic moment in Bohr magnetron (µb ) units
(µb =eh/ 4m = 9.27x10-24 Am2 or Jt -1)
µ increase with the increasing number of unpaired electrons or in other words observed magnetic
moment guise a useful indication about the number of unpaired electrons.
Eg- Calculate the magnetic moment of Mn2+ if the at no. =25 ,
Z=25 So, d5 has 5 unpaired electrons , n=5
µ= 5(5+2) =5.92µb

Formations of complex compounds :- The transition metals form a large no. of complex compo
undo due to
(i) the comparitively smaller sizes of the metal ions,
(ii) their high ionic charges and
(iii) the availability of d-orbital for bond formation
Eg .[PtCl4]2- , [Cu(NH3)4], [Fe(CN)6]4– etc.

Catalytic Properties-
a) Transition metals show catalytic property because of their ability to adopt multiple oxidation
states. Catalysts at a solid surface involve the formation of bond 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 & the
activation energy is lowered, moreover transition metals can change their oxidation states.
Eg- Fe3+ catalyses the reaction between I2& persulphate ions.
b) The catalytic property of transition metals is due to their tendency to form reaction intermediates
with suitable reactants. These intermediates give reaction paths of low activation energy and
therefore increase the rate of reaction. The reaction intermediates decompose yielding products
and regenerating the original substance. The transition metals form reaction intermediates due to
the presence of vacant orbitals & tendency to form variable oxidation state.

Formation of interstitial compounds :-

Many of the transition metals form interstitial compounds which are formed when small atoms like
B,H, N or C are trapped inside the crystal lattices of metals. They are usually non-stoichiomatric and
are neither typically ionic nor covalent. There small atoms enter into the void sites, eg In, Ti. If C
enters the void going the composition TiC or TiH1.7, VH0.56 etc.

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Physical & Chemical characteristics of these compounds
(i) High m.pt, higher than pure metals.
(ii) Very hard , some borides approach diamond in hardness
(iii) Retain metallic conducting
(iv) Chemically inert

Alloy Formation:-
Alloys are homogenous solid solutions in which the atoms of one metal are distributed randomly
among the atoms of the other but the metals should have similar metallic radii within 15% of each
other.
The alloys formed have high m.pt & are hard.
Eg. Alloys of Cr, V, W, Mo, Mn etc, stainless steel is or alloys of Fe, Ni, Cr
Alloys of transition metals with non-transition metals, such as Brass (Cu-Zn), Bronze (Cu- Sn).

Some important compounds of transition elements

Oxides & oxo metals ions
Metal + O2high/ temp MxOy
The higher oxidation state in the oxides coincides with the group no. eg, Sc2O3(Sc is +3),
Mn2O7(Mn is +7).
Beyond group 7- no higher oxides. Eg- Fe2O3(Fe is +3)
Besides the oxides, oxocations, stabilize V(V) as VO2+, V(IV) as VO2+ and Ti(IV) as TiO2+.
As the oxidation number of metal increases, ionic character decreases .
Mn2O7 is a covalent compound which is green oil.
Mn2O7 gives HMnO4 Acids in high oxidation state.
CrO3 gives H2Cr2O7
V2O5 is amphoteric, V2O3 is basic and V2O4 is less basic.
When dissolved in acidic salts it gives VO2+ salts.
CrO is basic and Cr2O3 is amphoteric. (High oxidation states are more covalent and more
acidic).

Potassium Dichromate , K2Cr2O7 –

Preparation - From Chromite ore
Chromates in turn are formed by fusion of Chromite ore( FeCr2O4) with Na2CO3 or K2CO3.
4 FeCr2O4 + Na2CO3 + 7O2 8Na2CrO4 + 2Fe2O3 + 8CO2.
Excess

Na2CrO4 is filtered and treated with H2SO4 to obtain orange crystals of Na2CrO7.2H2O
Sodium dichromate is more stable than pot. dichromate
Na2Cr2O7 + 2KCl K2Cr2O7 + 2NaCl
Chromates and dichromates are interchangeable in aqueous solution depending upon pH of the
solution.
The O.S. of Cr in CrO42- and Cr2O72- is same.
2CrO42- + 2H+ Cr2O72- + H2O
(+6) (+6)

Cr2O72- + 2OH- 2CrO42- + H2O

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Chromate ion

Dichromate ion

Na2Cr2O7& K2Cr2O7 : Strong oxidizing agents

Used in Used in volumetric

Organic analysis
Chemistry

Chemical properties of K2Cr2O7 :-

In acidic solution, its oxidizing action can be represented as follows –
Cr2O72- + 14H+ + 6e-→ 2Cr3+ + 7H2O
Acidified K2Cr2O7 oxidises iodides to iodine, sulphide to S, Sn(II) to Sn(IV), Fe(II) to Fe(II) to Fe(III)
6I- 3I2 + 6e-
3 H2S 6H+ +3S + 6e-
3Sn2+ 3Sn4+ +6e-
6Fe2+ 6Fe3+ 6e-

The full ionic equation can be obtained by adding half equation for potassium dichromate to half
equation for the reducing agent, for eg.,
Cr2O72- + 14H+ + 6Fe2 2Cr3+ + 6Fe3+ + 7H2O
Uses – In leather industry, preparation of azo dyes.

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Potassium Permanganate (KMnO4)
Preparation – KNO3
2MnO2 + 4KOH + O2 ---------→ 2K2MnO4 + 2H2O
Oxidizing agent

3MnO42- + 4H+ 2MnO4- + MnO2 + 2H2O

(+6 (+7) (+4)

Commercially prepared by alkaline oxidative fusion of MnO2 followad by the electrolytic oxidation
of Manganate (VI)
Fuse with KOH
MnO2---------------------→ MnO42-
Oxidize with air or KNO3
Electrolytic oxidation

MnO42- -----------------------→ MnO4-

in alkaline solution
In the laboratory manganese (II) ion salt is oxidized by peroxodisulplate to permanganate
2Mn+2 + 5S2O82- + 8H2O 2MnO4- + 10SO42- 16H+

Properties-
1. Forms dark purple crystals
2. Not very soluble in water
3. Decomposes when heated at 513K
2KMnO4 K2MnO4 + MnO2 + O2

Tetrahedral manganese (green)ion Tetrahedral Permanganate (purple) ion

MnO4 is a strong oxidizing agent, both in neutral & acidic medium

Acidified KMnO4 oxidises oxalates to CO2, Iron(II) to iron (III), nitrites to nitrates and iodides to free
iodine The half reaction of reductants are-
COO-
5 10CO2 + 10e-
COO-
Fe+2 5Fe+3 + 5e-
5NO2- + 5H2O 5NO3- + 10H+ + 10e-
10I- 5I2 + 10e-

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Full reactions can be written by adding the half reactions of KMnO4 to half reactions of the reducing
agents and balancing them.

Acidic solutions –
10I- + 2MnO4- + 16H+ 2Mn+2 + 8H2O + 5I2 (iodides to KI)
5C2O42- + 2MnO4- + 16H 2Mn+2 + 8H2O + 10CO2 (oxalate ions to CO2)
5Fe2+ + MnO4- + 8H Mn+2 + 4H2O + 5Fe2+ (Fe2+(green) to Fe3+ (yellow))
5S2- + 2MnO4- + 16H+ 2Mn+2 + 8H2O + 5S (H2S to S)
5SO32- + 2MnO4- + 6H+ 2Mn+2 + 3H2O + 5SO42 (sulphites to sulphates)
5NO2- + 2MnO4- + 6H+ 2Mn+2 + 3H2O + 5NO3- (nitrites to nitrates)

Neutral medium -
2MnO4- + H2O + I- MnO2 + OH- + IO3- (iodides to iodates)
8MnO4- + H2O + 3S2O32- 8MnO2 + 2OH- + 6SO42 (thiosulphates to sulphates)
MnO4- + 2H2O + 3Mn2+ 4H+ + 5MnO2 (Manganese salt to Mn2+)

Uses- Used as uxidant, used for bleaching wool, cotton, silk and decolorization of oils.

The Inner Transition elements (f- Block)

Consists of two series:-
Lanthanoides (Ln; general Symbol) Actinoides
(14elements following La ) (14 elements following Ac)

Lanthanoides :-

Electronic Configuration:-

Atomic No. Name Symbol E.C.

57 Lanthanum La 5d1 6s2

58 Cerium Ce 4f15d16s1
59 Praseodymium Pr 4f3 6s2
60 Neodymium Nd 4f4 6s2
61 Promethium Pm 4f56s2
62 Samarium Sm 4f6 6s2
63 Europuim Eu 4f7 6s2
64 Gadolinium Gd 4f7 5d1 6s2
65 Terbium Tb 4f9 6s2
66 Dysprosium Dy 4f10 6s2
67 Holmium Ho 4f11 6s2
68 Erbium Er 4f12 6s2
69 Thulium Tm 4f13 6s2
70 Ytterbium Yb 4f14 6s2
71 Lutetium Lu 4f14 5d1 6s2

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Atomic & Ionic Series :- Decrease from La to Lu is due to Lanthanoid Contraction (The shielding of
one 4f electron by another less than that by one d electron by another & the increase in nuclear
charge along the series.

The almost identical radii of Zr (160pm.) and Hf (159), a consequence of the lanthanoid contraction,
account their occurrence together in nature and for the difficulty faced in their separation.

Colour and Para magnetism

Ln3+ are coloured both in solid and in aqueous solution due to the presence of f electrons.
La3+ and Lu3+ do not show any colour. However absorption bands are narrow probably because of
the excitants with in f level.
Ln3+ are paramagnetic except La3+& Ce4+ (f0 type) &f14 type ( Yb2+& Lu3+ ). Paramagnetism rises to
maximum in Neodymium .

Ionization enthalpies
I.E. depends on the degree of stability of empty, half filled and completely filled f-level.
This is indicated from the abnormally low values of the third ionization enthalpy of La, Gd, Lu.

Oxidation states.:-
Ln3+ compounds are predominant species. +2 & +4 ions in solution or in solid compounds are also
obtained occasionally.
Ce (IV) formation is favoured due to extra stability of noble gas configuration, but it is a strong
oxidant reverting to the common +3 state .
Pr, Nd, Tb and Dy also exhibit +4 state but only in oxides, MO2
Eu2+ is formed by losing the two s- electrons & its +7 configuration.

Properties and use :-

Ln are silvery white soft metals and tarnish rapidly in air. Hardness increases with increasing atomic
number. M.pt. ranges from 1000K – 1200K
Sm is steel hard. (m.pt 1623K)

Chemical behavior
In general earlier members of the series are quite reactive similar to Ca, with increasing atomic
number they behave more like aluminum.

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Ln3+ (aq) +3e- Ln(s)

Ln2O3 H2

Burns in with acids

O2

Heated with S with halogens

Ln2S3 Ln LnX3

Heated with H2O

N2
Ln (OH)3 +H2
Ln N C 2773K

Ln C2
∆
Ln + C ---------→ Ln3C + Ln2C3 + LnC2 (carbides)

Use :- Used in the production of alloy steels for plates & pipes eg. Mischmetal is an alloy which
consists of a lanthanoid metal (95%) and iron (5%) and trace of S, C, Ca and Al. A good amount of
this allay is used in Mg-based alloy to produce bullets, shell & lighter flint.

● Mixed oxides of Ln are used as catalyst in petroleum cracking .

● Ln oxides are used as phosphors in television screens & similar fluorescing surfaces.

The Actinoids:-

Atomic no. Name Symbol E.C.

89 Actinium Ac 6d17s2
90 Thorium Th 5f06d27s2
91 Protactinium Pa 5f26d17s2
92 Uranium U 5f36d17s2
93 Neptunium Np 5f46d17s2
94 Plutonium Pu 5f67s2
95 Americium Am 5f77s2
96 Curium Cm 5f76d17s2
97 Berkelium Bk 5f97s2
98 Californium Cf 5f107s2
99 Einsteinium Es 5f117s2
100 Fermium Fm 5f127s2
101 Mendelevium Md 5f137s2
102 Nobelium No 5f147s2
103 Lawrencium Lr 5f146d17s2

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Although the naturally occurring elements & the earlier member have relatively long half lives, the
latter members have values ranging from a day to 3 minutes for Lr (Z=103) These facts and high
radioactivity renders their study more difficult.

