salwan public school mayur vihar chemistry.pdf · chemistry, inorganic chemistry and physical...

1

SALWAN PUBLIC SCHOOL

MAYUR VIHAR

CHEMISTRY RECKONER

SESSION: 2020-2021

NAME: ________________________________

CLASS: _____XII_____ SECTION: ____________

2

Preface The structure of chemistry syllabus prescribed by CBSE for class 12 broadly covers the topics: Organic

Chemistry, Inorganic chemistry and physical chemistry. The content of the reckoner corresponding to

these topics has been organized to flow exactly as per the syllabus requirement.

Students will find each chapter enriched with the theoretical explanation along with variety of

questions. The reckoner has been so designed that these contain every necessary tool required for

clear understanding and conceptual development on each topic. The summary of the topic along with

important points to remember are provide for a quick revision. Conceptual problems are also

provided in each unit.

The reckoner includes NCERT textbook examples and questions with solutions and also very short

answer type, short answer type, long answer type question with answers, exemplar problems and

Value Based questions.

The numerical and examples illustrated for each unit/chapter shall enable students to understand

better the hidden aspects of the topic.In order to enhance the spectrum of knowledge already

acquired by the students, Practice Problems are given at the end of each unit. The pedagogy of the reckoner is enriched to extend the breadth of knowledge to inculcate investigatory skills and creative thinking promote analytical and problem-solving abilities

Evaluate progress of conceptual and application-based understanding of content.

Chemistry Faculty, Salwan Public School, Mayur Vihar

3

INDEX SNO. CONTENT Page No.

Syllabus and Mark Division 1-8

Design Of Question Paper 9-13

Unit-I

Unit-II

Unit-III

Unit-IV

Unit-V General Principles And Processes Of Isolation Of

Unit-VI

Unit-VII

Unit-VIII

Unit-IX

Unit-X Alcohols, Phenols And Ethers

Unit-XI

Unit-XII

Unit-XIII

Unit-XIV

Unit-XV

Sample Papers 191-213

4

CLASS XII (THEORY (2018-19)

Total Periods (Theory 160 + Practical 60) Time: 3 Hours 70 Marks

Unit No. Title No. of Periods Marks

Unit-I Solutions 10 23

Unit-II Electrochemistry 12

Unit-III Chemical Kinetics 10

Unit-IV Surface Chemistry 08

Unit-V General Principles and Processes of 08 19 Isolation of Elements

Unit-VI p-Block Elements 14

Unit-VII d -and f -Block Elements 12

Unit-VIII Coordination Compounds 12

Unit-IX Haloalkanes and Haloarenes 12 28

Unit-X Alcohols, Phenols and Ethers 12

Unit-XI Aldehydes, Ketones and Carboxylic 14

Acids

Unit-XII Organic Compounds containing 12

Nitrogen

Unit-XIII Biomolecules 12

Unit-XIV Polymers 06

Unit-XV Chemistry in Everyday Life 06

TOTAL 160 70

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 loweringof vapour pressure, Raoult's law, elevation of boiling point, depression of freezingpoint, 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 Gibbs energy change and EMF of a cell, fuel

cells, corrosion.

5

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, Arrhenious 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 multi-molecular 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: Some p -Block Elements 14 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).

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, magneticproperties, 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).

6

Unit IX: Haloalkanes and Haloarenes 12 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 12 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, electrophillic substitution reactions, uses of phenols. Ethers: Nomenclature, methods of preparation, physical and chemical properties, uses. Unit XI: Aldehydes, Ketones and Carboxylic Acids 14 Periods Aldehydes and Ketones: Nomenclature, nature physical and chemical properties, mechanism hydrogen

in aldehydes, uses.of carbonyl group, methods of preparation, of nucleophilic addition, reactivity of

alphaCarboxylic Acids: Nomenclature, acidic nature, methods of preparation, physical and chemical

properties; uses. Unit XII: Organic compounds containing Nitrogen 12 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 of proteins -

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.

7

Unit XIV: Polymers 06 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 XV: 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.

8

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 Micro-chemical methods are available for several of the practical experiments. Wherever possible, such techniques should be used.

60 Periods

9

Surface Chemistry Preparation of one lyophilic and one lyophobic sol Lyophilic sol - starch, egg albumin and gum

Lyophobic sol - aluminium hydroxide, ferric hydroxide, arsenous sulphide.

Dialysis of sol-prepared in (a) above.

Study of the role of emulsifying agents in stabilizing the emulsion of different oils.

Chemical Kinetics Effect of concentration and temperature on the rate of reaction between Sodium Thiosulphate and

Hydrochloric acid.

Study of reaction rates of any one of the following: Reaction of Iodide ion with Hydrogen Peroxide at room temperature using different concentration of

Iodide ions.

Reaction between Potassium Iodate, (KIO3) and Sodium Sulphite: (Na2SO3) using starch solution as

indicator (clock reaction). Thermochemistry

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

Chloroform.

Electrochemistry

Variation of cell potential in Zn/Zn2+|| Cu2+/Cu with change in concentration of electrolytes (CuSO4 or ZnSO 4) at room temperature. Chromatography Separation of pigments from extracts of leaves and flowers by paper chromatography and determination of Rf values. Separation of constituents present in an inorganic mixture containing two cations only (constituents

having large difference in Rf values to be provided).

Preparation of Inorganic Compounds Preparation of double salt of Ferrous Ammonium Sulphate or Potash Alum. Preparation of Potassium Ferric Oxalate. Preparation of Organic Compounds

Preparation of any one of the following compounds Acetanilide Di -benzal Acetone p-Nitroacetanilide 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.

10

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

food stuffs.

Determination of concentration/ molarity of KMnO4 solution by titrating it against a standard solution of: Oxalic acid,ss 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 - Pb2+, Cu2+, Al3+, Fe3+, Mn2+, Zn2+, Cu2+, Co2+, Ni2+, Ca2+, Sr2+, Ba2+, Mg2+,[NH4]+ Anions – [CO3]2-, S2-, [SO3]2-, [SO4]2-, [NO2]-, Cl- ,Br-, I-, [PO4]3-, [C2O4]2-, CH3COO-

(Note: Insoluble salts excluded)

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.

12

CHEMISTRY (Code No. 043) QUESTION PAPER DESIGN CLASS - XII (2019-20)

S.No. Typology of Questions VeryShort Short Long Long Total %

Answer- Answer Answer- Answer- II Marks Weightag

e Objective -I I (LA-II) type (SA) (LA-I) (5 marks) (VSA) (2Mark (3

(1 Mark) s) marks)

1 Remembering: 2 1 1 - 7 10% Exhibit memory of previously learned

material by recalling facts,

terms, basic concepts and

answers.

2 Understanding : 6 2 2 1 21 30% Demonstrate

understanding of facts and ideas by

organizing, comparing, translating, interpreting,

giving descriptions and

stating main ideas.

3 Applying: 6 2 2 1 21 30% Solve problems to new

situations by applying acquired knowledge, facts,

techniques and rules in a

different way.

4 Analysing : 6 1 2 - 14 20% Examine and break

information into parts

by

identifying motives or causes. Make inferences and find evidence to

support generalizations.

13

Evaluating: Present and defend opinions by making

judgements about

information, validity of

ideas or quality of work

based on a set of

criteria.

Creating: - 1 - 1 7 10%

Compile information

together in a different

way

by combining elements

in

a new pattern or

proposing

alternative solutions.

TOTAL 20x1=20 7x2=14 7x3=21 3x5=15 70(37) 100%

QUESTION WISE BREAK UP

Type of Question Mark per Question Total No. of Total Marks Questions

VSA/ Objective 1 20 20

SA 2 7 14

LA-I 3 7 21

LA-II 5 3 15

Total 37 70

No chapter wise weightage.Care to be taken to cover all the chapters.

Suitable internal variations may be made for generating various templates keeping the overall weightage to different form of questions and typology of questions same.

Choice(s): There will be no overall choice in the question paper. However, 33 % internal choices will be given in all the sections

14

UNIT- I SOLUTIONS

SYNOPSIS

SL.No. Topic Concepts Degree of importance

1 SOLUTIONS Types of solutions **

2 Expressing conc. of solutions ***

3 Henry’s law and solubility

**

4 Vapour pressure of liquid *** and Raoult’s law

5 Colligative properties and ***

determination of molar mass

6 Abnormal molar mass

**

Solutionsare homogeneous mixtures of two or more non reacting substances (components).

Binary solution: The solution consisting of two components i.e. Solute + solvent. Expressing concentration of solutions:

Mass percentage (w/w): Mass% of a component = W of the component x 100 /W solution

Volume percentage (v/v): Volume% of a component = V of the component x 100 /V of the solution.

Mass by volume percentage (w/v): = W solute x 100/ V solution.

Parts per million (ppm): No of parts of the component x 106 / Total number of parts of in

The solution.

Mole Fraction (x): Mole fraction of a component (x) = No. of moles of the Component /Total no

of moles of all the components.

No. moles of the component (n) = mass (W) / Molar mass (M)

For binary solution, eg. xA = nA / nA + nB ; xB = nB/ nA + nB and xA + xB = 1

Molarity (M): The no. of moles of solute per litre of solution.

Molality (m): The no. of moles of solute per kg of solvent. Molality = n solute / W solvent in kg,Unit

mol kg-1.

Effect of temperature: mass %, mole fraction and molality do not change with temperature wherea

Molarity changes with temperature because volume of solution (liquid) changes with temperature.

Solubility: Solubility of a substance is defined as the maximum amount of substance that can be

Dissolved in a specified amount of solvent to a given temperature.

15

Solubility of a solid in liquid:

Effect of temperature: the solubility of a solid in a liquid changes with temperature.

If the dissolution process is endothermic, the solubility should increase with the rise in temperature

and if Itis exothermic the solubility should decrease with rise in temperature.

Effect of pressure: Solubility of a solid in a liquid does not change with pressure as solids and liquids

are highly incompressible.

Like dissolves like: Polar solutes dissolve in polar solvents and non-polar solutes dissolves in non

polar solvents. In general, a solute dissolves in a solvent if the intermolecular attractions are similar in

both.

Solubility of a gas in liquid:

Increases with increase of pressure.

Decreases with increase of temperature. As dissolution of gas in liquid is an exothermic process, the

solubility should decrease with increase in temperature.

Henry’s law: At constant temperature, the solubility of a gas in a liquid is directly proportional to the

pressure of gas over the solution. Or The partial pressure of the gas in vapour phase (P) is directly

proportional to the mole fraction of the gas(X) of the solution.

P = kH.X where kH = Henry‘s law constant. It depends on the nature of gas.

Greater the kH value, the lower will be the solubility and vice versa. K H increases with increase in

temperature so solubility of a gas in a liquid decreases with increase of temperature.

Raoult’s law: Raoult’s law in its general form can be stated as, ― For any solution, the partial vapour

pressure of each volatile component in the solution is directly proportional to its mole fraction in the

solution.According to Raoult’s law for a solution of two volatile liquids,

P1α x 1 , P 2α x2. P1 = Po1 x1, P2 = Po

2 x2.

According toRaoult’s law for a solution containing volatile solvent and non- volatile solute.

P solvent = Posolvent x solvent

Since P solution = P solvent ,

P solution = Posolvent. x solvent.

16

When a non- volatile solute is dissolved in a pure solvent, the vapour pressure of the pure solvent decreases.

P solution < Posolvent because the non-volatile solute particles are present on surface of the solution so the

rate of evaporation of molecules of the volatile solvent from the surface of solution decreases.

Liquid- liquid volatile solutions are classified into two types on the basis of Raoult’s law asfollows:

Ideal solution:

The solution which obey Raoult‘s law at all concentration and at all temperature i.e. PA = PoA.xA,PB = Po

B.xB

If the intermolecular attractive forces between the solute –solvent(A – B interaction) are nearly equal to those

between the solvent – solvent (A – A) andsolute – solute (B – B) it forms ideal solutions. Enthalpy of

mixing, mixing H = 0. Volume change on mixing, mixing V = 0.

Examples: n - hexane and n-heptane, Bromoethane and chloroethane and Benzene and to luene.

Non Ideal solution:

The solution which do not obey Raoult‘s law at all concentration and at all temperature i.e.

PA ≠ PoAxA , PB ≠ Po

BxB

If the intermolecular attractive forces between the solute – solvent (A – B interaction) are not equal (either

stronger or weaker) to those between the solvent – solvent (A – A) and solute – solute (B – B) it forms non-

ideal solutions.

Enthalpy of mixing, ΔHmixing is not equal to 0.

Volume change on mixing, ΔVmixing is not equal to 0.

Examples: Ethanol and acetone, Ethanol and water, Chloroform and acetone Nitric acid and water.

Types of Non-Ideal solutions:

Non-ideal solution showing positive deviations. In this case, intermolecular attractive forces between the

solute and solvent(A – B) are weaker than those between the solvent(A – A) and the solute (B – B) i.e. PA>

PoAxA ,PB> PoBxB

17

mixing H = +ve , mixing V = +ve

Dissolution is endothermic so, heating increases solubility. Ethanol and acetone, Ethanol and

water, CS2 and acetone.

Non-ideal solution showing negative deviations

In this case, intermolecular attractive forces between the solute and solvent(A – B) are stronger than

those between the solvent(A – A) and the solute (B – B) i.e. PA<

PoAxA, PB< Po

BxB . mixing H = -ve, mixing V = -ve

Dissolution is exothermic so, heating decreases solubility.

Examples: Chloroform and acetone, Nitric acid and water, Phenol and aniline.

Azeotropes are binary solutions (liquid mixtures) having the same composition in liquid

andvapour phase and it is not possible to separate the components of an azeotrope by fractional

distillation.

18

Types of azeotropes:

Minimum boiling azeotrope

The non- ideal solutions showing positive deviation form minimum boiling azeotrope at aspecific

composition.

Example; 95% ethanol and 5% water (by volume)

Ethanol = 351.3K , Water = 373 K, Azeotrope = 351.1K

Maximum boiling azeotrope

The non- ideal solutions showing negative deviation form maximum boiling azeotrope at aspecific

composition.

Example : 68% Nitric acid and 32% water (by mass)

Nitric acid = 359K , Water = 373 K, Azeotrope = 393.5K

Colligative Properties:

The properties of solutions which depends on the no. of solute particles but irrespective ofnature of solute

particles.

Example: Relative lowering of vapour pressure , Elevation of boiling point ,Depression offreezing point ,

Osmotic pressure

Application: By measuring a colligative property of a solution molar mass of the non volatile solute can be

calculated Conditions for getting accurate value of molar mass.

Solute must be non volatile.

Solution must be dilute.

c) Solute particles must not undergo any association or dissociation in the solution.

Relative lowering of vapour pressure:

When a non volatile solute is added to a pure solvent its vapour pressure decreases.

Psolution< P0solvent , according to Raoult‘ s lawP solvent = P0

solvent x solvent Where, P solution

P solvent (Since solute is non volatile). P solution = vapour pressure of solution.

P solvent = vapour pressure of solvent in the solution.

P0solvent = vapour pressure of pure solvent.

19

Therefore, P solution = P0solvent.x solvent (Since xsolute+ xsolvent =1) .

0 0

(P0solvent – P solution) = P0

solvent x solute (Lowering of vapor pressure).

Therefore relative lowering of vapour pressure =

( P0solvent – P solution) / P0

solvent = x solute ----- (1)

Since, it depends on mole fraction of the solute (Xsolute), It is a colligative property.

To calculate molar mass : xsolute = n solute /( n solute + n solvent). Since the solution is dilute, n solute is neglected

from denominator.

x solute = n solute/ n solvent = ( Wsolute x Msolution ) /( Msolute x Wsolvent).

Elevation of boiling point :

Boiling point : The temperature at which vapour pressure of liquid becomes equal to atmospheric

pressure.The vapour pressure of the pure solvent decreases when non volatile solute is added to it and the

boiling point is inversely proportional to vapour pressure. Therefore, boiling point of the pure solvent

increases when a nonvolatile solute is added.

i.e. Tb> T0b(solvent) The difference in boiling point of solution and pure solvent is calledelevation of boiling

point (ΔTb). ΔTb = Tb (solution) – ΔT0b (solvent).

For dilute solution, the elevation of boiling point is directly proportional to molality (molal concentration) of

the solution.ΔTb = Kb * molality… … … (1) Where Kb= Boiling point elevation constant or molal elevation

Constant or ebullioscopic constant. Molality (m) = nsolute / W solvent in kg =

Wsolute/(MsolutexWsolvent in kg) … (2)

By substituting (2) in (1), the molar mass of non volatile solute(M solute) can be calculated.

Depression of freezing point

Freezing point: Temperature at which the vapour pressure of the substance in its liquid phase is equal to its

vapour pressure in its solid phase. Since the vapour pressure of the pure solvent decreases in the presence of

non volatile solute and freezing point is directly proportional to vapour pressure, the freezing point of pure

solvent decreases when non volatile solute is added to it Tf(solution)< T0f(solvent).

The difference in freezing point of pure solvent and solution is called depression of

freezing point (ΔTf) ΔTf =T0f(solvent)-Tf(solution).

For dilute solution, the depression of freezing point is directly proportional to molality (molal concentration)

of the solution.

ΔTf = Kf molality …………….(1)

20

Where Kf = freezing point depression constant or molal depression constant or cryoscopic constant.

Molality(m)= W solute / (Msolute xWsolvent in kg ). …………….(2).

Osmosis: The process of flow of solvent molecules from pure solvent to the solution through semi permeable

membrane. Osmotic pressure (Π) : Minimum pressure that must be applied to the solution to prevent the flow of solvent

into the solution through a semipermiable membrane It‘ s a colligative property as it is directly proportional

to molarity (molaroncentration) of the solution at the given temperature.

Π = Molarity × R × T (since Π = (n/V) ×R×T) ………………..(1).

Isotonic solutions: two different solutions having the same osmotic pressure at same temperature.

Hypertonic and hypotonic solutions: a solution which has more Osmotic pressure than other solution is

hypertonic and similarly the solution having less osmotic pressure than the other solution is called hypotonic.

Reverse osmosis: When a pressure larger than osmotic pressure is applied to the solution side, the pure

solvent flows out of solution to solvent side through semi permeable membrane this process is called reverse

osmosis.

Use: Reverse osmosis is used to desalination of sea water (into drinking water).

Abnormal molar mass and van’t Hoff factor (i):

When the non volatile solute particles undergo association (Eg: acetic acid) or dissociation (Eg: electrolytes

like KCl, NaCl etc), the abnormal molar mass is observed.

Formula to calculate i; i = Total number of moles of particles after association or dissociation /Total number

of moles of particles before association or dissociation.

Always’i‘ value is less than unity (i < 1) for association and’i‘ is greater than unity(i >1) for dissociation, the

modified equation for colligative properties for association or dissociation:

1. relative lowering of vapour pressure = i*Xsolute

2. ΔTb = i × Kb × molality

3. ΔTf = i × Kf × molality

4. Π = i × C× R x T

IMPORTANT FORMULAE

Molarity (M) = WB/ MBV(L)

Molality (m) = WB x 1000 / MBWA (g).

Δp/p0 = (p0-p) /p = xB.

ΔTb = kb x WB x 1000 / MBWA.

ΔTf = kf x WB x 1000 / MBWA

π = WBRT / MBV

21

CONCEPT BASED SOLVED QUESTIONS VERY SHORT ANSWER QUESTION (1 MARK )

1. What type of liquids form ideal solutions?

Hints: Liquids having similar structures and polarities.

2. Why does chemist prefer to refer concentration of solution in terms of molality?

Hints: Molality does not change with temperature where as molarity changes with temperature

because volume changes with temperature. Therefore molality is better.

3. What is van’tHoff factor for KCl in aqueous solution?

Hints: i = 2

4. What happens when blood cells are placed in pure water?

Hints: Due to osmosis water enters the blood cells through the cell wall, as a result it swells and

May even burst.

5. Two liquids say X and Y boil at 380 K and 400 K respectively. Which of them is more volatile? Why?

Hints: Liquid – X because lower the boiling point more will be volatile (evaporation).

6. What is the effect of rise in temperature on solubility of gases?

Hints: Dissolution of gas is exothermic process. Hence according to Le- Chatelier‘s principle, the solubility

of gas should decrease with rise in temperature.

7. Mention a large scale use of the phenomenon called ‘reverse osmosis’.

Hints: Desalination of Sea water.

8.Why it is advised to add ethylene glycol to water in a car radiator while driving in a hill station?

Hints:As an antifreeze to lower the freezing point of water.

9. What is the expected Van’t Hoff’s factor ‘i’ value for K3[Fe(CN)6] ?

Hints: Therefore, i = 4.

10. The molar mass of bio molecules is determined by osmotic pressure and not by other colligative

properties. Why?

Hints: Osmotic pressure is measured around room temperature whereas bio molecules are generally

unstable at higher temperatures.

22

SHORT QUESTION (2& 3* MARKS)

1.Why is the freezing point depression of 0.1M NaCl solution nearly twice that of 0.1M glucose solution?

Hints:NaCl is an electrolyte and dissociates completely whereas glucose being a non-electrolyte does

notdissociate.Hence, the number of particles in 0.1M NaCl solution is nearly double than that in 0.1M

glucose solution.Depression in freezing point being a colligative property is nearly twice for NaCl solution

than for glucose solution of same molarity.

2.What is the effect of temperature on Henry’s law constant (KH) and on solubility of a gason liquid?

Hints:Solubility of a gas in liquid decreases with increase in temperature. KH value increases with the

increase in temperature.

3.Calculate the no. of moles of methanol in 5L of its 2m solution, if the density of the solution is 0.981KgL-

1[molar mass of methanol = 32g mol-1]

Hints:Molality = n solute / W solvent in kg 2 mol Kg-1 = nsolute /(W solution – W solute ) = nsolute /

0.981×5 – n solute × 32 /1000.

n solute = 9.21 mol.

4.What type of deviation from ideal behaviour will be shown by a solution of cyclohexane and ethanol?

Why?

Hints:Positive deviation because ethanol has intermolecular hydrogen bond, but the addition of Cyclohexane

will break some of the hydrogen bonds. Therefore solute-solvent molecular attractive force will be weaker

than solute-solute and solvent- solvent molecular attractive force.

5.What care is generally taken during intravenous Injections and why?

During intravenous injection, the concentration of the solution should be same as that of blood so that they

are isotonic. Because if the solution concentration is hypertonic than blood cell will shrink and if it is

hypotonic than blood cell will swells / burst.

6.When X and Y are mixed the solution becomesWarmer and when Y and Z are mixed the solution becomes

cooler. Which of this solution will exhibit positive Deviation and which will be the negative deviation?

Hints:X and Y Show the negative deviation, Y and Z Shows the positive deviation.

23

7 . Calculate the concentration of sugar solution with osmotic pressure of 2.46atm at 300K. Hints:Π = 2.46atm T= 300K R =0.0821LatmK-1 mol-1

= CRT

C = Π /RT = 2.46atm /0.0821LatmK-1 mol-1x 300K = 0.1M = 34.2gL-1.

8.The mole fraction of helium in a saturated solution at 20°C is 1.2 x 10-6. Find the Pressure of helium

above the solution.Given Henry’s constant at 20°C is 144.97 kbar.

Hints: pHe = KH x XHe = (144.97 x 10 3bar)( 1.2 x 10-6 = 0.174 bar.

LONG ANSWER QUESTIONS (5 MARKS)

Q1. Calculate the mass percentage of benzene (C6 H 6) and carbon tetrachloride (CCl 4) if 22 g of benzene

is dissolved in 122 g of carbon tetrachloride.

Solution:

Given,

Mass of solute (WB ) = 22g

Mass of solvent (W B) = 122 g

Mass of solution = 22g + 122 g = 144 g

Mass % of benzene = ?

Mass % of carbon tetrachloride = ? Now, Mass percentage of benzene

And, Mass percentage of carbon tetrachloride

Thus, Mass percentage of benzene = 15.27%

And, Mass percentage of carbon tetrachloride = 84.72%

24

Q2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Solution:

Given, % of benzene by mass = 30%

Thus, % of solvent (carbon tetrachloride) = 70% Mole fraction of benzene in the given solution = ?

This means, 30g of benzene (C6 H 6) is dissolved in 70 g of CCl 4

Now, Molecular mass of benzene (C6 H6 ) = (12 x 6) + (1 x 6) = 78 g mol – 1 Molecular mass of carbon

tetrachloride (CCl4 ) = 12 + (35.5 x 4 ) = 154 g mol – 1

Number of moles

Therefore, Number of moles of C 6H 6

And, number of moles of CCl4

Thus, mole fraction of benzene in the given solution

Thus, mole fraction of benzene in the given solution = 0.457

Q3. Calculate the molarity of each of the following solutions:

30 g of Co(NO 3) 2. 6H2 O in 4.3 L of solution

30 mL of 0.5 M H 2SO4 diluted to 500 mL. Solution:

Given,

Mass of solute (W B) = 30 g

Molar mass of given solute Co(NO 3) 2.6H 2O = 58.7 + 2[14 + (16 x 3)] + 6 ( 2 + 16) = 58.7 + (2 x 62) + (6

x 18)

= 58.7 + 124 + 128 g mol – 1 = 290.7 g mol – 1

Now, Number of moles of Co(NO 3)2 .6H 2O

25

Now, we know that, Molarity

Thus, molarity of given solute = 0.24 M

Given, 30 mL of 0.5 M H2SO4 diluted to 500 mL. Thus,

Thus, required molarity = 0.3 M

Alternate method:

Number of moles present in 1000 ml of 0.5 M H 2SO4 solution = 0.5 mol

Therefore, number of moles present in 1 ml of solution = 1 / 1000 mL

Therefore, number of moles present in 30 mL of solution

Now, we know that, Molarity

Q4. Calculate the mass of urea (NH2 CONH 2) required in making 2.5 kg of 0.25 molal aqueous solution.

Solution: Given,

Molality (m) = 0.25 m

Molar mass (M B) of urea (NH2 CONH 2) = 14 + 2 + 12 + 16 + 14 + 2 = 60 g mol – 1 Mass of solution = 2.5

kg = 2500 gm

Mass of solute (W B) = ?

Now, we know that, 0.25moles=0.25*60=15 g

Mass of solution=1000+15=1015g

1015 g of solution contain=15 g urea

2500g----------------------------=15*2500/1015=37

Thus, mass of given solute urea = 37 g

26

ENRICHMENT EXERCISE

1. Why a person suffering from high blood pressure is advised to take minimum quantity of common salt ?

2. What happens to vapour pressure of water, if a table spoon of glucose is added to it ?

3. Equimolar solutions of glucose and sodium chloride are not isotonic. Why ?

4. Two liquids A and B boil at 145 oC and 190 oC respectively. Which of them has a higher vapour

pressure at 80 oC.

5. Semipermeable membrane of cupric ferrocyanide is not used for studying osmosis in non aqueous

solutions. Why ?

6. Why is camphor preferred as a solvent in the determination of ΔTf ?

7. Addition of HgI2 to aq KI solution shows an increase in the vapour pressure. Why ?

8. What do you mean by mole fraction?

9. Why sugar readily dissolves in H2O despite its covalent nature ?

10. What is the units of molel elevation constant.

11. when is the value of vant Hoff factor more than one.

12. Before giving intravenous injection. What care is generally taken and Why?

13. Explain why a solution of Ethyl alcohol and water cannot be separated into pure components by

fractional distillation.

14. Osmotic pressure measurements are preffered over other colligative properties for molecular mass

determination of polymers. Why?

15. 4% NaOH solution (weight/volume) and 6% urea solution (weight / volume) are equimolar but not

isotonic. Explain.

16. What is the difference between molel elevation constantmolecula elevationconstant?

27

17. Why does vapour pressure of a liquid decrease when a non-volatile solute is added to it?

18. Why is molality of a solution preferred for expressing concentration over molarity?

19. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in litre

of water at 250C, assuming that it is completely dissociated .

20.(a) What is the effect of temp. on the vapour pressure of a liquid? Name one compound which

can be used as semipermeable membrane.

21.What concentration of nitrogen should be present in a glass of water at room temperature?

Assume a temperature of 250C, a total pressure of 1atm and mole fraction of nitrogen in air of 0.78. (KH

for nitrogen = 8.42 X10-7 M/mm Hg).

22. Henry’s law constant for CO2 dissolving in water is 1.67 X 108 Pa at 298K. Calculate the quantity of

CO2 in 1 litre of soda water when packed under 2.5atm CO2 pressure at 298K.

23.The partial pressure of ethane over a saturated solution containing 6.56 X 10-3g of ethane is 1bar. If the

solution were to contain 5.0 X 10-2g of ethane, then what will be the partial pressure of the gas?

24.Determine the osmotic pressure of a solution prepared by dissolving 2.5 X 10-2g of K2SO4 in 2L of

water at 250C, assuming that it is completely dissociated. (R=0.0821L atm K-1 mol-1, molar mass of K2SO4

= 174gmol-1).

25.1g of a non-electrolyte solute when dissolved in 50g of benzene lowered the freezing

point of benzene by 0.40K. Find the molar mass of the solute. (Kf for benzene = 5.12 K kg mol-1).

26.Calculate the amount of KCl which must be added to 1Kg of water so that the freezing point is

depressed by 2K.( Kf for water = 1.86 K kg mol-1).

27.At 250C, the saturated vapour pressure of water is 3.165k Pa (23.75mm Hg). Find the saturated vapour

pressure of a 5% aqueous solution of urea (carbamide) at the same temperature. (Molar mass of urea =

60.05g mol-1).

28. 15g of an unknown molecular material is dissolved in 450g of water. The resulting solution freezes at -

0.340C. What is the molar mass of the material? (Kf for water = 1.86 K kg mol-1).

28

29.A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500g of water.

This solution has a boiling point of 100.420C, what mass of glycerol was dissolved to make this solution?

(Kb for water = 0.512 K kg mol-1).

30.Calculate the freezing point of an aqueous solution containing 10.50g of MgBr2 in 200g of water

(Molar mass of MgBr2 = 184g, Kf for water 1.86 K kg mol-1).

31Calculate the boiling point of a solution prepared by adding 15g of NaCl to 250g of water. (Kb for

water = 0.512 K kg mol-1, molar mass of NaCl = 58.44 g mol-1)

32.A solution prepared by dissolving 8.95g mg of a gene fragment in 35ml of water has an osmotic

pressure of 0.335 torr at 250C. Assuming the gene fragment is non-electrolyte, determine its molar mass.

33.What mass of NaCl must be dissolved in 65g of water to lower the freezing point of water by 7.500C?

The freezing point depression constant (Kf) for water is 1.860C/m. Assume van’t Hoff factor for NaCl is

1.87. (Molar mass of NaCl = 58.5 g mol-1).

34.An aqueous solution containing 12.48g of barium chloride in 1 kg of water boils at

373.0832K. Calculate the degree of dissociation of barium chloride. (Given, Kb for H2O = 0.52 Km-1,

molar mass of BaCl2 = 208.34 g mol-1).

35.At 300K, 36g of glucose, C6H12O6 present per litre in its solution has an osmotic pressure of 4.98bar. If

the osmotic pressure of another glucose solution is 1.52 bar at the same temperature, calculate the

concentration of the other solution.

36.Calculate the boiling point of one molar aqueous solution. Density of KBr solution is

1.06 g ml-1. (Kb for water = 0.52 K kg mol-1, atomic mass; K=39, Br=80).

37.A solution prepared by dissolving 1.25g of oil of winter green (methyl salicylate) in 99g of benzene has

a boiling point of 80.310C. Determine the molar mass of this compound.(Boiling point of pure benzene =

80.100C and Kb for benzene = 2.530C kg mol-1).

38.What mass of ethylene glycol must be added to 5.50kg of water to lower the freezing point of water

from 00C to -100C? (Kf for water is = 1.86 K kg mol-1)

39.Calculate the amount of NaCl which must be added to one kg of water so that the freezing point is

depressed by 3K. (Given, Kf = 1.86 K kg mol-1)

29

KEY POINTS / STUDY TIPS:

1. Difference between molarity and molality

2. Henry’s Law and its and its applications

3. Raoult’s law for solutions containing volatile and non-volatile solute

4. Difference between Ideal and non-ideal solutions

5. Colligative properties

6. Relative lowering in vapour pressure

7. Elevation in boiling point

8. Depression in freezing point)Osmotic pressure

9. Abnormal molecular mass Van’t Hoff factor

MULTIPLE CHOICE QUESTIONS 1. The molality of 98% H2SO4 (density = 1.8 g/mL) by weight is: (a) 6 m (b) 18 m

(c) 10 m (d) 4 m 2. Which of the following does not show positive deviation from Raoult's law?

(a) benzone + chloroform (b) benzene + acetone

(c) benzene + ethanol (d) benzene + CCl4 3. Which solution will have least vapour pressure?

(a) 0.1 M BaCl2 (b) 0.1 M Uxa

(c) 0.1 M Na2SO4 (d) 0.1 M Na3PO4 4. Which condition is not satisfied by an ideal solution?

(a) ∆Hmix = 0 (b) ∆Vmix = 0

(c) ∆Pmix = 0 (d) ∆Smix = 0 5. Azeotrope mixture are:

a) mixture of two solids

b) those will boil at different temperature

c) those which can be fractionally distilled

d) constant boiling mixtures

6. If Kf value of H2O is 1.86. The value of ∆Tf for 0.1 m solution of non-volatile solute is (a) 18.6 (b) 0.186

(c) 1.86 (d) 0.0186

30

7. Solute when dissolve in water

(a) increases the vapour pressure of water

(b) decreases the boiling point of water

(c) decrease the freezing point of water

(d) All of the above 8. The plant cell will shrink when placed in: (a) water (b) A hypotonic solution (c) a hypertonic solution (d) an siotonic solution 9. The freezing point of 11% aquous solution of calcium nitrate will be: (a) 0°C (b) above 0°C (c) 1°C (d) below 0°C

10.The Van’t Hoff factor for 0.1 M Ba(NO3)2 solution is 2.74. The degree of dissociation is: (a) 91.3% (b) 87%

(c) 100% (d) 74% 11. Which of the following solutions would have the highest osmotic pressure:

(a)

M

NaCl (b)

M

Urea

M M

12. 0.5 M aquous solution of Glucose is isotonic with:

(a) 0.5 M KCl solution (b) 0.5 M CaCl2 solution (c) 0.5 M Urea solution (d) 1 M solution of sucrose 13. Which of the following is true for Henry's constant

(a) It decreases with temperature (b) It increases with temperature

(c) Independent on temperature (d) It do not depend on nature of gases.

14.Which one is the best colligative property for determination of molecular mass of polymer? (a) osmotic pressure (b) depression in freezing point (c) elevation in boiling point (d) osmosis

31

CONCEPT MAP OF SOLUTION

32

UNIT- II

ELECTROCHEMISTRY

SYNOPSIS

SL.No. Topic Concepts Degree of

importance

ELECTRO Galvanic cell and electrode **

II CHEMISTRY cell

Nernst equation ***

Conduction of electrolyte ***

solution

Electrolytic cell and ***

electrolysis

Batteries and fuel cells ***

Corrosion **

Types of conductors:

Metallic Conductors or Electronic Electric Conductors or Ionic Conductors

Conductors

Metallic substances which conduct Substances which conduct electricity in

Electricity

through them due to movement of

33

CH3COOH

Metallic Conductance decreases with Electrolytic conductance increases with

increase in Temperature increase in Temperature because

Movement of ions increases.