Electronic Configuration
The irregularities in the electronic configurations of the actinoids like those of in the lanthanoids are
related to the stability of fo ,f7 and f14 occupancies of the 5f orloitals.
Eg.Am : [Rn] 5f77s2
Cm : [Rn] 5f76d17s2

5f orbital can & do participate in bonding.

Common oxidation state is +3
The maximum oxidation state increases from +4 in Th, +5 in Pa, +6 in U and +7 in Np but decreases
in succeeding elements.

Magnetic Properties:- the variation of magnetic properties with the no. of unpaired 5f electrons is
similar to that of Ln.

Ionic Sizes: Decrease in size due to increase in the effective nuclear charge on the outermost shell
and poor shielding by 5f electrons. This is referred to as actinoid contractions.

Ionization Enthalpy :
The I.E. of early actinoids is lesser than that of early Ln as when 5f orbitals are beginning to be
occupied, they will penetrate less into the inner core of electrons. The 5f electrons, will therefore, be
more effectively shielded from the nuclear charge than are the 4f electrons of the corresponding Ln.
Because the outer electrons are less firmly held, they are available for bonding in the actinoids.

Physical and Chemical Reactivity

The actinoids are highly reactive when they are finely divided.

Actinoid Boiling water MxOy + MHn

moderate temp
Actinoid + Non metal Corresponding compound
Actinoid + HCl MxOy oxide layer
Actinoid + HNO3 MxOy
Actinoid + Alkali No reaction

Metallic radii of actinoids is more as compared to lanthanoids.

Comparison With Lanthanoids

1. Structural variability in actinoids is obtained due to irregularities in metallic radii which are
greater then lanthanoids.
2. Magnetic properties in actinoids are more complex than lanthanoids.

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3. Ionization enthalpies of early actinoids, though not accurately known are lower than early
lanthanoids. This is because 5f electrons penetrate less into the inner core and hence the
outer electron are less firmly held, they are available for bonding in actinoids.

Applications of d-and f Block Elements

1 Iron and steel are important construction materials. Their production is based on reduction
of iron oxides, removal of impurities, and addition of carbon and alloying metals such as Cr,
Mn, and Ni.
2 TiO is used in pigment industry.
3 MnO2 is used in battery cell. Also Zn and Ni/Cd.
4 Elements of group II are coinage metals.
5 V2O5 catalyses oxidation of SO2 in contact process,
6 Iron catalyst is used in Haber’s process.
7 TiCl4 and Al(CH3)3 forms Ziegler-Natta Catalyst.
8 Ni complexes are used in polymerization of alkynes.

Hands –on / IT Enabled work:

Whole chapter is done with the help of a presentation.

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Assignment

Chapter 7: d -and f -Block Elements

1. Complete and balance the following equations:

(i) Fe2+ + MnO4- + H+ →

(ii) MnO4- + H2O + I- →

2. Give reason:

(i) E0value for the Mn3+/Mn2+ couple is highly positive as compared to Fe3+/Fe2+ .
(ii) Iron has high enthalpy of atomization than that of copper.
(iii) Sc3+ is colourless in aqueous solution whereas Ti3+ is coloured.
(iv) Actinoid contration is greater from element to element than lanthanoid
contraction.

3. Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation
state equal to its group number.
4. Account for the following :
(i) Transition metals form large number of complex compounds.
(ii) The lowest oxide of transition metal is basic whereas the highest oxide is
amphoteric or acidic.
(iii) E° value for the Mn3+/Mn2+ couple is highly positive (+1.57 V) as compare to
Cr3+/Cr2+.

5. (i) How is the variability in oxidation states of transition metals different from
that of the p-block elements ?
(ii) Out of Cu+ and Cu2+, which ion is unstable in aqueous solution and why ?
(iii) Orange colour of Cr2O72– ion changes to yellow when treated with an alkali.Why?

6. Describe the general characteristics of transition elements with special reference to the
following :
(i) Variable oxidation states
(ii) Formations of coloured ions.

7. (i) What are interstitial compounds? Why are such compounds well known for
transition elements?
(ii) What are alloys? Name an alloy which contains a lanthanoid metal.
8. How is it that several transition metals act as catalysts? Give two examples of
reactions catalyzed by them?
9. Complete the following reactions:
(a) Cr2O72-(aq.) + I-(aq.) + H+
(b) MnO4_ (aq.) + Fe2+(aq.) + H+(aq.)
(c) MnO4_ (aq.) + S2O32-(aq.) + H2O(l)
(d) 5NO2-(aq.) + 2MnO4-(aq.) + 6H+(aq.)

(e) Cr2O72-(aq.) + H2S(g) + H+(aq.)

(f) Cr2O72-(aq.) + Fe2+(aq. )+ H+(aq.)
(g) MnO2(s) + KOH(aq.) + O2

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(h) 2CrO42- + 2H+
(i) Cr2O72-(aq.) + 3Sn2+ + 14 H+

10. (a)Compare the chemistry of actinoids with that of lanthanoids with reference to
(i) electronic configuration (ii) oxidation state
(iii) atomic sizes (iv) chemical reactivity
(b) Chemistry of actinoids is complicated as compared to lanthanoids. Give two
reasons.
(c) Write one similarity and one difference between the chemistry of lanthanoid and
actinoid elements.

11. When chromite ore FeCr2O4 is fused with NaOH in presence of air, a yellow coloured
compound (A) is obtained which on acidification with dilute sulphuric acid gives a
compound (B). Compound (B) on reaction with KCl forms a orange coloured
crystalline compound (C).
(i) Write the formulae of the compounds (A), (B) and (C).
(ii) Write one use of compound (C).
12. What may be the possible oxidation states of the transition metals with the following
d electronic configurations in the ground state of their atoms:
3d34s2, 3d54s2 and 3d64s2. Indicate relative stability of oxidation states in each case.

13. Calculate the number of unpaired electrons in following gaseous ions: Mn3+, Cr3+, V3+
and Ti3+. Which one of these is the most stable in aqueous solution?

14. How would you account for the following:

(a) The metallic radii of the third(5d) series of transition metals are virtually the same
as those of corresponding group members of the second (4d) series.
(b) Among lanthanoids, Ln(III) compounds are predominant. However, occasionally
in solutions form solid compounds,+2 and+4 ions are also obtained.
(c) The E0M2+/M for copper is positive (0.34V), copper is the only metal in the first
series of transition elements showing this behavior.
(d) The higher oxidation states are usually exhibited by the members in the middle of
the series of transition elements.
(e) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it
shows the highest oxidation state of +4.
(f) Actinoids show irregularities in their electronic configurations.
15. a) Which metal in the first transition series (3d series) exhibits +1 oxidation state most
frequently and why?
b) Which of the following cations are coloured in aqueous solution and why?
Sc3+, V3+, Ti4+ , Mn2+

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More Practice

Chapter 7: d- and f -Block Elements

1. What is Lanthanide contraction? What effect does it have on the chemistry of the
elements which follow lanthanoids.

2. Why is HCl not used to acidify a permanganate solution in volumetric estimations of

Fe2+ or C2O42- ?

3. Why are Zn , Cd and Hg normally not regarded as transition metals ?

4. Which of the following ion is paramagnetic: Sc3+ (Z=21), Cu+ (Z=29).

5. Why is the third ionization energy of Manganese (Z=25) unexpectedly high?

6. Evaluate the magnetic moment of a divalent ion in aqueous solution if its atomic no. is
25.

7. Explain why:
(1) Ce3+ can be easily oxidized to Ce4+ ( At. no. of Ce =58).
(2) Zr (Z=40) and Hf (Z=70) have very close value of atomic radii.
(3) The lowest oxidation state of manganese is basic while the highest is acidic.
(4) CrO42- is a strong oxidizing agent while MnO42- is not.

8. Predict which of the following will be coloured in aqueous solution?

Ti3+,V3+,Cu+,Sc3+,Mn2+ ,Fe3+,Co2+

9. Explain the following observations:

(i) In general the atomic radii of transition elements decrease with atomic number in
a given series.
(ii) The Eo value for Mn3+/Mn2+ couple is much more positive than for Cr3+/Cr2+ or
Fe3+/Fe2+ couple.
(iii) Cu+ ion is unstable in aqueous solutions.
(iv) Although Co2+ ion appears to be stable, it is easily oxidised to Co3+ ion in the
presence of a strong ligand.
(v) With the same d4 d-orbital configuration Cr2+ ion is reducing while Mn3+ ion is
oxidising.
(vi) The enthalpies of atomisation of the transition elements are quite high.
(vii) Transition metals form compounds which are usuallly coloured.
(viii) Transition metals exhibit variable oxidation states.
(ix) The actinoids exhibit a greater range of oxidation states than the lanthanoids.
(x) There occurs much more frequent metal-metal bonding in compounds of heavy
transition elements (3rd series).
(xi) There is in general an increase in density of element from titanium (Z=22) to
copper (Z=29).
(xii) The gradual decrease in size (actinoid contraction) from element to element is
greater among the actinoids than that among the lanthnoids. (lanthanoid
contraction).
(xiii) The greatest numbers of oxidation states are exhibited by the members in the
middle of a transition series.

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(xiv) With the same d-orbital configuration (d4) Cr2+ ion is a reducing agent but Mn3+
ion is an oxidising agent.

10. What is meant by ‘disproportionation’ ? Give two examples.

11. Why Ce4+is oxidizing and Sm2+, Eu2+ are reducing in nature?

12. A mixed oxide of iron and chromium FeOCr2O3 is fused with sodium carbonate in the
presence of air to form an yellow coloured compound (A) On acidification the
compound the compound forms an orange coloured compound (B), which is a strong
oxidizing agent .
(i) Identify the compounds (A) and (B)
(ii)Write balanced chemical equation for each step.

13. Explain the following facts:

(a) transition metals act as catalysts.
(b)Chromium group elements have the highest melting points in their respective series.
(c) The enthalpies of atomization of transition elements are high.
(d) From element to element the actinoid contraction is greater than the lanthanoid
contraction.
(e) The Eo value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+
(f) Scandium (Z=21) does not exhibit variable oxidation states and yet it is regarded as a
transition element.

14. An element ‘A’ exists as a yellow solid in standard state . It forms a volatile hydride ‘B’
which is a foul smelling gas and is extensively used in qualitative analysis of salts. When
treated with oxygen ,’B’ forms an oxide ‘C’ which is a colourless , pungent smelling
gas.This gas when passed through acidified KMnO4 solution , decolorizes it.’C’ gets
oxidized to another oxide’D’ in the presence of a heterogenous catalyst . Identify
A,B,C,D, and also give the chemical equation of reaction of ‘C’ with acidified KMnO4
solution and for conversion of ‘C’ to ‘D’.

15. (a) A blackish brown coloured solid ‘A’ when fused with alkali metal hydroxides in the
presence of air, produces a dark green compound ‘B’, which on electrolytic oxidation
in alkaline medium gives a dark purple coloured compound C. Identify A, B and C
and write the reactions involved.
(b) What happens when acidic solution of green compound (B) is allowed to stand for
some time? Give the equations involved .What is this type of reaction called?

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Assignment

Chapter 8: Coordination Compounds

1. Write the coordination number and oxidation state of Platinum in the complex
[Pt(en)2Cl2]
2. (a) Write the formula of the following coordination compound:
Iron(III)hexacyanoferrate(II).
(b) What type of isomerism is exhibited by the complex [Co(NH3)Cl]SO4?
(c) Write the hybridization and number of unpaired electrons in the complex [CoF6]3-

3. (i) What type of isomerism is shown by the complex [Co(NH3)6] [Cr(CN)6] ?

(ii) Why a solution of [Ni(H2O)6]2+ is green while a solution of [Ni(CN)4]2– is colourless ?
(At. no. of Ni = 28)
(iii) Write the IUPAC name of the following complex : [Co(NH3)5(CO3)]Cl.

4. Write IUPAC name of the following:

1) [Pt(NH3)2Cl2] 2) [CoBr2(en)2]+ 3) [Ni(NH3)6]Cl2
4) K4[Fe(CN)6] 3+ 5) [NiCl4]2- 6) [CrCl2(en)2]Cl
5. Write the formulas in the following cases.
1) Tetrahydroxozincate(II), 2) Hexaamminecobalt(III) sulphate
3) Hexaammineplatinum(IV), 4) Pentaamminenitrito-N-cobalt(III)

6. Give the formula of each of the following coordination entities:

(a) Co3+ ion is bound to one Cl-, one NH3 molecule and two bidentate ethylene
diamine(en) molecules.
(b) Ni2+ ion is bound to two water molecules and two oxalate ions.
Write the name and magnetic behavior of each of the above coordination entities.

7.

8. In the ring test for identification of nitrate ion, what is the formula of the compound
responsible for the brown ring formed at the interface of two liquids?

9. a) For the complex [Fe(CN)6]3–, write the hybridization type, magnetic character and spin
nature of the complex. (At. number : Fe = 26).
b) Draw one of the geometrical isomers of the complex [Pt(en)2Cl2]2+ which is optically
active.
10. Show the possible isomers of the following coordination entities?
(i) [Cr(C2O4)3]3- (ii) [Co(NH3)3Cl3] (iii) [Co(en)2Cl2]Cl

11. Name the isomerism exhibited by the following pair of coordination compounds:
[Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br
Give one chemical test to distinguish between these two compounds.
12. Using valence bond theory, predict the geometry and hybridization of [Cr(NH3)6]3+ ion,
[Fe(CN)6]3- (paramagnetic due to single unpaired electron) and [FeF6]3- (paramagnetic due
to 5 unpaired electron) [ Cr = 24, Fe=26].

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13. How many ions are produced from the complex [Co(NH3)6]Cl2 in aqueous phase.

14. What is spectrochemical series? Explain the difference between a weak field ligand and a
strong field ligand.

15. (i) Draw a sketch to show the splitting of d- orbitals in an octahedral crystal field. State
for a d6 ion how the actual configuration of the split d- orbitals in an octahedral crystal
field is decided by the relative values of Δo and P.
(ii) On the basis of CFT, write the electronic configuration of d4 ion if ∆0>P.

16. Give reasons:

1) K3[Fe(CN)6] is weakly paramagnetic whereas K3[FeF6] is highly paramagnetic.
2) Though CO is a weak lewis base yet it forms a number of stable metal carbonyls .
Explain.
17. Compare the following complexes with respect to their shape, magnetic behaviour and
the hybrid orbitals involved: (i) [CoF4]2-, (ii)[Cr(H2O)2(C2O4)2]-
(Atomic number Co = 27, Cr =24 )

18. Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems (ii) medicinal chemistry
(iii) analytical chemistry (iv) metallurgy of metals.

19. (a) What is a ligand? Give an example of a bidentate ligand.

(b) Explain as to how the two complexes of nickel, [Ni(CN)4]2- and Ni(CO)4, have different
structures but do not differ in their magnetic behaviour. (Ni =28)

20. Explain the following:

(a) The π-complexes are known for transition elements only.
(b) Nickel(II) does not form low spin octahedral complexes.
(c) [Fe(CN)6]4- and [Fe(H2O)6]2+ are of different colours in dilute solutions.

21. Hard water does not form leathers with soap. Rita uses a washing powdercontaining
sodium metapolyphosphate and ethylenediamine tetracetate(EDTA) while Sita is using
ordinary washing power.
(a) Which washing powder is move effective for washing clothes in hard waterand why?
(b) Name the values associated with the above passage.

22. (i) Name the ligand (compound) used for treatment of Lead poisoning.
(ii) Write the reaction involved for removal of lead from living organism.
(iii) Write the name of coordination compound used as a chemotherapeutic agent to curb
the growth of tumours.

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More Practice

Chapter 8: Coordination Compounds

1. Write IUPAC name of the following:

1) [Pt(NH3)2Cl(NH2CH3)]Cl
2) linkage isomer of [Co(NH3)S(ONO)]2+
3) [Co(NH3)6]Cl3
4) [Co(NH3)4Cl(NO2)]Cl
5) [Mn(H2O)6]2+
6) [Co)(en)3]3+
7) [Cr(H2O)4Cl2]Cl
8) [Co(NH3)4(H2O)2]Cl3
9) [Pt(NH3)4][NiCl4]

2. Give the electronic configuration of the

(a) d-orbitals of Ti in [Ti(H2O)6]3+ ion in an octahedral crystal field.
(b) Why is this complex coloured? Explain on the basis of distribution of electrons in the
d-orbitals.
(c) How does the colour change on heating [Ti(H2O)6]3+ ion?

3. A metal ion Mn+ having d4 valence electronic configuration combines with three
didentate ligands to form a complex compound. Assuming Δo > P
a. draw the diagram showing d orbital splitting during this complex formation.
b. Write the electronic configuration of the valence electrons of the metal Mn+
ion in terms of t2g and eg.
c. What type of hybridization will Mn+ ion have?
d. Name the type of isomerism exhibited by this complex.

4. Write the shape of Fe(CO)5 , Mn2(CO)10 , Co2(CO)8, Ni(CO)4 molecule

5. For the complex [Fe(en)2Cl2]Cl, (en = ethylene diamine), identify

1) The oxidation number of iron.
1) The hybrid orbitals and the shape of the complex
2) The magnetic behaviour of the complex
3) The number of geometrical isomers
4) Whether there is an optical isomer
5) Name of the complex. (At. No. of Fe =26)

6. Write the formulas in the following cases.

1) Tetrabromidocuprate(II)
2) Diamminedichloridoplatinum(II)
3) Amminebromidochloridonitrito-N-platinate (II)
4) Dichlorodobis (ethane-1,2-diamine) platinum (IV) nitrate

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Chapter 9- Haloalkanes and Haloarenes
POLYHALOGEN COMPOUNDS

1. CH2 CL2 (Methylene Chloride) / (Dichloromethane)

Uses: Used as a solvent as a paint remover, propellant in aerosol, as a process solvent in the
manufacture of drugs and as a metal cleaning and finishing solvent.
Harmful effects: Harms human central nervous system. Higher level in air causes dizziness, nausea,
tingling and numbness in the fingers and toes. Direct contact with skin causes intense burning and
can burn cornea of eyes.
2. CHCl3 (chloroform/trichloromethene)
Uses: Solvents, fats, alkaloids, iodine and other substances, in production of Freon refrigerant R-22.
Inhaling chloroform vapours depresses the central nervous system. Was used as an anesthetic but
now has been replaced by less toxic anesthetics.
Light
CHCl3 + O2 2COCl2 + 2HCl
(Carbonyl chloride)
Chloroform is slowly oxidized by the presence of light to an extremely poisonous gas, phosgene
which when inhaled may cause damage to liver, kidneys, and some people develop sores when the
skin is immersed in closed dark coloured bottles completely filled so that air is kept out.

3. CHI3 (lodoform/ Triiodomethane)

It has strong unpleasant smell .It was used as an antiseptic but the antiseptic properties are due to
the liberation of I2 and not due to CHI3. Due to its objectionable smell objectionable smell, other
formulations containing I2 are used.
4. CCL4(carbon tetrachloride/ Tetra chloromethane)
Uses: Used in synthesis of chlorofluorocarbons, manufacture of refrigerants and propellants or
aerosol cans, s a solvent, cleaning fluid, fire extinguishers.
Harmful effects: Exposure to CCl4 may cause liver cancer, dizziness, lightheadedness, nausea and
vomiting which can cause permanent damage to nerve cells. When CCl4 is released into the air , it
rises to the atmosphere and depletes the ozone layer which increases human exposure to UV rays,
leading to increased skin cancer, eye diseases and disorders and possible disruptions to the immune
system.

5. Freon (Chlorofluoro carbon compounds of CH4 and C2H6)

Freon is stable, unreactive, non toxic, non corrosive and easily liquefiable gases. Eg. Freon 12
(CCl2F2)
Preparation- CCl4 + 2 AgF / SbF2 CCl2F2 (Swartz reaction)
Uses- Aerosol propellants, refrigeration and air conditioning.
Freons, eventuall diffuse unchanged into the stratosphere here it initiates radical chain reaction that
can upset O3 balance.
6. DDT (p,p’-Dichlorodiphenyltrichloromethane) - It is the first organic chlorinated nsecticide.
Uses- as used against mosquito that spreads malaria and lice that carry typhus. Later, many species
of insects developed resistance to DDT, and it as also discovered to have toxicity towards fish. DDT
is not metabolized very rapidly by animals; instead it is deposited and stored in fatty tissues.

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Assignment

Chapter 9: Haloalkanes and Haloarenes

1. Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous
NaOH and why?

2. (a) Identify the chiral molecule in the following pair:

(b) Write the structure of the product when chlorobenzene is treated with methyl chloride in
the presence of sodium metal and dry ether.
(c)Write the structure of the alkene formed by dehydrohalogenation of
1-bromo-1-methylcyclohexane with alcoholic KOH.

3. Write the structure of 1-Bromo-4-chlorobut-2-ene.

4. Following compounds are given to you :

2-Bromopentane, 2-Bromo-2-methylbutane, 1-Bromopentane
(i) Write the compound which is most reactive towards SN2 reaction.
(ii) Write the compound which is optically active.
(iii)Write the compound which is most reactive towards β-elimination reaction.

5. Name the following halides according to IUPAC system:

(a) CH3CH(Br)CH=C(CH3)CH2Cl (b) CH2CH2Br

(c) CH3CH(CH3)CH(Br)CH3 (d) ClCH2C=CCH2Br

(e) Cl Cl (f) CH3CH ≡ C-I

6.

7. What happens when bromine attacks CH2=CH-CH2-C ≡ CH?

8. Write the structures of the following organic compounds:

(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 2-(2-Chlorophenyl)-1-iodo octane
(iv) 4-tert-Butyl-3-iodoheptane
9. Answer the following questions:
(i) What is meant by chirality of the compound? Give an example.

(ii) Which of the following compounds is more easily hydrolysed by KOH and why?

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CH3CH(Cl)CH2CH3 or CH3CH2CH2Cl

(iii) Which one undergoes SN2 faster and why?

I Cl

Or
10. Which one of the following reacts faster in an SN1 reaction and why?
Cl
Cl or

11. State one use of DDT and iodoform. Why chloroform is kept in dark coloured bottles
completely filled?

12. What are ambident nucleophiles? Explain with the help of an example.

13. Write short notes on:

(a) Fittig reaction (b) Swartz reaction

14. Account for the following:

a) tert-Butyl chloride reacts with aqueous NaOH by SN1 mechanism while n-butyl chloride
reacts by SN2 mechanism.
b) Among HI, HBr and HCl, HI is most reactive.
c) Alkyl halides though polar, are immisible with water.
d) Chlorobenzene is extremely less reactive towards nucleophillic substitutionreaction.
e) C–Cl bond length in chlorobenzene is shorter than C–Cl bond length in CH3–Cl.
f) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
g) SN1 reactions are accompanied by racemization in optically active alkyl halides.

15. Carry out the following conversions:

i) 1-Chlorobutane to n-octane
ii) Toluene to benzyl alcohol
iii) Benzyl chloride to benzyl alcohol

16. What will be the mechanism for the substitution of -Br by –OH in (CH3)2C(Br)CH2CH3?

17. Identify the following compounds from A to N:

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More Practice
Chapter 9: Haloalkanes and Haloarenes

1. Name the following halides according to IUPAC system:

(i) (CCl3)3CCl
(ii) CH3C(p-ClC6H4)2CH(Br)CH3
(iii) (CH3)3CCH=C(Cl)C6H4I-p

H3C H
H
H Br
H
2. Arrange the following compounds in an increasing order of their acid strengths:
(CH3)2CHCOOH, CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH

3. Write the structures of the following organic compounds:

4. Arrange the following compounds in increasing order of reactivity towards

nucleophillic substitution reaction:
1,4-Dichlorobenzene , 4-Methoxy chlorobenzene, benzene , 2,4,6- Trinitro chlorobenzene

5. Account for the following:

a) The dipole moment of chloro benzene is lower than that of cyclohexyl chloride.
b) Vinyl chloride is unreactive towards nucleophillic substitution reactions.
c) Grignard reagent should be prepared under anhydrous conditions.
d) Haloarenes are much less reactive than haloalkanes towards nucleophilic
substitution reactions.