Types of electrolytes

Strong Electrolytes Weak Electrolytes

The electrolysis which are The electrolyte which do not dissociate

completely dissociated into ions in completely in solution. e.g. Weak Acids:

solution. e.g. Strong Acids: HCl, CH3COOH, H2CO3, HCOOH.

H2SO4, HNO3.

Strong Base: NaOH, KOH. Salts: Weak Base: NH4OH

NaCl, KCl, ZnCl2.

The degree of dissociation (α) = 1. The degree of dissociation (α) < 1.

Resistance (R): The obstruction to the flow of current. S.I. unit of R = Ohm (Ω). R l/A or R = ρ l/A where L

= length of the conductor, A= area of cross section ρ = resistivity or specific resistance.

Resistivity (ρ): The resistance of a conductor of 1 m length and 1 m2 area of Cross

section. Its S.I. unit: Ohm m-1.

Conductance (G): The reciprocal of resistance (R) C=1/R S.I unit of C=S (Siemens).

Conductivity or specific conductance (ҡ) (kappa): It is the conductance of solution

kept between two electrodes with 1 m2 area of cross section and distance of 1 m. It is the reciprocal of

resistivity (ρ).κ = 1 / ρ. S.I unit of κ = S-m-1

R = ρ L/A 1 / ρ = 1/R * L/A κ = G x G*

Conductivity ( κ ) = Conductance (G) x Cell constant (G*)

Cell constant (G*): It is defined as the ratio of the distance between the two electrodes (l) to their area of

cross section A. Its S.I unit is m-1

34

Molar conductivity (^m): It is defined as the conductance of the solution containing one mole of

electrolyte and kept between the two electrodes with unit (1 m2) area of cross section and distance of unit

length (1m).

^m ( S cm2 mol-1) = K ( Scm-1) X 1000(cm3 L-1) /Molarity (mol L-1)

Limiting molar conductivity ( ^m0): The molar conductivity at infinite dilution or zero

concentration.Variation of conductivity (K) and molar conductivity (^m) with conc.(c)

For strong electrolytes: ^m increases slowly with dilution due to increase in movement of ions on

dilution. ^m = ^0 – A c ½ Where A is a constant which depends on the nature of electrolyte i.e. the

charges on the cation and anion.

For weak electrolyte: Λ m increases steeply on dilution due to increase in the number of ions (or the

degree of dissociation).

Kohlrausch law of independent migration of ions:

The limiting molar conductivity of an electrolyte (Λ0m) is the sum of limiting molar conductivities of

cation (λ0+) and anion (λ 0

-). Λ0 m = ٧+ λ0+ + ٧- λ0 Where ٧+ and ٧- are no. of cation and no of anions

formed on respectively.

Applications:

(1)This law is used to calculate ٨0 m for any electrolyte from the λ 0 ofindividual ions

to determine the degree of dissociation (α) α = Ʌm / Ʌ0m where Λ m = the molar conductivity at the

concentration C and Λ0 m = the limiting molar conductivity.

For weak electrolyte like acetic acid, Ka = Cα2 / 1-α.

35

Difference between:

Electrochemical cell or Galvanic Electrolytic Cell

cell orVoltaic Cell

A device in which electrical energy is used

A device in which electrical tobring about a chemical reaction

energy isproduced from chemical

Reaction.

E.g. Daniel cell, dry cell, leads E.g. Electrolysis of molten NaCl,

storage battery. Electrolysis

of dil. Aq. H2SO4 sol. using Pt electrodes.

Daniel cell : Zn (s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu (s) E0cell = 1.10 V

Electrode potential: the potential difference between electrode (metal) and the electrolyte (metal ion

solution).

Cell potential: the potential difference between the two electrodes of a galvanic cell.

Ecell = Ecathode– Eanode

EMF (electromotive force) of cell: the potential difference between the two electrodeswhen no current is

drawn (flowing) through the cell.

S H E (standard hydrogen electrode): It is used as a reference electrode to measure thestandard electrode

potential of the other electrode by assigning standard electrode potential ofS H E is zero. [E0H+ = zero]. It

consists of a platinum electrode (foil) coated with finely divided platinum dipped in an acidic solution

with 1 M( H + (aq) ion )and pure hydrogen gas(at 1 bar)is bubbled through the solution.H+ (aq) | H2 (g) |

Pt (s)

Nernst equation: It shows the relationship between the electrode potential (electrode)and concentration of

the solution.

For the electrode reaction Mn+ (aq) + ne - M (s)

36

E.g. for cell reaction: Cu (s) + 2 Ag +( aq) Cu2+ (aq) + 2 Ag (s)

Electrolysis: .Decompsition of electrolyte into ions by passage of electricity.

Faradays laws:

First law: the amount of substance formed (m) or chemical reaction occurs at any electrode is directly

proportional to quantity of electricity (Q) passed through the electrolyte. m = z Q Where

Z=electrochemical equivalent. Q = I t where Q = quantity of electricity in coulombs (C), I = current in

ampere (A) and t = time in (s).

Second law: When the same quantity of electricity is passed through the different electrolytic solutions,

the amounts of different substance liberated are directly proportional to their chemical equivalent weights.

Faraday: The charge on one mole of electrons is called one faraday (1F) and its value is 96478 C mol-1

or approximately 96500 C mol-1

Application of electrochemical series

1. The substance which has higher standard electrode potential (E0 value) is stronger oxidizing agent or

has greater tendency to get reduced.

The substance which has lower standard electrode potential (E0 value) is stronger reducing agent or

hasgreater tendency to undergo oxidation.

Primary Cells: In these the reaction occurs only once and battery then becomes dead after use over a

period of time. It cannot be recharged and reused again.

Dry cell (leclanche cell) (1.) Anode- Zn container.

(2.) Cathode- graphite rod surrounded by powered MnO2 and carbon.

Electrolyte- a moist paste of NH4Cl and ZnCl2.

Anode: Zn(s) → Zn 2+ (aq) + 2e- ,Cathode: MnO2 + NH4+ + 1e- → MnO (OH) + NH3

Cell potential: nearly 1.5V.

Mercury cell

37

Anode – Zn-Hg amalgam,Cathode - A paste of HgO and carbon

Overall cell reaction: Zn (Hg) + HgO(s) → ZnO(s) + Hg (l)

Cell potential is 1.35V

Secondary batteries: After use, they can be recharged by passing current through it in opposite

direction and so they can be reused again. E.g. Lead storage battery, nickel- cadmium cell.

Lead storage battery:

Anode – Pb plate, Cathode – grid of lead packed with PbO2.

Electrolyte – 38% solution f sulphuric acid (1.3 g / ml)

Anode: Pb(s) + SO42- (aq) → PbSO4 (s) +2e-

Cathode: PbO2 (s) + SO42- (aq) + 4 H +(aq)+2e- → 2 PbSO4 (s) + 2 H2O (l)

Overall cell reaction: Pb(s) + PbO2 (s) + 2 H2SO4 → 2 PbSO4 (s) + 2H2O(l)

used in automobiles and inverters.

On recharging the battery, the above reaction is reversed.

Nickel- Cadmium cell: Anode: Cd,Cathode: Ni(OH)3

Overall cell reaction, Cd (s) + 2 Ni (OH) 3 (s) → CdO (s) + 2Ni (OH) 2 (s) + H2O (l) Nickel cadmium cell

has longer life than the lead storage cell but more expensive to manufacture.

Fuel Cells: galvanic cells which convert the energy of combustion of fuels like hydrogen, methane

directly into electrical energy. E.g. Hydrogen – Oxygen Fuel Cell.

Hydrogen and oxygen are bubbled through porous carbon electrodeinto concentrated aqueous sodium

hydroxide solution. To increase the rate of electrode reaction catalysts like palladium or platinum are

used.

Corrosion: formation of oxides or other salt of metal on the surface of metallic objects when exposed to

air or water. E.g. rusting of iron, tarnishing of silver.

Rusting Of Iron: Corrosion of iron is known as rusting. Rust is hydrated ferric

38

oxide, Fe2O3.X H2O. It is an electrochemical phenomenon. At a particular spot oxidation takes place,

which acts as

Anode: 2Fe (s) → 2Fe2+ +4e- E0 red = - 0.44V Electrons released at anode move through the iron metal

and go to another spot and reduce the oxygen in presence of H + ion.

Cathode: O2 (g) + 4 H + (aq) + 4e- →2H2O (l) E0 red = +1.23V.

Overall reaction: 2Fe (s) + O2 (g) + 4 H + (aq) → 2Fe2+ (aq) + 2H2O (l)

E0 cell = +1.67V.

Methods Of Preventing Corrosion:

Galvanization: the process of coating Zinc over iron.(2). Cathodic protection or sacrificial electrode: In

this method more reactive metal like Mg (or) Zn are made as sacrificial anodeand are connected to iron

pipe or tank.

CONCEPT BASED SOLVED QUESTIONS

VERY SHORT ANSWER QUESTIONS (1 MARK)

(1)Name two metal which can be used for cathodic protection of iron.

Hints:(Mg, Zn)

(2)What is the relationship between E0 cell and equilibrium constant at 298 K?

Hints:E0 cell = 0.059 log KC /n.

(3)Can a nickel spatula be used to stir a solution of copper sulphate? Support your answer with reason.

Hints:[E0Ni

2+ / Ni = - 0.25V, E0Cu

2+/Cu = + 0.34V]

No, since nickel has lower E0 value than copper it undergoes oxidation.

(4) A Leclanche cell is also called dry cell. Why?

Hints:Leclanche cell consists of zinc anode (container) and carbon cathode. The electrolyte is a oist paste

of MnO2, ZnCl 2, NH4Cl and carbon black. Because there is no free liquid in the cell, it is called dry cell.

(5) What are fuel cells?

Hints:A fuel cell is a galvanic cell for converting the energy of a fuel directly into electrical energy

without use of a heat engine.

(6) What is meant by Faraday‘s constant?

Hints:Faraday‘s constant is the quantity of electricity carried by one mole of electrons. 1 F = 96500 C/mol

39

(7) Define the term – Resistivity?

Hints:The resistively of a substance is its resistance when it is one meter long and its area of cross Section

is one m2 .

(8) State the factors that affect the value of electrode potential?

Hints:Factors affecting electrode potential values are – a) Concentration of electrolyte b) Temperature.

(9) Define the term – standard electrode potential?

Hints:When the concentration of all the species involved in a half-cell is unity, then the electrode

potential is called standard electrode potential.

(10) What does the positive value of standard electrode potential indicate?

Hints:The positive value of standard electrode potential indicates that the element gets reduced more

easily than H+ ions and its reduced form is more stable than Hydrogen gas.

SHORT QUESTIONS (2 MARKS)

(1) Give reasons:

Rusting of iron is quicker in saline water than in ordinary water.

Aluminium metal cannot be produced by the electrolysis of aqueous solution of aluminium salt.

Hints: (a) Because the conductivity of saline water is more than ordinary water.

Al is highly reactive and cannot be reduced easily as compared to Al 3+ ions; water is reduced easily

since E0 reduction for water is higher.

(2) Predict the products of electrolysis obtained at the electrodes in each case when the electrodes used

areplatinum.

An aqueous solution of AgNO3(b) A dilute aqueous solution of H2SO4. Hints: (a) Silver is deposited at

cathode and oxygen is anode.

Cathode: Ag+ (aq) + 1e- → Ag (s),

Anode: 2 H2O (l) → O2 (g) + 4H+ + 4e-

Hints:(b) H2 gas at cathode and O2 gas at anode. Cathode: H2O (l) + 1e- → ½ H2 (g) + OH-, Anode: 2 H2O

(l) → O2 (g) + 4H+ (aq) + 4e-

40

Fill in the blanks-

1 F is the charge present on __________ of electrons.

1 F passed through CuSO4 sol deposits _________ of Cu. Hints:(i) 1 mole (ii) ½ mole.

Q. Name the cell used for low current devices like hearing aids, watches etc. Also give the half-cell

reactions for such a cell?

Hints:. This cell is mercury cell – Half cell reactions are

Anode: Zn (Hg) + 2OH- ZnO + H2O + 2e- and

Cathode: HgO + H2O +2e- Hg (e) + 2OH-

Q. The conductivity of metals decreases while that of electrolytes increases with increases in

temperature.Why?

Hints:With increase in temperature, the K.E. of metal cation increases and obstructs the free flow of

electrons decreasing the conducts of metal while in case of electrolytes, increased temperature increases

the mobility of ions this increases the conductance of ions.

Q. Can an electrochemical cell act as electrolytic cell? How?

Hints: Yes, An electrochemical cell can be converted into electrolytic cell by applying an

externalopposite potentialgreater than its own electrical potential.

Q. What is an electrochemical series? How does it predict the feasibility of a certain redox reaction?

Hints:The arrangement of metals and ions in increasing order of their electrode potential values is known

as electrochemical series. The reduction half reaction for which the reduction potential is lower than the

other will act as anode and one with greater value will act as cathode .Reverse reaction will not occur.

Q. Give some uses of electrochemical cells?

Hints:Electrochemical cells are used for determining the a) pH of solutions b) solubility product and

equilibrium constant c) in potentiometric titrations

Q.How is standard electrode potential of a cell related to :- 1) Equilibrium constant? 2) Gibbs free energy

change.

Hints: (i) Standard electrode potential and equilibrium constant

E0cell = log kc Where E0 cell = standard electrode potential of cell, R = Gas constant, T = temperature in

Kelvin,n= no .of electrons, F = Faraday‘s constant and Kc = Equilibriumconstant

Standard electrode potential and Gibbs free energy change- ΔG0 = - n F E0 cell

Where ΔG0 = Change in Gibbs ‘free energy n = No. of electrons.

41

Q. What are the factors on which conductivity of an electrolyte depend?

Hints:The conductivity of an electrolyte depends upon

The nature of electrolyte added.

Size of the ions produced and their solvation

Concentration of the electrolyte

Temperature

Q. What do you mean by primary and secondary battery?

Hints:In the primary batteries, the reaction occurs only once and after the use over a period of time battery

becomes dead and cannot be reused again. A secondary batteryafter used, can be recharged by passing

current through it in the opposite direction so that it can be used again.

ENRICHMENT EXERCISE

1. What is electronic concept of oxidation and reduction?

2. What are weak and strong electrolytes?

3. Which type of reaction is corrosion –oxidation reduction?

4. Give expression of equilibrium constant (k) for a reaction

Zn(s) + Cu2+ (aq) Zn2+ (aq) +Cu(s)

5. Dissociate the following electrolytes: NaCl, CuCl2, AgNO3

6. Conductivity of a solution is because of ______________________.

7. How many electrons are needed for the reduction of Ag+ to Ag and MnO4- to Mn2+?

8. Can youstir a copper sulphate solution with a nickel spoon?

9. Which electrode is considered positive terminal in a cell?

10. What is the relation between resistance and conductance?

11. What are the signs of G, K and E0cell for a spontaneous reaction?

42

12. For a reaction: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

a). Name the positive and negative terminals.

b) What is the function of salt bridge?

c) What is the direction of flow of current?

13. Two metals A and B with electrode potential of -0.76 V and +0.80 respectively which of the two will

liberate hydrogen from dil. H2SO4?

14. A current of 0.50A was passed through an electrolyte solution containing AgNO3 solution with inert

electrodes. The weight of Ag deposited was 1.50g .How long did the current flow?

15.Why does the conductivity of a solution decreases with dilution?

16 .Given the Eo of cell for K+/K = -2.93V, Hg2+/Hg= 0.79V Cr3+/Cr=-0.74V, Ag+/Ag=0.80V Mg2+/Mg= -

2.37. Arrange these metals in their increasing order of reducing power.

17. Give the products of electrodes of aqueous solution of CuCl2 with platinum electrodes.

18. Why Kc(equilibrium const.) is related only to Eocell and not E cell

19. Which type of metal can be used in cathodic protection of iron against rusting?

20. Conductivity of 0.00241 M acetic acid is 7.896*10-5 Scm-1.Calculate its molar conductivityand if ^0

for acetic acid is 390.5 Scm2/mol; what is its dissociation constant?

21. Calculate the equilibrium constant Kc for the reaction at 298K,

Zn(s) + Cu2+(aq) ↔ Zn2+(aq) + Cu(s) ∆G0 = -212300 kJ mol-1.

[Given, E0Zn2+/Zn = -0.76V, E0

Cu2+/Cu = +0.34V]

22. (i) Write the formulation for galvanic cell in which the reaction

Cu(s) + 2Ag+ (aq) --→ Cu2+(aq) + 2Ag(s)

takes place. Identify the cathode and the anode reactions in it.

43

(ii) Write Nernst equation and calculate the emf of the following cell

Sn(s)/Sn2+(0.04M) II H+(0.02M)/H2(g)/Pt(s) [given E0 Sn2+/Sn = -0.14V]

23. Calculate the cell emf and ∆G0 for the cell reaction at 250C.

Zn(s)/Zn2+(0.1M) II Cd2+(0.01M)/Cd(s)

[Given, E0 Zn2+/Zn=-0.763V, E0

Cd2+/Cd= -0.403V, 1F = 96500 C mol-1, R = 8.314 JK-

1mol-1]

24. Calculate the standard electrode potential of Ni2+/Ni electrode if emf of the cell, Ni(s)/Ni2+(0.01M) II

Cu2+(0.1M)/Cu(s) is 0.059V. [Given, E0Cu2+/Cu = 0.34V]

What will happen when?

Eext =1.10V

Eext <1.10V

Eext > 1.10 V is applied on the cell Zn / Zn2+ || Cu2+/ Cu ? 26. What is SHE? How will you determine the

Potential of Zn electrode Cu electrode using SHE?

27. Account for the following:

Zn displaces hydrogen from dilute HCl while Cu can not.

Li is the strongest reducing agent while Fluorine is strongest oxidising agent.

copper sulphate solution can not be stored in zinc vessel.

E0 Cu2+ / Cu = + 0.34 V E0 Zn2+ / Zn = - 0.76 VE0 Li+ / Li = - 3.05 V E0 F2 /2 F- = + 2.87 V)

28. Explain the terms

1. resistance

2. resistivity

3. conductance

4. conductivity

5. molar conductivity.

Q. Write the Kohlrausch equation and draw a graph to show the variation of molar conductivity with

dilution for

KCl

CH3COOH.

Q.State Kohlrausch law of independent migration of ions.. Mention two applications of the law.

44

Q. Give the products of electrolysis of

NaCl (molten)

NaCl (aq)

CuSO4(aq) using inert electrode like Pt

CuSO4(aq) using Cu electrodes

AgNO3(aq) using Ag electrode

Q. Write the reactions involved at each electrode in mercury cell. Why does the cell potential of this cell

remain constant?

Q. What are fuel cells? Write the reactions involved at each electrode in H2-O2 fuel cell.

Q. What are secondary cells? Write the reactions involved at each electrode in lead storage cell when

battery is in use

battery is not in use.

Q. Explain corrosion of iron as a electrochemical process. Mention the methods to prevent corrosion.

KEY POINTS / SUBJECT TIPS:

Working of galvanic cell and the effect of external EMF on its working.

Representation and Calculation of EMF of Cell

Nernst Equation and its derivation for electrode potential, Emf of Cell, Kc and Gibbs free energy

Calculation of conductivity and molar conductivity

Variation of molar conductivity with dilution for strong and week electrolyte.

Electrolysis of aqueous and molten NaCl solution

Faraday’s laws of electrolysis

Primary and secondary storage cells and their working

Fuel cell and Corrosion

45

MULTIPLE CHOICE QUESTIONS (1 Mark) 1. When a lead storage battery is discharged: (a) SO2 is evolved (b) lead is formed (c) H2SO4 is consumed (d) PbSO4 is consumed 2. How many coulomb are required for the oxidation of 1 mol of H2O2 to O2?

(a) 9.65 ×104C (b) 93000 C

(c) 1.93 × 105C (d) 19.3 × 102C 3. KCl is used in salt bridge because:

(a) It forms a good jelly with agar-agar

(b) It is a strong electrolyte

(c) It is a good conductor of electricity (d) Migration factor of K+ and Cl– ions are almost equal 4. The nature of curve of E° cell against log KC is: (a) a straight line (c) a hyperbola (b) parabola (d) an elliptical curve

5. For a spontaneous reaction the ∆G, equilibrium constant (K) and E°cell will be respectively. (a) – ve, < 1, – ve (b) – ve, > 1, + ve (c) – ve, > 1, – ve (d) + ve, > 1, – ve

6. Determine the value of E°cell for the following reaction, cu2+ + Sn+2 → Cu + + Sn+4, equilibrium constant is 106.

(a) 0.1773 (b) .01773

(c) 0.2153 (d) 1.773 7. Which is the best reducing agent?

(a) F– (b) Cl–

(c) Br– (d) I– 8. If a salt bridge is removed between the half cells, the voltage: (a) drops to zero (c) increase gradually (b) does not chang (d) increases rapidly

46

9. Faraday's law of electrolysis are related to the:

(a) Atomic number of the cation

(b) atomic number of the anion

(c) equivalent weight of the electrolyte

(d) speed of the cation 10. A current of 9.65 amp flowing for 10 minutes deposits 3.0 g of a metal. The equivalent wt. of the

metal is: (a) 10 g (b) 30 g

(c) 50 g (d) 96.5 g

47

UNIT- III

CHEMICAL KINETICS

SYNOPSIS

SL.No. Topic Concepts Degree of Importance

III CHEMICALS Expression of rate of *

KINETICS Reaction

Rate law, order & ***

molecularly of a reaction

Integrated rate method ***

alongwith half life

Arrhenius equation and **

temperature dependence

Effect of catalyst and **

Collision Theory of

chemical reactions

Rate of chemical reaction:-The change in molar concentration of the species taking part in a reaction per

unit time.

Unit is mol L-1S-1 or mol L-1 min-1

For gaseous reactions atm s-1 or atm min-1

Average rate:-obtained by dividing the change in concentration of any one of the reactant

or product by the time taken for the change ie Δx/Δt

For a reaction of the type

aA +bB→ cC+dD ,the average rate expressions are

r ave = - [A]/ t×1/a = - [B]/ t ×1/b = + [C]/ t ×1/c = + [D]/ t ×1/d

The negative sign implies that the change in concentration of reactant is negative Average

rate does not give the true rate of the reaction

Instantaneous rate of a reaction:- The rate of a reaction at a particular moment of time.

Average rate expression becomes instantaneous rate expression as Δt→0

Instantaneous rate rinst= dx/dt where dx- infinitesimally small change in concentration and

dt- infinitesimally small interval of time.

Instantaneous rate is experimentally obtained by taking the tangent at any instant on the curve obtained by

plotting concentration vs time.

48

QUESTIONS

Q1. Write the average and instantaneous rate expressions of the following reaction.

5Br-(aq)+BrO3-(aq) +6H +(aq) → 3Br2(aq) +3H2O(l)

Rav = - [Br-]/ t × 1/5 = - [BrO3-]/ t = - [H+]/ t ×1/6 = + [Br2]/ t×1/3 = +

[H2O]/ t×1/3

rinst = - d [Br-]/dt ×1/5 = -d [BrO3-]/dt

= +d[Br2]/dt ×1/3= +d[H2O]/dt ×1/3

Q2. In a reaction 2A→Products, the concentration of A decreases from 0.5 mol L-1 to 0.4 molL-1 in 10

minutes. Calculate the rate during this interval.

= ½ (0.4-0.5) mol L-1/10 min = 0.005 mol L-1 min-1

Q3. A chemical reaction 2A→ 4B + C in gas phase occurs in a closed vessel. The concentration of B is

found to increase by 5×10-3 mol L-1 in 10 seconds, calculate

(a) the rate of appearance of B (b) the rate of disappearance of A.

Sol: - rate = -1/2 x Δ[A]/Δt= +1/4 x Δ[B]/Δt = +Δ[C]/Δt

Rate of appearance of B = 5×10-3 mol L-1/10 s= 5×10-4 mol L-1 s-1

Rate of disappearance of A,I .e. Δ[A]/Δt = 2/4×Δ[B]/Δt

1/2×5×10-4 mol L-1S-1 =2.5×10-4 mol L-1s-1

Rate law: -The expression in which reaction rate is given in terms of molar concentrations of reactants

with each term raised to some power, which may or may not be same as the stoichiometric coefficient of

the reacting species in a balanced chemical equation.

For a general reaction

aA + bB→ cc + dD The rate law expression isRate =k [A]x[B]y

Where k- rate constant, ’x‘ may or may not be equal to ‘a‘ , ‘y‘ may or may not be equal to ‘b‘.

Order of a reaction:-The sum of exponents of the concentration of the reactants in the rate law expression.

In the rate law expression; rate = [A]x [B]y

‘x‘--- order of the reaction w.r.t reactant A ,‘y‘ order of the reaction w.r.t reactant B x+y is the overall

order of the reaction.

For nth order reaction the unit of rate constant ‘k’ is (mol L-1)1-n s-1.

49

Q4. The rate constant of a reaction is 2.3x 10-5mol-1Ls-1. What is the order of the

reaction?

Here 1-n = -1, so n=2 i.e., the reaction is of second order.

Molecularity of a reaction: -The number of reacting species (atoms,ions or molecules) taking part in an

elementary reaction which must collide simultaneously in order to bring about a chemical reaction.

Complex reactions involving more than three molecules in the stoichiometric equation, must(is assumed

to be) be taking place in more than one step (elementaryreaction).

The overall order of the reaction is controlled by the slowest step ,called rate determining step

The rate can be written from the slowest step

For single step reactions, the order and molecularity will be the same.

Q5. Suggest a probable mechanism for the reaction

The rate law is rate = k[H2O2][I-].

H2O2 + I-→ H2O + IO-------- slow (2) H2O2 + IO-→ H2O + I- +O2----- fast

Q6. Write any 3 important differences between order &Molecularity

ORDER

MOLECULARITY

1. Theoretical concept. Determined experimentally.

2.Can be zero or fractional number. Always whole numbers.

3.For complex reactions, order is

applicable for both the elementary Applicable only for elementary

and overall reaction. reactions and no meaning for overall

complex reaction.

50

Q7.For a reaction A + B → Products, the rate law is given by Rate =k [A]1/2 [B]2, what is the order of the

reaction?

Order = 1/2 + 2= 2.5,Integrated rate expression for zero order reactions k= [R]0 – [R]/ t where k—rate

constant for zero order reaction

[R]0--- initial concentration of the reactant [R]---- concentration of the reactant at time ‘t‘

variation in the concentration Vs time graph for zero order reaction

For first order reaction K = 2.303/t log [R]0/[R]

Slope= -k/2.303

[logR0]

Plot of variation of log[R] Vs time for a first order reaction

For first order gas phase reaction of the type A(g) → B(g) +C(g)

K = 2.303/t log pi/(2pi-pt)

Pi---- initial pressure of gas A ,Pt----- partial pressure of gas A at time ’y‘.

Half- life of a reaction:- Time taken for reducing the concentration of a reactant to one half of it‘s

Concentration.

For zero order reaction, half- life t1/2 = [R]0/2k

For first order reaction t1/2 = 0.693/k

For nth order reaction t1/2α [R]01-n

51

Q7. The rate constant of a reaction w.r.t.the reactant A is 6 min-1. If we start with [A]=0.8 moL-1, when

would [A] reach the value of 0.08 molL-1?

For first order reaction, t = 2.303/k log [R]0/[R] ,[R]0 = 0.8 mol L-1, [R] = 0.08 mol L - 1, k = 6 min-1,t =

2.303/6 min-1 log [0.8mol L-1/0.08 molL-1] = 2.303/6 min-1 x log 10 = 0.38 min.

Q8. If half- life period for a first order reaction in A is 2 min. How long will it taketo reach 25% its initial

concentration? For first order reaction k = 0.693/t1/2

k =0.693/2 min-1.

k = 0.3465 min-1, [R]0 = 100 ,[R] = 25

t= 2.303/k log [R]0/[R] = 2.303/0.3465 min-1. log 100/25 = 2.303/0.3465 min-1 log 4 = 6.65min.

Pseudo first order reaction: - reaction which appears to be a second order ,butactually first order are

called pseudo first order reaction.Eg:- hydrolysis of esters

CH3COOC2H5 +H2O ---→ CH3COOH + C2H5OH Rate law for this reaction is; rate =

k[CH3COOC2H5][H2O]

But the concentration of water does not change during the course of the reaction

So; [H2O] is constant Therefore rate = k1[CH3COOC2H5] . The reaction behaves as a first orderreaction

Arrhenius equation for temperature dependence of rate constant:-

k= Ae-Ea/RT Or log k = logA – Ea/2.303RT, log k2/k1 = Ea/2.303[T2-T1/T1T2] where A = pre- exponential

factor; Ea— Activation energy, R=gas constant, T=temperature in kelvin scale. K1 and k2 are rate

constant.

slope = -Ea/2.303R

Activation energy :- The energy required to form the reaction intermediate {refer

NCERT Text Book page no: 112}

52

Effect of catalyst on reaction rate :- A catalyst alters the rate of a reaction as it provides an alternate

pathway or reaction mechanism by reducing the activation energy between reactants and products and

hence lowering the potential energy barrier [Refer NCERT Text Book fig.4.11 page no ; 115]

A catalyst (a) does not alter the Gibb‘s energy ΔG of a reaction.

Catalyses only spontaneous reactions (c) does not change the equilibrium constant of a reaction (d) is

highly specific

Collision theory of chemical reactions

collision frequency :-a)The number of collisions per second per unit volume of the reaction mixture

{Refer NCERT Text Book page no 115&116}

According to collision theory; rate = p ZAB e –Ea/RT P --- probability factor; Z-

---collision frequency of reactants A&B e-Ea/RT represents the fraction of molecules with energies equal

to or greater than Ea.



Q.The reaction A+B →C has zero order, write it‘s rate law equation.

Ans:- rate =k [A]o[B]0

Q.Give one example of a fractional order reaction.

Ans:- H2+ Br2 → 2HBr rate = k[H2][Br2]1/2 order of the reaction = ½ 3.

Q.What is an elementary reaction?

Ans:- the reaction place in a single step eg :- N2(g)+ O2(g) → 2NO(g)

Ā Ā ᜀ Ā ᜀ Ā ᜀ Ā ᜀ Ā ᜀ Ā ᜀ Ā ᜀ Ā ᜀ Ā ЀĀ Ā Ā ᜀ Ā ᜀ Ā ᜀ Ā ᜀ Ā ᜀ Ā ᜀ Ā ᜀ Ā I

Q. In some chemical reactions , it is found that a large number of colliding molecules have energy more

than threshold energy value , yet the reaction is quite slow.Why? Ans:- The reactant molecules may not

be properly oriented.

53

SHORT ANSWER QUESTIONS (2/3* MARKS)

1. What is activation energy? How is it related to rate constant K?

2. Give the unit of rate constant for

zero order

first order

3.Draw the graph of reaction coordinate Vs potential energy for a reaction

a) with out catalyst

b) with catalyst.

4. Generally rate of a reaction doubles when the temperature is raised by 10K. Explain this statement with

the help of the Maxwell Boltzmann distribution curve.

5. Draw the graphs

a) concentration Vs time for reactant.

b) Concentration Vs time for product

c) potential energy diagram of a catalysed reaction.

ENRICHMENT EXERCISE

1. Write rate law for zero order reaction.

2. What is the order of reaction for the following rate law expression: Rate =k[A]3/2[B]1/2

3.Express the relation between the half- life period of a reactant and its initial conc. for a reaction of nth

order.

4.How does the value of rate constant vary with reactant conc.?

5.How does rate constant vary with activation energy?

6.A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of reaction?

54

7. The rate constants for a first order reaction is 0.005 minute-1 . Calculate its half -life.

If slope of line obtained by plot of log [N2O5] vs Time is -2.147 X 10-4 s-1. Calculate the value of ‘k’.

8.Define pseudo unimolecular reaction. Give two examples.

9.The rate of decomposition of NH3 on platinum surface is zero order.

10. What is the rate of production of N2 and H2, If k = 2.5 X 10-4M s-1?

11.The half- life of radioactive decayof C -14 is5730years. An archeological artifact contained wood had

only 80%of the C -14 found in living tree .Estimate the age of the sample.

12.At 300 k, a certain reaction is 50% complete in minutes. At 350 k, the same reaction is 50% complete

in 5 minutes. Calculate the activation energy of the reaction?

13.Explain- (1) rate law (2) order of reaction (3) molecularity.

14.What is the effect of catalyst on rate of reaction?

15.A first order reaction is 20 % complete in 10 minutes .Calculate the time for 75 % completion of the

reaction.

16.State and explain Arrhenius equation. How can we determine the activation energy of areaction using

this equation?

17.For the assumed reaction X2 + 3Y2 →2XY3 , write the rate equation in terms of rate of disappearance of

Y2 and X2.

18.Decomposition of NH3 (g) on surface of catalyst 2NH3↔N2 (g) + 3H2(g) .Under low pressure follows

first order kinetics while at high pressure it is zero order reaction. Why?

19.In the reversible reaction 2N 2↔N2O4.Find out the rate of disappearance of NO2..

20.Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t1/2)

of the reaction.

21.Describe the role of activated complex in the reaction and give its stability with activation energy?

55

22.The reaction, N2(g) + O2(g) ↔ 2NO(g) contributes to air pollution wherever a fuel is burnt in air at a

high temperature. At 1500K, equilibrium constant, K for it is 1.0 x 10-5. Suppose in a case [N2] = 0.80

mol L-1 and [O2] = 0.20molL-1 before any reaction occurs. Calculate the equilibrium concentrations of the

reactants and products after the mixture has been heated to 1500K.

23.A first order reaction takes 100 min for completion of 60% of the reaction. Find the time when 90% of

the reaction will be completed.

24.The thermal decomposition of HCOOH is a first order reaction with a rate constant of 2.4 x 10-3 s-1 at

a certain temperature. Calculate how long will it take for third-forth of the reaction to complete. (log 0.25

= -0.6021)

25.A first order reaction has a rate constant value of 0.00510 min-1. If we begin with 0.10 M concentration

of the reactant will remain after 3 hours?

26.The decomposition of phosphine, PH3 proceeds according to the following equation

4PH3(g) → P4(g) + 6H2(g)

27.It is found that the reaction follows the following rate equation, Rate = k [PH3]. The half life of PH3 is

37.9 sec at 1200C.

28.How much time is required for 3/4th of PH3 to decompose?

29.What fraction of the original sample of PH3 remains behind after 1min?

30.The decomposition of a compound is found to follow a first order rate law. If it takes 15 min for 20%

of original material to react, calculate

31.The rate constant The time at which 10% of the original material remains unreacted?

32.The half life of a first order reaction is 5 x 104s. What percentage of the initial reactant will react in 2

h? Hydrogen peroxide, H2O2 (aq) decomposes to H2O(l) and O2(g) in a reaction that is of first order in H2O2

and has a rate constant, k = 1.06 x 10-3 min-1.

33.How long will it take 15% of a sample of H2O2 to decompose?

34. How long will it take 85% of a sample of H2O2 to decompose?

56

35.The half life for decay of radioactive 14C is 5730years. An archaeological artifact containing wood has

only 80% of the 14C found in a living tree. Estimate the age of the sample.

36.The decomposition of A into products has a value of K as 4.5 x 103S-1 at 100C and energy of activation

60kJmol-1. At what temperature would K be 1.5 x 104s-1?

37.The rate of a reaction becomes four times when the temperature changes from 293K to 313K.