6. Carry out the following conversions:

(i) Propene to propyne
(ii) Ethanol to but-1-yne

7. Identify the following compounds from A to K:

(i) (CH3)3CBr + H2O heat A

C2H5ONa, heat
(ii) (CH3)2CHCH(Br)CH2CH3 B

(iii) CH2=CHCH2Br + CH3C ≡ CNa liq NH3 C

(iv) C6H4CH2Cl + C2H5ONa D

(v) CH3CH2CH2OH + SOCl2 E

(vi) CH3CH=C(CH3)2 + HBr F

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(x) Cl

+ C2H5ONa J

NO2 NO2

(xi) CH(CH3)2 + Br2 heat K

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Assignment

Chapter 10: Alcohols, Phenols and Ethers

1. Write the structures of the main products in the following reactions:

i. O CH2 ─ C ─ OCH3
|| || NaBH4

ii. CH = CH2
+ H2O H+

iii. |OC2H5

+ HI

2. (a) Write the product(s) in the following reactions :

(i)

CH3
|
(ii) CH3 –CH– O – CH2 – CH3 HI ?+?
(iii) CH3 – CH = CH – CH2 – OH PCC ?
(b) Give simple chemical tests to distinguish between the following pairs of compounds :
(i) Ethanol and Phenol
(ii) Propanol and 2-methylpropan-2-ol

3. (a) Write the formula of reagents used in the following reactions :

(i) Bromination of phenol to 2,4,6-tribromophenol
(ii) Hydroboration of propene and then oxidation to propanol.
(b) Arrange the following compound groups in the increasing order of their property
indicated :
(i) p-nitrophenol, ethanol, phenol (acidic character)
(ii) Propanol, Propane, Propanal (boiling point)

(c) Write the mechanism (using curved arrow notation) of the following reaction :
CH3 – CH2 – O+H2 CH3CH2OH CH3 – CH2 – O+ – CH2 – CH3 + H2O
|
H

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4. Give the I.U.P.A.C. name of the following:

CH3

(i) CH3CH2-CH-CH-CH2OH

OH

CH3
(iii) CH3-- C----- CH—CH3
C2H5 OH

(iii ) CH3—CH2-CH2-OCH2-CH-CH-CH2CH3

CH3

( iv) C6H5OCH2CH2CH2CH2CH—CH2CH3

CH3

(v) H3C-CH-CH2-C----CH-CH2OH

CH3 OH CH3

5. Arrange the following as

(i) Decreasing order of boiling points:

2-methyl-2-propanol, 1-butanol,2-methyl –1-propanol and 2-butanol.
(ii) Decreasing order of their acidic character:

OH OH OH

(a) (b) (c)

OCH3 NO2
(iii ) Increasing reactivity towards Lucas reagent:
1-butanol , 2-methyl-2- propanol , 2-butanol.

6. Write the chemical equation when 1-propanol react with

(i) excess of HBr (ii) H2SO4 at 410 K
(iii ) H2SO4 at 443K (iv) acidified KMnO4

7. How does phenol react with the following?

(i) Acetyl chloride.
(ii) Bromine in water.
(iii) Chloroform in presence of NaOH.

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8. Convert
(1) Phenol to picric acid
(2) Cumene to phenol.
(3) Phenol to Salicyaldehyde
(4) Phenol to anisole
(5) Propan-2-ol to 2-Methylpropan-2-ol

9. Explain giving reasons:-

(i) Alcohols are generally soluble in water but alkyl halides are not.
(ii) Phenols exhibit an acidic character.
(iii) Phenols has a smaller dipole moment than methanol.
(iv) 2,3- dimethylbutanol has got lower boiling point than hexanol .
(v) 2-Nitrophenol is more volatile than 4-Nitrophenol.

10.

11.

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More Practice
Chapter 10: Alcohols, Phenols and Ethers

1. Complete the following reactions:

(a) CH3CH2--C-CH3 NaBH4

||
O
CH3
|
(b) CH3CH2-CH-CH2-COCH3 + CH3MgBr A H2O B
CH3
|
(c ) CH3-C-OH Cu
| 573K
CH3
CH3
|
(d) C2H5Br + NaO-C-CH3
|
CH3

CH3
|
(e ) CH3-C-Br + NaOC3H7
|
CH3

OH
Br2 water
(f)

Cu-ZnO – Cr2O3
(g) CO + 2H2
200-300 atm, 573-673 K
OCH3
AlCl3
(h) + CH3COCl

2 Convert:
(i) Chlorobenzene to Phenol
(ii) Aniline to phenol
(iii) Propanone to Propene

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Assignment

Chapter 11: Aldehydes, Ketones and Carboxylic Acids

1 (A), (B) and (C) are three non-cyclic isomers of a carbonyl compound with molecular formula
C4H8O. Isomers (A) and (C) give positive Tollen’s test whereas isomer (B) does not give
Tollen’s test but gives positive Iodoform test. Isomers (A) and (B) on reduction with
Zn(Hg)/conc.HCl gives the same product D.

(i) Write the structures of (A), (B), (C) and (D).

(ii) Out of (A), (B) and (C) isomers, which one is most reactive towards addition of NaHSO3
and why?

2 How do you convert the following:

(a) Ethanal to propanone
(b) Toluene to benzoic acid

(c) Benzoic acid to benzaldehyde

(d) Benzoic acid to benzaldehyde

(e) Prapanone to Propene

3 Account for the following:

(a) Aromatic carboxylic acids do not undergo Friedal-Crafts reaction.
(b) pKa value of 4-nitrobenzoic acid is lower than that of benzoic acid.

4 Write structures of compounds A, B and C in each of the following reactions :

(i) C6H5Br Mg/dry ether A (a) CO2(g) B PCl5 C
(b) H3O+
(ii) CH3CN (a) SnCl2/HCl A dil. NaOH B Δ C
(b) H3O+
5 Write IUPAC names for the following :

CH3
(a)

O
(b) CH2=CHCH2CHO
(c) (CH3)2C=CHCOCH2CH3

6 a) Arrange the following compounds as directed:

b) Acetaldehyde, acetone, Methyl tert-butyl ketone (reactivity towards HCN)
c) Benzoic acid, 3,4-Dinitrobenzoic acid, 4-methoxybenzoic acid
d) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH (acid strength)
e) a) CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3(increasing order of their
boiling point)
f) b) Ethanal, propanal, propanone, butanone (increasing order of their reactivity
towards nucleophilic addition).
7 Give brief description with suitable example:
a) Cannizzaro Reaction

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b) Hell-Volhard Zelinsky Reaction.
c) Clemmensen Reduction
d) Wolff-Kishner Reduction.
e) Cross aldol condensation
f) Etard reaction

8 Account for the following:

a) Carboxylic acids have higher boiling points than alcohols of comparable molecular masses.
b) Electrophilic substitution in benzoic acid takes place at meta-position.
c) Monochloroethanoic acid has a higher pKa value than dichloroethanoic acid.
d) Ethanoic acid is a weaker acid than benzoic acid.
e) The boiling points of aldehydes and ketones are lower than of the corresponding acids.

9 How will you convert the following?

a) Acetaldehyde into 2-Butenal
b) Acetic acid to Acetic anhydride
c) Butanol to butanoic acid
d) 4-Methylacetophenone to benzene-1,4-dicarboxylic acid
e) Butan-2-one to butan-2-ol
f) Phenol to 2,4,6- tribromophenol
10 Distinguish between the following
a) CH3CHO and C6H5CHO
b) C2H5OH and CH3CH2COCH2CH3
c) C6H5COOH and C6H5OH
d) C6H5COCH3 and C6H5COC6H5
e) CH3COCH3 and C2H5OH
f) CH3COCH3 and C3H7OH

11 Complete the following reactions:

(a) CH3CO CH3 +NH2NH2 KOH/glycol

(b) C6H5COCH3+ C6H5OH NaOH/ I2

(c) C6H5NH2 NaOH/CaO

12

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13

14 Complete the following reaction statements by giving the missing starting material, reagent
or product as required:

(i) ? O3 2 =O
Zn-H2O

(ii) CH2 ? CHO

(iii) CH2CH3 KMnO4, KOH, heat ?

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More Practice

Chapter 11: Aldehydes, Ketones and Carboxylic Acids

1 Account for the following:

a) Formaldehyde gives Cannizzaro’s reaction whereas acetaldehyde does not
b) Carboxylic acids do not give the characteristic reactions of carbonyl group.
c) Aldehydes are more reactive than ketones towards nucleophilic addition reactions.
d) Chloroacetic acid has lower pKa value than acetic acid.
e) The aldehydes an ketones undergo a number of addition reactions.
f) Ethanoic acid is a weaker acid than benzoic acid.

2 How will you convert the following?

i) Acetophenone to Ethyl benzene
ii) Acetone to tert-butyl alcohol
iii) Benzyl alcohol to phenyl ethanoic acid
iv) Bromobenzene to benzoic acid
v) p-methyl acetophenone to benzene 1,4 –dicarboxylic acid
vi) Benzoic acid to benzyl amine
vii) p-nitrobenzamide to p-nitroaniline
viii) A primary alcohol to an aldehyde
ix) Ethanol to acetone
x) Benzene to acetophenone
xi ) Benzoic acid to benzaldehyde

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5 An organic compound contains 69.77% carbon, 11.63% hydrogen and the rest is oxygen.
The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms
an addition compound with sodium hydrogen sulphite and gives a positive iodoform test.
On vigorous oxidation it gives ethanoic and propanoic acids. Deduce the possible structure
of the organic compound.

6 An organic compound with molecular formula C9H10O forms 2,4-DNP derivative, reduces
Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1,2-
benzenedicarboxylic acid. Identify the compound.

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Assignment

Chapter 12: Organic Compounds containing Nitrogen

1. (a) Write the reactions involved in the following:

(i) Hofmann bromamide degradation
(ii) Diazotization
(iii)Gabriel phthalimide synthesis
(b)Give reasons:
(i) (CH3)2NH is more basic than (CH3)3N in an aqueous solution.
(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts.

2. (a)Write the structures of the main products of the following reactions:

(CH3CO)2O, Pyridine
(i) C6H5NH2
(ii) C6H5SO2Cl (CH3)2NH
(iii) C6H5N2+Cl- CH3CH2OH

(b)Give a simple chemical test to distinguish between Aniline and N,N-

dimethylaniline.
(c) Arrange the following in the increasing order of their pKb values:
C6H5NH2 , C2H5NH2, C6H5NHCH3

3. Write IUPAC name of the following compound :

(CH3CH2)2NCH3

4. Give reasons :
(i) Acetylation of aniline reduces its activation effect.
(ii) CH3NH2 is more basic than C6H5NH2.
(iii)Although –NH2 is o/p directing group, yet aniline on nitration gives a significant
amount of m-nitroaniline.

5. Write IUPAC names of the following:

a) CH3—CH—CH—CONH2 b) Br N(CH3)2
NH2 CH3

c) O2N—CH2—CH2—CH=CH—CHO d) (CH3)3CCN

e)

6. For an amine RNH2, write the expression for Kb to indicate its basic strength

7. Describe the following processes giving suitable examples of each:

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i) Diazotization
ii) Coupling reaction
iii) Carbyl amine reaction
iv) Hofmann’s bromamide reaction

8. How will you carry out the following conversions:

i) Nitrobenzene to acetanilide
ii) Methyl amine to ethyl amine
iii) Nitrobenzene to phenol
iv) Toluene to m-nitro benzoic acid
v) Acetic acid to ethyl amine

9. Write structures of the following:

i) P- Toluidine
ii) Picric acid
iii) Sulphanilic acid
10. Write a chemical reaction in which the iodide ion replaces the diazonium group in a
diazonium salt.

11. Write reactions for what happens when:

a) Phenol is treated with benzene diazonium chloride in presence of NaOH
b) Aniline is treated with benzaldehyde
c) Ethyl amine is treated with excess of methyl iodide.

12. Arrange in increasing order of boiling point: C2H5NH2, C2H5OH, (CH3)3N

13. Complete the following reaction equations:

i) C6H5N2Cl + H3PO2 + H2O
ii) C6H5NH2 + Br2(aq)
14.

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More Practice
Chapter 12: Organic Compounds containing Nitrogen

1. Account for the following:

a) Ammonolysis of alkyl halide does not give a corresponding pure amine
b) pKb for aniline is more than that for methylamine
c) Boiling point of methylamine is less than that of methanoic acid
d) Aniline cannot be prepared by Gabriel Pthalimide synthesis
e) Nitration of toluene is easier compared to nitrobenzene
f) Before nitration aniline is converted to acetanilide.
g) Aniline does not undergo Friedel-Crafts reaction.
h) Methylamine solution in water reacts with ferric chloride solution to give a
precipitate of ferric hydroxide.

2. A compound X having molecular formula C3H7NO, reacts with Br2 in presence of NaOH
to give another compound Y. The compound Y reacts with HNO2 to form ethanol and N2
gas. Identify the compounds X and Y and write the reactions involved.

3. A compound A of the molecular formula C3H7O2N on reaction with Fe and conc. HCl
gives a compound B of the molecular formula C3H9N. Compound B on treatment with
NaNO2 and HCl gives another compound C of the molecular formula C3H8O. The
compound C gives effervescence with Na. On oxidation with CrO3, the compound C
gives a saturated aldehyde having 3 carbon atoms. Deduce the structures of A, B and C
and write the reactions involved.
4. Describe a test to distinguish between each of the following pairs
a) Ethyl amine and aniline
b) N-methyl aniline and N,N-dimethyl aniline

5. How will you carry out the following conversions:

i) Aniline to N-phenyl ethanamide
ii) Aniline to benzoic acid
iii) Benzene to m-dichlorobenzene
iv) 2-nitropropane to acetone
v)Benzonitrile to acetophenone

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Assignment

Chapter 13: Biomolecules

1. Define the following with an example of each;

(i) Polysaccharides
(ii) Denatured proteins
(iii) Essential aminoacids

2. (i) Write the product when D- Glucose reacts with bromine water.
(ii) Amino acids show amphoteric behaviour. Why?
(iii) Write one difference between α-helix and β-pleated structures of proteins.

3. State two main differences between globular proteins and fibrous proteins. Give one example
of each.

4. a) Write the full forms of DNA and RNA. Write the names of the bases in them.
b) What are three types of RNA molecules which perform different functions?

5. Write chemical equations for the reactions of glucose with

(i)Acetic Anhydride (ii) NH2OH (iii) HNO3 (iv) HI

6. Define and classify vitamins. Name the main disease caused due to lack of vitamins and its
sources in each of the following ;A , B 6, E ,D, B 12 and K .

7. (a) Write any two reactions of glucose which cannot be explained by the open chain structure
of glucose molecule.
(b) Write the structure of the product obtained when glucose is oxidized with nitric acid.

8. Define enzymes .State the activity of enzyme. How do enzymes differ from ordinary
chemical catalysis? Comment on the specificity of enzyme action.

9. In what way is a nucleotide different from a nucleoside? Illustrate with examples?

10. What is essentially the difference between alpha-glucose and beta-glucose? What is meant by
pyranose structure of glucose?

11. (a) Name some biological functions of nucleic acids.

(b) What is the name given to the linkage which holds together two nucleotides

12. Exlain what is meant by

(i) a peptide linkage
(ii) a glycosidic linkage

13. a) Write the name of two monosaccharides obtained on hydrolysis of lactose sugar.
b) Why Vitamin C cannot be stored in our body ?
c) What is the difference between a nucleoside and nucleotide ?
14. What is glycogen? How is it different from starch? How is starch structurally different from
cellulose?

15. How do you explain amphoteric behaviour of amino acids?

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16. Define denaturation in proteins.

17. Which of the following is a disaccharide: Starch, Maltose, Fructose, Glucose?

Hands-on/ IT Enabled work:

 The entire chapter is done with the help of a presentation.
 A hand–out is given on proteins, vitamins and nucleic acids.

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Assignment
Chapter 14: Polymers
1. Write a difference between LDPE and HDPE.

2. What are biodegradable polymers? Give an example.

3. Write the structures of the monomers used for getting the following polymers :
(i) Dacron
(ii) Melamine – formaldehyde polymer
(iii)Buna-N
4. Write the names and molecular structures of the monomers of the following and classify
them as addition or condensation polymers :
(i) Natural rubber (ii) Glyptal (iii) ) Buna-S (iv) Bakelite
(v) PVC (vi) Nylon-6 (vii) Neoprene (viii) Polypropene

5. (i)What is the role of t-butyl peroxide in the polymerization of ethene? Write its mechanism.
(ii)Identify the monomers in the following polymer :
[NH – (CH2)6 – NH – CO – (CH2)4 – CO– ]n
(iii)Arrange the following polymers in the increasing order of their intermolecular
forces : Polystyrene, Terylene, Buna-S
(iv)What are elastomers? Give an example of it.

6. (a) Describe chain growth and step growth polymerization with the help of an example.
(b) Classify the following as addition or condensation polymers:
Nylon-66, Neoprene, Polythene
7. (a)What is the difference between nylon-6 and nylon-66?
(b)What does the part ‘6,6 mean in the name nylon-6,6 ?

8. What is Teflon? What are its uses?

9. Differentiate the following pair of polymers based on the property mentioned against each.
(i)Novolac and Bakelite (structure)
(ii)Buna-S and Terylene ( intermolecular forces of attraction)

10. What is the repeating unit in the condensation polymer obtained by combining
HO2CCH2CH2CO2H (succinic acid) and H2NCH2CH2NH2(ethylene diamine).
11. How is melamine polymer prepared? Give its two uses. What type of polymer is it?
12. Name the monomers of Nylon2-nylon6 polymer.
13. PHBV (Poly-β-Hydroxybutyrate-co-β-hydroxy valerate) is a biodegradablepolymer. It is a
copolymer of 3-hydroxy valerate acid and 3-hydroxy pentanicacid.
(a) How PHBV has found utility in medicines as Capsule?
(b) Write the name of polymer used in artificial limb popularly known as Jaipurfoot.

Hands-on/ IT Enabled work:

 Different samples of polymers will be shown in the class.
 A hand out will be given.

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CHEMISTRY IN EVERYDAY LIFE

DRUGS

Drugs are chemicals of low molecular masses which interact with macromolecular targets and
produce as biological response. When biological response is therapeutic and useful, these
chemicals, are called medicines and if taken in higher doses, they behave as poisons. Use of
chemicals for therapeutic effect is called Chemotherapy.

Classification of Drugs:
a) On the basis of pharmacological effect: It is useful for doctors because it provides them the
whole range of drugs available for treatment of a particular problem. For e.g.: analgesics for
pain killing effect, antiseptics kill or arrest growth of microorganisms.
b) On the basis of drug action: It is based on the action of a drug on a particular biochemical
process. Eg- antihistamines which inhibit the action of histamines which causes inflammation in
the body.
c) On the basis of Chemical structure: Some drugs share a common feature and often have similar
Pharmacological activity. Eg; Sulphonamides have H2N-C6H4-SO2-NHR structural feature in
common.
d) On the basis of molecular targets: Drugs usually interact with biomolecules such as
carbohydrates, lipids, proteins & nucleic acid. These are called target molecules. Drugs
possessing some common structural feature have the same mechanism of action on targets.

DRUG –TARGET INTERACTION

Macromolecules of biological origin perform various functions in the body. For eg- Proteins which
perform role of biological catalyst in the body are called ENZYMES & those which are crucial to
communication system are called RECEPTORS.
Hormones are biological chemical messengers secreted by endocrine glands. Example- Insulin,
noradrenalin.

ENZYMES AS DRUG TARGETS:

a) Catalytic action of enzymes : For understanding interaction between drug and enzyme we first
study the function of enzymes.
Enzymes hold the substrate for a chemical reaction. Active sites of enzymes hold substrate
molecule in a suitable position, so that it can be attacked by the reagent effectively. Substrate
binds themselves to the active sites by ionic bonding, hydrogen or by vanderwaals interaction.

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It provides functional groups that will attack the substrate and carry out chemical reaction.

b) Drug-enzyme interaction: Drugs inhibit any of the above mentioned activities of enzymes. These
can block the binding site of enzyme and prevent the binding of substrate or can inhibit catalytic
activity of enzyme. These are called enzyme inhibitor.
These can occur in two different ways---
(i) Drugs compete with the natural substrate for their attachment on active sites of enzymes.
These are called competitive inhibitors.

(ii) Some drugs do not bind to the enzyme’s active site. These bind to some different enzyme
site called allosteric site. This binding of inhibitor at allosteric site changes the shape of
the active site in such a way that substrate cannot recognize it.

If the bond between an enzyme and inhibitor is a strong covalent bond and it cannot be easily
broken, then the enzyme is blocked permanently. The body then degrades the enzyme-inhibitor
complex and synthesizes the new enzyme.

RECEPTORS AS DRUG TARGETS

Receptors are proteins that are crucial to body’s communication process. Receptor proteins are
embedded in cell membranes in such a way that their small part possessing active site projects out of
the surface of the membrane and opens on the outside region of the cell membrane.

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(c) Receptor regains structure after removal of chemical messenger

There are a large no. of different receptors in the body that interact with different chemical
messengers. These receptors show selectivity for one chemical messenger over the other because
their binding sites have different shape, structure and aminoacid composition.
Drugs that bind to the receptor site and inhibit its natural function are called antagonists. Drugs that
mimic the natural messenger by switching on the receptors are called agonists. These are useful
when there is lack of natural chemical messenger.

CHEMICALS IN MEDICINE

The chemical substances used for treatment of diseases and for reducing suffering from pain are
called medincines or drugs.
Chemotherapy- is a science in which suitable chemicals are used for treatment of diseases.
1) Antipyretics- The chemicals use to lower body temperature in high fever are called antipyretics.
Eg- Aspirin, paracetamol and phenacetin

2) Analgesics- The chemical substances used to relieve pains without causing impairment of
consciousness, mental confusion, incoordination or paralysis or some other disturbances of
nervous system are called analgesics. These are of two types.
a) Non-narcotic drugs or non-addictive drugs- Eg- aspirin, analgin, novalgin, naproxen,
ibuprofen & diclofenac sodium or potassium.

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Aspirin: Finds use as antipyretic, prevention of heart attack because of its anti-blood
clotting action. Aspirin is supposed to be toxic to liver which gets hydrolysed in stomach
giving salicylic acid which sometimes cause bleeding in stomach. Therefore, overdosage
and its use in empty stomach should be avoided.
b) Narcotic drugs or Addictive drugs- Which produce sleep and unconsciousness. These
can also be used as analgesics. Eg- morphine, codeine, heroin, marijuana. When used in
medicinal doses, they relieve pain and produce sleep. In excessive doses, they produce
stupar coma, convulsions and ultimately leading to death. These narcotics are called
opiates because they are obtained from opium poppy.

3) Antiseptics and disinfectants:

Antiseptics are chemical substances used either to kill or prevent the growth of micro-organisms.
These are not harmful to living tissues and can be applied on wounds, ulcers, diseased skin
surfaces. They are also used to reduce odours resulting from bacterial decomposition of the
body or in the mouth. Eg- Soframycin, Bithional is added to medicated soaps, tincture of Iodine
( 2-3% soln of iodine in alcohol-water mixture), Iodoform, Boric acid in dilute aqueous solution is
antiseptic for eyes etc.
Disinfectants are chemical substances which are used to kill micro-organisms but they cannot be
applied on living tissues. They play a major role in water treatment and public health sanitation.
These are commonly applied on inanimate objects like floors, drainage system etc. Eg- Cl2 at a
conc. of 0.2 to 0.4 ppm makes water fit for drinking, Phenol derivative, thymol.
Some substances act both as antiseptics and disinfectants. Eg- Dettol (a mixture of chloroxylenol
and terpineol ), 0.2% soln. of phenol acts as antiseptic & 1% soln acts as disinfectant.

4) Tranquilizers: The chemical substances used for treatment of stress, mild and severe mental
diseases are called tranquilizers. They release mental tension and reduce anxiety. These are
essential component of sleeping pills. These are also called psychotherapeutic drugs.
Noradrenaline, a hormone which induces feeling of well being and helps in changing mood. If
the level of nordrenaline is low for some reason, then signal sending activity becomes low, and
the person suffers from depression. In such situations, antidepressant drugs are required.
Eg: iproniazid and phenalzine are antidepressant drugs. They inhibit the enzyme which
catalyse the degradation of noradrenaline.
Chlordiazepoxide and meprobamate are used to relieve tension.
Equanil, diazepam,veronal and serotonin are used in controlling depression and hypertension
Barbiturates like veronal, amytal, membutal, seconal and luminal are hypotonic ie: sleep
producing agents.