38.Calculate the energy of the activation (Ea) of the reaction assuming that it does not change with

temperature. (R = 8.314 JK-1 mol-1, log 4 = 0.6020)

39.The activation energy for the reaction 2HI(g) → H2(g) + I2(g) is 209.5 kJ mol-1 at 581K. Calculate the

fraction of molecules having energy equal to or greater than activation energy. (R = 8.314JK-1mol-1)

40.Explain the following terms:

a) average rate of a reaction

b) instantaneous rate of a reaction.

c) rate constant of a reaction

d) rate law of a reaction

e) order of a reaction

41.What do you mean by pseudo first order reaction.Give one example.

42.Give one example each of

a) zero order reaction

b) first order reaction.

42.Derive integrated rate law for

a) zero order reaction

b) first order reaction. and Show that t1/2 for

a) zero order reaction is directly proportional to initial concentration of the reactant. b) first order reaction

is independant initial concentration of the reactant.

57

KEY POINTS /Subject Tips:-

1.Instantaneous and average rate of reaction

2.Factors affecting rate of a chemical reaction

3.Molecularity and order of reaction.

4.Units of rate constant of different orders of reactions

5.Integrated rate law expression for zero and first order reactions

6.Pseudo first order reactions

7.Temperature dependence of rate of a chemical reaction

8.Arrhenious equation and collision theory

MULTIPLE CHOICE QUESTIONS (1 Mark)

1. The half life period of a first order reaction is 100° seconds. Its rate constant is:

(a) 0.693 sec–1 (b) 6.93 × 10–3 sec–1

(c) 6.93 × 10–2 sec–1 (d) None of these

2. In Arrehenius equation if a graph is plotted between 10 gK and 1/T, the slope

of the curve will be:

−Ea

R

58

3. The rate low for a reaction 2C + D → A + E is

−d[D] = K[C]2[D]

dt

if C is present in large excess, the order of the reaction will be:

(a) zero (b) first

(c) second (d) third

4. What is the activation energy for the reverse of this reaction?

N2

O4(g)

→

2NO2(g)

Data for the given reaction is ∆H = 54 KJ/mol and εa = 57.2 KJ.

(a) –54KJ (b) 3.2 KJ

(c) 60.2 KJ (d) 111.2 KJ

5. The rate constant of a reaction becomes equal to the pre exponential factor when:

a. the absolute temperature is zero

b. the activation energy is infinity

c. the absolute temperature is infinity

d. the activation energy is zero

6.A second order reaction between A and B is elementary reaction: A + B → Product

rate law expression of this reaction will be:

(a) Rate = K[A][B] (b) Rate = K[A]0[B]2

(c) Rate = K[A]2[B]0 (d) Rate = K[A]3/2[B]1/2

7. The order and molecularity of the chain reaction, H2(g) + Cl2(g) hν→ HCl

(a) 2, 0 (b) 0, 2

(c) 1, 1 (d) 3, 0

59

8.Which of the following is pseudo first order reaction?

a. 2H2O2 → 2H2O + O2

b. 2O3 → 3O2

c. CH3COOC2H5 + NaOH → CH3COONa + C2H5OH

d. CH3COOC2H5 + H2O →H+ CH3COOH + C2H5OH

9.A large increase in the rate of reaction for rise in temperature is due to:

a. Increase in the number of collisions

b. Increase in the number of activated molecules

c. Lowering of activation energy

d. Shortening of the mean free path.

10.For a creactionj, the following data were obtained:

Concentration (mol/L) 0.1 0.05 0.025 0.0125

Half life in (sec) 30 29.9 30.1 30

the order of reaction is:

(a) 2 (b) 1

(c) 0 (d) fractional

60

Concept map of ChemicalKinetics

sources-Internet:www.studiestoday.com

61

UNIT- IV SURFACE CHEMISTRY

SYNOPSIS


IV SURFACE Type of Adsorptions and ***

CHEMISTRY characteristics,

Adsorption isotherms

Heterogeneous and

homogenous catalyst, **

zeolites and enzymes

Classification of Colloids

***

Preparation and ***

properties of Colloids

Adsorption: The accumulation of molecular species at the surface rather than in the bulk of a solid or

liquidAdsorbate:The molecular species or substance, which concentrate at the surface.Adsorbent: The material on the surface of which the adsorption takes place. Adsorption is essentially a

surface phenomenon.Desorption: The process of removing an adsorbed substance from a surface on which it is adsorbed.Factors featuring adsorption The extent of adsorption increases with the increase of surface area per unit mass of the adsorbent at a

givetemperature and pressure. Easily liquefiable gases(i.e., with higher criticaltemperatures) are readily

adsorbed.Adsorption isaccompanied by decrease in enthalpy aswell as decrease in entropy of the system.

Types of Adsorption

Physical adsorption Chemical adsorption

1. It arises because of van der 1. It is caused by chemical bond

Waals‘forces. formation.

2. It is not specific in nature. 2. It is highly specific in nature.

3. It is reversible in nature. 3. It is irreversible.

4. Enthalpy of adsorption is low (20- 4. Enthalpy of adsorption is high (80-

40 kJ mol–1). 240 kJ mol-1).

5. No appreciable activation energy is 5. High activation energy is sometimes

needed. needed.

6. It results into multimolecular layers 6. It results into unimolecular layer.

on adsorbent surface under high

. Freundlich adsorption isotherm: Empirical relationship between the quantity of gas adsorbed by unit mass of

solid adsorbent and pressure at a particulartemperature.

x/m= k.P1/n (n > 1) where x is the mass of the gas adsorbed on mass m of the adsorbent at pressure P, k and n

are constants which depend on the nature of the adsorbent andthe gas at a particular temperature.

62

Taking logarithm log x/m= log k + 1/n log P

When 1/n = 0, x/m = constant, the adsorption is independent of pressure. When 1/n = 1, x/m= k P, the

adsorption varies directly with pressure.

Catalysis:

Catalysts - Substances, which alter the rate of a chemical reaction and themselves remain chemically and

quantitatively unchanged after the reaction.

Promoters - Substances that enhance the activity of a catalyst. For example, inHaber‘s process for

manufacture of ammonia, molybdenum acts as a promoter for iron which is used as a catalyst.

Poisons – Substances that decrease the activity of a catalyst.

Homogeneous catalysis:When the reactants and the catalyst are in the samephase (i.e.liquid or gas).

Eg:- Oxidation of sulphur dioxide into sulphur trioxide with dioxygen in the presence of oxides of

nitrogen as the catalyst in the lead chamber process.

NO(g)

2SO2(g) + O2(g) 2SO3(g)

Heterogeneous catalysis: The catalytic process in which the reactants and thecatalyst are in different

phases. is known as heterogeneous catalysis. Eg:-Oxidation of sulphur dioxide into sulphur trioxide in the

presence of Pt.

2SO2(g)→ 2SO3(g).

The mechanism of heterogeneous catalysis involves five steps:

a) Diffusion of reactants to the surface of the catalyst.

b) Adsorption of reactant molecules on the surface of the catalyst.

c) Occurrence of chemical reaction on the catalyst‘s surface through formation of an intermediate.

d) Desorption of reaction products from the catalyst surface.

63

Diffusion of reaction products away from the catalyst‘s surface.

Important features of solid catalysts

Activity:The activity of a catalyst depends upon the strength of chemisorptions to a large extent. The

reactants must get adsorbed reasonably strongly on to the

catalyst (but not so strongly) to become active.

(b) Pt

Eg:- 2H2 (g) + O2 (g) 2 H2O(l)

Selectivity: The selectivity of a catalyst is its ability to direct a reaction to yield a particular product.

Shape-selective catalysis: The catalytic reaction that depends upon the pore structure of the catalyst and

the size of the reactant. Zeolites are good shape-selective catalysts. Eg:- ZSM-5 converts alcohols directly

into gasoline (petrol) by dehydrating them to give a mixture of hydrocarbons.

Enzymes:The enzymes are biochemical catalysts.

Inversion of cane sugar: The invertase enzyme converts cane sugar into glucose and fructose.

COLLOIDS: A colloid is a heterogeneous system in which one substance is dispersed (dispersed phase)

as very fine particles in another substance called dispersion medium. Range of diameters is between 1 and

1000 nm.

Classification of Colloids

(i)Based on Physical State of Dispersed Phase and Dispersion MediumSee table 5.4 NCERT Text- book

Solidsin liquids – sols eg:- starch sol

Liquids in solids – gels eg:- butter

Liquids in liquids – emulsions eg:- milk

(ii)Based on nature of interaction between dispersed phase and the dispersion medium

64

Lyophilic colloids Lyophobic colloids

1.Solvent liking 1.Solvent hating

2.Reversible sols 2.Irreversible sols

3.Quite stable 3. Unstable.Need stabilizing agents to

preserve.

4.Cannot be easily Coagulated 4.Can be coagulated easily by adding

small amount of electrolyte

(iii)Based on the type of the particles of the dispersed phase

Multimolecular Macromolecular Associated colloids (Micelles)

colloids Colloids

Atoms or molecules Solutions in which At low concentrations behave as

aggregate together to the size of the normal strong electrolytes, but at

form colloidal range macro molecules higher concentrations exhibit

species .Eg:- gold may be in the colloidal behaviour due to the

sol,sulphur sol Colloidal formation of aggregates.

range.Eg:- starch Eg:- soaps & detergents

Sol

Kraft temperature (Tk)- Temperature above which the formation of micelles takes place.

Critical micelle concentration (CMC) – Concentration above which the formation of micelles takes place.

Peptization_ Process of converting a precipitate into colloidal sol by shaking it with dispersion medium in

the presence of a small amount of electrolyte.

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

through a suitable membrane.

Electro-dialysis: Dialysis can be made fasterby applying an electric field if the dissolved substance in the

impure colloidal solution is only an electrolyte.

Properties of colloidal solutions

(i)Tyndall effect: - The scattering of light rays by colloidal particles due to whichthe path of light is

illuminated.

Tyndalleffect is observed only when The diameter of the dispersed particles is not much smaller than the

wave length of the light used.

The refractive indices of the dispersed phase and the dispersion medium differ greatlyin magnitude.

(ii)Brownian movement: - The zig-zag movement of colloidal particles due to the unbalanced

bombardment of the dispersed particles with the molecules of the dispersion medium.

65

(iii) Electrophoresis :- The movement of colloidal particles under an appliedelectric potential Charge on

colloidal particles :- The colloidal particles develop charge due to the following reasons:

Electron capture by sol particles during electro dispersion of metals Due to preferential adsorption of ions

from solutions Due to formulation of electrical double layer.

Coagulation or precipitation of colloidal particles or the process of settling ofcolloidal particles is due to

(a) addition of electrolytes (b) electrophoresis(c) boiling (d)mixing two oppositely charged sols.

Hardy – Schulze rule: -Greater the valence of the coagulating ion added to a sol, the greater is its power to

cause precipitation. The coagulation power of some of the cations is in theorderAl3+>Ba2+>Na+. The

coagulating power of some of the anions is in the order [Fe(CN)6] 4->PO43->SO4

2- >Cl- .

Coagulating value of an electrolyte: The minimum concentration of an electrolyte in millimols per litre

required to cause precipitation of a sol in two hours.

Emulsion :- colloidal system where a liquid is dispersed in another liquid. Typesof emulsionsoil in water

type (o/w) eg: milk, vanishing creamwater in oil type (w/o) eg: butter , cold cream Emulsions are

stabilized by emulsifyingagents. eg : soaps.


VERY SHORT ANSWER QUESTIONS (1MARK)

1.Adsorption is always exothermic. Why?

Hints:Adsorption decreases the surface energy of the adsorbent, which appears as heat according

tothermodynamicterm ∆G = ∆H –T ∆S.So it is exothermic.

2.The enthalpy of adsorption of chemisorptions is high. Why?

Hints:Chemisorptions involves chemical bond formation

3.When a solution of acetic acid in water is shaken with charcoal the concentration of the acid decreases

in the solution .Why?

Hints:Part of the acid is adsorbed by charcoal

66

4.What is meant by co-enzyme?

Hints:The non- protein substance present along with enzymes,which enhances the activity of enzymes.

5. 1 gm of activated charcoal adsorb more SO2(g) (critical temperature 630K) than

methane(criticaltemperature 190K).Why?

Hints:Easily liquefiable gases (higher critical temperature) are easily adsorbed by a solid, because

near the critical temperature the van der Waal‘s forces between the gas molecules and the solid adsorbents

is stronger.

6. Why is FeCl3 preferred over KCl in case of a cut leading to bleeding?

Hints:FeCl3 helps in coagulation of blood more effectively than KCl. Greaterthe valency of coagulating

ion, more will be coagulating power.

7.What are aerosols? Give an eg.

Hints:Colloids of a solid in gas. eg :- smoke

8.The sky appears blue. Why?

Hints:Due to scattering of blue light by dust particles.

9.Why are powdered substance more effective adsorbents than their crystalline forms?

Hints:Because powdered form provides more surface area due to which extent of adsorption increases.

10.Give 2 examples of positively charged colloids.

(1) Hemoglobin (blood) (2) Fe(OH)3 sol

SHORT QUESTIONS (2& 3* MARK)

1. Write the suitable adsorbent for the following adsorbates:

(a) chlorine gas (b) moisture (c) polluting gases like NO2& SO2 (d) gases like H2,O2,

Hints:(a) Charcoal (b) silica gel (c) charcoal (d) transition metals like Ni, Co etc 2. What happens when?

2.A beam of light is passed through As2S3 sol KCl is added to Fe(OH)3sol.

Hints: (i)The path of the light become visible due to Tyndall effect.Fe(OH)3 gets coagulated

3. What happens when?

Gelatin is added to gold sol Colloidal sol of Fe2O3 and As2S3 are mixed.

Hints:(i) Gold sol gets stabilized.

67

4.The oppositely charged colloids get neutralized and coagulated. 4. Explain the cleansing action of soap.

Hints:Soap molecules form micelles around the oil droplets in such a way that the hydrophobic part of

soappoints to the oil and hydrophilic part projects towards water. Thus soap helps in emulsification and

washingaway of oils and fats.

5.Explain why hydrophilic sols are relatively more stable than hydrophobic sols? Hints:This is because

hydrophilic sols are extensively hydrated and there is strong interaction between disperse phase and

dispersion medium.

6.What is electro dialysis?

Hints:The process of dialysis used for the purification of colloids can be accelerated by applying an

electric field. The oppositely charged ions of the electrolyte present in the colloids migrate towards the

respective electrodes easily.

7. Account for the following Fe(OH)3 sol is positively charged.

The extent of physical adsorption decrease with rise in temperature.

Hints:(i)A deep red sol of Fe(OH)3 is obtained by the hydrolysis of FeCl3 . The sol particles are positively

charged because of preferential adsorption of Fe3+ ions

Adsorption is an exothermic process. So the rate of physical adsorption decreases with the rise in

temperature in accordance with the Le-Chatlier principle.

8. Give the preparation of sulphur sol & gold sol. Hints:Oxidation of H2S with SO2 gas gives sulphur sol.

SO2(g) + 2H2S(g) → 3S(sol) + 2H2O

By reduction of auric chloride by formaldehyde gives gold sol. 2AuCl3 + 3HCHO + 3H2O → 2 Au (sol) +

3HCOOH + 6HCl

9.Name two industrial heterogeneous catalytic processes.

Hints:Haber‘s process for the manufacture of ammonia,iron is used as catalyst.

Contact process for the manufacture of sulphuric acid, V2O5is used as catalyst.

10.*(i) Why is alum added to water for purification?

(ii)Explain why deltas are formed where river & sea water meet.

Cottrell‘s smoke precipitator is fitted at the mouth of the chimney used in factories.

Hints: (i) For coagulating sand &soil particles. River water gets coagulated by electrolytes in sea water so

as to form deltas. It removes poisonous gases by adsorption & smoke free from poisonous gases

comes out.

68

ENRICHMENT EXERCISE Q1. What is meant by the terms Adsorption and Absorption? Give one e.g. of each.

Q2. Name the factors on which the adsorption of a gas on a solid depends.

Q3. What do x and m represent in the following expression? X /m = k p 1/n

Q4. The ziz-zag motion of colloidal particles is called ____________________.

Q5. How can we differentiate colloidal solution from true solution and suspension on basis of particle

size?

Q6. What are lyophilic and lyophobic sols?

Q7. Define the terms

a) Peptisation

b) Coagulation

c) Brownian Movement

Q.8.When a beam of light is passed through a colloidal sol. Electric current is passed through a

colloidal sol.

Q9. Give an e.g. of associated colloid.

Q10.Give 2 examples of heterogeneous catalysis.

Q11. What is the range of particle size in colloidal solution in nm?

Q12. What is collodion?

Q13. How does an increase in temperature affect both physical as well as chemical adsorption?

Q14. Explain why lyophilic sols are relatively more stable than lyophobic sols?

Q15. What are micelles? How do they differ from normal colloidal solution?

Q16. Explain the following terms:

a)Hardy- Schulze Rule

b)Electrophoresis

69

(c) Emulsification

Q17. What is the difference between multimolecular, macromolecular and associated colloids?

Give one example of each.

Q18. What do you understand by activity and selectivity of catalyst?

Q19. What is shape selective catalyst?

Q20. Adsorption of a gas on the surface of solid is generally accompanied by decrease in entropy;

still it is aspontaneous process.

Q21. Write three distinctive features of chemisorptions, which are not found in physisorption.

Q22. Explain how the phenomenon of adsorption finds application in each of the following

processes?

a)Production of vacuum

b)Heterogenious catalysis

c)Froth floatation process

Q23. What is adsorption isotherm? Describe Freundlich adsorption isotherm.

Q24. Discuss the effect of pressure and temperature on the adsorption of gases on solid.

Describe the application of adsorption in controlling humidity.

Q25. Describe the following terms with an example in each case.

a)Macromolecular sol

b)Peptisation

c)Emulsion

Q26. Differentiate among a homogeneous solution, a suspension and a colloidal solution, gave a

suitable example of each.

Q27. Explain the following terms giving a suitable example for each.

a)Aerosol

b)Emulsion

c)Micelle

70

Q28. Classify colloids where the dispersion medium is water. State their characteristics and write an example

of each of these classes.

Q29. Explain what is observed when

a)An electric current is passed through a sol?

b)A beam of light is passed through a sol?

Q30.An electrolyte (say NaCl or KCl) is added to ferric hydroxide sol.(or hydrated ferric oxide sol)?

a)Electrophoresis

b)Dialysis

c)Tyndall effect

Q31. Distinguish between multimolecular and macromolecular colloids? Give one example of each type. How

are associated colloids different from these two types of colloids?

Q32. How are the following colloids different from each other in respect of their dispersion medium and

dispersed phase? Give one example of each.

a) Aerosol

b) Emulsion

c) Hydrosol

KEY POINTS/ Subject Tips:-

Difference between physical and chemical adsorption Frendlisch adsorption isotherm Applications of adsorption Homogenious and heterogeneous catalysis Modern adsorption theory of catalysis Enzymes and their mechanism of action Colloids and their classification Methods of preparation, physical and chemical properties of colloids Emulsions and its types

71

Queations 1. Rate of physisorption 1.Match the column and choose correct option:

(A) Smoke P. foam

(B) Butter Q. emulsion

(C) Hair cream R. aerosol

(D) Whipped cream S. gel

(a) A–P, B–S, C–Q, D–R (b) A–R, B–Q, C–S, D–P

(c) A–R, B–S, C–Q, D–P (d) A–S, B–P, C–R, D–Q

Matching Column Type

2. Column 1 Column 2

(A) Soap in water P. Associated colloid

(B) Starch gelatin Q. Lyoptrilic colloid

(C) Gold sol R. Collodion

(D) Cellulose nitrate in alcohol S. Lyophobic colloid

(a) A–R, B–S, C–Q, D–P (b) A–P, B–Q, C–S, D–R

(c) A–R, B–S, C–P, D–Q (d) A–P, B–Q, C–R, D–S

Fill in the blanks take question

3.Collodion is a 4% solution of ................... in a alcohol or ether.

4.Starch is an example of ................... colloids.

Assertion and Reason Type Question

5. Statement 1 : Lyophilic sols are more stable than Lyophobic sols.

Statement 2 : Lyophilic sols are highly hydrated in the solution.

(a) both statement are correct and statement 2 is correct explanation of statement 1

(b) both are correct explanation but statement 2 is not explain statement 1.

(c) statement 1 is true both statement 2 is false

(d) statement 2 is false but statement 2 is true

72

MULTIPLE CHOICE QUESTIONS (1 Mark)

1. Adsorption increases with:

(a) decrease in temperature

(c) decrease in pressure

(b) increase in temperature

(d) decrease in surface area

2. The colloidal system consisting of a liquid adsorbate in a solid adsorbent is termed as:

a) aerosol b) aerosol

(c) emulsion d) emulsion

3. Which of the following has least coagluating value for positive sol?

(a) Cl– (b) SO4 2–

(c) PO 4–3 (d) [Fe(CN)6]-4

4.Which can adsorb larger volume of hydrogen gas?

(a) Colloidal solution of platinum

(b) finely divided nickel

(c) finely divided platinum

(d) colloidal Fe(OH)3

5. What is the emulsifer in milk?

(a) albumin (b) soap

(c)gelatin (d) caesin

73

6. Which one of the following gases will be adsorbed most easily?

(a) N2 (b) H2

(c)O2 (d) CO2


(a) N2 (b) H2

(c)O2 (d) CO2


(a) N2 (b) H2

(c)O2 (d) CO2

9.Cottrell precipitator works otheprinciple

(a) distribution law (b) addition of electrolate

(c)Le-chattelier principle (d) Neutralisation of charge on collids

10. The formation of micelles takes place only above:

(a)inversion temperature (b) Boyle temperature

(c)critical temperature (d) Kraft temperature

11.A colloidal solutions show:

(a) very high osmotic pressure (b) high osmotic pressure

(c)low asmotic pressure (d) no osmotic pressure

12.Alums purify muddy water

by:

(a) dialysis (b) adsorption

(c) absorption (d) coagulation

13.Which of the following is an example of associated colloid?

(a) soap in water (b) protein in water

(c) rubber in benzene (d) AgNO3 in water

74

14.The coagulating power of an electrolyte for blood decrease in the order.

(a) Na+, Al+3, Ba+2 (b) PO4 –3, SO4 –2, Cl–

(c) Al+3, Ba+2, Na+ (d) Cl–,, SO4 2- ,PO4

3-

75

Concept map of SurfaceChemistry

sources-Internet: www.studiestoday.com

http://www.studiestoday.com/

76

UNIT-V

GENERAL PRINCIPLES AND PROCESSES

OF ISOLATION OF ELEMENTS

SYNOPSIS


V GENERAL Occurrence of metals **

PRINCIPLES and Concentration of

AND PROCESS Ores

OF ISOLATION

OF ELEMENTS

Extraction of crude ***

metal from ore

Refining ***

The chemical substances in the earth’s crust obtained by mining are called Minerals.

Minerals, which act as source for metal, are called Ore.

The unwanted impurities present in ore are called Gangue.

The entire process of extraction of metal from its ore is called Metallurgy.

Removal of gangue from ore is called Concentration, Dressing or Benefaction of ore.

Concentration by Hydraulic washing is based on the difference in gravities of ore and gangue particles.

Concentration by Magnetic separation is based on differences in magnetic properties of ore components.

If either of ore or gangue is capable of attracted by a magnet field, then such separation is carried out.

Concentration by Froth Flotation Process is based on the facts that sulphide ore is wetted by oil & gangue

particles are wetted by water.

Concentration by Leaching is based on the facts that ore is soluble in some suitable reagent & gangue is

insoluble in same reagent. e.g. Bauxite ore contains impuritiesof silica, iron oxide & TiO2 .The powdered

ore is treated with NaOH which dissolve Al & impurities remains insoluble in it.

Al2O3 +2NaOH + 3 H2O 2 Na [Al(OH)4].

Calcination involves heating of ore in absence of air below melting point of metal. In this process volatile

impurities escapes leaving behind metal oxide.

Fe2O3.xH2O

Fe2O3 +xH2O

ZnCO3

ZnO +CO2

CaCO3.MgCO3

CaO + MgO + 2CO2

77

Roasting involves heating of ore in presence of air below melting point of metal in reverberatory furnace.

In this process volatile impurities escapes leaving behind metal oxide and metal sulphide converts to

metal oxide.

ZnS + 3 O2 2ZnO+2SO2

2PbS + 3 O2 2 PbO +2 SO2

2 Cu2S + 3 O2 2Cu2O + 2 SO2

Reduction of metal oxide involves heating of metal in presence of suitable reagent Coke or CO2.

Reactions taking place at different zones of blast furnace in extraction of iron:-

Zone of reduction:- Temperature range 250oC-700oC

3Fe2O3+CO 2 Fe3O4+CO2

Fe3O4+CO 3 FeO+ CO2

FeO +CO Fe+ CO2

Zone of slag formation:- Temperature range 800oC-1000oC

CaCO3 CaO+CO2

CaO+SiO2 CaSiO3

SiO2+2C

Si+2CO,

MnO2+2C

Mn+2CO

(iii) Zone of fusion: - Temperature range 1150oC-1350oC

CO2 +C 2CO

(iv) Zone of fusion: - Temperature range 1450oC-1950oC:

C +O2

CO2

78

FLOW SHEET FOR EXTRACTION OF IRON: Iron ore (Magnetite Fe3O4)

Concentration is done by Gravity separation followed by magnetic separation

Calcination &Roasting i.e. Ore + Air +Heat → Moisture,CO2,SO2, As2O3 removed And FeO oxidized to Fe2O3

Smelting of mixture of Ore, coke & CaCO3 takes place in long BLAST FURNACE.

Pig iron is obtained which is remelted and cooled and then cast iron if formed. Iron: - It contains Fe 93-95%, Carbon 2.5-5%, and Impurities 3%.

Cast Iron: - It contains Fe 99.5-99.8%, Carbon 0.1-0.2% Impurities0.3%.

Spongy iron: - Iron formed in the zone of reduction of blast furnace is called spongy iron. It contains impurities

of C, Mn , Si, etc.

FLOW SHEET FOR EXTRACTION OF COPPER: Copper pyrites,CuFeS2 Concentration is done by Froth floatation process

Powered ore + water + pine oil +air sulphide ore in froth Roasting is done in presence of air.The reactions taking place:

S+O2

SO2, 4As +3 O2

2As2O3, 2 CuFeS2

Cu2S +FeS + SO2

Smelting in small blast furnace of a mixture of Roasted ore,coke and silica

2FeS +3O2

2FeS +2 SO2 ,

FeO +SiO2

FeSiO3 (slag)

79

A mixture of Cu2S,FeS & silica is obtained from blast furnace known as Copper matte Besserisation of

copper matte is done in Bessemer converter

2Cu2S +2 Cu2O 6 Cu + SO2

Melted copper is cooled and SO2 is evolved.Such copper is known asBlister Copper

FLOW SHEET FOR EXTRACTION OF ALUMINIUM:-

Bauxite Al2O3,2 H2O

Concentration of ore is done by leaching. Bauxite is treated withNaOH.

Following reactions takes place:-

Al2O3 +2 NaOH +3 H2O 2 Na[Al(OH)4] and impurities of Fe2O3,TiO2& SiO2 are removed.

Na[Al(OH)4 then treated with CO2 to get pure Al2O3.

Na[Al(OH)4] +2CO2 Al2O3+NaHCO3

Electrolytic reduction of pure alumina takes place with cryolite (Na3AlF6)& fluorospar to get aluminium

At cathode: Al3+ + 3 e- Al At anode: 2 O2- O2 +4 e-

Vapour phase refining is used for extraction of Nickel (MOND PROCESS) and Zirconium &Titanium

(VAN ARKEL PROCESS).

Zone refining is used for extraction of Si, Ge, Ga, etc.

Chromatography method is based on selective distribution of various constituents of a mixture between

two phases, a stationary phase and a moving phase. The stationary phase can be either solid or liquid on

solid support.

Column chromatography is based on adsorption phenomenon. This method is useful for those elements,

which are available in small amounts and the impurities are not much different in chemical properties

from the element to be purified.

80


VERY SHORT ANSWER QUESTIONS (1 MARK) Q.1- What is slag?

Hints:- It is easily fusible material, which is formed when gangue still present in roasted ore combines

with the flux. e.g. CaO (flux) + SiO2 (gangue) →CaSiO3 (slag)

Q.2- Which is better reducing agent at 983K, carbon or CO?

Hints:- CO, (above 983K CO being more stable & does not act as a good reducing agent but carbon does.)

Q.3- At which temperature carbon can be used as a reducing agent for FeO?

Hints:- Above 1123K, carbon can reduce FeO to Fe.

Q.4- What is the role of graphite rods in electrometallurgy of aluminium?

Hints:- Graphite rods act as anode, are attacked by oxygen to form CO2 and so to be replace timeto time.

Q.5- What is the role of cryolite in electrometallurgy of aluminium?

Hints:- alumina cannot be fused easily because of high melting point. Dissolving of alumina in cryolite

furnishes Al3+ions, which can be electrolyzed easily.

Q.6- What are depressants?

Hints:- It is possible to separate two sulphide ore by adjusting proportion of oil to water in froth flotation

process byusing a substance known as depressant.e.g. NaCN is used to separate ZnS and PbS.

Q.7- Copper can be extracted by hydrometallurgy but not Zn. Why? Hints:- The E0 of Zn is lower than

that of Cu thus Zn can displace Cu2+ ion from its solution. On other hand side to displace Zn from Zn2+

ion, we need a more reactive metal than it.

Q.8- Give name and formula of important ore of iron.

Hints:- Haematite – Fe2O3, Magnetite –Fe3O4, Iron pyrites FeS2.

Q.9- Give name and formula of important ore of Copper.

Hints:- Copper pyrites CuFeS2, Malachite CuCO3. Cu (OH) 2, Cuprite Cu2O.

Q.10- Give name and formula of important ore of Zinc.

Hints:- Zinc blende - ZnS, Calamine- ZnCO3, Zincite – ZnO

.

81


Q.1 Describe the method of refining of nickel.

Hints:- In the Mond Process, Ni is heated in a stream of CO forming a volatile complex, which then

decomposes at higher temperature to give Ni. At 330-350K: - Ni + 4CO → Ni (CO) 4

At 450-470K Ni (CO)4 → Ni + 4 CO

Q.2- What is Zone Refining? Explain with example. Hints:- Zone refining is a method of obtaining a

metal in very pure state. It is based on the principal that impurities are more soluble in molten state of

metal than solidified state.

In this method, a rod of impure metal is moved slowly over circular heater. The portion of themetal being

heated melts & forms the molten zone. As this portion of the rod moves out of heater, it solidified while

the impurities pass into molten zone. The process is repeated to obtain ultra-pure metal and end of rod

containing impure metal cutoff.

Q.3 Write the principal of electro-refining.

Hints:- In this method of purification impure metal is made Anode and pure metal is made the cathode.

On passing electricity, pure metal is deposited at the cathode while the impuritiesdissolve in solution as

anode mud. E.g. electro- refining of copper:-

At Cathode: - Cu2+ + 2e- → Cu

At Anode: - Cu → Cu2+ + 2e-

Q.4- Describe the method of refining of Zirconium and Titanium.

Hints:- Van Arkel process is used for obtaining ultrapure metal. The impure metal is converted into

volatile compound,which then decomposes electrically to get pure metal.

At 850K: - Zr impure) + 2 I2 → ZrI4

At 2075K:- ZrI4 → Zr (pure) + 2 I2

Q.5- Out of C & CO, which is better reducing agent for ZnO?

82

Hints: Since free energy of formation of CO from C is lower at temperature above 1120K whilethat of

CO2 from carbon is lower above 1323K than free energy of formation 0f ZnO. However, the free energy

of formation of CO2 from CO is always higher than that of ZnO. Hence, C isbetter reducing agent of ZnO.

Q6 The value of f G0 for Cr2O3 is -540kJ/mole & that of Al2O3 is -827kJ/mole. Is the reduction of Cr2O3

possible with aluminium?

Hints:- The desired conversion is

4 Al + 2Cr2O3 → 2Al2O3 + 4Cr

It is obtained by addition of following two reactions:-

4Al + 3O2 → 2 Al2O3 f G0=-827kJ/mole

2Cr2O3 → 4Cr + 3O2 f G0==+ 540 kJ/mole

Therefore, G0 for desired reaction is -827+540=-287, as a result reduction is possible.

Q.7:- Why copper matte is put in silica lined converter?

Hints::- Copper matte consists of Cu2S and FeS. When blast of air is passed through molten matte in

silica-lined converter,FeS present in matte is oxidized to FeO, which combines with silica to form slag.

2FeS + 3O2→2FeO +2 SO2, (ii) FeO + SiO2 →FeSiO3 (slag),

2Cu2S + 3O2 →2Cu2O+2SO2, (iv) 2Cu2O+2Cu2S→ 6Cu + SO2

Q.8- What is meant by term chromatography?

Hints:-Chromato means Colour and graphy means writing because the method was first used for

separation of coloured substance. It is based on selective distribution of various constituents of a mixture

between two phases, a stationaryphase and a moving phase. The stationary phase can be either solid or

liquid on solid support.

Q.9-Why is reduction of metal oxide easier if metal formed is in liquid state at temperature of reduction.

Hints:- The entropy of a substance is higher in liquid state than solid state. In the reduction of metal

oxide, the entropy change will be positive if metal formed is in liquid state. Thus, the value of G0

becomes negative and reduction occurs easily.

Q.10-Why is reduction of metal oxide easier if metal formed is in liquid state at temperature of reduction.

83

Hints:- The entropy of a substance is higher in liquid state than solid state. In the reduction ofmetal oxide,

the entropy change will be positive if metal formed is in liquid state. Thus, the value of G0 becomes

negative and reduction occurs easily.

ENRICHMENT EXERCISE Q1. Differentiate between a mineral and an ore.

Q2. Why is it that only sulphide ores are concentrated by froth floatation method?

Q3.What is flux? And write down the role of it in metallurgy of iron and copper.

Q4. State the role of depressant in froth floatation process.

Q5.Silver ores and native gold have to be leached with metal cyanides. Suggest a reason for this.

Q6. Name the method used for refining of (i) Nickel (ii) Zirconium (iii) Tin.

Q7. Write down the basic principle of Hydraulic washing and magnetic separation.

Q8.Why is the reduction of a metal oxide easier if the metal formed is in liquid state at thetemperature of

reduction?

Q9.How is Cast iron different from pig iron?

Q10.Copper matte is put in silica lined converter?

Q11.Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous

solution of NaCl is subjected to hydrolysis?

Q12.What is role of cryolite in the metallurgy of aluminium?

Q13. Write down the reactions taking place in different zones in the blast furnace during extraction of

iron.

Q14. State briefly the basic principles of following metallurgical operations. (i).Vapour phase refining

(ii).Electrolytic refining. (iii) Zone refining Q15.What is meant by the term ‘Chromatography’?

84

Q16Write the reactions involved in the extraction of Fe from hematite ore Cu from copper pyrite ore C)

Al from alumina

Q17. Mention the principles of refining the metals by

Electrolytic refining

Zone refining

Vapour phase refining

Q18. ∆fG MgO and ∆fG CO are given below

∆fG MgO(s) at 1273 K = - 941 KJ/mole

∆fG MgO(s) at 2273K = - 314 KJ/mole

∆fG CO (g) at 1273 K = - 439 KJ/mole

∆fG CO (g) at 2273 K&n = -628 KJ/mo

Q 19. Determine which temperature is suitable for reduction of MgO using C?