5) Antimalarials: These are chemical substances used for treatment of malaria. Eg- Chloroquine,
paraquine etc.

6) Antimicrobials: are chemical substances used to cure infections due to micro-organisms. The
disease in human beings may be caused due to variety of micro-organisms like virus, bacteria
etc. which are called microbes. They can be seen only by microscope. The disease causing
microbes are called pathogens. Our body possesses natural defense mechanism against the
pathogenic microbes. Skin is impervious to microbes. Our body secretions kill the microbes or
inhibit their growth. Some common examples are lysozyme in tears, nasal secretions, saliva,
lactic acid in sweat etc. The pathogens reach the tissues due to breach in defence mechanism and
cause infections.

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The control of microbial diseases can be achieved by:
(i) Drugs which kill organisms in the body (bactericidal)
(ii) Drugs which inhibit or arrest the growth of organisms (bacteriostatic)
(iii) Increasing immunity and resistance to infections of the body (immunity)
Antimicrobial substances may be synthetic chemicals like sulphonamides or antibiotics-
like tetracycline, penicillin, chloramphenicol etc. The common example of antimicrobial
drug is sulphanilamides which are effective in wide range of micro-organisms.

7) Antifertility Drugs: These chemical substances control pregnancy. Their basic aim is to prevent
conception or fertilization. The birth control pills are essentially a mixture of esterogen and
progesterone derivative. Both of these compounds are hormones. Progesterone suppresses
ovulation. Synthetic progesterone are more potent than progesterone.
The common pills used for a combination of progesterone, norethindrone and estrogen
(ethynylestradiaol) is novestrol.

8) Antacids: The chemical substances which neutralize excess acid in the gastric juices and give
relief from acid indigestion, acidity, heart burns, and gastric ulcers are called antacids. Baking
soda in water is a common antacid. Other example are magnesium hydroxide, calcium
carbonate, sodium bicarbonate, potassium bicarbonate, magnesium carbonate, potassium
bicarbonate, aluminium phosphate. Magnesium oxide is also used as an antacid ingredient since
it reacts with water to form Mg(OH)2. The antacids are available in the form of liquids, gels or
tablets. Generally, liquid antacids are more effective than tablets because of great surface area
available for interaction and neutralization of acid. An advancement in treatment of
hyperacidity came through the discovery that histamines stimulates the secretion of pepsin and
hydrochloric acid. To prevent interaction of histamines with the receptors present in the
stomach wall, the drug cimetidine has been designed. This resulted in release of lesser amount
of acid. The drug is now replaced by ranitidine. A more effective new class of drugs is
omeprazole and lansoprazole which prevents formation of acid in stomach.

9) Antihistamines: are chemical substances which diminish or abolish the main actions of
histamines release in the body and hence prevent the allergic reactions caused by antigens.
Histamines are responsible for nasal congestion associated with common colds, cough, allergic
response to pollens etc. Synthetic drugs such as bromopheniramine (Dimetapp) and terfenadine
(seldane) are used as antihistamines. Antihistamines are also called anti-allergic drugs. These
are used to treat allergy, eg, skin rashes, conjunctivitis etc. These drugs relieve sneezing , nasal
discharge, mild asthama, itching of eyes, nose and throat. The common antihistamine drugs are
Benadryl, avil, zeet, bromethazine, actidil, anistine, foristal etc.

10) Anaesthetics: are chemical sunstances which produce general or local insensibility to pains and
other sensations. Cocaine, novocaine are local anaesthetic chloroform, diethyl & vinyl ethers are
general anaesthetics.

11) Antibiotics: are chemical substances which are produced by micro-organisms (bacteria, fungi
and moulds) and can inhibit the growth or even destroy micro-organisms. Antibiotic refers to a
substance ( produced wholly or partly by chemical synthesis) which in low concentration
inhibits growth or destroys micro-organisms by intervening in their metabolic processes.
First antibiotic produced was penicillin by Alexander Fleming in 1929. Antibiotic can be either
bactericidal or bacteriostatic.
Bactericidal: Pencillin, Aminoglycosiders, Ofloxacin.

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Bacteriostatic: Erythromcin, Tetracycline, Chloramphenicol.
Pencillin is narrow spectrum. These can be used for curing sore throat, reheumatic fever, local
infections etc. Streptomycin, neomycin is used for treatment of tuberculosis, meningitis,
pneumonia etc.
Broad spectrum antibiotics are effective against several micr-organisms. Thereforem these are
for curing a variety of diseases. Eg- thetracycline, chloromycetin and chloramphenicol. Eg-
Chloramphenicol is a broad spectrum antibiotic which is used for curing typhoid, acute fever,
dysentery, whooping cough, pneumonia, eye infections, certain urine infections etc.
Sulphadrugs are used against pneumonia, tuberculosis, diphtheria etc. Some examples are
sulphadiazine, sulphathiazole, sulphaacetamide etc.

S.No Type of Medicine Used as Examples

.
1 Analgesics Relieve Pain Aspirin, Ibuprofen
2 Antipyretics Lowers body temperature Paracetamol, Phenacetin
3 Antiseptics & Kill or prevent growth of Phenol, Chlorine, dettol
Disinfectants microorganism
4 Tranquilizers Treatment of stress & mental Barbituric acid & its
diseases derivatives (Seconal,
Luminal, Veronal etc)
5 Antimicrobials Cure infections due to Sulphonamides
microorganisms
6 Antifertility drugs Birth control Novestrol (ethynylestradiol)
& Progestrone
(norethindrone),
mifepristone
7 Antacids Removes excess acid in stomach Magnesium hydroxide,
Magnesium trisilicate,
aluminium hydroxide gel
Ranitidine
8 Antihistamines Treatment of hyperacidity, Brompheniramine &
stimulates secretion of pepsin & terfenadine
HCl in the stomach. Also
responsible for nasal congestion
associated with common cold
9 Antibiotics Produced by microorganisms & Pencillin, Tetracycline,
can inhibit the growth of other Chloramphenicol
microbes

Chemicals in Food

Many chemicals are added to food for their preservation and enhancing their appeal. These include
flavourings, sweetness, antioxidants, fortifiers, emulsifiers and antifoaming agents.

1. Antioxidants:
Antioxidants are the important class of compounds which prevent oxidation of food materials.
These compounds retard the action of oxygen on the food and thereby help in preservation.
These act as sacrificial materials. i.e. they are more reactive towards oxygen than the materials
they are protecting. They also reduce the rate of involvement of free radicals in the aging

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process. Most important antioxidants used are butylated hydroxy anisole (BHA) and butylated
hydroxy toluene (BHT). The addition of BHA to butter increases its storage life.

OH OH
(CH3)3C C(CH3)3 C(CH3)3

OCH3
CH3
BHT BHA

Sometimes BHT and BHA are added in combination with citric or ascorbic acid to produce a
more active synergistic effect. SO2 and sulphate are useful antioxidants for wine and beers,
sugar syrups and cut peeled on dried fruits and vegetables.

2. Preservatives:
These are the chemical substances which are added to the food materials to prevent their
spoilage and to retain their nutritive value for long periods. These preservatives prevent
rancidity of food & inhibit the growth of microorganisms during storage. Example: Common
salt, sugar, oils, Sodium benzoate, salts of propanoic acid and ascorbic acid.
3. Artificial Sweetening agents:
The artificial sweetners are another type of food additives. Eg; Saccharin which is marketed as
soluble of calcium salt. It is 300 times sweet than cane sugar. It is life saver for diabetic patients
and is of great value to people who need to control intake of calories.
Aspartame: Unstable at cooking temperature, therefore it is used as a sugar substitute to cold
foods and soft drinks.
Alitame: more stable during cooking than aspartame
Sucralose: good artificial sweetener.

4. Edible colors:
Edible colour that are used for food are dyes; ex- dyes are used to dye orange peels so hat
oranges retain their colour. Colour is also added to fruit juices. Food colours do not have any
nutritional value. The use of some of the azodyes are dangerous for young children and asthma
patients.
Terazime, a widely used dye is harmful
Natural dyes like carotene are safe food edible colours.
PFA {Prevention of food Adulteration Act}- govt. has passed it for the protection of consumer
interests.

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SOAPS AND DETERGENT

Soaps are sodium or potassium salts of long chain fattyacids. Soaps containing sodium salts are
formed by heating fat (ie glyceryl ester of fatty acid) with aqueous sodium hydroxide solution and
potassium salts are prepared by using potassium hydroxide. This reaction is known as
saponification. Soap obtained remains in colloidal form and is precipitated from the solution by
adding NaCl.

CH2-O-C-C17H35 CH2OH
| |
CH2-O-C-C17H35 + 3NaOH 3C17H35COONa + CH2OH
| |
CH2-O-C-C17H35 CH2OH

Soaps cannot be used in hard water as hard water contains certain metal ions such as Ca 2+ and Mg2+
which form a curdy white precipitate of calcium and magnesium salt. This is called scum and is
hinderance to good washing because this insoluble ppt. adheres onto the fibre of the cloth as
gummy mass.

Synthetic detergents: They are sodium or potassium salts of sulphonic acid. Eg: sodium
alkylbenzene sulphonate which have a general formula: CH3(CH2)xC6H4SO3Na+

Advantages of detergents:

Detergents can be work in hard water. The anions of synthetic detergent do not precipitate in the
presence of Ca2+ and Mg2+. They can work will even in acidic water.

Types of detergents;

There are three types of detergents;

(a) Anionic detergents are synthesized from long chain alcohol. The long chain alcohols are treated
with conc. H2SO4to form alkyl hydrogen sulphate of high molecular mass and finally alkyl
sulphate are neutralized with alkali to form salts. It is called anionic detergent because large part
of the molecule is anion. The anionic detergent is largest in use as household detergents. E.g.-
Alkylbenzenesulphonate . They are effective in acidic solutions to form an alkyl hydrogen
sulphate which is soluble where as soap are not effective due to formation of insoluble fatty
acids.
CH3(CH2)11OSO3-Na+, CH3(CH2)11C6H4SO3-Na+

(b) Cationic detergent: These are mostly acetates or chlorides of quaternary ammonium salt. They
are more expensive therefore are used to limited extent. Such detergent possess germicidal
properties and are extensively used as germicides. e.g [CH3(CH2)11N+(CH3)3]Br-

(c) Non-Ionic detergent: Some of the detergent are non- ionic , like the esters of high molecular
mass formed by reactions between polyethylene glycol and stearic acid. They do not possess
any ion.

CH3(CH2)16COOH+HO(CH2CH2O)nCH2CH2OH
CH3(CH2)16COO(CH2CH2O)nCH2CH2OH

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Some liquid dishwashing detergents are of non-ionic type branched hydrocarbon chain detergents
are non-biodegradable and cause water pollution. The hydrocarbon side chain stops bacteria from
attacking and breaking the chain. These molecules degrade slowly leading to water pollution.
Unbranched or linear alkyl chain detergents do not create pollution as they are more prone to attack
by bacteria, thus can be biodegraded.

Hands-on/ IT Enabled work:

Whole chapter is done with the help of a presenation with animations.

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Assignment

Chapter 15: Chemistry in Everyday Life

1. (i) Why is bithional added to soap?

(ii) What is tincture of iodine? Write its one use.
(iii) Among the following, which one acts as a food preservative?
Aspartame, Aspirin, Sodium Benzoate, Paracetamol

2. Define the following and give one example :

(i) Anionic detergents
(ii) Antimicrobials
(iii)Antioxidants
(iv)Broad spectrum antibiotics
(v) Artificial sweetening agents
(vi) Foodpreservatives

3 What type of medicines are omeprazole and lansoprazole?

4 Give an example of drug used in case of mental depression.

5 For which disease chloramphenicol is used?

6 (a)Name the sweetening agent used in preparation of a sweet for a diabetic patient.
(b) What problem arises in using alitame as artificial sweeteners?
(c) Why is use of aspartame limited to cold foods?

7 Name a broad spectrum antibiotic and diseases for which it is prescribed.

8 (a)How are antiseptics distinguished from disinfectants? Give 2 examples of each.

(b) Name a substance that can be used as an antiseptic as well as disinfectant.

9 Name the action of the following on the human body:

(i) Equanil (ii) Morphine (iii) Norethindrone (iv)Aspirin (v) Penicillin
(vi) Luminal (vii) Seconal

10 What are the essential components of dettol?

11 What are detergents? Give their scheme of classification. Why are detergents preferred over
soaps?

12 Why is ethanol added to soaps?

13 What are biodegradable and non-biodegradable detergents? What are the consequences of
using the latter kind? Give one example of each kind.
14 Why soaps do not act on hard water?
15 Explain the term ‘chemotherapy’.

16 Describe the function of the following with one example for each :
a. Tranquilizers b. Antifertility drugs c. Antihistamines
d. Analgesics e. Antioxidants f. Antacids

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17 Explain the cleansing action of soaps.

18 Account for the following:

a) Aspirin drug helps in the prevention of heart attack.
b) Diabetic patients are advised to take artificial sweetener instead of natural
sweeteners.
c) Detergents are non-biodegradable while soaps are biodegradable.

19 Except for vitamin B12, all other vitamins of group B, should be supplied regularly in diet.
Why?

20 In order to wash clothes with water containing dissolved calcium hydrogen carbonate
which cleaning agent will you prefer and why, soaps or synthetic detergents? Give one
advantage of soap over detergents

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Practice paper (Summer Vacation)

(Electrochemistry, Chemical Kinetics, Solutions,

p-block elements, Salt Analysis)

Time: 1 hr 45 min Max. Marks –50

No. of printed pages: 2

General Instructions:
(i) All questions are compulsory.
(ii) This paper has two parts.
(iii) Part A is theory from question nos. 1 to 14 carrying 35 marks.
(iv) Part B is practical related questions numbering 15 to 19 carrying 15 marks.
(v) Use log tables if necessary, use of calculators is not permitted.

PART A

1 For a reaction, A+ B → P, the reaction is of first order in reactant A and second order 1
in reactant B.
(i) How is the rate of this reaction affected when the concentration of B doubled.
(ii) What is the overall order of rection if A is present in large excess.

2 Write the anode and cathode reactions occurring in a mercury cell. 1

3 Why ZnO turns yellow and shows enhanced conductivity on heating? 1

4 Draw the structural formulae of molecules of following compounds: 1

a) ClF5 b)XeF4

5 Complete the following chemical equations: 2

a) NaOH(aq) (Hot and Conc) + Cl2(g)
b) XeF6(s) + H2O (l)

6 How many grams of chlorine can be produced by the electrolysis of molten NaCl with 2
a current of 1A for 15 minutes?
(given atomic mass of Cl=35.5u, 1F=96500 C/mol)

7 State Kohlrausch law of independent migration of ions. Write and expression for the 2
molar conductivity of acetic acid at infinite dilution according to Kohlrausch law.

8 Non ideal solutions exhibit either positive or negative deviations from Raoult’s law. 2
What are these deviations and why are they caused? Explain with one example for
each type.

9 (i) Why is the freezing point depression of 0.1M sodium chloride solution nearly 2
twice that of 0.1M glucose solution?
(ii) A solution containing 8g of a substance in 100g of diethyl ether boils at 36.860C ,
whereas pure ether boils at 35.60C. Determine molecular mass of solute. [For
ether, Kb = 2.02 K Kg mol-1]

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10 The decomposition of phosphine, PH3, proceeds according to the following equation: 2
4PH3 (g) → 3P4 (g) + 6H2(g)
It is found that the reaction follows the following rate equation: Rate = k[PH3]
The half-life of PH3 is 37.9s at 1200C. How much time is required for 3/4th of PH3 to
decompose? [log2=0.3010]

11 a) What is meant by the ‘rate constant, k’ of a reaction? If the concentration be 2

expressed in mol L-1 units and time in seconds, what would be the units for k (i) for a
zero order reaction and (ii) for a first order reaction?
b) What type of collisions is known to be effective?

12 A well known mineral fluorite is chemically calcium fluoride. It is known that in one 3
unit cell of this mineral there are 4Ca2+ ions and 8F- ions and that Ca2+ ions are
arranged in a fcc lattice. The F- ions fill all the tetrahedral holes in the face centered
cubic lattice of Ca2+ ions. The edge of the unit cell is 5.46 x 10-8 cm in length. The
density of the solid is 3.18 g/cm3. Use this information to calculate Avogadro’s
number.
(Molar mass of CaF2 = 78.08 gmol-1)

13 Write the Nernst equation and calculate the e.m.f of the following cell at 298K 3
Zn(s)|Zn2+(0.1 M) || Cd+2(0.01M)|Cd(s)
(E0Zn2+/Zn= -0.761V, E0Cd2+/Cd = 0.40V ) . Furthur show:
(i) The carriers of current within the cell.
(ii) E0 values for the electrode 2 Zn+2/2Zn .
(iii) Which electrode is negatively charged?
(iv) Individual reactions at each electrode.

14 a) From the graph: 3

[A] What is the order of the reaction?

b) A first order decomposition reaction takes 40 min for 30% decomposition.

Calculate its t1/2 value. (log10=1, log7=0.8451)

15 a) 2g each of two solutes A and B (molar mass of A > B) are dissolved separately in 3
200g each of the same solvent. Which will show greater elevation in boiling point?

b) The molal elevation constant for H2O is 0.52 K/m. Calculate the boiling point of
solution made by dissolving 6 g of urea (NH2CONH2) in 200 g of H2O.
a) Complete the following chemical reaction equations: 5
16 i. XeF4 (s) + H2O (l)
ii. ClF + H2O
b) Explain the following observations giving appropriate reasons:
i. Halogens are strong oxidizing agents.
ii. Bleaching action of chlorine is permanent where as that of SO2 is temporary.

PART B

17 Why is it that tests for Barium, Strontium and Calcium to be done in order? 2

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18 Explain one confirmatory test for Ni2+. 2

19 (a) Give the formula of reddish yellow vapours evolved during chromyl chloride test? 1,2
(b) Explain the Cl2 water test for iodide with equation.

20 Explain the indicatory and confirmatory tests for sulphide. 3

21 (a) How can one distinguish between sulphite and sulphate using BaCl2 test. 5
(b) What is the colour seen in flame for strontium salt.
(c) What is the yellow ppt. in K2CrO4 test for lead due to?
(d) What is the canary yellow ppt. in Ammonium Molybdate test for Phosphate due
to?
(e) What is the formula for brown ring?

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Academic Session- 2018-19
Preboard Examination
Subject – Chemistry
M/1/1
Time: 3 hrs Max. Marks – 70
No. of printed pages: 7

General Instructions:
(a) All questions are compulsory.
(b) Section A: Q. nos. 1 to 5 are very short answer questions and carry 1 mark each.
(c) Section B: Q. nos. 6 to 12 are short answer questions and carry 2 marks each.
(d) Section C: Q. nos. 13 to 24 are also short answer questions and carry 3 marks each.
(e) Section D: Q. nos. 25 to 27 are long answer questions and carry 5 marks each.
(f) There is no overall choice. However an internal choice has been provided in two questions
of one mark, two questions of two marks, four questions of three marks and all the three
questions of five marks weightage. You have to attempt only one of the choices in such
questions.
(g) Use log tables if necessary, use of calculators is not allowed.

Section- A

1 Analysis shows that FeO has a non-stoichiometric composition with formula Fe0.95O. 1
Give reason.
OR

The intermetallic compound LiAg crystallizes in cubic lattice in which both lithium
and silver have coordination number of eight. What is the type of crystal lattice?

2 What is the reason for stability of lyophilic sols? 1

3 What type of isomers are [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br. 1

OR
Write the structures of geometrical isomers of complex ion [Co(en)2Cl2]+.

4 In the following pairs of halogen compounds, which would undergo SN2 reaction 1
faster? Explain.

5 Identify the monomers in the following polymer: 1

-[NH-(CH2)6—NH—CO—(CH2)4—CO--]n
Section – B

6 The rate constant for a first order reaction is 60 s-1. How much time will it take to 2
reduce 1 g of the reactant to 0.625 g? (log 1.6 = 0.204)

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7 Draw the molecular structures of the following : 2
(a) Noble gas species which is isostructural with BrO3-.
(b) Dibasic oxoacid of phosphorus.

8 Calculate the freezing point of a solution containing 8.1 g of HBr in 100 g of water, 2
assuming the acid to be 90% ionized. [ Given : Molar mass HBr = 81 g/mol, Kf water
= 1.86 K Kg/mol]
OR

Calculate the molality of ethanol solution in which the mole fraction of water is 0.88.

9. (a) Arrange the following in increasing order of Acidic character: 2

MnO2, Mn2O3, MnO
(b) What happens when KMnO4 is heated? Give chemical equation.

10. Explain the mechanism of the following reaction: 2

2C2H5OH conc H2SO4 C2H5OC2H5

413 K

11. How will you bring about the following conversions? 2

(a) Propanone to propane
(b) Ethanal to but-2-enal
OR

Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Ethanal and Propanal
(b) Benzoic acid and Phenol

12. Name the type of polymerization involved in the formation of the following polymers 2
from their respective monomers. Draw the structure of the polymer.
(i)PVC
(ii) Nylon 6

Section- C

13. Observe the graph in diagram and answer the following questions 3

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(a) What is the order of the reaction?
(b) If slope is equal to -2.0 x 10-6 sec -1, what will be the value of rate constant?
(c) Give the relationship between half life and its rate constant?

OR

For a certain chemical reaction

A + 2B 2C + D
The experimentally obtained information is tabulated below.
Experiment [A]o [B]o Initial Rate of
Moles/ L Moles/L reaction
Moles/L/s

1 0.30 0.30 0.096

2 0.60 0.30 0.384
3 0.30 0.60 0.192
4 0.60 0.60 0.768

For this reaction

(a) Derive the order of reaction w.r.t both the reactants A and B.
(b) Write the rate law.

14. Give a reason for the following: 3

(a) Rough surface of catalyst is more effective than smooth surface.
(b) Smoke passed through charged plates before allowing it to come out of
chimneys in factories.
(c) Ne gets easily adsorbed over charcoal than He.

15. Niobium crystallises in body-centred cubic structure. If the atomic radius is 143.1 pm, 3
calculate the density of Niobium. (Atomic mass of Niobium = 93 u).

16 (a) What process takes place when fruits are preserved by adding concentrated sugar 3
solution to protect against bacterial action.
(b) Give Reasons:
(i) When 2 g of benzoic acid is dissolved in 25 g of benzene , the experimentally
determined molar mass is always greater than the true value.
(ii) Mixture of chloroform and acetone shows negative deviation from
Raoult’s Law.

17. Account for the following facts: 3

(a) The reduction of a metal oxide is easier if the metal formed is in the liquid state
at the temperature of reduction.
(b) Limestone is used in the manufacture of pig iron from haematite.
(c) Pine oil is used in the froth floatation process used to concentrate sulphide
ores.
OR

Describe how the following steps can be carried out?

(a) Recovery of Gold from leached gold metal complex.

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(b) Conversion of Zirconium iodide to pure Zirconium,
(c) Formation of slag in the extraction of copper.
(Write the chemical equations also for the reactions involved)

18. (a) Mn2+ compounds are more stable than Fe2+ towards oxidation to +3 state. 3
(b) In 5d-transition series, which element is not regarded as transition metal and why?
(c) Which bivalent cation in 3d-transition series is most paramagnetic and why?

19. (a) Write IUPAC name of [Co(en)3][Cr(C2O4)3]. 3

(b) Discuss the hybridization, shape and magnetic behaviour of [Mn(CN)6]4-.
( Atomic number of Mn = 25)

20. Give reasons: 3

(a) C-Cl bond length in chlorobenzene is shorter than C-Cl bond length in CH3-Cl.
(b) Grignard reagent is kept under anhydrous conditions.
(c) SN1 reactions are accompanied by racemisation in optically active alkyl halide.

OR

Give reasons:
(a) n-Butyl bromide has higher boiling point than t- butyl bromide.
(b) Racemic mixture is optically inactive.
(c) The presence of nitro group (-NO2) at o/p positions increases the reactivity of
haloarenes towards nucleophilic substitution reactions.