KEY POINTS/Subject Tips:-

Methods of concentration of ore mainly leaching, froth floation and magnetic separation

Isolation of iron, aluminium and copper from its ore in detail

Reactions of blast furnance and reverbeatery furnance

Difference between calcinations and roasting

Purification of metals obtained by zone refining, gas phase refining, electro refining and

chromatographic method.

85

Fill in the blanks type questions 1.Cresols and aniline are used as .............. in froth floatation process: 2.Haematite is an ore of ...............

Match the following type questions 1.Match the column

Column 1 Column 2

(A) cyanide process P. ultrapure Ge

(B) zone refining Q. extraction of Cu

(C) Froth floatation process R. pine oil

(D) electrolytic refining S. extraction of Au

(a) A–S, B–P, C–R, D–Q (b) A–R, B–S, C–Q, D–P

(c) A–P, B–Q, C–R, D–S (d) A–S, B–R, C–P, D–Q

MULTIPLE CHOICE QUESTIONS

1. The main function of roasting is:

(a) Oxidation (b) reduction

(c) to remove volatile matter (d) to make slag

2. Which is not a mineral of Al?

(a) Diaspora (b) bauxite

(c) Corundum (d) galena

3. Brass contains:

(a) Cu + Sn (b) Cu + Ni

(c) Cu + Zn (d) Mg + Al

4. Flux used in the smelting of copper arc is:

(a) coke (b) magnesia

(c) silica (d) lime stone

5. The type of iron obtained from blast furnace is:

(a) pig iron (b) cast iron

(c) wrought iron (d) mild steel

6. Which one is leached with cyanide process?

(a) Cu (b) Al

(c) Ag (d) Zn

7. Which one is not a process of purification of metals:

(a) chromatrography (b) zone refining

(c) froth floatation (d) distillation

8. Which is a copper matte?

(a) Cu2O + FeS (b) Cu2S + FeO

86

(c) Cu2S + FeS (d) Cu2O + FeO 9.In metallurgical process of Al, cryolite is mixed in its molten state, because it

(a) decreases the amount of alumina (b) oxidises the alumina

(c) increases the melting point of alumina (d) decreases the melting

point of alumina

10. Froth floatation process is used for:

(a) cuprite (b) zincite

(c) copper pyrites (d) bauxite

11. The anode mud obtained during electro refining of Cu contains:

(a) Ag (b) Fe

(c) Au (d) Zn 12. SiO2 is a :

(a) flux (b) Gangue

(c) Slag (d) catalyst

87

Concept Map of GeneralPrinciples and isolation of elements

Sources-www.ncerthelp.com

88

UNIT-VI

The p- BLOCK ELEMENTS

SYNOPSIS

Sl.No.

Topic Concepts Degree of

importance

VI Preparation,properties of oxygen **

and Ozone

Oxoacids of sulphur ***

Preparation,properties of **

Chlorine

Oxoacids of halogensand **

Interhalogens

Chemistry of Noble gases ***

GROUP-16 ELEMENTS (CHALCOGENS)

Group 16 Elements:O,S,Se,Te,Po

General electronic configuration:ns2np4

Element Occurrence

Oxygen Comprises 20.946% by volume of the atmosphere.

Sulphur As sulphates such as gypsum CaSO4.2H2O,Epsom salt MgSO4.7H2O and Sulphides Such as

galena PbS, Zinc Blende ZnS, Copper Pyrites CuFeS2 As metal selenides and tellurides are in sulphide

ores.

Se & Te as a decay product of thorium and uranium minerals.

ATOMIC & PHYSICAL PROPERTIES

Ionisation enthalpy decreases from oxygen to polonium.

Oxygen atom has less negative electron gain enthalpy than S because of the compact nature of the oxygen

atom.However from the S onwards the value again becomes less negative upto polonium.

Electronegativity gradually decreases from oxygen to polonium, metallic character increases from oxygen

to polonium.

Oxygen & S are non-metals, selenium and tellurium are metalloids. Po is a radioactive metal.

Oxygen is a diatomic gas while S,Se&Te are octa- atomic S8,Se8&Te8 molecules which has puckered ’

ring’ structure.

89

CHEMICAL PROPERTIES

Common oxidation state:- -2,+2,+4 &+6.

Due to inert effect,the stability of +6 decreases down the group and stability of +4 increases.

Oxygen exhibits +1 state in O2F2,+2 in OF2.

Anomalous behavior of oxygen-due to its small size,high electronegativity and absence of d-orbitals.

DIOXYGEN

Prepared by heating oxygen containing salts like chlorates, nitrates heat

2KClO3 2KCl+3O2

OXIDES

A binary compound of oxygen with another element is called oxide. Oxides can be

Classified on the basis of nature

Acidic Oxides:- Non-metallic oxides. Aqueous solutions are acids. Neutralize bases to form salts.Ex:SO2,

CO2, N2O5 etc.

Basic Oxides: Metallic oxides. Aqueous solutions are alkalies. Neutralize acids to form salts.Ex:Na2O,

K2O etc.

Amphoteric oxides:-some metallic oxides exhibit a dual behavior. Neutralize both acids & bases to form

salts.Ex:-Al2O3, SbO2, SnO,etc.

OZONE

PREPARATION

Prepared by subjecting cold, dry oxygen to silent electric discharge. 3O2→2O3

PROPERTIES

Due to the ease with which it liberates atoms of nascent oxygen, it acts as a powerful oxidizing agent. For

eg:- it oxidises lead sulphide to lead sulphate and iodide ions to iodine.

PbS +4O3→PbSO4+4O2

SULPHUR DIOXIDE PREPARATION

Burning of S in air : S+O2→SO2

Roasting of sulphide minerals

(Iron pyrites) 4FeS2+1102→2Fe2O3+8SO2

(Zinc blend)2ZnS+3O2→2ZnO+2SO2

90

PROPERTIES

Highly soluble in water to form solution of sulphurous acid SO2+H2O→H2SO3

SO2 reacts with Cl2 to form sulphuryl chloride SO2+Cl2→SO2Cl2

It reacts with oxygen to form SO3 in presence of V2O5 catalyst 2SO2+O2→2SO3

Moist SO2 behaves as a reducing agent. It converts Fe(III) ions to Fe(II) ions&decolorizes acidified

potassium permanganate (VII) solution( It is the test for the SO2 gas).

SULPHURIC ACID PREPARATION

It is manufactured by contact process which involves 3 steps

Burning of S or Sulphide ores in air to generate SO2.

Conversion of SO2 to SO3 in presence of V2O5 catalyst

Absorption of SO3 in H2SO4 to give oleum.

PROPERTIES

1. In aqueous solution it ionizes in 2 steps

H2SO4+H2O H3O++HSO4-

HSO4-+H2O H3O++SO4

2-

2. It is a strong dehydrating agent Eg:-charring action of sugar

H2SO4

C12H22O11 12C+11H2O

It is a moderately strong oxidizing agent. Cu+2H2SO4(conc.) →CuSO4+SO2+2H2O

C+2H2SO4(conc.)→CO2+2SO2+2H2O

GROUP 17 ELEMENTS(HALOGENS)

Group 17 elements: F,Cl,Br,I,At

General electronic configuration: ns2np5 and the elements are F, Cl,Br,I.

Occurrence: As insoluble fluorides (fluorspar CaF2, Cryolite and fluoroapattie).

Sea water contains chlorides,bromides and iodides of Sodium, potassium magnesium and calcium, but is

mainly sodium chloride solution (2.5% by mass). Certain forms ofmarine life (various seaweeds).

91

ATOMIC & PHYSICAL PROPERTIES

i). Atomic & ionic radii increase from fluorine to iodine.

ii). Ionization enthalpy gradually decreases from fluorine to iodine due to increase in atomicsize.

iii). Electron gain enthalpy of fluorine is less than that of chlorine. It is due to small size of fluorine &

repulsionbetween newly added electron &electrons already present in its small 2p orbital.

iv). Electronegativity decreases from fluorine to iodine. Fluorine is the most electronegative element in

the periodic table.

v). The color of halogens is due to absorption of radiations in visible region which results in the excitation

of outerelectrons to higher energy level.

vi). Bond dissociation enthalpy of fluorine is smaller than that of chlorine is due to electron-electron

repulsion among the lone pair in fluorine molecules where they are much closer toeach other than in case

of chlorine. The trend: Cl-Cl>Br-Br>F-F>I-I.

CHEMICAL PROPERTIES

OXIDATION STATES:-1.However, chlorine, bromine &iodine exhibit +1, +3, +5, +7 oxidation states

also.Fluorine forms two oxides OF2 and O2F2. These are essentially oxygen fluorides because of the higher

electronegativity of fluorine than oxygen.

Anomalous behavior of fluorine- due to its small size, highest electronegativity, low F-F bond dissociation

enthalpy and absence of d-orbitals.

TRENDS IN PROPERTIES

CHLORINE PREPARATION 1.MnO2 +4HCl MnCl2 +Cl2 +2 H2O

2.4NaCl+MnO2+4H2SO4 Mncl2 + 4 NaHSO4 + 2 H2O + Cl2

2KMnO4+16HCl +2KCl+2MnCl2 + 8H2O+5Cl2

DEACON’S PROCESS

CuCl2

4HCl+O2 2Cl2+2H2O

92

5. By electrolysis of brine solution. Cl2 is obtained at anode.

PROPERTIES

i). With cold and dilute Cl2 produces a mixture of chloride and hypochlorite but with hot and concentrated

alkalis itgives chloride and chlorate.

2NaOH+Cl2 NaCl+NaOCl+H2O

6NaOH+3Cl2 5NaCl+NaClO3+3H2O

ii).With dry slaked lime it gives bleaching powder.

2Ca (OH) 2+2Cl2 Ca (OH) 2+CaCl2+2H2O

iii). It is a powerful bleaching agent; bleaching action is due to oxidation

Cl2+H2O 2HCl+ [O]

Colored substance+[O] colorless substance

iv). Action of concentrated H2SO4 on NaCl give HCl gas.

420 K

NaCl+H2SO4 NaHSO4+HCl

3:1 ratio of conc. HCl & HNO3 is known as aqua-regia & it is used for dissolving noble metals like Au

and Pt.

OXOACIDS OF HALOGENS

HOCl,HOCl2,,HOCl3,HOCl4

Interhalogen compounds are prepared by direct combination of halogens.

Ex: ClF, ClF3, BrF5, IF7.

GROUP 18 ELEMENTS

GROUP 18 ELEMENTS: He, Ne, Ar,Kr,Xe &Rn

General electronic configuration: ns2np6

Atomic radii- large as compared to other elements in the period since it corresponds to Vander Waal radii.

Inert – due to complete octet of outermost shell, very high ionization enthalpy & electron gain enthalpies

are almost zero.

93

The first noble compound prepared by Neil Bartlett was XePtF6&Xenon. O2+PtF6

-led to the discovery of

XePtF6 sincefirst ionization enthalpy of molecular oxygen (1175 kJmol-1) was almost identical with that

of xenon (1170kJmol-1).

PROPERTIES:

673K, 1bar

Xe+F2 XeF2

873k, 7bar

Xe + 2 F2 XeF4

573k, 60-70 bar

Xe + F2 XeF6

XeF6 + MF M+[XeF7]-

XeF6 +H2O XeOF4 +2 HF

XeF6 +2 H2O XeO2F2 (partial Hydrolysis)

94



1. Why H2S is acidic and H2O is neutral ?

Hints:The S – H bond is weaker than O – H bond because the size of S atom is bigger than that of Oatom.

Hence H2S can dissociate to give H+ Ions in aqueous solution.

2.Name two poisonous gases which can be prepared from chlorine gas? Hints:Phosgene (COCl2), tear gas

(CCl3NO2)

3.Name the halogen which does not exhibit positive oxidation state .

Hints:Flourine being the most electronegative element does not show positive oxidation state.

4.Iodine forms I3- but F2 does not form F3- ions .why?

Hints:Due to the presence of vacant d-orbitals , I2 accepts electrons from I-ions to form I3-ions , but

because of d-orbitals F2 does not accept electrons from F-ions to form F 3- ions.


1. Why is HF acid stored in wax coated glass bottles?

Hints:This is because HF does not attack wax but reacts with glass. It dissolves SiO2 present in glass

forminghydrofluorosilicic acid.

SiO2 +6HF H2SiF6+2H2O

2. How is ozone estimated quantitatively?

Hints:When ozone reacts with an excess of potassium iodide solution Buffered with a borate buffer (pH 9.2),

Iodide is liberated which can be titrated against a standard solution of sodium thiosulphate. This is a

quantitative method for estimating O3 gas.

3.Write the balanced chemical equation for the reaction of Cl2 with hot and concentrated NaOH .Is this

reaction a disproportionation reaction? Justify:

Hints: 3Cl2+6NaOH 5NaCl+NaClO3+3H2O

Yes, chlorine from zero oxidation state is changed to -1 and +5 oxidation states.

95

4.Account for the following. (i)SF6 is less reactive than SF4.

(ii) 0f the noble gases only xenon compounds are known.

Hints:. (i)In SF6 there is less repulsion between F atoms than In SF4.

(II)Xe has low ionization enthalpy & high polarizing power due to larger atomic size. 5.With what neutral

molecule is ClO- Isoelectronic? Is that molecule a Lewis base? Hints:ClF .Yes, it is Lewis base due to

presence of lone pair of electron.

ENRICHMENT EXERCISE

1.Why are halogens coloured.

2.What are chalcogens?

3.Which noble gas is Radioactive?

4.Explain why fluorine always exhibit an oxidation state of – 1 only.

5.Which compound led to the discovery of compounds of noble gas?

6.Name the most electronegative element.

7.Why is 6 OF compound not known?

8.Why do noble gases have comparatively large atomic sizes?

9.Arrange in decreasing order of Ionic character

M- F, M -Cl, M- Br, M- I

10.Arrange the following in the order of property indicated:

a) AS2O3, ClO2 GeO2 Ga2O3(Increasing acidity)

b) H2 O, H2 S, H2 Se, H2Te (Increasing acid strength).

11. Arrange in decreasing order of bond energy:

F2 , Cl2 , Br2 , I2

96

12. Complete the following:

i) IO3- + I- + H +

13. Give the chemical reactions in support of following observations: b) Sulphur exhibits greater tendency for

catenation than selenium.

14. How would you account for following:

Enthalpy of dissociation of F2 is much less than that of Cl2

Sulphur in vapour state exhibits paramagnetism.

15. Draw structures of following:

a) H2S2O5b) XeF4

16. Despite lower electron affinity of F2, it is stronger oxidizing agent than Cl2 . Explain.

17. Addition of Cl2 to KI solution gives if brown colour but excess at if turns if colourless. Explain.

18. Draw the structures of

(a) Sulphuric acid

(b) sulphurus acid

(c) Peroxo di sulphuric acid

(d) Pyro sulphuric acid (oleum)

(e) BrF3(f) XeOF4

(g) XeF2

(h) XeF4

(i) XeF6

(j) XeOF4

(k) XeO3

KEY POINTS/Subject tips:-

Trends of physical and chemical properties in various groups

Oswald’s process and contact process

Oxoacids of sulphur and chlorine with structures

Chemical properties of ozone, sulphuric acid, nitric acid, HCl and interhalogen compounds

All the oxo acids of S,Cl and structures of Xenon compounds given in text book.

97


1.On addition of conc. H2SO4 to a chloride salt, colourless fumes are evolved but in case of iodise salt,

violet fumes come out. This is because

(a) H2SO4 reduces Hl to l2 (b) Hl is of violet colour

(c) Hl gets oxidised to l2 (d) Hl changes to HlO3

2. Affinity for hydrogen decreases in the group from flourine to iodine which of the halogen acids

should have highest bond dissociation enthalpy?

(a) HF (b) HCl

(c) HBr (d) Hl

3.Arrange the following hydrides of group 16 elements in order of increasing stalility.

(a) H2S < H2O < H2Te > H2Se (b) H2O < H2Te < H2Se < H2S

(c) H2O < H2S < H2Se < H2Te (d) H2Te < H2Se < H2S < H2O

4. The hybridisation of sulphur in sulphur tetrafluroide is

(a) sp3d (b) sp3d2

(c) sp3d3 (d) sp3

5. On heating KClO3, we get

(a) KClO2 + O2 (b) KCl + O2

(c) KCl + O3 (d) KCl + O2 + O3

6.. The correct order of acidic strength is:

(a) K2O > CaO > MgO (b) CO2 > N2O5 > SO3

(c) Na2O > Mgo > Al2O3 (d) Cl2O7 > SO2 > P4O10

7.Which one is not a property of ozone?

(a) it acts an oxidising agent in dry state

(b) oxidation of K1 into KlO2

(c) PbS is oxidised to PbSO4

(d) Hg is oxidised to Hg2O

98

8.What is the correct operation when Br2 is treated with NaF, NaCl and Nal

taken in three test tukes lavelled (X), (Y) and (Z)?

Br2 Br2 Br2

NaF NaCl Nal

(a) F2 is liberated in (X) and Cl2 in (Y)

(b) Only l2 is liberated in (Z).

(c) Only Cl2 is liberated in (Y)

(d) Only F2 is liberated in (X)

More than one correct Response

9.Which of the following statements are correct?

(a) Among halogens, radius ratio between iodine and fluorine is maximum

(b) Leaving F–F bond, all halogens have weaker X—X bond than X—X' bond in interhalogens

(c) Among interhalogen compounds maximum number of atoms are present in iodine fluoride.

(d) Interhalogen compounds are more reactive than halogen compounds.

10.Which of the following statements are correct for SO2 gas?

a) It act as bleaching agent in moist conditions

b) It's molecule has linear geometry

c) It's dilute solution is used as disinfectant.

d) It can be prepared by the reaction of dilute H2SO4 with metal sulphide.

11. Which of the following statements are correct?

(e) All the three N—O bond lengths in HNO3 are equal.

(f) All P–Cl bond lengths in PCl5 molecule in gaseous state are equal.

(g) P4 molecule in white phosphorous have angular strain therefore white phosphours is very

reactive.

(h) PCl5 is ionic in solid state in which cation is tetrahedral and anion is octahedral.

99

12. Which of the following statements are true?

a) Only type of interactions between particles of noble gases is due to weak dispersion forces.

b) Ionisation enthalpy of molecular oxygen is very close to that of numon.

c) Hydrolysis of XeF6 is a redox reaction.

d) Xenon fluorides are not reactive.

13. Match the items of column 1 and column 2 and mark the correct option

Column 1 Column 2

(A) H2SO4 (1) Highest electron gain enthalpy

(B) CCl3NO2 (2) Chalcogen

(C) Cl2 (3) Tear gas

(D) Sulphur (4) Storage batteries

(a) A–4, B–3, C–1, D–2 (b) A–3, B–4, C–1, D–2

(c) A–4, B–1, C–2, D–3 (d) A–2, B–1, C–3, D–4

100

Concept map the p-Block elements

Gr.16 ns2np4. Gr.17 ns2np5 Gr.18 ns2np6

Oxygen family: halogen family inert gases

Atomic and ionic radii: down the gr.

Ionisation energy: down the gr.

Oxidation states:+3,+5 Oxidation states:+2,+4 oxidizing power :F2 > Cl2 > Br2 > I2

“ O” has less E.G than S B.D.E.:Cl2 > Br

2 > F2 > I

2

Acidic nature: H2O < H2S < H2Se <H2Te Acidic strength: HF < HCl < HBr < HI

Stability: HF > HCl > HBr > HI

Xe +F2 XeF

Xe +2F2 XeF4

Xe +3F2 XeF6

Thermal stability: H2O < H2S < H2Se < H2Te <H2Po

Reducing character: H2O < H2S < H2Se < H2Te <H2Po

Ionic character: MF > MCl > MBr >MI

Acid strength: HOCl < HClO2 < HClO3 < HClO4

XX’ (CIF, BrF, BrCl, ICl, IBr, IF) (Linear shape)

XX’3 (CIF3, BrF3, IF3, ICl3) (Bent T-shape)

XX’5 – CIF5, BrF5, IF5, (square pyramidalshape)

XX’7– IF7 (Pentagonal bipyramidalshape) Sources- Self made and internet: www.ncerthelp.com

101

UNIT- VII

THE d-AND f-BLOCK ELEMENTS

SYNOPSIS

Sl.No. Topic Concepts Degree of Importance

VII The d- & f- Position & electronic **

BLOCK configuration.

ELEMENTS

General properties like physical ***

properties ,variation in atomic

sizes,

Ionization, oxidation states ,

electrode potential , stability of

higher oxidation states , chemical

reactivity, magnetic properties ,

coloured compounds , complex

compounds , catalytic properties ,

Interstitial compounds , Alloy

formation.

II) Oxides of transition metal, and ***

their oxidizing properties

(i) K2Cr2O7

(ii) KMnO4

III)Inner Transition **

elements(Lanthanoids& Actinides)

1.The elements of periodic table belonging to group 3 to 12 are known as d-Block elements.

2.The general electronic configuration of these elements is (n -1)d1-10 ns 1-2 d- Block elements are

collectively known as Transition Elements because properties of these elements vary in between s-Block

and p-Block elements.

A transition element should have partially filled (n-1) d orbital.

Group 12 elements i.e. Zn, Cd, Hg have completely filled (n-1) d-orbital in atomic& ionic state& thus

these elements are considered as Typical Transition Elements.

5.All these elements are metals. They are less electropositive than s-block elements &more

electropositivethan p-block elements.

6.The atomic radii decreases from group 3 to 6 (i.e. Sc to Cr) because of increase in effective nuclear

Charge gradually.

102

The atomic radii of group 7,8 9 &10 elements (i.e. Fe,Co,Ni) is almost same because pairing of electrons

takeplace in (n-1)d orbital causing repulsion i.e. shielding of (n-1)d orbital.

Group 11 &12 elements i.e. Cu & Zn have bigger size due to strong shielding of completely filled (n-

1)dorbital.

The transition elements show variable oxidation state due to small energy

difference between(n-1)d &ns orbital as a result both (n-1)d &ns electrons takepart in bond formation.

The highest oxidation state of an element is equal to number of unpaired electrons present in (n-1)d

&nsorbital.

Transition elements have high enthalpy of atomization/ sublimation Because of large number of unpaired

electrons in their atoms, they have stronger interatomic interaction and hence strong metallic bonding

ispresent between atoms.

Most of transition elements are paramagnetic due to presence of unpaired electronsin (n-1) d orbital.

Most of transition elements are used as catalyst. It is due to

(i) partially filled (n-1) d orbital

(ii)Variable oxidation state

(iii) Ability to change oxidation state frequently.

Most of transition elements form coloured compounds due to presence of unpaired electrons in(n-1)

dorbital & thus they can undergo d-d transition.

Most of transition elements form complex compounds due to (i) small size (ii) high charge (iii) presence

of vacant d-orbital of suitable energy.

Transition elements have lower value of Reduction Potential due to high ionization potential, high heat

ofsublimation & low enthalpy of hydration.

Transition elements form interstitial compounds because size of interstitial voids issimilar to size of non-

metals C, N, O, H.

Transition elements form alloys due to similar ionic radii.

The oxides of transition metals in lower oxidation state are BASIC, intermediate oxidation state are

AMPHOTERIC, highest oxidation state are ACIDIC.

LANTHANOIDS: --

The 14 elements after Lanthanum having atomic number 58 to 71 are collectively known as Lanthanoids.

The general electronic configuration of these elements is [Xe] 4f(1-14) , 5d(0-1) ,6s2 .

Most common oxidation state of these elements is +3, but Ce shows +4, Eu +2, because they acquirestable

configuration.

The size of Lanthanoids and its trivalent ion decreases from La to Lu due to poor shielding of 4f

electrons. It is known as lanthanoids contraction.

103

ACTINOIDS:-

The 14 elements after Actinium having atomic number 90 to 113 are collectively known as Actinoids.

The general electronic configuration of these elements is [Rn] 5f(1-14), 6d(0-1), 7s2.

The size of actinoids and its trivalent ion decreases from Ac to Lw due to poor shielding of 5f electrons. It

is known as actinoids contraction.

The elements after U (92) are man-made known as transuranic elements.

POTASSIUM DICHROMATE:-

Preparation: - It takes place in three steps-

Conversion of chromite ore to sodium chromate.

Conversion of sodium chromate to sodium dichromate.

Conversion of sodium dichromate to potassium dichromate

Following reaction take place:--

3FeCr2O4+ 4 Na2CO3 +7O2

2 Na2CrO4+ 2Fe2O3 +8 CO2 2Na2CrO4 + 2 H+

Na2Cr2O7 + 2 KCl

Na2Cr2O7

K2Cr2O7

+ 2 Na+ + H2O

+ 2 NaCl

POTASSIUM PERMANGNATE:--

Preparation: --

Conversion of pyrolusite ore into potassium magnate

Conversion of potassium mangnate to potassium permaganate Following reactions take place:-

3MnO2 + 4 KOH + O2 3MnO42- +4H+

2K2MnO4 + 2H2O

2 MnO4- +MnO2 +2H2O

104

Compare the chemistry of Lanthanoids and Actinoids

Lanthanoids Actinoids

(a) Electronic configuration is 4f1-14 (a)Electronic configuration is 5f1-14 6d0-1

5d0-1 6s2 7s2.

(b) common oxidation state is +3 (b) common oxidation states are +3&+4.

(c) the size difference between two

(c) The size difference between the successive Actinoids is larger.

successive Lanthanoids is small.

(d) the ionization enthalpies are (d) the ionization enthalpies are higher

Higher



1.Ti2+, V2+, &Cr2+ are strong reducing agents .Why?

+4 is the stable oxidation state for Vanadium and +3 is the stable oxidation state for Chromium in their

aqueous solutions.

2.What does the E0 value of M3+/M2+ show for Mn3+& Co3+ (+1.57 V & +1.97V respectively) A:The high

E0 value of M3+/M2+ for Mn3+ and Co3+ shows that they are the strongest oxidizing agents in aqueous

solutions among all the rest of transition metal Series .

105

3.Why Na2Cr2O7 is not used in volumetric analysis?

A:Because it is deliquescent.

4.Why is that orange solution of K2Cr2O7 turns yellow on adding NaOH?

A: It changes to CrO4 2- ion.Cr2O7

2-+ 2 OH- 2 CrO4 2-

Orange Yellow

5.Arrange CrO, CrO3 and Cr2O3in increasing order of acidic strength.

A:CrO < Cr2O3< CrO3.

6. Which metal in the first series of transition metal exhibits +1 oxidation state most frequently and why?

A:Cu, due to stable d10 e- configuration.

Write an ionic equation representing the oxidizing property of acidified KMnO4 solution.

+ 16H+ + 10 I- 2 Mn2+ + 5 I2 + 8H2OA:2MnO4

7.Why do Zr and Hf show similar properties?

It is due to similar ionic size, which is due to Lanthanide contraction.

8.Why are Cd2+ salts white?

Cd2+ ion has no unpaired electron.

9.Which element of the first transition series shows the highest number of oxidation states?

Mn

11.What is meant by disproportionation reaction?

A:It is a transformation of a substance into two or more substances by simultaneous oxidation and

reduction.

12. Why does Ti4+ ion show diamagnetic nature? A:Ti4+ has no unpaired ē in it.

SHORT QUESTIONS (2& 3 MARK)

Q.1-Explain briefly how +2 oxidation state becomes more and more stable in the first half

of the first row transitioelements with increasing atomic number.

A.1-In M2+ ions, 3d-orbitals get occupied gradually as the atomic number increases. Since, the number of

empty d-orbitals decreases, the stability of cations increases from Sc2+ to Mn2+.Mn2+ is most stable as all d-

orbitals are singly occupied.

106

Q.2- Explain why transition elements have many irregularities in their electronic configurations?

A.2-In the transition elements, there is a little difference in the energy of (n-1) d-orbitals and ns-orbitals.

Thus,incoming electron can occupy either of shell. Hence, transition elementsexhibit many irregularities in

their electronic configurations.

Q.3-What are different oxidation states exhibited by Lanthanides?

A.3-The common stable oxidation state of lanthanides is +3.However some members also show oxidation

states of +2 & +4.

Q.4-How is the variability in oxidation states of transition metals different from that of the non-transition

metals? Illustrate with examples.

A.4-The transition elements use its (n-1)d, ns and np orbital and the successive oxidation states differ by

unity. Forexample, Mn shows all the oxidation states from +2 to +7. On other hand non transition elements

use its ns, np and nd orbitals and the successive oxidation statesdiffer by two units e.g. Sn2+, Sn4+ etc.

Q.5- Why do transition elements show variable oxidation states?

107

A.5- The transition elements show variable oxidation state due to small energy difference between (n-1) d

&ns orbital as a result both (n-1)d &ns electrons take part in bond formation.

Q.6-Why are Mn2+ compounds more stable than Fe2+ compounds towards oxidation to +3 state?

A.6-The electronic configuration of Mn2+ is [Ar] 3d5, i.e. all five d-orbitals are singly occupied. Thus this

is stableelectronic configuration and further loss of electron requires high energy .on other hand the

electronic configurationof Fe2+is [Ar] 3d6, i.e. Loss of one electron requires low energy

Q.7-To what extent do the electronic configuration decide the stability of oxidation state in the first series

of thetransition elements? Illustrate your answer with an example.

A.7-In a transition series, the oxidation state which lead to exactly half filled or completely filled orbitals

are morestable.e.g. the electronic configuration of Fe is [Ar] 3d6 ,4s2. It shows various oxidation state but

Fe(III) is more stablethan Fe(II).

Q.8-What is meant by disproportionation? Give two examples.

A.8-Those reactions in which same substance undergoes oxidation as well as

reduction are called disproportionationreactions.e.g.

2Cu+→ Cu2+ + Cu

3MnO42- +4 H+ → 2 MnO4

- +MnO2 + 2 H2O

Q.9- Which metal in the first series of transition metals exhibits +1 oxidation state mostfrequently and

why?

A.9- Copper with configuration [Ar] 3d10 4s1 exhibits +1 oxidation state. Copper loses 4s1 electron easily

and achieved a stable configuration 3d10 by forming Cu+.

Q.10- What are inner transition elements?

A.10- The f-block elements in which the last electron accommodated on (n-2) f-subshell are called inner

transitionelements. These include atomic numbers 58 to 71 and from 90 to 103.

Q.11- The paramagnetic character in 3d-transition series elements increases upto Mn and then decreases.

Explain why?

108

A.11- In the 3d-transition series as we move from Sc (21) to Mn (25) the number of unpaired electrons

increases and hence paramagnetic character increases. After Mn, the pairing of electrons In the d-orbital

starts and the number of unpaired electrons decreases and hence, paramagnetic character decreases.

Q.12- Comment on the statement that elements of the first transition series possess many properties

different from those of heavier transition metal

A.12-The following points justify that the given statement is true:-

Ionization enthalpies of heavier transition elements are higher than the elements of 3d series.

Consequently, heavier transition elements are less reactive in comparison to 3d-elements.

Melting points of heavier transition elements are higher than 3d-elements.

Higher oxidation states of heavier transition elements are stable whereas lower oxidation states are stable

in 3d-elements.

Q.13-What are transition elements? Which d-block elements are not regarded as transition elements and

why?

A.13- An element which has partially filled (n-1) d orbital is known as transition elements. Group 12

elements i.e. Zn,Cd, Hg have completely filled (n-1) d-orbital in atomic & ionic state & thus

theseelements are not considered asTransition Elements.

Q.14-What are interstitial compounds? Why are such compounds well known for transitionmetal?

A.14- Compounds of transition metal with relatively smaller non-metals are known as

interstitialcompounds. These compounds are well known for transition metals because size of C, N, O,

and B is similar to size of interstitial voids of transition metal.

Q15. What is Lanthanoid contraction? What are its consequences?

A.It is filling up of 4f orbital before 5d orbital results in a regular decrease in atomicradii are called

Lanthanoidcontraction.

Consequences of lanthanoid contraction

There is similarity in the properties of second and third transition series.

Separation of lanthanoids is possible due to lanthanide contraction.

It is due to lanthanide contraction that there is variation in the basic strength of lanthanidehydroxides.

Basicstrength decreases from La(OH)3 to Lu(OH)3.)

Q16.Why does Cerium act as strong oxidizing agent?

A16.Ce shows +4 oxidation states due to stable noble gas configuration, but it is a Strong oxidizing agent

and reduced to Ce3+. Its E 0 value for Ce4+ /Ce3+ is 1.74v It oxidizes water, but the rate of reaction is slow

and thus can be studied in detail and analyzed. Hence Ce acquires great use in analytical chemistry.

109

ENRICHMENT EXERCISE

Q1: Write the outer electronic configuration of Chromium atom (z=24)

Q2: What is the effect of increasing pH on Potassium dichromate solution.

Q3. Write the general configuration of d block elements.

Q4. Why are Zn and Cd are not regarded as transition elements.

Q5. Why do Zr and Hf exhibit similar properties?

Q6. Why is solution of Sc+3 colourless?

Q7. Name an ore of Manganese and Chromium

Q8. Calculate the Magnetic moment of a divalent ion in aqueous solution if its atomic number is 25

Q9. Complete the reactionNa2Cr2O7 + 2KCl

Q10. Name an alloy of lanthanoid metal.

Q11. Which is more basic La(OH)3 or Lu(OH)3 ? Why?

Q12. Why are Mn2+ compounds more stable than Fe+2 towards oxidation to their +3 state.

Q13. Why is HCl not used to acidify a permanganate solution in volumetiric estimation of Fe+2with

C2O42- ?

Q14.Which is stronger reducing agent Cr+2 or Fe+2 ? Why ?

Q15. Explain why Cu+ ion is not stable in aqueous solutions.

Q16.Write the steps involved in the preparation of KMnO4 from pyrolusite ore .

Q17. Why are enthalpies of atomization of transition elements high.

Q18.Write down the number of 3d electron in Ti+2,V+2,Fe+2 and Ni+2.

Q19.Pt (iv)compounds are relatively more stable than Ni(iv)compounds.

Q20.Out of zinc and cobalt salts which is attracted in magnetic field. Why?

Q21. Why is E0 value for Mn+3/Mn+2 couple much more positive than that for Cr+3/Cr+2 or Fe+3/Fe+2.

Explain.

Q22. What happens when potassium dichromate solution is heated with conc. Sulphuric acid and a

soluble metal chloride. Write the structure of Orange Compound produced during thereaction.

110

Q23. Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using

diphenylamine as indicator. What are the number of moles of Mohr’s salt required per mole of

dichromate?

Q24. Write the balanced Redox reaction between KMnO4 (molecular weight = 158 ) and oxalic acid.

Q25. State reasons for the following

Ce(III) is readily oxidized to Ce(IV).

Actinoids have a stronger tendency to form complexes than lanthanoids.

111


General trends of atomic size, ionization enthalpy, hydration enthalpy, oxidation potential and

reduction potential in transition elements.

Prepration of KMnO4 and K2Cr2O7 and their chemical properties.

Lanthanoid contraction and its consequences

Difference between lanthanoids and actinoids.

MULTIPLE CHOICE TYPE QUESTIONS

1. One of the characteristics of transition metals to form the complex ion is:

(a) Having unpaired electron in d-subshell

(b) Having paired electrons in d-subshells

(c) Providing empty d-orbitals

(d) Having small charge/size ratio

2. The correct electronic configuration of copper atom is:

(a) 3d104s1 (b) 3d104s2

(c) 3d94s2 (d)

3d54s24p

4

3.CrO3 dissolves in aqueous NaOH to give:

(a) CrO 2– (b) Cr(OH)

3

4

(c) Cr O 2– (d) Cr(OH)

2 7

112

4.. The electronic configuration of gadolinium (At. No 64) is:

(a) [Xe] 4f 8 5d0 6s2 (b) [Xe] 4f 7 5d2 6s2

(c) [Xe] 4f 3 5d5 6s2 (d) [Xe] 4f 6 5d2 6s2

5. Electronic configuration of a transition element in + 3 oxidation state is [Al]

3d5. What is is atomic number?