21. Complete the following equations: 3

22. Complete the following reactions: 3

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23. (a) Give one example of reducing sugar and one example of non-reducing sugar. 3
(b) Give two differences between RNA and DNA.

OR

(a) Which of the following biomolecule is insoluble in water? Justify.

Insulin, Haemoglobin , Keratin
(b) Draw the Haworth structure for α-D-Glucopyranose.
(c) Write chemical reaction to show that glucose contains aldehyde as carbonyl
group.

24. Explain the following terms with one suitable example for each: 3
(a) A sweetening agent for diabetic patients
(b) Enzymes
(c) Analgesics

Section- D

25. (a) Given are the conductivities and molar conductivities of NaCl solutions at 298 K 5
at different concentrations:

Concentration, Conductivity, Molar Conductivity

M Scm-1 S cm2 mol-1
0.100 106.74 x 10-4 106.7
0.05 55.53 x 10-4 111.1
0.02 23.15 x 10-4 115.8

Compare the variation of conductivity and molar conductivity of NaCl

on dilution. Give reason.

(b) Silver is electrodeposited on a metallic vessel of total surface area 900 cm2 by
passing a current of 0.5 A for two hours. Calculate the thickness of silver
deposited. [Given: density of Ag = 10.5 gcm-3, Atomic mass of Ag = 108 u,
1F= 96500 C/mol]
OR

(a) On the basis of the standard reduction potential values for following solution,
predict whether Ti4+ species may be used to oxidise Fe2+ to Fe3+.
Ti4+ + e- Ti3+; Eo = +0.01 V
Fe3+ + e- Fe2+; Eo = +0.77 V
(b) Why is rusting of iron quicker in saline water than in ordinary water?
(c) A copper-silver cell is set up. The copper ion concentration in it is 0.10 M. The
conc. of Ag+ is not known. The cell potential measured is 0.422 V. Determine
log [Ag+]. [EoAg+/Ag = +0.80 V, EoCu2+/Cu = +0.34 V]

26. (a) Write down the equations for hydrolysis of XeF4 and XeF6. Which of these two 5
reactions is a Redox reaction?
(b) Account for the following:
(i) F2 is strongest oxidising agent among halogens.

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(ii) Fluorine exhibits only -1 oxidation state whereas other halogens exhibit
higher +ve oxidation state.
(iii) Acidity of oxoacid of chlorine is HOCl < HOClO < HOClO2 < HOClO3

OR

(a) Why does chlorine water lose its yellow colour on standing? How chlorine
water has both oxidising and bleaching properties. Give reason
(b) What happens when Cl2 reacts with cold dilute solution of sodium hydroxide?
Write equations only.
(c) Account for the following:
(i) H3PO2 and H3PO3 act as good reducing agents while H3PO4 does not.
(ii) ICl is more reactive than I2

27 An aromatic compound ‘A’ of molecular formula C7H7ON undergoes a series of

reactions as shown below. Write the structures of A, B, C, D and E in the following
reactions:

A Br2+ KOH C6H5NH2 NaNO2 +HCl B CH3CH2OH C

(C7H7ON)
CHCl3 KI
+
KOH
D E
OR

(a) Write the structures of main products when aniline reacts with the following
reagents:
(i) Br2 Water
(ii) HCl
(iii) (CH3CO)2O/pyridine
(b) Arrange the following in the increasing order of their boiling point:
C2H5NH2, C2H5OH, (CH3)3N
(c) Give a simple chemical test to distinguish between the following pair of
compounds:
(CH3)2NH and (CH3)3N

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Academic Session: 2017-18
Pre Board Examination
Subject: Chemistry
M/1/1

Time : 3 Hrs Max marks : 70

General Instructions:

● All questions are compulsory.

● Question nos. 1 – 5 are very short answer questions and carry 1 mark each.
● Question nos. 6 – 10 are short answer questions and carry 2 mark each.
● Question nos.11 – 22 are also short answer questions and carry 3 mark each.
● Question no. 23 is a value based question and carry 4 marks.
● Question nos.24 – 26 are long answer questions and carry 5 mark each.
● Use log tables, if necessary. Use of calculators is not allowed.
● This paper has 5 printed sides.

Q1. In the Arrhenius equation, what does the factor e-Ea/RT corresponds to ? (1)

Q2. Write Reimer -Tiemann reaction. (1)

Q3. What happens when Phenol is treated with Br2 in carbon disulphide as medium. (1)

Q4. Give the structure and IUPAC names of the product expected from the catalytic (1)
reduction of butanal.

Q5. Name the reagent used in dehydrogenation of benzyl alcohol to benzaldehyde. (1)

Q6. Complete the following reactions: (2)

(a) Ca3P2 + H2O

(b) P4 + KOH + H2O
Q7. Which of the following solutions has higher freezing point? (2)

0.05 M Al2(SO4)3 , 0.1 M K3[Fe(CN)6]. Justify.

Q8. Calculate the emf of the following cell at 298 K: (2)

Cr(s) |Cr3+ (0.1M)|| Fe2+ (0.01 M)| Fe(s)

[given: E0cell = +0.30V]

OR

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The conductivity of 10-3mol/L acetic acid at 250C is 4.1 x 10-5 S cm-1. Calculate
its degree of dissociation, if Λ0m for acetic acid at 250C is390.5 S cm2mol-1.

Q9. Give reason: (2)

(i) (CH3)3N is less basic than (CH3)2NH.
(ii) C2H5NH2 has lower boiling point than C2H5OH.
Q10. (i) Give the Haworth projection of α- D(+)glucopyranose. (2)
(ii) Glucose is oxidized and a dicarboxylic acid is obtained. Write the
relevant equation and name the acid formed.

Q11. The freezing point of benzene decreases by 2.12 K when 2.5 g of benzoic acid (3)
(C6H5COOH) is dissolved in 25 g of benzene. If benzoic acid forms a dimer in
benzene, calculate the van’t Hoff factor and the percentage association of
benzoic acid (Kf for benzene = 5.12 K kg mol-1)

Q12. (i) What are the linkages present in polynucleotide chain. (3)
(ii) Name a heterocyclic base present in RNA but not in DNA.
(iii) What are essential amino acids? Give one example.

Q13. Starting from elemental sulphur , how would you prepare H2SO4? Draw the (3)
structure of Sulphuric acid and predict its basicity.

OR

Write equations for the preparation of Nitric acid by Ostwald process? Draw
the structure of Nitric acid and predict its basicity.

Q14. The rate constant for a first order reaction is 60 s-1. How much time will it take (3)
to reduce 1 g of the reactant to 0.0625 g? [ log 16 = 1.204]

Q15. (i) Name and draw the monomer of Nylon-6. (3)

(ii) What are biodegradable polymers? Give one example of polyamide
which is biodegradable.
(iii) Give one use of ethylene glycol in the manufacture of polymers.
Q16. (i)Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar (3)
conductivity of ‘ B’ increases 1.5 times while that of ‘A’ increases 25 times.
Which of the two is a strong electrolyte? Justify your answer.
(ii) The products of electrolysis of aqueous NaCl at the respective
electrolytes are: Cathode : H2; Anode : Cl2 and not O2. Explain ca

Q17. (i) What type of detergents create water pollution. (3)

(ii) What is antifertility drug. Give one example.
(i) What is the cause of feeling of depression in human beings. Name
the class of drugs used for treating depression.

Q18. (i) Which type of isomerism is shown by [ Cr(H2O)6]Cl3, (3)

[CrCl(H2O)5]Cl2.H2O, [CrCl2(H2O)4]Cl.2H2O?

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(ii) Why is [Co(en)3]3+ more stable than [Co(NH3)6]3+?
(iii) How many unpaired electrons are present in [CoF6]3-.
(Atomic number of Co is 27).

Q19. (i) Write the principle for the refining of Germanium. (3)
(ii) What happens when silver ore is treated with sodium cyanide?
Write the chemical reaction involved.
(iii) Can Magnesium be used as a reducing agent for extracting
Aluminium from Alumina at all temperatures. Why/ Why not?

Q20. Give reasons for the following: (3)

(i) Acidic character increases from H2O to H2Te
(ii) Oxygen does not form OF6 whereas sulphur forms SF6.
(iii) Fluorine does not show higher positive oxidation state.

Q21. Write the structures of compounds A, B, C in the following reaction: (3)

(i) C6H5COOH NH3 , ∆ A Br2/KOH(aq) B NaNO2/HCl C

0-5C

(ii) C6H5N2+BF4- Cu/NaNO2 A Sn/HCl B Br2(aq) C

Q22. (i) Write the reaction between 30 haloalkanes containing assymetric (3)
carbon with methyl amine. Will it form optically active or inactive
product.
(ii) Give a chemical reaction to illustrate Williamson Synthesis to
prepare aromatic ether.

Q23. It was observed that due to rising air pollution in Delhi , when the AQI levels (4)
reached the level of 1000, every individual got concerned to protect themselves.
Through the day smog could be seen all over. Air was also filled with
poisonous gases. People adopted to various means to breath in pure air.
Sujata, who is a class 12 science student , advised the family to plant various
forms of plants , use nasal masks, and put air purifiers. Curious brother asked
his sister Sujata, what is contained in air purifier? Sujata said it contains filters
and adsorbents, that helps to clean the air.
(i) What kind of colloid is smog? Name the dispersed phase and
dispersion medium.
(ii) Majority of the particulate matter is filtered at the initial stage of
filters in the air purifier , what is then the role of adsorbents.
Name one adsorbent.
(iii)What values are discharged by Sujata.

Q24. (i) Following is the schematic alignment of magnetic moments : (5)

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Identify the type of magnetism. What happens when these
substances are heated?
(ii) If the radius of the octahedral void is ‘r’ and radius of the atoms in
close packing is’ R’. What is the relation between ‘r’ and ‘ R’?
(iii) Tungsten crystallizes in body centred cubic unit cell. If the edge of
the unit cell is 316.5 pm. What is the radius of tungsten atom?

OR
(i) Identify the type of defect shown in the following figure:

What type of substances show this defect?

(ii) A metal crystallizes in a body centred cubic structure . If ‘a’ is the

edge length of its unit cell, ‘r’ is the radius of the sphere. What is
the relationship between ‘r’ and ‘a’?
(iii) An element with molar mass 63 g/mol forms a cubic unit cell with
edge length of 360.8 pm. If its density is 8.92 g/cm3. What is the
nature of the cubic unit cell?

Q25. (i) Identify the Oxoanion of chromium which is stable in acidic medium. (5)
(ii) What happens when potassium permanganate is heated. Write the
equation for the same.

(iii) The magnetic moments of few transition metal ions are given below:
Metal ion Magnetic moment (BM)
a) Sc 3+ 0.00
b) Cr3+ 4.90
c) Ni2+ 2.84
d) Ti3+ 1.73
(at.no. Sc=21, Ti =22, Cr =24, Ni =28)

Which of the given metal ions:

a. has the maximum number of unpaired electrons?

b. forms colourless aqueous solution?
c. exhibits the most stable +3 oxidation state?

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OR

(i) Write the reaction between dichromate ion and iodide ion in acidic
medium.
(ii) The lanthanoid element that exhibits +4 oxidation state.
(iii) Consider the standard electrode potential values (M2+/M) of the
elements of the first transition series.
Ti V Cr Mn Fe Co Ni Cu Zn

-1.63 -1.18 -0.90 -1.18 -0.44 -0.28 -0.25 +0.34 -0.76

Explain:

(a) E0 value for copper is positive.

(b) E0 value of Mn is more negative as expected from the trend.
(c) Cr2+ is a stronger reducing agent than Fe2+.

Q26. (i) Complete the following reactions: (5)

(a) CH3COOH P2O5, heat

(b) C6H5COOH + Br2 FeBr3

(ii) Write the structural formulae and names of all four possible
aldol condensation products from ethanal and propanal.
(iii) What is the use of formalin?

OR

(i) Distinguish between following pairs of compound by suitable

chemical test:
(a) Benzoic acid and Propanol
(b) Formaldehyde and Acetaldehyde
(ii) Convert :

(a) Benzene to m-nitrobenzoic acid

(b) Benzaldehyde to Phenyl acetic acid
(c) Toluene and Methylbenzoate

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Sample Question Paper Chemistry class XII
2017-18
TIME 3 HRS MM:70

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