(a) 25 (b) 26

(c) 27 (d) 24

6. CrO3 dissolves in aqueous NaOH to give:

(a) CrO 2– (b) Cr(OH)

3

4

(c) Cr O 2– (d) Cr(OH)

2 7

7. The electronic configuration of gadolinium (At. No 64) is:

(a) [Xe] 4f 8 5d0 6s2 (b) [Xe] 4f 7 5d2 6s2

(c) [Xe] 4f 3 5d5 6s2 (d) [Xe] 4f 6 5d2 6s2

8.Which of the following oxidation state is common for all lanthanoids?

(a) + 2 (b) + 3

(c) + 4 (d) + 5

9.Which of the following is amphotric oxide?

Mn2O7, CrO3, Cr2O3, CrO, V2O5, V2O4.

(a) V2O5, Cr2O3 (b) Mn2O7, CrO3

(c) CrO, V2O5 (d) V2O5, V2O4

113

10. Which of the following ions show higher spin only magnetic moment value?

(a) Ti3+ (b ) Mn+2

(c) Fe2+ (d) Co3+

11.. Which of the following will not act as oxidising agents?

(a) CrO3 (b) MnO3

(c) WO3 (d) CrO24

–

12. Match the components/elements given in column I with uses given in column II.

Column 1 Column 2

(A) Lanthanoid oxide (1) Production of iron alloy

(B) Lanthanoid (2) Television screen

(C) Misch metal (3) Petroleum cracking

(D) Magnesium based alloy (4) Lanthanoid metal + iron

(E) Mixed oxides of lanthanoids (5) Bullets

are employed (6) In X-ray screen

(a) A–4, B–3, C–1, D–2 (b) A–3, B–4, C–1, D–2

(c) A–4, B–1, C–2, D–3 (d) A–2, B–1, C–3, D–4

13. Assertion: Cu2+ iodide is not known.

Reason: Cu2+ oxidises I– to iodine

14. Assertion: Separation of Zr and Hf is difficult.

Reason : Because Zr and Hf lie in the same graph of the periodic table

15. Which of the following ions show higher spin only magnetic moment value?

(a) Ti3+ (b) Mn2+

(c) Fe2+ (d) Co3+

16. Which of the following will not act as oxidising agents?

(a) CrO3 (b) MnO3

(c) WO3 (d) CrO24

–

114

Assertion and Reason Type

Note : In the following questions a statement of assertion followed by a statement of reason is given.

Choose the correct answer out of the following choice.

(a) Both assertion and reason are True, and reason is the correct explanation of the assertion.

(b) Both assertion and reason are True, but reason is not the correct explanation of the assertion.

(c) Assertion is not True, but reason is True.

(d) Both assertion and reason are False.

17.Assertion: Cu2+ iodide is not known.

Reason: Cu2+ oxidises I– to iodine.

18. Assertion: Separation of Zr and Hf is difficult.

Reason : Because Zr and Hf lie in the same graph of the periodic table.

19 Match the components/elements given in column I with uses given in column II.

Column 1 Column 2

(A) Lanthanoid oxide (1) Production of iron alloy

(B) Lanthanoid (2) Television screen

(C) Misch metal (3) Petroleum cracking

(D) Magnesium based alloy (4) Lanthanoid metal + iron

(E) Mixed oxides of lanthanoids (5) Bullets

are employed (6) In X-ray screen

(a) A–4, B–3, C–1, D–2 (b) A–3, B–4, C–1, D–2

(c) A–4, B–1, C–2, D–3 (d) A–2, B–1, C–3, D–4

20. Match the solutions given in Column 1 and the colours given in column II.

Column 1 Column 2

(i) FeSO4.7H2O (a) Green

(ii) NiCl2.4H2O (b) light pink

(iii) MnCl2.4H2O (c) Blue

(iv) COCl2.6H2O (d) Pale green

(v) Cu2Cl2 (e) Pink

(f) Colourless

115

Concept Map of d and f- BlockElements

Sources-Internet:www.studietoday.com

116

UNIT- VIII

CO-ORDINATION COMPOUNDS

SYNOPSIS


VIII CO- 1.Werner’s theory and **

ORDINATION some important Terms

COMPOUND

2.Nomenclature& ***

isomerism of co-ordination

compounds

3.Valence Bond theory ***

4.Crystal field theory in ***

octahedral complexes

5.Bonding in metal *

carbonyls

1. Coordination compounds

Coordination compounds are compounds in which a central metal atom or ion is linked to a number of

ions or neutral molecules by coordinate bonds or which contain complex ions.Example:

K4[Fe(CN)6]; [ Cu(NH3)4]SO4; Ni(CO)4

The main postulates of Werner’s theory of coordination compounds

2.In coordination compounds metals show two types of linkages or valencies- Primary and Secondary.

The primary valencies are ionisable and are satisfied by negative ions.

The secondary valencies are non- ionisable and are satisfied by neutral molecules or negative

ions.The secondary valence is equal to the C.N and is fixed for a metal.

The ions or groups bound by secondary linkages to the metal have characteristic spatial arrangements

corresponding to different coordination nos.

3.Difference between a double salt and a complex:

Both double salts as well as complexes are formed by the combination of two or more stable compounds

in stoichiometric ratio. However, double salts such as Mohr‘s salt,

117

FeSO4.(NH4)2SO4.6H2O, potash alum, K2SO4.Al2(SO4)3.24H2O, etc. dissociate into simple ions

completely when dissolved in water. However, complex ions such as [Fe(CN)6]4– of K4[Fe(CN)6], do not

dissociate into Fe2+ and CN– ions.

IMPORTANT TERMINOLOGY

(i)Coordination entity: It constitutes the central metal ion or atom bonded to a fixed number of ions or

moleculerepresented within a square bracket.

(ii)Central atom/ ion: In a coordination entity, the atom/ion to which a fixed number of ions/groups are

bound in a definite geometrical arrangement around it, is called the central atom or ion.

(iii)Ligands: The neutral or negative ions bound to the central metal or ion in the coordination entity.

These donate a pair/s of electrons to the central metal atom /ion. Ligands may beclassified as-

a)Monodentate /Unidentate: Ligands bound to the central metal atom/ion through a single donor atom.

Ex- Cl- ;H2O ; NH3 ; NO 2-.

Didentate: The Ligands which bind to the central metal atom/ion through two donor atoms.Ex- C2O42-

(ox); H2NCH2CH2NH2(en)

Polydentate: The Ligands which bind to the central metal atom/ion through two or more donor atoms

present in asingle ligand. Ex- (EDTA)4-

Chelating ligands: Di- or polydentate ligands that uses two or more donor atoms to bind to a single metal

ion toform ring- like complexes. (Ox); (EDTA)

Ambidentate ligand: A ligand that can ligate through two different atoms, one at a time.

Ex-

NO2- ; SCN-

Coordination number: The no. of ligand donor atoms to which the metal is directly bonded through sigma

bondsonly. It is commonly 4 or 6.

Counter ions: The ionisable groups written outside the square bracket. Ex- K+ in K4[Fe(CN)6] OR Cl- in

[Co(NH3)6]Cl3

Coordination Polyhedron: The spatial arrangement of the ligand atoms which are directly attached to the

central metal atom/ion. They are commonly Octahedral, Square-planar or Tetrahedral.

Oxidation number: The charge that the central atom would carry if all the ligands are removed along with

their pairs of electrons shared with the central atom. It is represented in parenthesis.

118

Homoleptic complexes: Complexes in which a metal is bonded to only one kind of donorgroups. Ex-

[Co(NH3)6] 3+

Heteroleptic complexes: Complexes in which a metal is bonded to more than one kind of donor groups.

Ex- [Co(NH3)4 Cl2]+

Nomenclature of mononuclear coordination compounds The following rules are used-

i) The cation is named first in both positively and negatively charged coordination entities. ii) The ligands

are named in an alphabetical order before the name of the central atom/ion.

iii) The name of the anionic ligands end in –o, those of neutral and cationic ligands are

the same except aqua for H2O, ammine for NH3, carbonyl for CO and nitrosyl for NOthese are placed

within enclosingmarks .

When the prefixes mono, di, tri, etc., are used to indicate the number of the individual ligands in the

coordination entity. When the names of the ligands include a numerical prefix, then the terms, bis, tris ,

tetrakis are used, the ligand to which they referbeing placed in parenthesis.

Oxidation state of the metal in cation, anion, or neutral coordination entity is indicated by roman numeral

in parenthesis. If the complex ion is a cation, the metal is same as the element. The neutral complex

molecule is named similar to that of the complex cation.

6.Names of some common ligands

NEGATIVE Neutral

LIGANDS ligands

Symbol Name Charge Symbol Name Charge

CN- Cyanido -1 NH3 Ammine 0

Cl- Chlorido -1 H2O Aqua 0

Br- Bromido -1 NO Nitrosyl 0

F- Fluorido -1 CO Carbonyl 0

SO42- Sulphato -2 NH2CH2- 1,2-ethane 0

CH2NH2 diamine

C2O42- Oxalato -2

ONO- O-Nitrito -1

NO2 - N-Nitrito -1 Positive

Ligands

NO3 - Nitrato -1 NO+ Nitrosonium +1

SCN- Thiocynato -1 NO2 + Nitronium +1

NCS- Isothionato -1

OH- Hydroxo -1

119

Isomerism in coordination compounds

Two or more substances having the same molecular formula but different spatial arrangements are called

isomers andthe phenomenon is called isomerism. Coordination compounds show two main types of

isomerism-a) Structural Isomerism b) Stereoisomerism

STRUCTURAL ISOMERISM:- It arises due to the difference in structures of coordination compounds.

It is furthersubdivided into the following types-

Ionization isomerism: This form of isomerism arises when the counter ion in a complex salt isitself a

potential ligand and can displace a ligand which can then become the counter ion. An example is provided

by the ionizationisomers [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4.

Hydrate or solvate isomerism: This form of isomerism is known as ‗hydrate isomerism‘ in case where

water isinvolved as a solvent. This is similar to ionizationisomerism. Solvate isomers differ by whether or

not a solventmolecule is directly bonded to the metal ion or merely present as free solvent molecules in

the crystal lattice. Anexample is provided by the aqua complex [Cr(H2O)6]Cl3 (violet) and its solvate

isomer [Cr(H2O)5Cl]Cl2.H2O (grey-green).

Linkage Isomerism: Linkage isomerism arises in a coordination compound containing ambidentate

ligand. A simpleexample is provided by complexes containing the thiocyanate ligand, NCS–, which may

bind through the nitrogen to give M–NCS or through sulphur to give M–SCN.

Coordination isomerism: It arises from the interchange of ligands between cationic and anionic entities of

different metal ions present in a complex .

Example [Co(NH3)6][Cr(CN)6] & [Cr(NH3)6][Co(CN)6]

STEREOISOMERISM: Stereo isomers have the same chemical formula and chemical bonds but they

have differentspatial arrangement. They are of two kinds

a. Geometrical isomerism

b. Optical isomerism

a).Geometrical Isomerism- This type of isomerism arises in heteroleptic complexes due to different

possible geometric arrangements of the ligands. Important examples of this behaviour are found with

coordination numbers 4 and 6.In a square planar complex of formula [MX2L2].(X and L are unidentate),

the two ligands X may be arranged adjacent to each other in acis isomer, or opposite to each other in a

trans isomer [MABXL]-Where A,B,X,L are unidentates Two cis- and one trans- isomers are possible.

120

Another type of geometrical isomerism occurs in octahedral coordination entities of the type [Ma3b3]

like[Co(NH3)3(NO2)3]. If three donor atoms of the same ligands occupy adjacent positions at the corners

of anoctahedral face, we have the facial (fac) isomer. When the positions are around the meridian of the

octahedron, we get the meridional (mer) isomer.

a)Optical Isomerism: Optical isomers are mirror images that cannot be superimposed on one another.

These arecalled as enantiomers. The molecules or ions that cannot be superimposed are called chiral. The

two forms are called dextro (d) and laevo (l) depending upon the direction they rotate the plane of

polarised light in a polarimeter (d rotates to the right, l to the left). Optical isomerism is common in

octahedral complexes involving

didentate ligands. In a coordination entity of the type [CCl2(en)2]2+, only the cis-isomer shows optical

activity.

121

Coordination Type of hybridization

number

Tetrahedral 4 SP3

Square planar 4 dsp2

Trigonal 5 sp3d

bipyramidal

Octahedral 6 sp3d2

Octahedral 6 d2sp3

CRYSTAL FIELD THEORY:

The metal-ligand bond is ionic arising purely from electrostatic interactions between the metal ion and the

ligand.

Ligands are treated as point charges or dipoles in case of anions and neutral molecules.

In an isolated gaseous metal atom or ion the five d-orbitals are degenerate.

Degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal /ion.

In a complex the negative field becomes asymmetrical and results in splitting of the d-orbital

A) CRYSTAL FIELD SPLLITING IN OCTAHEDRAL COORDINATION ENTITIES

1. For d4ions, two possible patterns of electron distribution arise:

If Δo < P, the fourth electron enters one of the eg orbitals giving the configuration t 2g3 eg

1

Ligands for which Δo < P are known as weak field ligands and form high spin complexes.

If Δo > P, it becomes more energetically favourable for the fourth electron to occupy a

t2gorbital with configuration t2g4 ego. Ligands which produce this effect are known as strong field ligands

and form low spin complexes.

122

B) CRYSTAL FIELD SPLLITING IN TETRAHEDRAL COORDINATION ENTITIES

1.The four surrounding ligands approach the central metal atom/ion along the planes between the axes

The t2g orbitals are raised in energy (2/5) t .

The two eg orbitals are lowered in energy (3/5) t

The splitting is smaller as compared to octahedral field splitting, t=(4/9) 0.

Pairing of electrons is rare and thus complexes have generally high spin configurations.

BONDING IN METAL CARBONYLS The metal-carbon bond in metal carbonyls possess both σ and π character. The M–C σ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The M–C π bond is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding π* orbital of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal .


VERY SHORT ANSWER QEUSTIONS (1 MARK)

1. What are ambidentate ligands? Give two examples for each.

ANS. Ambidentate ligands are ligands that can attach themselves to the central metal atom

through two different atoms. For example:

123

(a)

(The donor atom is N) (The donor atom is oxygen)

(b)

(The donor atom is S) (The donor atom is N)

Q2. Using IUPAC norms write the formula for the following: Tetrahydroxozincate(II) ANS. [Zn(OH)4]2-

Q3. Using IUPAC norms write the formula for the following: Hexaamminecobalt(III) sulphate

ANS. [Co(NH3)6]2 (SO4)3

Q4. Using IUPAC norms write the formula for the following: Pentaamminenitrito-O-cobalt(III)

ANS. [Co(ONO) (NH3)5]2+

Q5. Using IUPAC norms write the systematic name of the following: [Co(NH3)6]Cl3 ANS.

Hexaamminecobalt(III) chloride

Q6. Using IUPAC norms write the systematic name of the following:

[Pt(NH3)2Cl(NH2CH3)]Cl

ANS. Diamminechlorido(methylamine) platinum(II) chloride

Q7. Using IUPAC norms write the systematic name of the following: [Co(en)3]3+ ANS. Tris(ethane-1, 2-

diammine) cobalt(III) ion

Q8. Draw the structures of optical isomers of: c[Cr(C2O4)3]3-ANS .

124

Q9. What is meant by the chelate effect? Give an example.

ANS. When a ligand attaches to the metal ion in a manner that forms a ring, then the metal- ligand

association is found to be more stable. e.g. [Ni (en)3]2+


Q1. What is spectrochemical series? Explain the difference between a weak field ligand and a strong field

ligand.

ANS. A spectrochemical series is the arrangement of common ligands in the increasing order of their

crystal-field splitting energy (CFSE) values.

I−< Br−< SCN−< Cl−<S2-< F−< OH−< C2O42-∼ H2O < NCS−< NH3< en <CN-< CO

Q2. [Cr(NH3)6]3+is paramagnetic while [Ni(CN)4]2-is diamagnetic. Explain why?

ANS. Cr is in the +3 oxidation state i.e., d3 configuration. Also, NH3 is a weak field ligand

that does not cause the pairing of the electrons in the 3d orbital.

Cr3+ :

Therefore, it undergoes d2sp3 hybridization and the electrons in the 3d orbitals remain unpaired. Hence, it

is paramagnetic in nature.

In [Ni(CN)4]2−, Ni exists in the +2 oxidation state i.e., d8 configuration.

Ni2+:

CN− is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then, Ni2+ undergoes dsp2

hybridization.

Q3. A solution of [Ni(H2O)6]2+is green but a solution of [Ni(CN)4]2-is colourless. Explain. ANS.In

[Ni(H2O)6]2+, is a weak field ligand. Therefore, there are unpaired electrons in Ni2+. In this complex, the d

125

electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of d−d

transition is present. Hence, Ni(H2O)6]2+is coloured

In [Ni(CN)4]2-, the electrons are all paired as CN- is a strong field ligand. Therefore, d-d transition is not

possible in [Ni(CN)4]2-. Hence, it is colourless. As there are no unpaired electrons, it is diamagnetic.

Q2. Draw all the isomers (geometrical and optical) of: (i) [CoCl2(en)2]+

[Co(NH3)Cl(en)2]2+

[Co(NH3)2Cl2(en)]+ ANS. (i) [CoCl2(en)2]+ In total, three isomers are possible. Trans-isomers are optically inactive. Cis-isomers are optically active.

(iii) [Co(NH3)2Cl2(en)]+

Q3. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit

optical isomers?

ANS. [Pt(NH3)(Br)(Cl)(py)

From the above isomers, none will exhibit optical isomers. Tetrahedral complexes rarely show optical

isomerization. They do so only in the presence of unsymmetrical chelating agents.

Q4. What is meant by stability of a coordination compound in solution? State the factors which govern

stability of complexes.

126

ANS. The stability of a complex in a solution refers to the degree of association between the two species

involved in a state of equilibrium. Stability can be expressed quantitatively in terms of stability constant

or formation constant.

For this reaction, the greater the value of the stability constant, the greater is the proportion of

ML3 in the solution.

Q5. (a) Discuss the nature of bonding in the following coordination entities on the basisof valence bond

theory:

(i) [Fe(CN)6]4-(ii) [FeF6]3-(iii) [Co(C2O4)3]3-(iv) [CoF6]3-

ANS. (i) [Fe(CN)6]4-In the above coordination complex, iron exists in the +II oxidation state.Fe2+

: Electronic configuration is 3d6Orbitals of Fe2+ion:

As CN− is a strong field ligand, it causes the pairing of the unpaired 3d electrons.Since there are six

ligands around the central metal ion, the most feasible hybridization is d2sp3. The d2 sp3 hybridized orbital

of Fe2+ are:

127

6 electron pairs from CN− ions occupy the six hybrid d2sp3orbitals.Then, Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons). (ii) [FeF6]3−

In this complex, the oxidation state of Fe is +3. Orbitals of Fe+3 ion:

There are 6 F− ions. Thus, it will undergo d2sp3 or sp3d2 hybridization. As F− is a weak field ligand, it does not cause the pairing of the electrons in the 3d orbital. Hence, the most feasible hybridization is sp3d2.sp3d2 hybridized orbitals of Fe are: Hence, the geometry of the complex is found to be octahedral. (iii) [Co(C2O4)3]−

Cobalt exists in the +3 oxidation state in the given complex.Orbitals of Co3+ ion:Oxalate is a weak field ligand. Therefore, it cannot cause the pairing of the 3d orbital electrons. As there are ligands, hybridization has to be either sp3d2 or d2sp3 hybridization.sp3d2 hybridization of Co3+: The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp3d2 orbitals. Hence, the geometry of the complex is found to be octahedral.

(iv) [CoF6]3−Cobalt exists in the +3 oxidation state. Orbitals of Co3+ ion: Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons. As

128

result, the Co3+ ion will undergo sp3d2 hybridization.sp3d2 hybridized orbitals of Co3+ ion are: 81

Hence, the geometry of the complex is octahedral and paramagnetic.

ENRICHMENT EXERCISE Q.Why do tetrahedral complex not show geometrical isomerism?

Q.Why does the colour changes on heating [Ti(H2O)6]3+ .

Q.[Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3- is weakly paramagnetic. Explain.

Q.What happens when potassium ferrocyanide solution is added to a ferric salt solution?

A coordination compound has a formula (CoCl3. 4NH3). It does not liberate NH3but precipitates chloride

ion as AgCl. Give the IUPAC name of the complex and write its structural formula.

Q.Write the correct formula for the following co-ordination compounds. CrCl3 . 6H2O (Violet, with 3

Chloride ions/ Unit formula) CrCl3 . 6H2O (Light green colour with 2 Chloride ions/ unit formula)

Q.Give the electronic configuration of the d-orbitals of Ti in [Ti (H2O) 6]3+ ion in anoctahedral crystal

field.

Co(II) is stable in aqueous solution but in the presence of strong ligands and air, it can get oxidized to

Co(III). (Atomic Number of cobalt is 27). Explain.

Q.Give a chemical test to distinguish between [Co(NH3)5Br]SO4 and [Co(NH3)5Br]SO4Br. Name the type

of isomerism exhibited by these compounds.

Q.What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution

of copper sulphate?

129

Q.Why is that no precipitate of copper sulphate is obtained when H2S (g) is passed through this solution?

Q.Aqueous copper sulphate solution (blue in colour) gives a green precipitate with aqueous potassium

fluoride, a bright green solution with aqueous potassium chloride. Explain these experimental results. 2.

A metal complex having the composition Cr(NH3)4Cl2Br has been isolated in two forms, A and B. The

form A reacts with AgNO3 solution to give a white precipitate readily soluble in dilute aqueous ammonia

whereas B give a pale yellow precipitate soluble in concentrated ammonia solution. Write the formulae of

A and B and write their IUPAC names.

Q.Explain the following

i). All octahedral complexes of Ni2+must be outer orbital complexes.

ii). NH4+ ion does not form any complex.

iii). (SCN)-1 ion is involved in linkage isomerism in co-ordination compounds.

13. A metal ion Mn+ having d4 valence electronic configuration combines with three didentate ligands to

form complexes. Assuming o> P Draw the diagram showing d orbital splitting during this complex

formation. Write the electronic configuration of the valence electrons of the metal Mn+ ion in terms of t2g

and eg. What type of the hybridization will Mn+ ion have? Name the type of isomerism exhibited by this

complex.

The coordination no. of Ni2+ is 4. NiCl2 + KCN(excess) → A( a cyano complex ), A + Conc HCl(excess)

→ B ( a chloro complex )

Q.Write IUPAC name of A and B.

Q.Predict the magnetic nature of A and B

Q.Write hybridization of Ni in A and B

Q.Explain the following

i). Cu(OH)2 is soluble in ammonium hydroxide but not in sodium hydroxide solution. ii). EDTA is used to

cure lead poisoning

iii). Blue coloured solution of [CoCl4] 2- changes to pink on reaction with HgCl2.

130


Ligands and its types

Isomerism in co-ordination complexes

Warner’s theory of primary and secondary valencies

Valance bond theory

Crystal field splitting for octahedral and tetrahydral complexes

Bonding in carbonyls and the stability of complexe

MULTIPLE CHOICE QUESTIONS 1. Which of the following compounds formed by Cu2+ ions is most stable?

a) Cu2+ + 4NH3 → [Cu(NH3)4]2+; Log K = 11.6

b) Cu2+ + 4CN– → [Cu(CN)4]2–; Log K = 27.3

c) Cu2+ + 2 en → [Cu(en)2]2+; Log K = 15.4

d) Cu2+ + 4H2O → [Cu(H2O)4]2+; Log K = 8.9

2. The colour of the coordination compounds depends on the crystals field splitting. What will be

the correct order of absorption of warleinth of light in the visible region. for the complenes

[CO(NH3)6]3+; [CO(CN)0]3–; [CO(H2O)6]3+.

a) [CO (CN)6]3– > [CO(NH3)6]3+ > [CO(H2O)6]3+

b) [CO (NH3)6]3+ > [CO(H2O)6]3+ > [CO(CN)6]3–

c) [CO (H2O)6]3+ > [CO(NH3)6]3+ > [CO(CN)6]3–

d) [CO (CN)6]3– > [CO(NH3)6]3+ > [CO(H2O)6]3+

4. The correct IUPAC name of [Pt(NH3)2 Cl2] is

(a) Diamminedichloridoplatinum (II)

(b) Diamminedichlorideplatinum (IV)

(c) Diamminedichlorideplatinum (0)

(d) Dimminedichlorideplatinum (IV)

131

5. This stabilisation of coordination compounds due to chelation is called the chelate effect. Which

of the following is the most stable complen species?

(a) [Fe (CO)

5

] (b) [Fe(CN) ]3–

6

(c) [Fe(C O ) ]3+ (d) [Fe(H O)

]3

+

2 4 3 2 6

6.IUPAC name of [Pt(NH3)2Cl(NO2)] is:

(a) Platinum diaminechloritrite

(b) Chloronitrito-N-ammine platinum(II)

(c) Diamminechloridonitrite-N-platinum (II)

(d) Diamminechlornitrite-N-platinate(II)

7. Identify the optically acive compounds from the following:

(a) [CO(en) ]3+ (b) trans [CO(en) Cl ]+

3 2 2

(c) Cis [CO[en) Cl ]+ (d) [Cr(NH ) Cl]

2 2 3 5

8. Match the complen species given in column 1 with the possible isomerism given in column 1 and

assign the correct code:

Column 1 Column 2

(A) [CO(NCS) (NH3)5] (SO3) (1) + 4

(B) [CO(NH3)4 Cl2] SO4 (2) 0

(C) [CO (S2O3)3] Na4 (3) + 1

(D) [CO2(CO)8] (4) + 2

(5) + 3

(a) A–1, B–2, C–4, D–5 (b) A–4, B–3, C–2, D–1

(c) A–5, B–1, C–4, D–2 (d) A–4, B–2, C–2, D–3

Note: In the following questions a statement of assertion followed by a statement of reason is given.

Choose the correct answer out of the following choice.

132

(a) Both assertion and reason are True, and reason is the correct explanation of the assertion.

(b) Both assertion and reason are True, but reason is not the correct explanation of the assertion.

(c) Assertion is True, but reason is False.

(d) Both assertion and reason are False.

9. Assertion: Toxic metal ions are removed by the chelating ligands.

Reason: Chelate complens tend to be move stable.

10. Assertion: [(Fe(CN)6]3– ion shows magnetic moment corresponding to two unpaired electrons.

Reason: Because it has d2sp3 type hybridisation.

11.The CFSE for octahedral [CaCl6]4– is 18,000 cm–1. The CFSE for tetrahedral [COCl4]2– will be:

(a) 18,000 cm–1 (b) 16,000 cm–1

(c) 8,000 cm–1 (d) 20,000 cm–1

12. Match the complen ions given in column 1 with the hybridisation and number of unpaired

electrons given in column 2 and assign the correct code.

Column 1 Column 2

(A) [Cr(H2O)6]3+ (1) dsp2; 1

(B) [CO(CN)4 ]2– (2) sp3d2; 5

13. Identify the optically acive compounds from the following:

(a) [CO(en) ]3+ (b) trans [CO(en) Cl ]+

3 2 2

(c) Cis [CO[en) Cl ]+ (d) [Cr(NH ) Cl]

2 2 3 5

14. This stabilisation of coordination compounds due to chelation is called the chelate effect. Which of

the following is the most stable complen species?

(a) [Fe (CO)5]

(b) [Fe(CN)6 ]3–

(c) [Fe(C2O4)3] +3 (d) [Fe(H2O)6]+3

133

CONCEPT MAP OF CO-ORDINATIONCOMPOUNDS

Sources-Internet: www.studiestoday.com

134

UNIT- IX

HALOALKANES AND HALOARENES

SYNOPSIS


IX HALOALKANES 1. Classification *

AND &Nomenclature

HALOARENES

2. Methods of ***

preparation

3.Chemical Reactions. ***

4.Polyhalogen **

compounds.

1. Important terms and concept

R-H +X2 R-X+H–X

Halogen derivative of alkenes called halo alkenes e.g. R - CH3 (ii)Halogen derivative of arenes called

halo arenes e.g. R - C6H5

2- (a) Vicinal di halides – where the two halogens are attached on the adjacent carbon atom.

e.g. CH2Cl - CH2Cl

(b) Geminal dihalide- where two halogen atoms are attached to the same carbon atom eg CH3CHBr2

3. Important Reactions

(i) Nucleophilic substitution(ii) Eletrophilic substitution(iii) Elimination reaction

Carbylamine reaction(v) Reimer Tiemann reaction(vi) Wurtz reaction(vii) Wurtz fittig reaction.

Nucleophillic substitution -It involves the substitution of an atom or group by another atom or group.

A–B+C A–C+B

It must be remembered that A – B and A – C both are covalent compounds.

(i) In aliphatic system

R–X+OH- R–OH+X-

R–X+ O-C2H5 R-O-C2H5+X–

135

In general

R–X+:B- R–B+X-

B- OH-, RO-, CN-, RCOO-, NO3-, SH-, RS-Bimolecular Nucleophilic substitution SN2

(i) It takes place in one step.

Most of the SN2 reactions are second order but some time when nucleophilic reagent is present in excess

quantity the reaction is of Ist order but still proceeds by SN2

(iii)It is bimolecular It leads to inversion of configuration. The attack of nucleophilic occurs from

direction opposite to the leaving group.

SN1

It takes place in two steps.

All are Ist order. (iii)Unimolecular

It leads racemisation

Retention of configuration

The preservation of spatial arrangement of bonds at an asymmetric centre during the chemical reaction.

Stereochemistry of SN1 reaction

If an alkyl halide is optically active then the product is racemic mixture, here the attack of Nucleophile

from the both side [50:50 mix of the two enantiome

e.g. Nitration – how Electrophilic produce.

136

HNO3 + 2H2SO4 NO2 + + H3O+ +2HSO4-

OR

BF3+ HNO3NO2++ HOBF3

-

Elimination reaction

Two groups or atoms attached to two adjacent carbon atom and simultaneous formation of multiple bonds

between these carbon atom.[Reverse of addition]

Two types (i) β. Elimination - E1 Two step eliminate

(ii)α- elimination

Saytzaff’s Rule

Some important name reaction

1. Carbylamines reaction.

R – NH2 +CHCl3 +3 KOH RNC + 3 KCl + 3 H2O

C6H5NH2 + CHCl3 + 3 KOH C6H5NC + 3 KCl +3H2O

138



1. Give one example each of

Markwonikov’s addition.

Kharasch effect.

Sand Meyer reaction

Diazotisation reaction

Finkelstein reaction

Swarts

Wurtz reaction

Wurtz Fittig reaction

Fittig reaction

2. Explain the classification of halo alkanes based on

a) number of halogen atoms.

b) compounds having sp3 C-X bond

c) compounds having sp2 C-X bond

d) dihalides. Give one example each and their IUPAC names.

3. Carry out the following conversions:

Propene to a) Propan-1-ol b) Propan-2-ol

Ethanol to but-1-yne

1-Bromo propane to 2-Bromo propane and vice versa.

Toluene to benzyl alcohol.

Benzene to

4-bromonitro benzene

3-bromonitro benzene.

Benzyl alcohol to 2-phenyl ethanoic acid.

Ethanol to

Propane nitrle

Ethyl isocyanide.

139

8. Aniline

Chloro benzene

Bromo benzene

Iodo benzene.

2-Chloro butane to 3,4- dimethylhexane.

2-Methyl-1-propene to 2-chloro-2-methylpropane.

Ethyl chloride to propanoic acid.

But-1-ene to n-butyl iodide.

2-chloropropane to propan-1-ol

Isopropyl alcohol to iodoform.

SHORT ANSWER QUESTION ( 2 MARKS)

Q1. Why haloalkanes are more reactive than haloarenes?

Ans. In haloarenes,(i) there is double bond character b/w carbon and halogen due to resonance effect

which makes them less reactive. In benzene, carbon being sp2hybridisedwhich is smaller in size than sp3

present in haloalkanes. So C-Cl bond in aryl halides is shorter and stronger.

Q2. Why do haloalkenes undergo nucleophillic substitution whereas haloarenes undergo electophillic

substitution ?

Ans.Due to more electro negative nature of halide atom in haloalkanes carbon atom becomes slightly

positive and is easily attacked by nucleophillic reagents while in haloarenes due to resonance, carbon

atom becomes slightly negative and attacked by electrophillic reagents.

Q3.When an alkyl halide is treated with ethanolic solution of KCN, the major product is alkyl cyanide

where as if alkyl halide is treated with AgCN, the major product is alkyl isocyanide.

Ans.KCN is ionic they can attach through C or N but C-C bond is strong than C-N bond. So alkyl cyanide

is the major product but AgCN is covalent so more electronegative N can attach to C and forms

isocyanides.

Q4.How do 10, 20, 30 alcohols differ in terms of dehydrogenation?

Cu,3000c

Ans. 10alcohol aldehyde

Cu,300oc

2oalcohol -----------> ketone

Cu,300oC

30alcohol --------------->

Alkene

140

Q5. Why are the reaction of alcohol /phenol with acid chloride carried out in the presence of pyridine?

Ans. Because esterification reaction is reversible and presence of base (pyridine ) neutralizes HCl

produced during reaction thus promoting forward reaction .

Q6 Explain why o-nitrophenol is more acidic than o-methoxy phenol?

Ans. -NO2 group is electron with drawing group, e- density on O decreases and loss of H+ is easy

whereas –OCH3 group is electron releasing group, which increases e- density on O , which makes

difficult to the loss of H+, hence are less acidic .

Q7. Aryl halides cannot be prepared by the action of sodium halide in the presence H2SO4 .Why?

Ans. Due to resonance the carbon- oxygen bond in phenols has partial double bond and it is stronger than

carbon oxygen single bond.

Q8. Why Grignard reagent should be prepared under anhydrous conditions?

Ans. Grignard reagent react with H2O to form alkanes, therefore they are prepared under anhydrous

condition.

Q9. Why is Sulphuric acid not used during the reaction of alcohols with KI ?

Ans. It is because HI formed will get oxidized to I2 by conc. Sulphuric acid which is an oxidizing agent.

Q10. p- dichlorobenzene has highest m.p. than those of ortho and m-isomers.?

Ans. p- dichlorobenzene is symmetrical, fits into crystal lattice more readily and has higher melting point.

Q11. Although chlorine is an electron- withdrawing group, yet it is ortho and para directing in

electrophilic aromatic substitution reactions. Why?

Ans. Chlorobenzene is resonance hybrid, there is –ve charge at 0- and para positions, electrophilic

substitution reaction will take place at 0- and para position due to +R effect.+R effect is dominating over

– I effect..

Q12. The treatment of alkyl chlorides with aqueous KOH lead to the formation of alcohols but in

presence of alcoholic KOH alkenes are major products. Explain?

Ans. In aqueous KOH,OH- is nucleophile which replaces another nucleophile.

R-X +KOH R-OH +KX

Where as in alcoholic KOH

C2H5OH +KOH C2H5O- + K+

CH3CH2-Cl + alcoholic KOH CH2=CH2 + C2H5OH

141

Q13. Explain why vinyl chloride is unreactive in nucleophilic substitution reaction? Ans. Vinyl chloride

is unreactive in nucleophilic substitution reaction because of double bond character between C=Cl bond

which is difficult to break.

CH2 = CH – Cl CH2- –CH =Cl+

Q14. Arrange the following compounds according to reactivity towards nucleophillic substitution reaction

with reagents mentioned:-

4- nitrochlorobenzene, 2,4 dinitrochlorobemzene, 2,4,6, trinitrochlorobenzene with CH3ONa

Ans- 2,4,6, trinitrochlorobenzene>2,4 dinitrochlorobemzene>4- nitrochlorobenzene

Q15. Which compound will react faster in SN2 reaction with OH-?

Ans-(a) CH3Br and CH3I (SN2) CH3I will react faster than CH3Br.

(CH3)3C-Cl or CH3Cl (SN2) CH3Cl will react faster than 30halide.

Q16. Arrange in order of boiling points.

Bromobenzene, Bromoform, chloromethane,Dibromo-methane

1-chloropropane, Isopropyle chloride, 1-Chlorobutane.

Ans. (a) chloromethane < Bromobenzene < Dibromo-methane< , Bromoform (b), Isopropyl chloride <1-

chloropropane <1-Chlorobutane

(As Branching increases , boiling point decreases) Q17. Which compound undergoes SN1 reaction first ?

Ans. (a) 30 >20 >10 (b) 20

142

ENRICHMENT EXERCISE

1 . What happens when?

a. N-butylchloride is treated with alcoholic KOH

b. Bromobenzene is treated with Mg in the presence of dry ether.

c. Chlorobenzene is subjected to hydrolysis.

d. Ethyl chloride is treated with aq. KOH

e. Methyl bromide is treated with sodium in presence of dry ether.

2. An alkyl halide X of formula C6H13Cl on treatment with potassium tert-butoxide gives 2 isomeric

alkenes Y and Z (C6H12). Both alkenes on hydrogenation gives 2-3-Dimethylbutane predict the structures

of X, Y and Z?

3. Explain why chlorination of n-butane in presence of light at 298 K gives a mixture of 2-Chlorobutane

and 1-chlorobutane?

4. Describe the preparation of from ethanol using bleaching powder?

5. The Cl atom in Chlorobenzene is ortho and para directing explain why?

6. Why chlorobenzene cannot be hydrolised with aq NaOH at room conditions?

7. What is chloropicrin and how do you obtain it?

8. Nucleophilic substitution in aryl halides is facilitated by electron withdrawing groups.While

electrophilic substitution reaction is facilitated by electron releasing groups? Give reason?

9. Cholobenzene reacts with sodamide in the presence of liquid of liquid ammonia whereas, 2-bromo-3-

methyl anisole fails to react? Give reason?

10.. Which of the following has highest dipole moment? CCl4,CHCl3,CH2Cl2

11.. Differentiate between chiral and achiral molecules.

12.. What are enantiomers ? Draw the structures of possible enantiomers of 3-methylpent-1-ene.

13. Differentiate between retention and inversion.

143

14. Arrange the following in the order of increasing SN2 reactivity: CH3Cl, CH3Br,

CH3CH2Cl, (CH3)2CHCl 2. Arrange the following in the decreasing order of SN1 reactivity:

CH3CH2CH2Cl, CH2=CHCHClCH3 and CH3CH2CHClCH3.

15. RCl is hydrolysed to ROH slowly but the reaction is rapid if a catalytic amount of KI is added to the

reaction mixture.

16. Identify and indicate the center of chirality,if any in the following molecules ? How many

sterioisomers are possible for each?

2-Aminobutane

3-Bromopent-1-ene

1,2-Dichloropropane

17. Account for the following:

Halo alkanes have higher boiling point than the corresponding parent alkane.

Boiling point of halo alkanes RI>RBr>RCl> RF

Boiling point of 1-Bromo butane >2-Bromo butane> 1-Bromo- 2-methyl propane> 2-Bromo- 2-methyl

propane.

Melting point of p-Dichlo benzene is higher than its ortho and meta isomer.

Halo alkanes are polar in nature but sparingly soluble in water.

Iodo alkane can not be prepared by the reaction of alcohol with KI and sulphuric acid. Phosphoric acid is

used in place of sulphuric acid.

Order of reactivity of alcohol with HX is tert alcohol> sec alcohol > primary alcohol..

Halo arenes can not be prepared by treating phenol with HX or NaX in the presence of sulphuric acid.

Iodination of benzene is carried out in the presence of HIO3 or HNO3.

Propane on chlorination gives 2-chloro propane as a major product and not 1-chloro propane.

144

Kharasch effect is possible only with HBr and not with HCl and HI.

Alcohol reacts with thionyl chloride to give pure halo alkane.

Finkelstein reaction of halo alkane is carried out in the presence of dry acetone.

Order of reactivity of halo alkanes as per substitution bimolecular nucleophilic

is primary halide > secondary halide>tertiary halide.

Order of reaction as per substitution unimolecular is tertiary halide>secondary halide >primary halide.

Bezylic halides and allylic halides are more reactive towards nucleophile than halo alkanes.

Chloro ethene is less reactive towards nucleophile than chloro ethane.

Halo arenes are less reactive towards nucleophile than halo alkanes.

SN1 mechanism is ruled out in the reaction of halo arenes with nucleophile.

Electron with drawing groups like NO2 at ortho and para position with respect to halogen facilitates

nucleophillic substitution reaction.

Electron with drawing groups like NO2 at meta position with respect to halogen has no effect on

nucleophillic substitution reaction.

Halo arenes are less reactive towards electrophile than benzene.

Although chlorine atom has electron with drawing effect electrophillic substitution occur at ortho and

para position.

Order of reactivity of alkyl halide RI>RBr>RCl>RF

Halo alkanes react with KCN to give alkyl cyanide as a major product while it gives alkyl isocyanide as a

major product with AgCN.

Halo alkanes give nitrito alkane with KNO2 while nitro alkane with AgNO2.

CH3I undergoes SN2 reaction faster than CH3Cl.

2. Explain the following with suitable examples:

chiral and chirality

enantiomers

racemic mixture

retention of configuration

Inversion of configuration.

3. Give the products and explain the mechanisms of the following reactions:

a) CH3 CH2 Br + OH-

b)( CH3)3C-Br + OH-

c)n-BuBr + KOH

145

4. What happens when

a) n-butyl chloride is treated with alcoholic KOH.

b) bromobenzene is treated with Mg in the presence of dry ether.

c) chlorobenzene is subjected to hydrolysis.

d) ethyl chloride is treated with aqueous KOH.

e) methyl bromide is treated with Na in the presence of dry ether.

g) methyl chloride is treated with

KCN

AgCN

KNO2

AgNO2

5. Write the structure of the major organic product in each of the following reactions:

acetone

CH3CH2Cl + NaI heat

ethanol

(CH3)3 C-Br +KOH heat

ethanol

CH3CH2Br + KCN

C6H5ONa +CH3Br

CH3CH2OH +SOCl2

peroxide

CH3 CH=CH2 +HBr

CH3 CH=CH2 +HBr

(CH3)2C=CH2+ HBr

CH3CH=C(CH3)2 +HBr

CH3CH2CH2OH+SOCl2

CH3CH2Br +NaI

146

+ SOCl2

Br2/heat

Br2/heat

+HI

+Mg A

Dry ether

147

r) RBr +Mg A CH3CH (D)CH3

Dry ether D2O

(CH3)3C-C(CH3)3 RX A B Na/ether Mg/ether H2O

Arrange the compounds of each set in order of decreasing reactivity towards a) SN2 displacement. b) SN1

displacement.

2-bromo-2-methylbutane, 1-bromopentane, 2-bromo pentane

1-bromo-3-methylbutane, 2-bromo-2-methylbutane,3-bromo-2-methylbutane

1-bromo butane, 1-bromo-2-methyl propane, 1-bromo-2-phenyl propane.

Methyl chloride, Methyl bromide and Methyl iodide.

7. Distinguish chemically between

CH3Cl, CH3Br, CH3I

Chloro benzene and chloro methane

chloro benzene and benzyl chloride

CHCl3 and CCl4

Primary alkyl halide A (C4H9Br) react with alcoholic KOH to give B.B reacts with HBr to give C which

is an isomer of A. When A is treated with sodium in dry ether it gives a compound D C8H18 which is

different from the compound when n-butyl bromide is treated with sodium. Give the structural formula of

A and complete the reaction.

8.Which alkyl halide from the following pairs would you expect to react more rapidly by SN2

Mechanism? Explain your answer.

1-Bromo butane and 2-Bromo butane.

2-Bromo butane and 2-Bromo-2-methyl propane.

Cyclo hexyl chloro methane and chloro cyclo hexane.

Iodo butane and chloro butane.

148

10. Predict the order of reactivity of the following compounds in SN1 and SN2 mechanism.

C6H5CH2Br

C6H5CH(C6H5) Br

C6H5CH(CH3) Br

C6H5(CH3)(C6H5) Br. Explain your answer.


Do nomenclature

Prepare name reactions to solve:

Complete the reactions

Conversions

Distinguish between

Prepare mechanism of the reactions

Do NCERT exercise

149


1. Arrange the following compounds in increasing order of rate of reaction towards nucleophilic

substitution:

Cl

Cl Cl

NO2

(a) a < b < c (b) a < b < a

(c) a < c < b (d) c < a < b

2. Arrange the following compound in increasing order of rate of reaction towards nucleophilic

substitution.

Cl Cl Cl

CH3 NO2 NO2

(a) i < ii < iii (b) i < iii < ii

(c) ii < i < iii (d) iii < ii < i

3. Which of the following undergoes nucleophilic substitution exclusively by SN 1 mechanism?

(a) Benzyl chloride

(b) Ethyl chloride

(c) Chlorobenzene

(d) Isopropyl chloride

4. The increasing order of nucleophilicity would be

(a) Cl– < Br– < I–

(b) I– < Cl– < Br–

(c) Br– < Cl– < F–

(d) I– < Br– < Cl–

150

5. m-Xylene reacts with Br2 in presence of FeBr3, what are products formed

6. Name the product of C6H5—CH==CH—CH3 with HBr

7. Which of the following compound will undergo racemisation when reacts with aq. KOH?

(a) (i) and (ii)

(b) (ii) and (iv)

(c) (iii) and (iv)

(d) (iv)

8. Match the items of column 1 and column 2

Column1 Column 2

(A) CH2Cl2 P. Antiseptic

(B)

CCl4 Q. Insecticide

(C) (p-Cl C6H4)CHCCl3 R. Pyrene

(D)

CHI3 S. Refrigerend

(a) A–R, B–Q, C–S, D–P (b) A–S, B–R, C–Q, D–P

(c) A–Q, B–P, C–S, D–R (d) A–P, B–S, C–R, D–Q

151

Assertion and Reasoning

8.Assertion : SN2 reaction proceeds with inversion of configuration.

Reason : SN2 reactionoccurs in one step

9. Assertion : Treatment of chloroethane with saturated solution of AgCN give ethyl isocyanide as major

product.

Reason : Cyanide ion (CN–) is an ambident nucleophile.

152

CONCEPT MAP OF HALOALKANES ANDHALOARENES

Source-internet:www.studiestoday.com

153

UNIT- X

ALCOHOLS , PHENOLS AND ETHERS

SYNOPSIS

SL. No. Topic Concepts Degree of Importance

X ALCOHOLS 1. Classification & **

PHENOLS AND Nomenclature

ETHERS

2. Preparation of Alcohols ***

&Phenols

3. Chemical Reactions ****

4.Important commercial **

alcohols

5.Preparation of Ethers and ***

its chemical reactions

156

NAME REACTIONS

1. Kolbe’s Reaction:

Sodium phenoxide when heated with Carbon dioxide at 400K under a pressure of 4-7 Atmosphere

followed by acidification gives 2-Hydroxybenzoic acid (Salycilic acid) as the main product along with a

small amount of 4-Hydroxybenzoic acid.

2. Reimer- Tiemann Reaction:

Treatment of phenol with chloroform in presence of aq.NaoH or KOH at 340 K followed by hydrolysis

of the resulting product gives 2-Hydroxybenzaldehyde (salicylaldehyde) as the major product.

3. Williamson Synthesis:

It involves the treatment of an alkyl halide with a suitable sodium alkoxide to prepare symmetrical and

unsymmetrical ethers. The reaction involves SN2 attack of an alkoxide ion on primary alkyl halide.

R – X + R1 – O – Na

R – O – R1 + NaX

CH3I + CH3CH2ONa

CH3OCH2 – CH3 + NaI

CH3CH2 – I + PhONa

PhOH + NaI

Both simple and mixed ether can be produced.

157

Depending upon structure and cleavage of unsymmetrical ethers by halogen acid may occur either by SN2

or SN1 mechanism

CH3

CH3

SN2

e.g. (i) CH3 CH O CH3 + HI

CH3–I+CH3–CH-OH

158

CH3 SN1

(ii) CH3C O CH3 + H I (CH3)3 - C – I +CH3 - OH

CH3

Limitations of williamson synthesis

(CH3)3-C-ONa + CH3Br

CH3-O-C (CH3)3

(CH3)3-C-Br + Na-O-CH3

( CH3 )2-C= CH2

+ NaBr +CH3OH

159

Friedel-Craft’s Reaction: Anisole undergoes Friedel-Crafts reaction,i.e., the alkyl and acyl groups are

introduced at ortho and para positions by reaction with alkyl halide and acyl halide in the presence of

anhydrous aluminium chloride (a Lewis acid) as catalyst.

160

DISTINCTION BETWEEN PAIRS OF COMPOUNDS

Lucas Test: This test consists of treating an alcohol with Lucas reagent (mixt. of conc. HCl

andanhydrous ZnCl2) at room temp, when turbidity due to the formation of insoluble alkylchloride

is observed

i) If the turbidity appears immediately, the alcohol is tertiary.

ii)If the turbidity appears in about 5 minutes, the alcohol is secondary

A primary alcohol, however does not react with Lucas reagent at room Temp. and hence no

turbidity is formed. ROH +HX

RX + H2O

(a) Phenol and Benzoic acid

Test Phenol Benzoic acid

With FeCl3 Gives violet colour No characteristic colouration

With aq Does not give Gives brisk effervescence of CO2

NaHCO3 brisk

Effervescence of CO2

CO2

B. 1-Butanol and Diethyl ether

Test Ethanol Diethyl ether

With sodium Gives Brisk effervescence Of H2 Does not gives brisk

metal

CH2-CH2-CH2ONa + H2

161

Ethanol and Phenol

Test Ethanol Phenol

With litmus paper No action Turns blue litmus to red

With I2 and NaOH Gives yellow ppt of No Iodoform reaction

Iodoform

With FeCl3 No characteristic colour Gives violet colouration

2-Pentanol and 2-methyl – 2 Propanol

Test 2-Pentanol 2-methyl – 2 Propanol

Lucas test: Reaction with Does not react at room Gives turbidity

anhydrous Zinc chloride temperature. immediately.

and conc. HCl

Ethanol and 2-Propanol

Test Ethanol 2-Propanol

Heat with I2 solution And Forms yellow ppt of No yellow ppt.

NaOH Iodoform

CH3-CH2-OH CHI3+

HCOONa

Ethanol and 1-Propanol.

Test Ethanol 1-Propanol

Reaction with Iodine Gives yellow ppt of No yellow ppt of

Solution and NaOH, on Iodoform Iodoform

heating.

162

IMPORTANT MECHANISMS

Hydration of ethane to yield ethanol

Alkenes react with water in the presence of acid as catalyst to form alcohols. In case of

unsymmetrical alkenes, the addition reaction takes place in accordance with

164

Important reactions:

373 K CH3CH2HSO4 H2SO4 413 K CH3CH2OH CH3CH2 OCH2CH3 433 K to 444 K

CH2 = CH2+ H2O

(2) Preparation of phenol from Cumene

165


SHORT QUESTIONS ( 2 MARKS)

1.Explain phenol is acidic? Phenoxide ion is resonance stabilised

Note:If electron with drawing group are attached into the benzene ring it enhances acidic character and

vice versa.

2,4,6-Trinitrophenol > 2,4,-Dinitrophenol > 4-Nitrophenol > phenol Phenol > m- cresol > P- cresol > O- cresol m-methoxyphenol > phenol> O- methoxy phenol > P- methoxy phenol. O- chloro phenol > O- bromophenol > O- iodo phenol > O- fluoro phenol

FORMATION OF PICRIC ACID

HNO3 Phenol Br2 (aq)

Picric acid (2,4,6-Trinitrophenol)

2,4,6-Tribromophenol(white ppt)

166

Q2) Phenols do not give protonation reactions readily. Why?

Ans:- The lone pair on oxygen of O-H in phenol is being shared with benzene ring through

resonance.Thus,lone pair is not fully present on oxygen and hence phenols do not undergo protonatian

reactions.

Q3. Explain why propanol has higher boiling point than that of the hydrocarbon, butane?

ANS. The molecules of Butane are held together by weak van der Waal‘s Forces of attraction while those

of propanol are held together by stronger intermolecular hydrogen bonding.

Ans. The nitro-group is an electron-withdrawing group. The presence of this group in the ortho position

decreases the electron density in the O−H bond. As a result, it is easier to lose a proton. Also, the o-

nitrophenoxide ion formed after the loss of protons is stabilized by resonance. Hence, ortho nitrophenol is

a stronger acid. On the other hand, methoxy group is an electron-releasing group. Thus, it increases the

electron density in the O−H bond and hence, the proton cannot be given out easily.Therefore ortho-

nitrophenol is more acidic than ortho-methoxyphenol.

167

Q5. Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular

masses. Explain this fact.

ANS. Alcohols can form hydrogen bonds with water and break the hydrogen bonds already existing

between water molecules Therefore they are soluble in water. Whereas hydrocarbons cannot form

hydrogen bonds with water and hence are insoluble in water.

Q6. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer

which will be steam volatile. Give thane. O-nitrophenol is steam volatile due to intramolecular hydrogen

bonding and hence can be separated by steam distillation from p-nitrophenol which is not steam volatile

because of inter-molecular hydrogen bonding.

Q7. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitablemethod.

Give reason.

ANS. The formation of ethers by dehydration of alcohol is a bimolecular reaction (SN2) involvingthe

attack of an alcohol molecule on a protonated alcohol molecule. In the method, the alkylgroup should be

unhindered. In case of secondary or tertiary alcohols, the alkyl group ishindered. As a result, elimination

dominates substitution.

Q8. What is meant by hydroboration-oxidation reaction? Illustrate it with an example

ANS. Diborane (BH3)2 reacts with alkenes to give trialkyl boranes as addition product. This givesalcohol

by hydrogen peroxide in the presence of aqueous sodium hydroxide.

Q9. Which out of propan-1-ol and propan-2-ol is stronger acid?

ANS Propan-1-ol is stronger acid than propan-2-ol. The acidic strength of alcohols is in the order

10>20>30.

Q10. What is denaturation of an alcohol?

ANS. The commercial alcohol is made unfit for drinking by mixing in it some copper sulphate (togive it a

colour) and pyridine (a foul smelling liquid). It is known as denaturation of alcohol.

Q11. Give IUPAC name of CH3OCH2OCH3

ANS. Dimethoxymethane

168

Q12. Diethyl ether does not react with sodium. Explain.

ANS. Diethyl ether does not contain any active hydrogen.

Q13. Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of

ethanol.

ANS. The acidic nature of phenol can be represented by the following two reactions: (i) Phenol reactswith

sodium to give sodium phenoxide, liberating H2.

Q14. How does phenol react with dilute and conc. HNO3?

ANS. (i) With dilute nitric acid at low temperature (298 K), phenol yields a mixture of ortho and para

Nitrophenols.

Q15.With concentrated nitric acid, phenol is converted to 2,4,6-trinitrophenol. The product is commonly known as picric acid

Q16. How does phenol react with Br2 in CS2 and Bromine water?

ANS. (i) When the reaction is carried out in solvents of low polarity such as CHCl3 or CS2 and at low

temperature, monobromophenols are formed.

169

Q17.When phenol is treated with bromine water, 2,4,6-tribromophenol is formed as whiteprecipitate.

Q18. How do you account for the fact that unlike phenol, 2, 4-dinitrophenol and 2, 4, 6-trinitrophenol are

soluble in aqueous solution of sodium carbonate?

ANS. 2, 4-Dinitrophenol and 2, 4, 6-trinitrophenol are stronger acids then carbonic acid (H2CO3) due to

the presence of electron withdrawing – NO2 groups. Hence, they react with Na2CO3 to form their

corresponding salts and dissolve in aq. Na2CO3 solution.

Q19. Explain why phenols do not undergo substitution of the –OH group like alcohols. ANS. (i) Due to

electron withdrawing effect of phenyl group, the C—O bond in phenol is less polar, whereas in case of

methanol the methyl group has electron releasing effect and hence

C—O bond in it is more polar.

(ii)C—O bond in phenols has partial double bond character due to resonance and hence is difficult to

cleave.

Q20. Account for the following:

a). Boiling point of the C2H5OH is more than that of C2H5Cl

b). The solubility of alcohols in water decreases with increase in molecular mass.

ANS. A). Because of hydrogen bonding.

b). With increase in molecular mass the non-polar alkyl group becomes more predominant.

Q21. Answer the following

a). What is the order of reactivity of 10, 20 and 30 alcohols with sodium metal?

b). How will you account for the solubility of lower alcohols in water?

170

Ans: a). 10>20>30.

b). Here—OH group is predominant and the alcohol molecules can form hydrogen bonds with

watermolecules.

Q22. Give reasons:

i)Nitration of phenol gives ortho- and para- products only.

ii)Why do alcohols have higher boiling points than the haloalkanes of the same molecular mass?

ANS (1) –OH group increases the electron density more at ortho and para positions through itselectron

releasing resonance effect. Alcohols are capable of forming intermolecular H-bonds.

Q23. Account for the following:

(i)Phenols has a smaller dipole moment than methanol

Phenols donot give protonation reactions readily.

ANS i). In phenol the electron withdrawing inductive effect of –OH group is opposed by electron

releasing the resonance effect of –OH.

ii). The lone pair on oxygen of –OH in phenol is being shared with benzene ring through resonance. Thus,

lone pair is not fully present on oxygen and hence phenols do not undergo protonation reactions.

Q24. Explain the following facts that in aryl alkyl ethers (i) The alkoxy group activates the benzene ring

towards electrophilic substitution and (ii) It directs the incoming substituents to ortho and para positions

in benzene ring.

171

Thus, benzene is activated towards electrophilic substitution by the alkoxy group.

(ii)It can also be observed from the resonance structures that the electron density increases more at the

ortho and para positions than at the meta position. As a result, the incoming substituents are directed to

the ortho and para positions in the benzene ring.

Q25. How are primary, secondary and tertiary alcohols prepared from Grinard Reagents?

ANS.

The reaction produces a primary alcohol with thanei, a secondary alcohol with other aldehydes and

tertiary alcohol with ketones.

Q26. Give the equations of oxidation of primary, secondary and tertiary alcohols by Cu at

573 K.

Q27. Give equations of the following reactions:

(i)Oxidation of propan-1-ol with alkaline KmnO4 solution.

Bromine in CS2 with phenol. (iii)Dilute HNO3 with phenol.

172

Q28) Ortho- nitrophenol is more acidic than ortho –methoxy phenol. Why?

Ans:- Nitro group is electron with drawing which increases acidic charcter.

ENRICHMENT EXERCISE

1.Classify the following alcohols as primary10, secondary 20 and tertiary alcohols 30 :-Ethanol, 2-methyl-

1-propanol, 2-methyl-2-propanol, 2-proponal

2.Explain the following reactions- (I) Reimer-Tiemann Reaction (Ii) Williamson’s Synthesis of Ethers.

3.Name an electron releasing and electron withdrawing group.

4.Mention two important uses of methanol.

5.Write the general formula for mixed ETHER.

6.Write IUPAC name of the compound- CH3-CH2-CH(OH)-CH(OH)-CH3

7.Write one distinction test of – Phenol and alcohol

8.Give the preparation of phenol from CUMENE

9.Which Of the Following Will Have Higher B.P? CH3OH OR CH3-CH2-CH2-OH

173

10.Give Reaction of Phenol with Chloroform in Presence Of aqueous NaOH

11.Name Reagent Used In Oxidation of 10 Alcohols to Aldehyde.

12.Phenol Is more acidic than Alcohol. Why?

13Write One chemical Testto distinguish Ethyl Alcohol and 2-Propanol.

14.Why are ethers Sparingly Soluble in Water?

15.Give Balance Chemical Equation for the Reaction-(i)Oxidation of ethanol in presence of copper

catalyst at 573 K. (ii) Dehydration of ethanol at 443k.

16.Explain the mechanism of the acid catalyzed dehydration of alcohol at high temp.

Carry the following conversions:

(i)phenol to aniline.(ii)phenol to picric acid. (iii) 2-propanol to 1-bromopropane. 17.Show how will you

synthesize: (i) 1-phenylethanol from a suitable alkene. (ii) cyclohexylmethanol using an alkyl halide by

an SN2 reaction. (iii) pentan-1-ol using a suitable alkyl halide?

18.How are the following conversions carried out? (i) Propene → Propan-2-ol (ii) Benzyl chloride →

Benzyl alcohol (iii) Ethyl magnesium chloride → Propan-1-ol.

19.Name the reagents used in the following reactions: (i) Oxidation of a primary alcohol to

carboxylic acid. (ii) Oxidation of a primary alcohol to aldehyde. (iii) Bromination of phenol

to 2,4,6-tribromophenol.


Do nomenclature

Prepare name reactions to solve:

Complete the reactions

Conversions

Distinguish between

Prepare mechanism of the reactions

174

MULTIPLE CHOICE QUESTIONS 1. Arrange the following compound in decreasing order of boiling point

(i) propan-1-ol (ii) butane-1-ol

(iii) butan-2-ol (iv) pentan-1-ol

(a) i > iii > ii > iv (b) i > ii > iii > iv

(c) iv > iii > ii > i (d) iv > ii > iii > i

2. What is the correct order of reactivity of alcohols in the dehydration reaction?

2

(a) 1° > 2° > 3° (b) 1° < 2° > 3°

(c) 3° > 2° > 1° (d) 3° > 1° > 2°

3. Iodoform canbe prepared from all except

(a) butan-2-one (b) acetophenone

(c) propan-2-ol (d) propan-1-ol

4.Which of the following alcohol on dehydration with conc H2SO4 gives but-2-enc?

(a) 2-methylpropan-2-ol (b) Butan-1-ol

(c) 2-methyl propan-1-ol (d) Butan-2-ol

5. Which of the following alcohol give iodoferm test?

(a) Butan-1-ol (b) Propan-1-ol

(c) Propan-2-ol (d) Ethanol

6. Which of the following is a weaker acid than phenol?

(a) 4-Methoxy phenol (b) 3, 5-dinitrophenol

(c) 4-Methyl phenol (d) 4-Nitrophenol

175

7. The ether O—CH2 when treated with HI produces.

(a) CH2I (b) CH2 OH

(c) I (d) OH

8.Correct statements in case of n-butanol and t-butanol are:

(a) both are having equal solubility in water

(b) t-butanol is more soluble in water than n-butanol

(c) boiling point of t-butanol is lower than n-butanol

(d) boiling point of n-butanol is lower than t-butanol

9. The major product obtained on interactionof phenol with NaOH and CO2 is

(a) Benzoic acid (b) Salicaldehyde

(c) Salicylic acid (d) Pthalic acid

176

CONCEPT MAP OF ALCOHOLS, PHENOLS ANDETHERS

Sources-Internet:www.studies.com

177

UNIT-XI

ALDEHYDES, KETONES AND CARBOXYLIC ACIDS

SYNOPSIS


XI ALDEHYDES, 1.Nomenclature and Structure **

KETONES AND of Carbonyl group.

CARBOXYLIC

ACIDS

2.Preparation of aldehydes and ***

ketones.

3. Nucleophilic Addition ****

Reactions (Mechanism)

4. Methods of preparation of **

carboxylic acids

5.Chemical reactions

**

The π Electron cloud of >C=O is unsymmetrical. On the other hand, due to same electronegativity of the

two carbon atoms, the π-electron of the >C=C< bond is symmetrical.

Nature of carbonyl group:- The Pi electron cloud of >C=O is unsymmetrical therefore, partial positive

charge develop over carbon of carbonyl group while negative charge develop over oxygen of carbonyl

group and dipole moment is approximate 2.6D.

Aldehydes, Ketones and Carboxylic acids are important classes of organic compounds containing

carbonyl groups. They are highly polar molecules.

They boil at higher temperatures than the corresponding hydrocarbons and weakly polar compounds such

as ethers. Lower members are soluble in water because they can form H-bond with water. Higher

members are insoluble in water due to large size of their hydrophobic group.

Aldehydes are prepared by-

a. Dehydrogenation of primary alcohols

b. Controlled oxidation of primary alcohols.

c. Controlled and selective reduction of acyl halides. Aromatic aldehydes can be prepared by-

178

a.Oxidation of toluene with chromyl chloride or CrO3 in the presence of acetic anhydride b. Formylation

of arenes with carbon monoxide and Hydrochloric acid in thepresence of anhydrous aluminiumchloride /

Cuprous chloride

c. Hydrolysis of benzal chloride. Ketones are prepared by-

a. oxidation of secondary alcohols b. Hydration of alkenes

c. Reaction acyl chlorides with dialkylcadmium d. By Friedel crafts reaction.

Carboxylic acids are prepared by –

a. oxidation of primary alcohols, aldehydes and alkenes b. hydrolysis of nitriles

c. Treatment of thanei reagent with carbondioxide.

NAME REACTIONS

1.ROSENMUND REDUCTION: Acyl chlorides when hydrogenated over catalyst, palladium on barium

sulphate yield aldehydes. Benzoyl chloride Benzaldehyde 2.STEPHEN REACTION - Nitriles are reduced to corresponding imines with stannous chloride in the

presence of Hydrochloric acid, which on hydrolysis give corresponding aldehyde.

H3O+

RCN + SnCl2 +HCl RCH =NH RCHO

ETARD REACTION- On treating toluene with chromyl chlorideCrO2Cl2, the methyl group is oxidized

to a chromium complex, which on hydrolysis gives corresponding benzaldehyde.

179

4.GATTERMAN-KOCH REACTION-When benzene or its derivative is treated with CO and HCl in

the presence of anhyd. Aluminium chloride or cuprous chloride, it gives benzaldehyde or substituted

benzaldehyde.

CO, HCl - CHO

Anhyd.AlCl3 /CuCl

5.FRIEDEL CRAFTS ACYLATION REACTION When benzene or substituted benzene is treated

with acid chloride in presence of anhydrous aluminium chloride , aromatic ketones are obtained.

6.CLEMMENSEN REDUCTION The carbonyl group of aldehydes and ketone is reduced to –CH2

group on treatment with zinc amalgam and conc. Hydrochloric acid.

7.WOLFF- KISHNER REDUCTION

On treatment with hydrazine followed by heating with sodium or potassium hydroxide in high boiling

solvent like ethylene glycol

8.ALDOL CONDENSATION - Aldehydes and ketones having at least one α-hydrogen condense in the

presence of dilute alkali as catalyst to form β-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol).

180

9-CROSS- ALDOL CONDENSATION When aldol condensation is carried out between two different

aldehydes and / or ketones,a mixture of self and cross-aldol products are obtained.

10.CANNIZARO REACTION Aldehydes which do not have an α-hydrogen atom, undergo self-

oxidation and reduction (disproportionation) reaction on treatment with concentrated alkali,to yield

carboxylic acid salt and an alcohol respectively.

11.HELL-VOLHARD-ZELINSKY REACTION (HVZ) Carboxylic acids having an α – hydrogen are

halogenated at the α –position on treatment with chlorine or bromine in the presence of small amount of

red phosphorus to give α –halocarboxylic acids.

12.ESTERIFICATION Carboxylic acids react with alcohols or phenols in the presence of a

mineral acid such as conc.H2SO4 as catalyst to form esters. 144

13.DECARBOXYLATION: Carboxylic acids lose carbon dioxide to form hydrocarbons when their

sodium salts are heated with sodalime NaOH and CaO in the ratio 3: 1.

181

DISTINGUISH

Benzophenone Acetophenone

It doesn’t give iodoform test It gives iodoform test

Propanol /3-Pentanone Ethanol /2-Pentanone

It doesn’t give iodoform test. It gives iodoform test.

20 alcohol 30 alcohol

With Lucas reagent it gives turbidity It gives turbidity immediately.

after 5 minutes

Benzoic acid Benzene

add NaHCO3 Effervescence No effervescence comes out.

obtained(CO2)

Phenol Benzoic acid

It gives thane colour with FeCl3 test It doesn’t give thane colour

with FeCl3

It doesn’t give effervescence of CO2 Effervescence of CO2

evolve when NaHCO3

is added

Aniline Ethylamine

It gives azo-dye test (orange dye) It doesn’t give azo-dye test.

Phenol Alcohol

It gives FeCl3 test(violet colour) It doesn’t give this test

Benzaldehyde Propanal

It gives tollen’s reagent test It also give tollen’s reagent test

It doesn’t give fehling’s solution test It gives fehling’s solution test

Acetic acid Formic acid

It doesn’t gives tollen’s reagent It gives tollen’s reagent test

It doesn’t give fehling’s test It gives fehling’s test

182

SOME IMPORTANT TERMS

Cyanohydrin: When aldehydes and ketones react with HCN, hydrogen cyanide add across

the >C=O to yield cyanohydrins.

H+ HO

C =O +:CN- C NC

Semicarbazone: Nucleophile such as ammonia derivative H2N-Z like Semicarbazide H2N-NH-CO-

NH2add to the carbonyl group of aldehydes and ketones,in the presence of acid to yield addition product

called semicarbazone.

R

H

+

R

C =O +H2N-NH-CONH2

C = NNHCONH2

+ H2O

H H

Ketal: One mole of a dihydric alcohol adds to ketones in the presence of dry HCl to yield cyclic products

called ketals.

R CH2OHH+ R OCH2

R

CH2OH

R OCH2

Imine: Nucleophile such as ammonia H2N-Hadds to the carbonyl group of aldehydes and ketones,in the

presence of acid to yield addition product called IMINES

R

H+

R C =O + H2N-H

C =NH + H2O

H H

183

(v) 2,4-DNP-derivative:

Nucleophile such as ammonia derivative H2N-Z like 2,4-dinitrophenylhydrazine H2N-NH-C6H5add to the

carbonyl group of aldehydes and ketones,in the presence of acid to yield addition product called 2,4-

Dinitrophenylhyrazone.

NO2

NO2

R R

H H

Schiff’s base: Nucleophile such as ammonia derivative H2N-Z like Amine H2N-Radd to the carbonyl

group of aldehydes and ketones,in the presence of acid to yield addition product

called substituted imine Schiff‘s base

H+

RCHO + RNH2 RCH = NR + H2O


SHORT QUESTIONS ( 2 MARKS) 1.Name the reaction and the reagent used for the conversion of acid chlorides to the corresponding

aldehydes.

Name : Rosenmund‘s reaction

Reagent : H2 in the presence of Pd (supported over BaSO4) and partially poisoned by addition of Sulphur

or thaneiz.

2.Suggest a reason for the large difference in the boiling points of butanol and butanal, although they have

same solubility in water.

A. The b. pt. of butanol is higher than that of butanal because butanol has strong intermolecular H-

bonding while butanal has weak dipole-dipole interaction. However both of them form H-bonds with

water and hence are soluble.

184

3. What type of aldehydes undergo Cannizaro reaction ?

A. Aromatic and aliphatic aldehydes which do not contain α- hydrogens.

Out of acetophenone and benzophenone, which gives iodoform test ? Write the reaction involved. (The

compound should have CH3CO-group to show the iodoform test.)

A. Acetophenone (C6H5COCH3) contains the grouping (CH3CO attached to carbon) and hence given

iodoform test while benzophenone does not contain this group and hence does not give iodoform test.

I2/NaOH

C6H5COCH3

+ 3 I2 + 4 NaOH

CHI3 + C6H5COONa + 3 NaI + 3 H2O

I2/NaOH

C6H5COC6H5

No reaction

185

z

5. Give Fehling solution test for identification of aldehyde gp (only equations). Name the aldehyde which

does not give Fehling‘s soln. test.

A. R — CHO + 2 Cu2+ + 5OH– RCOO– + Cu2O + 3 H2O Benzaldehyde does not give Fehling soln.

test.

(Aromatic aldehydes do not give this test.)

6. What makes acetic acid a stronger acid than phenol?

A. Greater resonance stabilization of acetate ion over phenoxide ion.

Why HCOOH does not give HVZ (Hell Volhard Zelinsky) reaction but CH3COOH does? A. CH3COOH

contains alpha hydrogens and hence gives HVZ reaction but HCOOH does not contain alpha-hydrogen

and hence does not give HVZ reaction.

During preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst,

water or the ester formed should be removed as soon as it is formed.

A. The formation of esters from a carboxylic acid and an alcohol in the presence of acid catalyst in a

reversible reaction.H2SO4

RCOOH + ROH RCOOR + H2O

To shift the equilibrium in the forward direction, the water or ester formed should be removed as fast as it

is formed.

9.Arrange the following compounds in increasing order of their acid strength.

Benzoic acid, 4-Nitrobenzoic acid, 3, 4-dinitrobenzoic acid, 4-methoxy benzoic acid.

A. 4-methoxybenzoic acid < benzoic acid < 4-nitrobenzoic acid < 3,4, dinitrobenzoic acid. 10. How will

you distinguish between methanol and ethanol?

A. By Iodoform test: Ethanol having alpha-methyl group will give yellow ppt. of iodoform whereas

methanol does not have alpha methyl group will not give ppt. of iodoform.

10.Distinguish between: (i) Acetaldehyde and acetone (ii) Methanoic acid and Ethanoic acid.

A. (i) Acetaldehyde will give positive tests with Tollen‘s reagent and Fehling solutions whereas acetone

will not give these tests.

186

11.Methanoic acid gives Tollen‘s reagent test whereas ethanoic acid does not due to difference in their

boiling points.

12.Why are aldehydes more reactive than ketones ?

A. It is because of two reasons: The carboxyl compounds (both aldehydes & ketones) undergo

nucleophilic addition reaction. (i) + I effect: The alkyl group in ketones due to their e– releasing character

decrease the electrophilicity (+ ve) charge on c-atom and thus reduce its reactivity.

(Ii) Steric hindrance: Due to steric hindrance in ketones, they are less reactive.

13. Give the composition of Fehling A and Fehling B?

A. Fehling A = aq. CuSO4, Fehling B = alkaline sodium potassium tartarate (Rochelle salt). 14. Name

one reagent which can distinguish between 2-pentanone and 3-pentanone ?

14. 2-pentanone has a CH3CO-group, hence gives positive iodoform test. 3-pentanone does not have a

CH3CO-group, hence does not give positiIe iodIform test.

15. Why PCC cannot oxidise methanol to methane and while KMnO4 can?

This is because PCC is a mild oxidising agent and can oxide meththaneizethanal only. While KMnO4

being sthaneiidising agent oxidises it to metthaneizid.

16. Would you expect benzaldehyde to be more reactive or less reactive in nucleophlic addition reaction

than propanal? Explain.

A. C-atom of carbonyl group of benzaldehyde is less electrophilic than C-atom of carbonyl group in

propanal. Polarity of carbonyl group is in benzaldehyde reduced due to resonance making it less reactive

in nucleophilic addition reactions.There is no such resonance effect in propanal and so the polarity of

carboxyl group in it is more than in benzaldehyde.This makes propanal more reactive than benzaldehyde.

17. Why does methanal not give aldol condensatithanei ethanol gives?

A. This is because only those compounds which have alpha hydrogen atoms can undergo aldol reaction.

Ethanol possesses alpha -hydrogen and alpha undergoes aldo– condensation.Methanal has no alpha

hydrogen atoms, hence does not undergo aldol condensation.

187

18.Why does methanal undergo Cannizaro‘s reactthaneiBecause it does not possesses alpha hydrogen

atom. Only those aldehydes can undergo Cannizaro reaction which does not possess alpha hydrogen

atoms.

Arrange the following in order of increasing boiling points:

CH3CH2CH2OH, CH3CH2CH2CH3, CH3CH2 — OCH2CH3, CH3CH2CH2CHO A. CH3CH2CH2CH3<

C2H5OC2H5< CH3CH2CH2CHO < CH3 (CH2)2 OH

19.Why does solubility decreases with increasing molecular mass in carboxylic acid? A. Because of

increase in alkyl chain length which is hydrophobic in nature.

20.Although phenoxide ion has more no. of resonating structures than carboxylate ion,carboxylic acid is a

stronger acid. Why?

A. Conjugate base of phenol i.e.phenoxide ion has non-equivalent resonance structures in which –ve

charge is at less electronegative C-atom and +ve charge is at more electronegative O-atom i.e. Resonance

is not so effective.In carboxylate ion, – ve charge is delocalised on two electronegativethaneizatnce

resonance is more effective.

21.There are two — NH2 group in semicarbazide. However, only one is involved information of

semicarbazones. Why ?

A. Although semicarbazide has two — NH2 groups but one of them is involved in resonance.

As a result, e– density on one of the — NH2 group is reduced and hence it does not act as

nucleophile.Lone pair of other — NH2 group is not involved in resonance and is available for

nucleophilic attack.

22.Arrange the following carboxyl compounds in increasing order of their reactivity in nucleophilic

addition reactions. Explain with proper reasoning:

Q.Benzaldehyde. p-tolualdeyde, p-nitrobenzaldehyde, Acetophenone.

A.Acetophenone is a ketone while all others are aldehydes, therefore it is least reactive. In p-

tolualdehyde, there is methyl group (CH3) at para position w.r.t. to the carboxyl gp, which increases the

electron density on the carbon of the carboxyl gp by hyperconjugation effect therebymaking it less

reactive than benzaldehyde.

188

23.Which acid of each pair shown here would you expect to be stronger? CH3CO2H or FCH2CO2H

A. FCH2CO2H

24.Arrange the following compounds in increasing order of their reactivity towards HCN. Explain it with

proper reasoning. Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone.

A. Addition of HCN to the carboxyl compounds is a nucleophilic addition reaction.The reactivity towards

HCN addition decreases as the + I effect of the alkyl gps. increases and/or the steric hindrance to the

nucleophilic attack by CN– at the carboxyl carbonincreases. Thus the reactivity decreases in the order:

Ditert-butyl Ketone < tert-Butyl methyl Ketone < Acetone < Acetaldehyde.

ENRICHMENT EXERCISE

1.Account for the following:

Aldehyde and ketone are polar in nature.

Aldehyde and ketones have higher boiling point than hydro carbons of comparable molar mass.

Aldehyde and ketones have lower boiling point than alcohols of comparable molar mass.

Ketone has higher boiling point than aldehyde of comparable molar mass.

Oxidation of primary alcohol to aldehyde is carried out using PCC as an oxidizing agent.

Rosenmund's reduction of acid chloride to aldehyde is carried out using quinoline and sulphur.

Aldehyde is more reactive than ketone towards nucleophile.

Butanone is less reactive than propanone.

2,2,6-Tri methyl cyclo hexanone is less reactive towards nucleophile than cyclo hexanone.

Para nitro benzaldehyde is more reactive towards nucleophile than benzaldehyde.

Para methyl benzaldehyde is less reactive towards nucleophile than benzaldehyde.

Reaction of aldehyde with alcohol to give acetal is carried out in the presence of HCl(g).

Formaldehyde and benzaldehyde undergoes cannizaro reaction and not aldol condensation.

Acetaldehyde undergoes aldol condensation and not cannizaro reaction.

Aromatic aldehyde and ketones undergoes electrophillic substitution at meta position.

Carboxylic acid do not show the reactions of aldehyde and ketone though it has >C=O group.

189

Carboxylic acid has higher boilimg point than aldehyde, ketone and alcohol of comparable molar mass.

In semi carbazide, only one NH2 group is involved in the formation of semi carbazone.

Aldehyde, ketone and carboxylic acid are soluble in water.

In oxidation of primary alcohol to carboxylic acid is not carried out using acidified potassium dichromate.

Carboxylic acid is more acidic than alcohol.

Carboxylic acid is more acidic than phenol.

Acidity of CCl3COOH>CHCl2COOH>CH2ClCOOH>CH3COOH.

Acidity of FCH2COOH>ClCH2COOH>BrCH2COOH>I CH2COOH.

a chloro propanoic acid is more acidic than ß chloro propanoic acid.

An organic compound A(C8H8O) gives orange red precipitate with 2,4-DNP reagent. It responds to

iodoform test. It does not respond to Tollen’s reagent test. It does not decolorise bromine water. A on

oxidation using CrO3 give B. Identify the compounds and write thw equations of the reactions involved.

An organic compound C9H10O forms 2,4-DNP derivative, reduces Tollen’s reagent andUndergoes

cannizaro reaction. On vigorous oxidation it gives Benzene-1,2-dicarboxylic acid. Identify the compound.

An organic compound contain 69.77% C and 11.63% H and the remaining O. Molecular mass of the

compound is 86 u. It does not reduce Tollen’s reagent, gives positive iodoform test and respond to sodium

bisulphate test. On oxidation it gives acetic acid and propanoic acid. Give the structure of the organic

compound.

Arrange the following in the increasing order of property mentioned:

Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone( reactivity with HCN)

2-Bromo butanoic acid, 3-Bromo butanoic acid, Butanoic acid, 3-Methyl propanoic acid (acid strength)

Benzoic acid, 4-Nitro benzoic acid, 3,4-Dinitro benzoic acid, 4-Methoxy benzoic acid (acid strength)

Ethanal, Propanal, Propanone, Butanone( increasing order of reactivity towards nucleophile)

190

Benzaldehyde, p-Tolualdehyde, p-Nitro benzaldehyde, Acetophenone(increasing order of reactivity

towards nucleophile)

Butanal, Butanol, Ethoxy ethane and Butane (increasing order of boilimg point)

Suggest chemical tests to distinguish between

Propanal and propanone

Acetaldehyde and Benzaldehyde

Phenol and Acetic acid or Benzoic acid

2 – pentanone and 3 – pentanone

Formaldehyde and Acetaldehyde

7.Conversions

Propanoic acid to 1 – propanol 2. Ethanal to Acetone 3. Benzoic acid to Benzyl Chloride

Benzoic acid to aniline 5. Toluene to Benzaldehyde 6. Benzoic acid to Benzaldehyde 7. Acetamide to

ethanol 8. Acetaldehyde to Isopropyl alcohol

1. Why do Carboxylic acids not give the characteristic reactions of a carbonyl group? 9.Cyclohexanone

forms cyanohydrin in good yield but 2,2,6 trimethyl cyclo-hexanone does not. Why?

2. Why is carboxyl group in benzoic acid meta directing?

3. Treatment of Benzaldehyde with HCN gives a mixture of two isomers which cannot be separated even

by careful fractional distillation. Explain why?

4. Sodium bisulphite is used for the purification of aldehydes and Ketones. Explain.

5. Why pH of reaction should be carefully Controlled while preparing ammonia derivatives of carbonyl

compound?

6. Why formic acid is stronger acid than acetic acid?

7. Why is oxidation of alcohols to get aldehydes carried out under controlled conditions?

8. Why the oxidation of toluene to benzaldehyde with CrO3 is carried out in the presence of acetic

anhydride.

191

KEY POINTS/SUBJECT TIPS:-

1.Prepare nomenclature

2.Corelate physical properties and chemical reactions of aldehydes, ketones and carboxylic acids with

their structures.

3.Mechanism of selected reactions of aldehydes and ketones.

4.Understand various factors affecting the acidity of carboxylic acids and their reactions.


1. Which one is most acidic

(a) Ethanoic acid (b) Methanoic acid

(c) Ethanol (d) Phenol

2. Which has highest boiling point

(a) Ethanal (b) Ethanoic acid

(c) methoxy methane (d) Propane

3. Compound A and C in the following reaction are ............

CH3CHO + CH 3MgBr ------> A --B2H6---> B

(a) identical (b) position isomer

(c) functional isomer (d) optical isomer

4.Reduction of aldehydes and ketones into hydrocarbons using zinc amalgam and cone. HCl is called:

(a) Cope reduction

(b) Dow reduction

(c) Wolff Kishner reduction

(d) Clemensen reduction

192

5. CH3CHO and C6H5 CH2CHO can be distinguished chemically by

(a) Benedict’s test

(b) Iodoform test

(c) Tollen’s reagent test

(d) Fehling’s solution test

6. Increasing order of rate of HCN addition to compound (I-IV) is

(i) HCHO (ii) CH3COCH3

(iii) PHCOCH3 (iv) PHCOPH

(a) iv < ii < iii < i (b) iv < iii < ii < i

(c) iii < iv < ii < i (d) i < ii < iii < iv

7. Among the following which has the lowest pka value:

(a) CH3COOH (b) HCOOH

c) CH3—CH2—OH (d) (CH3)2CHCOOH

8. The correct order of acidity in given compounds

(i) FCH2COOH (ii) ClCH2COOH

(iii) NO2CH2COOH (iv) CH3COOH

(a) i > ii > iii > iv (b) iv > iii > ii > i

(c) iii > iv > i > ii (d) iii > i > ii > iv

dilNaOH HCN

Heat ----> A+ H3O-->B

(a) CH2 CH—CH—COOH (b) CH2 CH—CH—OH

OH CN

(c) CH3—CH2—CH—COOH (d) CH3—CH—COOH

OH OH

193

CONCEPT MAP OF ALDEHYDES, KETONES AND CARBOXYLICACIDS

Source-Internet: www.ncerthelp.com

194

UNIT- XII AMINES

SYNOPSIS

SL.No. Topic Concepts Degree of Importance

XII AMINES 1. Nomenclature and ***

preparation of amines.

2. Basic character of **

Amines(pKb) and

comparisons in gaseous and

aqueous phase.

3. Carbylamine Reaction ***

,Hinsberg’s Test.

4. Electrophilic substitution. **

5. Chemical reactions ***

involving Diazotization.

IUPAC NOMENCLATURE

NAME REACTIONS

1.Gabriel phthalimide synthesis

195

Gabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment with ethanolic

potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by

alkaline hydrolysis produces the corresponding primary amine. Aromatic primary amines cannot be

prepared by this method because aryl halides do not undergo nucleophilicsubstitution with the anion

formed by phthalimide.

2.Hoffmann bromamide degradation reaction

Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an

aqueous or ethanolic solution of sodium hydroxide. The amine so formed contains one carbon less than

that present in the amide.

3.Carbylamine reaction

Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide

form isocyanides or carbylamines which are foul smelling substances. Secondary and tertiary amines do

not show this reaction. This reaction is known as thaneizati reaction or isocyanide test and is used as a test

for primary amines.

4.Hinsberg Test

Benzenesulphonyl chloride (C6H5SO2Cl), which is also known as Hinsberg‘s reagent, reacts with

primary and secondary amines to form sulphonamides.

196

The reaction of benzenesulphonyl chloride with primary amine yields N-ethylbenzenesulphonyl amide.

The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron

withdrawing sulphonyl group. Hence, it is soluble in alkali.

(b) In the reaction with secondary amine, N,N-diethylbenzenesulphonamide is formed.

Since N, N-diethylbenzene sulphonamide does not contain any hydrogen atom attached to nitrogen atom, it

is not acidic and hence insoluble in alkali.

ITertiary amines do not react with benzenesulphonyl chloride. This property of amines reacting with

benzenesulphonyl chloride in a different manner is used for the distinction of primary, secondary and

tertiary amines and also for the separation of a mixture of amines.

5.Sandmeyer Reaction

The Cl–, Br– and CN– nucleophiles can easily be introduced in the benzene ring of diazonium salts in the

presence of Cu(I) ion.

6.Gatterman Reaction

Chlorine or bromine can be introduced in the benzene ring by treating the diazonium salt solution with

corresponding halogen acid in the presence of copper powder.

197

7.Coupling reactions

The azo products obtained have an extended conjugate system having both the aromatic rings joined

through the –N=N– bond. These compounds are often coloured and are used as dyes.

Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is

coupled with the diazonium salt to form p-hydroxyazobenzene. This type of reaction is known as coupling

reaction.Similarly the reaction of diazonium salt with aniline yields p-aminoazobenzen

198

DISTINCTION BETWEEN PAIRS OF COMPOUNDS

Give one chemical test to distinguish between the following pairs of compounds. (i)Methylamine and

dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline

(iv)Aniline and benzylamine (v) Aniline and N-methylaniline.

ANS. (i) Methylamine and dimethylamine can be distinguished by the thaneizati test. Carbylamine test:

Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide

form foul-smelling isocyanides or carbylamines. Methylamine (being an aliphatic primary amine) gives a

positive thaneizati test, but dimethylamine does not.

Secondary and tertiary amines can be distinguished by allowing them to react with Hinsberg‘s reagent

(benzenesulphonyl chloride, C6H5SO 2Cl). Secondary amines react with Hinsberg‘s reagent to form a

product that is insoluble in an alkali. For example, N, N−diethylamine reacts with Hinsberg‘s reagent to

form N, N−diethylbenzenesulphonamide, which is insoluble in an alkali. Tertiary amines, however, do

not react with Hinsberg‘s reagent.

(iii)Ethylamine and aniline can be distinguished using the azo-dye test. A dye is obtained when

147aromatic amines react with HNO2 (NaNO2 + dil.HCl) at 0-5°C, followed by a reaction with the

alkaline solution of 2-naphthol. The dye is usually yellow, red, or orange in colour. Aliphatic amines give

a brisk effervescence due (to the evolution of N2 gas) under similar conditions.

OH

199

00–50C

CH3NH2 + HNO2 C2H5 OH + N2 ↑ +H2O

Aniline and benzylamine can be distinguished by their reactions with the help of nitrous acid, which is

prepared in situ from a mineral acid and sodium nitrite. Benzylamine reacts with nitrous acid to form

unstable diazonium salt, which in turn gives alcohol with the evolution of nitrogen gas.

On the other hand, aniline reacts with HNO2 at a low temperature to form stable diazonium salt. Thus,

nitrogen gas is not evolved.

Aniline and N-methylaniline can be distinguished using the Carbylamine test. Primary amines, on heating

with chloroform and ethanolic potassium hydroxide, form

foul-smelling isocyanides or carbylamines. Aniline, being an aromatic primary amine, gives positive

thaneizati test. However, N-methylaniline, being a secondary amine does not.

200



Q1. Account for the following: i)Pkb of aniline is more than that of methylamine. (ii) Ethylamine is soluble in water whereas aniline is not. (iii)Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

(iv)Although amino group is o– and p– directing in aromatic electrophilicsubstitution reactions, aniline on

nitration gives a substantial amount of m-nitroaniline.

v)Aniline does not undergo Friedel-Crafts reaction.

Diazonium salts of aromatic amines are more stable than those of aliphatic amines. Ans. (i) pKb of aniline

is more than that of methylamine:

Aniline undergoes resonance and as a result, the electrons on the N-atom are delocalized over the benzene

ring. Therefore, the electrons on the N-atom are less available to donate.

129

On the other hand, in case of methylamine (due to the +I effect of methyl group), the electron density on

the N-atom is increased. As a result, aniline is less basic than methylamine. Thus, pKb of aniline is more

than that of methylamine. (ii)Ethylamine is soluble in water whereas aniline is not:

Ethylamine when added to water forms intermolecular H−bonds with water. Hence, itis soluble in water.

But aniline does not undergo H−bonding with water to a very large extent due to the presence of a large

hydrophobic –C6H5 group. Hence, aniline is insoluble in water. (iii)Methylamine in water reacts with

ferric chloride to precipitate hydrated ferricoxide: Due to the +I effect of –CH3 group, methylamine is

more basic than water. Therefore, in water, methylamine produces OH− ions by accepting H+ ions from

water.Ferric chloride

201

(FeCl3) dissociates in water to form Fe3+ and Cl− ions.Then, OH− ion reacts with Fe3+ ion to form a

precipitate of hydrated ferric oxide.

(iv)Although amino group is o, p− directing in aromatic electrophilic substitution reactions, aniline on

nitration gives a substantial amount of m-nitroaniline:

Nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to give anilinium

ion (which is meta-directing).For this reason, aniline on nitration gives a substantial amount of m-

nitroaniline.

(v)Aniline does not undergo Friedel-Crafts reaction:

A Friedel-Crafts reaction is carried out in the presence of anhydrous AlCl3. But AlCl3 is acidic in

nature(Lewis acid), while aniline is a strong base. Thus, aniline reacts with AlCl3 to form a salt. Due to

the positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated. Hence,

aniline does not undergo the Friedel-Crafts reaction.

vi)Diazonium salts of aromatic amines are more stable than those of aliphatic amines:

The diazonium ion undergoes resonance as shown below:

This resonance accounts for the stability of the diazonium ion. Hence, diazonium salts of aromatic amines

are more stable than those of aliphatic amines.

Q2. Why aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis? Ans. Gabriel

phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic

substitution (SN2) of alkyl halides by the anion formed by the phthalimide. But aryl halides do not

undergo nucleophilic substitution with the anion formed by the phthalimide.

202

Hence, aromatic primary amines cannot be prepared by this process.

Q3. Give possible explanation for each of the following:

(i)Why are amines less acidic than alcohols of comparable molecular masses?

(ii)Why do primary amines have higher boiling point than tertiary amines?

iii)Why are aliphatic amines stronger bases than aromatic amines?

Ans. (i) Amines undergo protonation to give amide ion. Similarly, alcohol loses a proton to give alkoxide

ion.In an amide ion, the negative charge is on the N-atom whereas in alkoxide ion, the negative charge is

on the O-atom. Since O is more electronegative than N, O can accommodate the negative charge more

easily than N. As a result, the amide ion is less stable than the alkoxide ion. Hence, amines are less acidic

than alcohols of comparable molecular masses.

In a molecule of tertiary amine, there are no H−atoms whereas in primary amines,two hydrogen atoms

are present. Due to the presence of H−atoms, primary amines undergo extensive intermolecular

H−bonding.As a result, extra energy is required to separate the molecules of primary amines. Hence,

primary amines have higher boiling points than tertiary amines.

(iii)Due to the –R effect of the benzene ring, the electrons on the N- atom are less available in case of

aromatic amines. Therefore, the electrons on the N-atom in aromatic amines cannot be donated

Q4. Write short notes on ammonolysis.

When an alkyl or benzyl halide is allowed to react with an ethanolic solution of ammonia, it undergoes

nucleophilic substitution reaction in which the halogen atom is replaced by an amino (−NH2) group. This

process of cleavage of the carbon-halogen bond is known as ammonolysis.When this substituted

ammonium salt is treated with a strong base such as sodium hydroxide, amine is obtained.

Though primary amine is produced as the major product, this process produces a mixture of primary,

secondary and tertiary amines, and also a quaternary ammonium salt.

Q5. Write short notes on acetylation.

Acetylation is the process of introducing an acetyl group into a molecule. Aliphatic and aromatic primary

and secondary amines undergo acetylation reaction by nucleophilic substitution when treated with acid

chlorides, anhydrides or esters. This reaction involves the replacement of the hydrogen atom of –NH2 or

203

group by the acetyl group, which in turn leads to the production of amides. To shift the equilibrium to the

right hand side, the HCl formed during the reaction is removed as soon as it is formed. This reaction is

carried out in the presence of a base (such as pyridine) which is stronger than the amine.

Pyridine

C2 H5NH2 +CH3COCl

Q6.Why are amines basic in character?

C2H5NHCOCH3+ HCl

ANS. Like ammonia, the nitrogen atom in amines RNH2is trivalent and bears an unshared pair of

electrons. Thus it acts like a Lewis base and donates the pair of electrons to electron- deficient species

which further increases due to +I effect of alkyl radical.

Q7. Arrange the following in decreasing order of their basic strength: C6H5NH2, C2H5 NH2, (C2H5)2NH,

NH3.

The decreasing order of basic strength of the above amines and ammonia follows the following

order:(C2H5)2NH > C2H5 NH2 > NH3> C6H5NH2

Q8. Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

204

ANS. (i) Aromatic amines react with nitrous acid (prepared from NaNO2 and a mineral acid such as HCl)

at 273 – 278 K to form stable aromatic diazonium salts, NaCl and H2O. (ii)Aliphatic primary amines react

with nitrous acid (prepared from NaNO2 and a mineral acid such as HCl) to form unstable aliphatic

diazonium salts, which further produce alcohol and HCl with the evolution of N2 gas.

ENRICHMENT EXERCISE

1. Write IUPAC Name of C6H5N(CH3)3Br ?

2. Which reaction is used for preparation of pure aliphatic & aromatic primary amine?

3. Name one reagent used to distinguish between primary, secondary & tertiary amines?

4. What is the directive influence of amino group in arylamines?

5. Why are benzene diazonium salts soluble in water?

6. Which is more basic: CH3NH2& (CH3)3N?

7. Which is more basic, aniline or ammonia?

8. Write the IUPAC name of C6H5NHCH3?

9. What product is formed when aniline is first diazotized and then treated with Phenol in alkaline

medium?

10. Primary amines have higher boiling points than tertiary amines why?

11. Why is it necessary to maintain the temperature between 273 K & 278 K during diazotization?

12. Arrange the following in order of decreasing basic strength: Ethyl amine, Ammonia, Triethylamine ?

13. Why aniline is acetylated first to prepare mono bromo derivative?

14. Arrange the following in decreasing order of their basic strength. C6H5NH2, C2H5NH2, (C2H5)2NH,

NH3 in gaseous state.

15. Write chemical equation for the conversion CH3-CH2-Cl into CH3–CH2-CH2-NH2.

205

16. Write the equation involved in Carbylamines reactions?

17. How will you distinguish the following pairs?

18. Methanamine and N-methyl methanamine(ii) Aniline and ethylamine

19. Write chemical reactions involved in following name reactions.

20. Hoffmann Bromoamide reaction.(ii) Diazotisation reaction.

21. Direct nitration of aniline is not carried out. Give reason.

22. The presence of base is needed in the ammonolysis of alkyl halides. Why?

23. How will you convert

i)Benzene into aniline (ii) Benzene into N, N-dimethylaniline

iii)Aniline to Sulphanilic acid

24. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’

which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the

structures and IUPAC names of compounds A, B and C.

25. How will you carry out the following conversions

Aniline to Phenol (b)Acetamide to Ethylamine I Aniline to p-nitroaniline.


1.Prepare classification and nomenclature of amines

2.Methods of prepration of amines.

3.Name reactions

4.Distinguish between primary, secondary and tertiary amines.

5.Describe the methods of prepration of diazonium salts and their importance in the synthesis of aromatic

compounds including azo

206


1.In the nitration of benzene using a moisture of Conc H2SO4 and conc. HNO3 the species which

initiates the reaction is:

(a) NO2+ (b) NO2

-

(c) NO2 (d) NO

2. The correct IUPAC name of Cu2 = CH—CH2NHCH3 is:

(a) Allymethyl amine (b) 2-amino-4-pentene

(c) 4-aminopent-lene (d) N-methylprop-2-en-anine

3. Which is the weakest base:

(a) CH3NH2 (b) ( CH3)2 NH2

(c) (CH3)3NH2 (d)C6H5NH2

4 The most acidic among the following compounds is:

(a) CH3NH2 (b) ( CH3)2 NH2

(c) (CH3)3NH2 (d)C6H5NH2

5. Which of the following statement about primary amine is false:

(a)Aryl amines react with nitrous acid to produce phenol

(b)Alkylamines are stronger base than ammonia

(c)Alkyl amines are stronger base than aryl amines

(d)Alkyl amines react with nitrous acid to produce alcohol

6. CH3CH2COOH

The structure of ‘c’ would be

(a) CH3CH2CO NH2 (b) CH3CH2NHCH3

(c) CH3CH2NH2 (d) CH3CH2CH2NH2

207

NH2 N

7. NaNO2 B CH3 C

HCl Cold


CH3 CH3

N N CH

3

CH3

(b) N N N

CH3

(c)

CH3

N N N

(d) N N CH2 N N

CH3

8.Which of the following is most stable diazoniom salt?

(a) CH N+ X– (b) C H N+X–

3 2 6 5 2

(c) CH CH N + X– (d) C H CH N + X–

3 2 2 6 5 2 2

9.Method by which aniline can not be prepared is:

a) reduction of nitrobenzene with H2/Pd in ethanol.

b) potassium salt of phthalimide treated with chlorobenzene

c) hydrolysis of phenyl isocyanide with acidic solution

d) degradation of benzamide with bromine in alkaline medium solution.

208

10. Methylamine reacts with HNO2 to form

(a) CH3—O—N==0 (b) CH3OCH3

(c) CH3OH (d) CH3CHO

11. CH3CH2COOH


(a) CH3CH2CO NH2 (b) CH3CH2NHCH3

(c) CH3CH2NH2 (d) CH3CH2CH2NH2

12. The correct IUPAC name of Cu2 = CH—CH2NHCH3 is:

(a) Allymethyl amine (b) 2-amino-4-pentene

(c) 4-aminopent-lene (d) N-methylprop-2-en-anine

Assertion and Reasoning

Assertion : n-Propylamine has higher boiling point than trimethylamine.

Reason : Among n-propylamine molecules, there is hydrogen bonding but there is not hydrogen bonding in

trimethylamine.

Assertion : Aniline does not undergo Friedel Crafts reaction.

Reason : Friedel Crafts reaction is an electrophilic substitution reaction.

209

CONCEPT MAP OF AMINES

Sources-Internet: www.studiestoday.com

210

UNIT-XIII

BIOMOLECULES SYNOPSIS

SL.No. Topic Concepts Degree of

importance

XIII BIOMOLECULES 1. Classification of carbohydrates *

with examples

2.Reactions of Glucose, Open & ***

cyclic str. Of glucose

4. Glycosidic Linkage, invert sugar **

5. Polysaccharides – starch( **

Amylose and amylopectin) and

cellulose

6. Peptide Linkage **

7. Proteins – Classifications ***

Denaturation

8. Vitamins ***

9. DNA & RNA **

Carbohydrates- These are optically active polyhydroxy aldehydes or ketones due to presence of chiral `C’

or the compounds which produce these on hydrolysis except dihydroxy acetone which is not optically

active.

Classification-

Monosaccharides – Those carbohydrates which cannot get hydrolysed e.g. glucose, fructose, galactose etc.

Oligosaccharides- Those carbohydrates which give two or more monosaccharide’s on hydrolysis e.g.

sucrose on hydrolysis gives glucose and fructose. Raffinose on hydrolysis gives glucose, fructose and

galactose.

(iii)Polysaccharides- Those carbohydrates which on hydrolysis give large number of monosaccharide’s.

e.g. starch, cellulose, glycogen.

3.Sugar-

(i)Reducing Sugars- Those which reduce Fehling’s or Tollen’s reagent. They have free aldehydic groups,

eg , glucose, galactose

(ii)Non Reducing Sugars- Those which do not reduce Fehling’s or Tollen’s reagent. They do not have free

functional group, e.g. sucrose

211

4.Glucose- It is a monosaccharide’s with molecular formula C6H12O6

5. Preparation (i)From Sucrose

C12H22O11 + H2O ----- C6H12O6 + C6H12O6 ( Only from sucrose)

(ii)From Starch

(C6H10O5)n + Nh2O ----- C12H22011 + H20 ---- 2C6H12O6

6.Structure

(i)Fischer structure –

CHO — (CHOH)4 — CH2OH

(+) Glucose has `D’ configuration as shown

`D’ means —OH group on first chiral `C’ from the bottom is on right hand and + means it is dextro rotatory,

i.e. it rotates plane polarized light towards right. Reactions of D-glucose

a)On prolonged heating with HI it yields n-hexane.

HI,Δ

OHC-(CHOH)4-CH2OH ( D-glucose) CH3-(CH2)4-CH3(n-hexane)

With hydroxylamine it forms an oxime and adds a molecule of HCN to give cyanohydrin. NH2OH

OHC-(CHOH)4-CH2OH ( D-glucose) HO-N=CH-(CHOH)4-CH2OH

HCN

NC-CH(OH)-(CHOH)4-CH2OH

On reaction with a mild oxidizingagent like bromine water, it gets oxidized to a six carbon carboxylic

acid,gluconic acid.

Br2 (aq)

OHC-(CHOH)4-CH2OH HOOC-(CHOH)4-CH2OH (Gluconic acid)

212

On oxidation with HNO3 glucose yields a dicarboxylic acid, saccharic acid. HNO3

OHC-(CHOH)4-CH2OH HOOC-(CHOH)4-COOH(saccharic acid) (ii)Cyclic Structure OF

Glucose: the straight chain is unable to explain the following reactions.

(a)It does not give the 2, 4-DNP test, Schiff’s Test and does not form the hydrogensulphite product with

NaHSO3 .

The pentaacetate of glucose does not react with NH2OH, indicating the absence of free aldehydic group.

Glucose exist in 2 different crystalline forms α and β forms. These are called anomers.

They differ in optical rotation, they also differ in melting point.

Anomers are isomers which have a different configuration across C-1 (first chiral ‘C’ atom).

CH2OH CH2OH

O O

1 1

OH OH OH H

H OH

H

7.Glycosidic Linkage: The linkage between two monosaccharide units through oxygen is called the

glycosidic linkage.

Proteins: These are macro molecules made up of amino acids joined via a peptide link (-(-CO-NH-)- is the

peptide linkage). These are required for growth and development of the body.

Amino Acids: These contain an amino (-NH2) and an acidic (-COOH) group and are therefore amphoteric in

nature. In solution they exist in the form of zwitter ion.

213

11.Structure and Shape of Protein

Primary Structure Secondary structure Tertiary structure Quarternary

structure

The specific sequence It is the shape in Represents overall Protein can be

of amino acids in the which the long folding of the composed of two or

polypeptide chain. polypeptide chain polypeptide chain. It more polypeptide

Change in amino can exist. It is of two gives rise to the chains called sub

acids sequence types : α- helix and fibrous or globular units. The spatial

changes the protein. β- pleated. These molecular arrangement ofthese

They have covalent structures arise shapes.Forces sub units with

bonds. dueto regular folding stabilizing the 2o respect to each other

of the backbone of and 3o structures quaternary structure

the polypeptide are hydrogen bonds, of the protein.

chain due to H- disulphide linkages,

bonding between the van der waal’s and

C=o and –NH- electrostatic forces of

groups of the peptide attraction.

bond.

12.Denaturation of Protein: The protein in native state, when subjected to a physical change like

temperature, Ph etc undergoes uncoiling and looses it’s biological activity. The 2o and 3o structures are

destroyed, only 1o structure is retained.

Renaturation of Protein:

Some proteins regain their biological activity by reversible process it is called Renaturation of Proteins. In

such a cases, when temperature in Ph of a denatured proteins is brought back to conditions in which the

native protein is stable, secondary and tertiary structures of proteins are restored to which leads to recovery

of biological activity.

Enzymes: These are biocatalyst which catalyse biochemical reactions and generally are globular proteins

e.g., invertase, zymase, phenylalaninehydroxylase, urease etc.

Vitamins: They are organic compounds required in the diet in small amounts to perform specific biological

functions for maintenance of optimum growth and health of the organism. They are classified as follows

(i) Fat Soluble Vitamins: Vitamin A, D, E and K. They are stored in liver and adipose tissues.

(ii) Water Soluble Vitamins: B complex vitamins and vitamin C. They need to supplied regularly in diet as

they are excreted in urine and cannot be stored (except vitamin B12) in our body.

Their deficiency causes diseases. Biotin (Vit H) is however neither fat nor water soluble. It’s deficiency

leads to loss of hair.

214

Nucleic Acids: These are biomolecules which are long chain polymers of nucleotides. They are: (i)

Deoxyribonucleic acid (DNA) (ii) Ribonucleic acid (RNA)

They are responsible for protein synthesis and transfer of genetic characteristics to offspring’s.

Composition of Nucleic Acid:

They are made up of pentose sugar (β-D-2-deoxyribose in DNA and β-D-ribose in RNA), phosphoric acid

and nitrogen containing heterocyclic compound (base). DNA- Bases present are Adenine (A), Thymine

(T), Guanine (G) and Cytosine(C).

RNA- contains Adenine (A), Guanine (G), Cytosine(C) and Uracil (U).

Nucleoside: The unit formed by the attachment of a base to 1’-position of sugar (Base + Sugar).

Nucleotide: Nucleoside and phosphoric acid at 5’-position. Nucleotides are bonded by phosphodiester

linkages between 5’ and 3’ carbon atoms of pentose sugar (Base+ Sugar+

Phosphoric Acid).

DNA : has a double helical structure with AT and GC linked together through 2 and 3 hydrogen bonds

respectively. It is responsible for transfer of genetic characteristics.

RNA: is of three types- messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA).

RNA helps in protein synthesis.

Biological Functions of Nuclei Acid: DNA is chemical basis of hereditary and have the coded message for

proteins to be synthesized in the cell. RNA carry out the protein synthesis in the cell.

215



Q1 – Which functional groups are present in monosaccharides?

Ans - ―OH and ―CHO or ―OH and >CO

Q2 – Name an aldopentone, aldohexone and ketohexone.

Ans – Ribose,glucose and fructose respectively.

Q3 – What is animal starch?

Ans – Glycogen.

Q4 – Which types of bonds are present in a protein molecule?

Ans – Peptide bonds, hydrogen bonds, sulphide bonds, ionic bonds etc.

Q5 – Which α-helix or β-helix is more stable?

Ans – α-helix is right handed and is more stable due to intermolecular H bonding between first and fourth

amino acid.

Q6 – Name the vitamin whose deficiency causes rickets?

Ans – Vitamin D.

Q7 – Name the purine bases present in DNA.

Ans – Adenine and guanine.

Q8 – Give an example of Vitamin which is Water soluble (b) Fat soluble Ans – (a) Vitamin C (b)Vitamin

D.

Q9 – Name a protein which is insoluble in water.

Ans – Keratin.

Q10 – Name polysaccharides that make up starch and what is the difference between them. Ans – Amylose

which is linear polymer of α-glucose and amylopectin which is branched polymer of α-glucose. Amylose is

water soluble whereas amylopectin is water insoluble.

216

Q12 – What are anomers?

Ans – Monosaccharide which differ only in the orientation of the ―OH group at C-1.e.g, α-glucose and β-

glucose.

Q13 – Where does the water present in the egg go after boiling the egg?

Ans – On boiling during denaturation process water gets adsorbed/absorbed in the denatured proteins.

Q14 – What do you understand by glycosidic linkage?

Ans – During condensation of two monosaccharides, a water molecule given out and two monosaccharides

get linked together by an oxide or ethereal linkage (―O―) called as glycosidic linkage.


Q1– What are essential and non- essential amino acid? Give two examples of each type. Ans – Essential

amino acids are those which are not produced in our body.e.g., valine, leucine.

Non-essential amino acids are those which are produced by our body. E.g. glycine and alanine.

Q 2 – How do you explain the amphoteric behavior of amino acids?

Ans – Amino acids have both acidic as well as basic group and they react both with acids as well as bases,

therefore they are amphoteric in nature.

Q 3 – What is the difference between a nucleoside and a nucleotide?

Ans – Nucleoside = sugar + base

Nucleotide = sugar + base + phosphoric acid

Q4 – What is invert sugar?

Ans – An equimolar aqueous solution of glucose and fructose is called invert sugar.

217

Q5 – Give three differences between DNA and RNA.

Ans – DNA RNA

1. It has deoxyribose as sugar. 1. it contains ribose as sugar.

2. It contains thymine along with adenine, 2.it contains uracil in place of thymine with

cytosine and guanine as bases. other bases.

3. It is responsible for maintaining heredity 3. it is responsible for protein synthesis.

traits from generation to generation.

Q6 – Give reactions with support cyclic structure of glucose.

Ans – (a)Glucose does not give 2,4-DNP test, Schiff`s test and sodium hydrogen suphide test.

(b)The pentaacetate of glucose does not react with NH2OH indicating absence of free ―CHO group.

IGlucose exists in two crystalline form α and β.

Q7. Mention structural differences between amylopectin and cellulose.

Amylopectin Cellulose

1. It is linear polymer of α- 1. It is linear polymer of β-

glucose. glucose.

2. It consists of branched 2. In cellulose, the chains are

chains of α-glucose. arranged to form bundles and

held together by hydrogen

bond between glucose and

adjacent strands.

Q8. What deficiency diseases are caused due to lack of vitamins B1, B6 and K in human diet?

Ans. Vitamins Deficiency Disease

B1 Beri beri (loss of appetite)

B6 Convulsions

K Increased blood clotting time

218

ENRICHMENT EXERCISE

1. What are biomolecules?

2. What is meant by RNA and DNA?

3. What are carbohydrates?

4. How are vitamins classified?

5. Name the disease caused by the deficiency of vitamin A and vitamin D?

6. What are polysaccharides? Name one of them.

7. What are the expected products of hydrolysis of lactose?

8. Why cannot vitamin C be stored in our body?

9. What is the basic structural difference between starch and cellulose?

10. Define the following-

a. Peptide linkage.

b. Primary structure of proteins and

c. Denaturation

11. Explain briefly-Essential and non- essential amino acids with one example each. 12.How do you

explain the amphoteric thanei of amino acids?

12.What happens when (D)-glucose is treated with the following reagents-

HIb. Bromine waterc. HNO3

a). Despite having an aldehyde group glucose does not give 2, 4-DNP test. What does this indicate?

b). Draw the Haworth structure of alpha –(D)-glucopyranose. c). What is the significance of D and (+)?

15.Name the only vitamin which can be synthesized in our body. Name one disease caused due to the

deficiency of this vitamin.

16.What is the difference between a nucleocide and a nucleotide?

17.How are vitamins classified? Name the vitamin responsible for coagulation of blood.

219

MULTIPLE CHOICE QUESTIONS 1.Which of the following statement is not true about glucose?

(a) it is an aldohexose

(b) on heating with HI it forms n-hexane

(c) it is present in furanose form

(d) it does not give 2, 4-DNP test

2.Which of the following acids is a vitamin?

(a) aspartic acid (b) ascorbic acid

(c) aoiphic acid (d) saccharic acid

3. Which of the following base is not present in RNA?

(a) Adenine (b) uracil

(c) Thymine (d) cytosine

4. Which of the following vitamins can be stored in our body?

(a) vitamin B1 (b) vitamin B2

(c) vitamin B6 (d) vitamin B12

5. Which one given below is non-reducing sugar?

(a) Glucose (b) sucrose

(c) Maltose (d) lactose

6. In a protein molecule amino acids are linked together by:

(a) peptide bond (b) dative bond

(c) glycosidic bond (d) phospodiestes bond

7. In DNA1 the complementary bases are:

(a) adenine and thynine ; guanine and eytosine

(b) adenine and thymine; guanine and uracil

(c) adenine and guanine; thymine and cytosine

(d) uracil and adenine; cytosine and guanine

220

8. Deficiency of vitamin B1, cause the disease:

(a) convulsions

(c) cheilosis

(b) beri-beri

(d) sterility

221

CONCEPT MAP OFBIOMOLECULES

sources-internet: www.studiestoday.com

222

UNIT-XIV POLYMERS

SYNOPSIS

SL.N

o. Topic Concepts Degree of Importance

XIV Polymers 1. Classification of polymers- **

addition and condensation,

Elastomers, fibres thermoplastics

and thermosetting. Step growth and

chain growth polymerization

2. Preparation of Polymers, their ***

monomers and the importance of

polymers in daily life.

Macromolecules – Large molecules having high molar mass e.g. proteins, nucleic acids. Polymers –

Macromolecules which consists of a very large number of single repeating structural units joined through

covalent bonds in a linear fashion.

Monomers – The repeating structural units of a polymers that are derived from simple reactive molecules.

Polymerization – The process of formation of polymers from respective monomers.

Classification of Polymers.

Based on source – Natural – found in plants and animals, e.g. proteins, cellulose, starch. Semi synthetic

polymers – Cellulose derivatives as cellulose acetate (rayon) and cellulose nitrate.

Synthetic – Polythene, synthetic fibers (nylon – 6,6) and rubbers (Buna-S).

Based on structure – Linear polymers – consists of long straight chain e.g. polythene and polyvinyl

chloride.

Branch chain polymers – contains linear chains having some branches e.g. low density polythene (LDP).

Cross linked or network polymers – formed from bi-functional and tri-functional monomers contains

strong covalent bonds between linear polymer chains e.g. bakelite, melamine.

Based on mode of polymerization – Addition polymers – Obtained by the addition of repeated units of

monomers (note – monomers must be unsaturated hydrocarbon or its derivatives)

223

Homopolymers – Additon polymers obtained from one type of monomer are called homopolymers e.g.

polythene.

Copolymer – Additon polymers obtained from two different type of monomers are called copolymers e.g.

Buna-S.

Condensation polymers – Obtained by repeated condensation reaction between two different bi functional

or tri functional monomeric unit e.g. nylon-6,6 is formed by condensation of hexamethylene diamine with

adipic acid..

Based on molecular forces –(Molecular forces Hydrogen bonding-strong force, vander waal forces-weak

forces.)

Elastomers – These are rubber like solids with elastic properties,containing weakest vander waalforces

e.g.Buna-S Buna-N,neoprene etc.

Fibres – are the polymers containing strong intermolecular forces like hydrogen bonding e.g. nylon and

terylene, etc.

Thermoplastic – which can be moulded into desired shapes by repeatedly heating? These polymers possess

intermolecular forces of attraction intermediate between elastomers and fibers e.g. polythene, polystyrene,

PVC, etc.

Thermosetting polymers – cannot be soften on heating e.g. bakelite, urea – formaldehyde etc.

Based on growth thaneization – Chain growth or Addition polymers – free radical mechanism.

Step growth polymerization or condensation. Free radical addition polymerization – in presence of free

radical generating initiator like benzoyl peroxide in presence of light thane on polymerization gives

polythene in following steps. Chain intiation step–

C6H5 – CO – O- O – CO – C6H5 2C6H5COO *2C6H5*

C6H5* + CH2= CH2 C6H5– CH2 – CH2*

Chain propogating step – in this step the formed new radical adds to the monomer

repeatedly to form a bigger free radical.

224

C6H5– CH2 – CH2* + CH2 ═ CH2 C6H5– CH2 – CH2 CH2 – CH2 * C6H5– (CH2 –

CH2 – CH2)n – CH2*

Chain terminating step – in this step the long chain free radicals can combine in different ways to give a

desired polymer.

2C6H5–(CH2 – CH2 – CH2)n – CH2 * C6H5 – (CH2 – CH2)n– CH2 – CH2 – CH2 –

(CH2 – CH2)n – C6H5

Biodegradable Polymers

PHBV (poly β-hydroxybutyrate – co- β-hydroxy valerate – it is obtained by copolymerization of 3-

hydroxybutanoic acid and 3- hydroxypentanoic acid. It is used in specially packaging orthopaedic devices

and in controlled release of drugs. It undergoes bacterial degradation in the environment.

Nylon -2-nylon 6 – polyamide copolymer of glycine and amino caproic acid. Commercially Important

Polymers

Name of Polymer Monomer Structure Uses

Polypropene Propene -[-CH2- Manufacture of

225

CH(CH3)-]n- ropes,toys,pipes

etc.

Polystyrene Styrene -[-CH2- As insulator,

CH(C6H5)n]- manufacture of

radio and

television

cabinets.

Polyvinyl chloride(PVC) Vinyl chloride -[-CH2- Manufacture of

CH(Cl)n-]- rain coats hand

bags, water

pipes.

Polythene Ethene -[-CH2-CH2- Insulation of

]n- wires, toys,

manufacture of

dustbins etc.

PolytetraFluroethene(Teflon) Tetrafluoroethene -[-CF2-CF2- Oil seal and

]n- Gasket and

thanei kitchen

wares

Polyarcylonitrile Acrylonitrile -[-CH2- Substitute for

CH(CN)-]n- wool

Terylene or Decron Ethylene glycol + Ropes, safety

Terephthalic acid - belts,tyre –cord ,

sails of

boats,saree and

dress material

Nylon-6,6 Hexamethylenediamine Stocking, socks,

+ Adipic acid - ropes,

Parachutes,

fabrics,

bristles of tooth

brush

Nylon-6 Caprolactum -[-CO- Tyre-cords,

(CH2)5-NH- Ropes, fabrics

]n-

Novolac(straight chain) Phenol + Used for binding

Formaldehyde - glue, laminated

C6H5OH + HCHO wooden planks

(x) Buna-S (Copolymer) 1,3-Butadiene + - Auto tyres floor tiles,

226

Styrene foot-wear

components

xi) Natural rubber 2-methyl-1,3- - Used for tyres

butadiene

xii) Neoprene 2-chloro-1,3- - Conveyor belts,

butadiene gasket , hoses

(xiii) Buna-N 1,3-butadiene + Resistance to action

(Copolymer) acrylonitrile - of petrol.

Make oil seals, tank

linings etc

3-hydroxybutanoic Packaging

(xiv) (PHBV) Poly acid and 3- Orthopaedic

hydroxypantanoic Devices

(biodegradable)

(xv) Nylon-2-nylon-6 Glycine + It is biodegradable

aminocaproic acid step growth

Polymer

(xvi)Glyptal Eyhylene glycol and Manufacture of

phthalic acid paints and

lacquers.

(xvii)PhenolFormaldehyde Formaldehyde + Combs, records,

resin(Bakelite) Phenol switches boards

227



1Name a natural elastomer.

Ans. Natural rubber.

2.Write name of a synthetic polymer which is an ester.

Ans. Nylon 6 or Nylon 6,6.

3.Name of monomer of Nylon 6.

Ans. Caprolactam

4.Write the monomer units of Bakelite.

Ans. Phenol and formaldehyde.

5.Define a copolymer.

Ans. The polymers made by addition polymerization from two different monomers are termed as

copolymers, e.g., Buna-S, Buna-N, etc.

6.Write one use of PVC.

Ans.: In manufacture of rain coats& vinyl flooring.

7.Define Polymer.

Ans.: Polymer is defined as very large molecules having molecular mass (103-107u). These are also

referred to as macromolecules.

8.Give an example of thermoplastics.

Ans: Thermoplastics are polythene, polystyrene, polyvinyls, etc.

9.To which class of polymers does Nylon-66 belong?

Ans: Polyamides

10.Name the type of monomers in terylene?

Ans: Ethylene glycol and terephthalic acid.

Arrange the following polymers in increasing order of their intermolecular forces.

(i) Nylon 6,6, Buna-S, Polythene.

(ii) Nylon 6, Neoprene, Polyvinyl chloride. Ans. (i) Buna-S < Polythene<Nylon 6, 6 (ii)Neoprene <

Polyvinyl chloride < Nylon 6.

12.Classify the following as addition and condensation polymers: Terylene, Bakelite, Polyvinyl chloride,

Polythene.

228

Ans. (i) addition polymers: Polyvinyl chloride, Polythene.

(ii)condensation polymers: Terylene , Bakelite.

What is meant by PTFE? Give its popular name. Ans.Polytetrafluoroethylene.it is called Teflon.

Write chemical name of (Ziegler-Natta catalyst). Ans.: Triethylaluminium and titanium tetrachloride

SHORT ANSWER QUESTIONS(2&3* MARKS)

1.What is a biodegradable polymer ? Give an example of a biodegradable aliphatic polyester.

Ans. Polymers which decomposes over a period of time due to environmental degradation by bacteria,

etc. are called biodegradable polymers. E. g. PHBV

2.How can you differentiate between addition and condensation polymerization ?

Ans.: In addition polymerization the molecules of the same monomer or different

.monomers add together on a large scale to form a polymer. The monomers used are

unsaturated compounds, e.g., alkenes, alkadienes and their derivatives.

Condensation thaneization generally involves a repetitive condensation reaction between

two di-functional monomers. These polycondensation reactions may result in the loss of

some simple molecules as water, alcohol, etc., and lead to the formation of high molecular

mass condensation

polymers. E.g. , Nylon 6,6.

3. What is the function of sulphur in vulcanization of rubber?

Ans: Sulphur introduces sulphur cross linkages between the chains. So it becomes more tensile

strength, elasticity and resistance to abrasion etc.

Write down the two differences between thermoplastic and thermosetting plastic and examples.

Ans.: Thermoplastic are the linear or slightly branched long chain molecules capable of repeatedly

softening on heating and hardeningon cooling. These polymers possess intermolecular forces of

attraction intermediate between elastomers and fibers. Some common thermoplastics are polythene,

polystyrene, polyvinyls, etc.

229

Thermosetting plastic polymers are cross linked or heavily branched molecules, which on heating

undergo extensive cross linking in moulds and again become infusible. These cannot be reused.

Some common examples are Bakelite, urea-formaldehyde resins, etc.

5.Differentiate Novolac and Bakelite on the basis of structure.

Ans: A linear product of Phenol – formaldehyde polymeris novolac, used in paints. Novolac on

heating with formaldehyde undergoes cross linking to form an infusible solid mass called bakelite.

It is used for making combs, phonograph records, electrical switches and handles of various

utensils.

6.Distinguish between the terms homopolymer and copolymer and give an example of each.

Ans: the addition polymers formed by the polymerization of a single monomeric species are known

as homopolymers, e.g., polythene.

The polymers made by addition polymerization from two different monomers are termed as

copolymers, e.g., Buna-S, Buna-N, etc.

7.How will you differentiate between LDP and HDP?

Ans: Low density polythene: It is obtained by the polymerization of thane under high pressure of

1000 to 2000 atmospheres at a temperature of 350 K to 570 K in the presence of traces of dioxygen

or a peroxide initiator (catalyst).

Low density polythene is chemically inert and tough but flexible and a poor conductor of

electricity.E.g., squeeze bottles, toys and flexible pipes.

High density polythene: It is formed when addition polymerization of thane takes place in a

hydrocarbon solvent in the presence of a catalyst Ziegler-Natta catalyst at a temperature of 333 K to

K and under a pressure of 6-7 atmospheres. HDP consists of linear.Molecules and has a high

density due to close packing. It is more tougher and harder. It is used for manufacturing buckets,

dustbins, bottles, pipes, etc.

8.* Write the names of monomers of the following polymers:

(i) Nylon 6,6 (ii) Neoprene(iii) Buna –N

Ans. (i) Hexamethylenediammine and adipic acid.

(ii)Chloroprene.

(iii)1, 3 – butadiene and acrylonitrile.

230

9*. How are polymers classified on the basis of structure?

Ans. On the basis of structure, the polymers are classified as below:

(i)Linear polymers such as polythene, polyvinyl chloride, HDP etc.

Branched chain polymers such as low density polythene, LDP, etc.

Cross linked polymers such as Bakelite, melamine, etc.

10*. Write the monomers of the following polymers:

Ans.(i) 1, 3 – butadiene and acrylonitrile (ii) tetrafluoroethene (iii)chloroprene.

11. Explain elastomeric polymers & Fibres

Ans: These are rubber – like solids with elastic properties. In these elastomeric polymers, the polymer

chains are held together by the weakest intermolecular forces. These weak binding forces permit the

polymer to be stretched. A few ‘crosslinks’ are introduced in between the chains, which help the polymer

to retract to its original position after the force is released as in vulcanized rubber. The examples are

buna-S, buna-N, neoprene, etc. Fibres arethe thread forming solids which possess high tensile strength

and high modulus. These characteristics can be attributed to the strong intermolecular forces like

hydrogen bonding. These strong forces also lead to close packing of chains and thus impart crystalline

nature. The examples are polyamides (nylon 6, 6), polyesters (terylene), etc.

ENRICHMENT EXERCISE

1.What are polymers?

2.What do you mean by the term polyamides?

3.Give examples of semisynthetic polymers.

4.What are biopolymers?

5.What are elastomers?

231

6.What are biodegradable polymers?

7. Mention one use of Buna-S.

8.Name the monomer of natural rubber.

9.Name the polymer used for making unbreakable crockery.

10.Name the polymer which is used for making non-stick utensils.

11.What is meant by polymerization? What type of polymer is Teflon and what is its chief use?

12.Give the classification of polymers based on molecular forces.

13.Arrange the following polymers in the increasing order of their intermolecular forces. Also classify

them as addition and condensation polymers: Nylon-6, Neoprene, PVC.

Neoprene < PVC < Nylon-6.

14.Why do the densities of low density polythene and high density polythene differ?

15.Could a copolymer be formed in both addition and condensation polymerization or not? Explain with

examples.

16.Give three differences between linear polymers and branched-chain polymers.

17.State the significance of numbers in the polymer names –Nylon-6 and Nylon-66. Write the monomers

used for making Nylon-66.

18.What is the difference between thermosetting and thermoplastic polymers?

19.Discuss the main purpose of vulcanization of rubber.

20.Write the names and structures of the monomers of the following polymers:

(i) Natural rubber (ii) Nylon-66 (iii) Terylene

21.Write the information asked for in the following polymers:

232

Bakelite: Materials used for preparation.

Synthetic rubber: Monomer unit.

PVC: Monomer unit.

Nylon-66: Materials required for preparation.

(a) What is the role of Benzoyl peroxide in polymerization of ethene?

(b) What are LDPE and HDPE? How are they prepared?

22. How does presence of double bonds in rubber molecules influence their structure and reactivity?


1.Explain the terms –monomers ,polymers and polymerization.

2.Distinguish between various classes of polymers and different types of polymerization processes.

3. Prepare Table for monomers and polymers.

233


1. An example of biopolymer is :

(a) Tefflon (b) Rubber

(c) Nylon-66 (d) DNA

2. Which of the following polymer do not involve cross linkage?

(a) Melamine (b) Bakelite

(c) Polythene (d) Vulcanised rubber

3. Polymer obtained by condensation polymerisation is:

(a) Polythene (b) Tefflon

(c) Phenol-formaldehyde (d) Nitrite rubber

4. Which is an example of thermosetting plastic?

(a) Polythene (b) PVC

(c) Neophene (d) Bakelite

5. Natural rubber is a polymer of:

(a) Butadine (b) Ethyne

(c) Styrene (d) Poly isophene

6. Terylene is a condensation product of ethylene glycol and

(a) Benzoic acid (b) Pnthalic acid

(c) Salicyclic acid (d) Terephthalic acid

7. The process involving heating of natural rubber with sulphur is known as:

(a) Vulcanisation (b) galvanisation

(c) Sulphonation (d) Bessemerisation

8. The interparticle forces present in Nylon-66 are:

(a) Vauder wall's forces

(b) Hydrogen bonding

(c) Dipole dipole interactions

(d) None of these

9. Which of the following polymers of glucose is stored by animals?

(a) Cellulose (b) Anylose

(c) Amylopectin (d) G-lycogen

10. The commercial name of polyacrylonitrite is .............

(a) Dacron (b) Orlon (acrilaw)

(c) PVC (d) Bakelite

234

CONCEPT MAP OF POLYMERS

Sources-Internet:www.studiestoday.com

235

UNIT- XV

CHEMISTRY IN EVERYDAY LIFE

SYNOPSIS

SL No. Topic Concepts Degree of importance

XV CHEMISTRY IN 1. Drug – target **

EVERYDAY interaction

LIFE

2. Classes of drugs – ***

Antacids, Antihistamines,

Tranquilizers, Analgesics,

antibiotics, Antiseptic &

Disinfectants

3. Chemicals in food – **

Antioxidants sweetening

agents, preservatives

4. Soaps & detergents

***

DRUGS – Drugs are chemical of low molecular masses, which interact with macromolecular targets and

produce a biological response.

CHEMOTHERAPY- The use of chemicals for therapeutic effect is called chemotherapy.

CLASSIFICATION OF DRUGS –

ON THE BASIS OF PHARMACOLOGICAL EFFECT-drugs for a particular type of problem e.g. as

analgesics-----for pain relieving.

ON THE BASIS OF DRUG ACTION-Action of drug on a particular biochemical process.

ON THE BASIS OF CHEMICAL ACTION-Drugs having similar structure .eg- sulpha drugs.

ON THE BASIS OF MOLECULAR TARGETS- Drugs interacting with biomolecules as lipids, proteins.

4. ENZYMES AS DRUG TARGETS

(i) CATALYTIC ACTION OF ENZYMES-

237

Enzymes have active sites which hold the substrate molecule .it can be attracted by reacting molecules.

Substrate is bonded to active sites through hydrogen bonds, ionic bonds, Vander Waal or dipole –dipole

interactions.

(ii) DRUG- ENZYME INTERACTIONS-

(a)Drug complete with natural substrate for their attachments on the active sites of enzymes .They are

called competitive inhibitors.

(b)Some drugs binds to a different site of the enzyme called allosteric sites which changes the shape of

active sites.

ANTAGONISTS- The drugs that bind to the receptor site and inhibit its natural function.

AGONISTS-Drugs mimic the natural messenger by switching on the receptor.

ANTACIDS-These are compounds which neutralize excess acid of stomach.eg-Aluminium hydroxide,

Magnesium hydroxide.

ANTI HISTAMINES-The drugs which interfere with the natural action of histamines and prevent the

allergic reaction. eg- rantidine, tegamet, avil.

TRANQULIZERS-The class of chemical compounds used for the treatment of stress, mild or even severe

mental diseases. Eg- luminal, seconal, equanil, idardil, iproniagid.

ANALGESICS-They reduce pain without causing impairment of consciousness, mental confusion or

some other disturbance of the nervous system. Eg - aspirin, seridon, phenacetin.

ANTIMICROBIALS-They tend to prevent/destroy or inhibit the pathogenic action of microbes as

bacteria ,virus ,fungi etc. They are classified as (i)ANTIBIOTICS-Those are the chemicals substances

which are produced by micro-organisms and use to kill the pathogenic micro-organism.

Eg- Pencillin , ofloxacin .NARROW SPECTRUM ANTI-BIOTICS-These are effective mainly against

gram positive or gram negative bacteria. Eg- Penicillin , streptomycin.

238

BROAD SPECTRUM ANTI-BIOTICS-They kill or inhibit a wide range of micro-organisms.

eg- chloramphenicol , tetracycline .

(ii)ANTISEPTICS OR DISINFECTANT-These are which either kill/inhibit the growth of micro-

organisms. Antiseptics are applied to the living tissues such as wounds, cuts, ulcers etc. eg-furacine,

chloroxylenol & terpinol (Dettol) .Disinfectant are applied to inanimate objects such as floors , drainage ,

system.

Eg- 0.2% solution of phenol is an antiseptic while 1% solution is disinfectant.

ANTIFERTILITY DRUGS- These is the chemical substances used to control the pregnancy. They are

also called oral contraceptives or birth control pills. Eg-Mifepristone, norethindrone.

ARTIFICIAL SWEETNING AGENTS-These are the chemical compounds which give sweetening effect

to the food without adding calorie.

They are good for diabatic people e.g.- aspartame, saccharin, alitame , sucrolose.

FOOD PRESERVATIVES- They prevents spoilage of food to microbial growth.eg-salt, sugar, and

sodium benzoate.

CLEANSING AGENTS-

SOAPS- They is sodium or potassium salts of long chain fatty acids. They are obtained by the

saponification reaction, when fatty acids are heated with aqueous sodium hydroxide. They do not work

well in hard water.

TOILETS SOAP-This is prepared by using better grade of fatty acids and excess of alkali needs to be

removed. colour & perfumes are added to make them attractive. (iv)MEDICATED SOAPS- Substances

of medicinal value are added.eg- Bithional, Dettol.ies of soaps, but actually contain no soap .They can

used in both soft and hard water .They are-

(i)ANIONIC DETERGENTS-They are sodium salts of sulphonated long chain alcohols or

hydrocarbons.eg-sodium lauryl sulphonate. They are effective in acidic solution.

H2SO4NaOH(aq)

-Na+ CH3 (CH2)10 CH2OH CH3 (CH2)10CH2OSO3H CH3 (CH2)10CH2SO3

Lauryl alcohol Sodium laurylsulphate

239

(ii)CATIONIC DETERGENTS- They are quaternary ammonium salts of amines with acetates , chlorides,

or bromides.They are expensive used to limited extent.eg-Cetyltrimethylammoniumbromide.

(iii)NON-IONIC DETERGENTS- They does not contain any ions. Some liquid dishwashing detergents

which are of non-ionic type.

BIODEGREDABLE DETERGENTS- The detergents which are linear and can be attacked by micro-

organisms are biodegradable.

Eg -sodium dodecylbenzenesulphonate.

NON-BIODEGREDABLE DETERGENTS- The detergents which are branched and cannot be

decomposed by micro-organisms are called non-biodegradable.It creates water pollution.



Q-1 Define the term chemotherapy?

Ans-1 Treatment of diseases using chemicals is called chemotherapy.

Q-2 Why do we require artificial sweetening agents?

Ans-2 To reduce calorie intake.

Q-3 what are main constituent of Dettol?

Ans-3 Choloroxylenol & Terpinol.

Q-4 Name the drug that are used to control allergy?

Ans-4. Antihistamines, Citrizine, Avil.

Q-5Why is the use of aspartame limited to cold food and drinks?

Ans-5 It is unstable at cooking temperature and decompose.

Q-6What is tranquilizers? Give an example?

Ans-6 They are the drug which are used to reduce the stress, mild and severe mental

disease.

Q-7 what type of drug chloramphenicol?

Ans-7 It is broad spectrum antibiotic.

Q-8 Why is bithional is added to the toilet soap?

Ans-8 It acts as antiseptic.

Q-9 what are food preservatives?

240

Ans-9 The substances that prevent spoilage of food due to microbial growth. eg- sodium

benzonate.

Q-10 Mention one important use of the following-

(i) Equanil (ii)Sucrolose

Ans-1 (i) Equanil- It is a tranquilizer.

(ii)Sucrolose-It is an artificial sweetener.

ENRICHMENT EXERCISE

Q-1. Define the following and give one example-(i)Antipyretics (ii) Antibiotics

Q-2. Name the medicines used for the treatment of the following-

(i) Tuberculosis (ii) Typhoid

Q-3. what are tincture of iodine?

Q-4. How is synthetic detergents better than soaps?

Q-5. what are sulpha drugs? Give two examples?

Q-6. what forces are involved in holding the active sites of the enzymes?

Q.7. Describe the following giving an example in each case- (i) Edible colours

(ii)Antifertility drugs

Q-8. Give two examples of organic compounds used as antiseptics? Q-9.What are Biodegredable and

non-biodegdredable detergents? Give one example of each.

Q-10.What are barbiturates? To which class of drugs do they belong? Give two examples. Q-11.What is

the use of –

Benadryl (ii) sodium benzoate (iii) Progesterone Q-12. Identify the type of drugs-

Ofloxacin (ii) Aspirin (iii) Cimetidine

Q-13. Describe the following with suitable example-

Disinfectant (ii) Analgesics

Broad spectrum antibiotics


1.Describe the basis of classification of drugs.

2.Various types of drug function in the body.

3.Artificial sweetening agents and food preservatives.

4.Chemistry of cleansing agents.

241

Match the class of compounds given in column with their functions given in column II.

Column I Column II

(i) Antagonists (a) Communicate message between two nuerons

and that between neurons to muscles

(ii) Agonists (b) Bind to the receptor site and inhibit its natural

function

(iii) Chemical messenger (c) Crucial to body's communication process

(iv) Inhibitors (d) Nimic the natural messenger

(v) Receptor (e) Inhibit activities of enzymes

Assertion and Reason Type Questions

Note: In the following questions a statement of assertion followed by a statement of reason is given.

Choose the correct answer out of the following choices.

Assertion and reason both are correct statement but reason does not explain assertion.

Assertion and reason both are correct and reason explains the assertion.

Both assertion and reason are wrong statement

Assertion is correct statement reason in wrong statement.

Assertion is wrong statement reason is correct statement.

Assertion : Enzymes have active sites that hold substrate molecule for a chemical reaction.

Reason : Drugs complete with natural substrate by attaching covalently to the active site of enzyme.

Assertion : Competitive inhibitors complete with natural substrate for this attachment on the active sites

of enzymes.

Reason : In competitive inhibition, inhibitor binds to the alhesteric site of the enzyme.

242


1.Which of the following statements are correct about barbiturates?

a.Hypnoties or sleep producing agents

b.These are tranquilizers

c.Non-narcotic analgesics

d.Pain reducing without disturbing the nervous system.

2.Which statement about aspirin is not true.

a.Aspirin belongs to narcotic analgesies

b.It is effective in relieving pain

c.It has antiblood clotting action

d.It is a neurologically active drug.

3.Which of the following is an example of antibiotic?

(a) Dimetapp (b) Sodium Benzoate

(c) sodium lauryl sulphate (d) Chloramphenicol

4. Which of the following is an example of analgesic?

(a) Opium (b) Sodium bicarbonate

(c) sodium lauryl sulphate (d) Amoxycillin

243

CONCEPT MAP OF CHEMISTRY IN EVERYDAYLIFE Sources-internet: www.syudiestoday.com

244

Unsolved Sample paper -1 Class: XII

Subject: Chemistry Time Duration: 3 Hours M.M: 70 General Instructions:

(i) All questions are compulsory. (ii) Marks of each question are indicated against it. (iii) Question nos. 1 to 20 are objective type questions and carry 1 mark each. (iv) Question nos. 21 to 27 are short- answer questions and carry 2 marks each. (v) Question nos.28 to 34 are also short -answer questions and carry 3 marks each. (vi) Question nos. 35 to 37 are long- answer question and carry 5 marks each. (viii) Use log tables if necessary. Use of calculators is not allowed.

Q1. Mixing of HNO3 and HCl is reaction:

(a) endothermic reaction (b) exothermic reaction (c) both exothermic and endothermic (d ) depend on entropy of reaction

Q2. Which can adsorb larger volume of hydrogen gas?

(e) Colloidal solution of platinum (f) finely divided nickel (g) finely divided platinum (h) colloidal Fe(OH)3

Q3. The correct IUPAC name of [Pt(NH3)2 Cl2] is

(a) Diamminedichloridoplatinum (II) (b) Diamminedichlorideplatinum (IV) (c) Diamminedichlorideplatinum (0) (d) Dimminedichlorideplatinum (IV)

Q4.Anyl halides are less reactive toward nucleophilic substitution reaction than alkyl halides due to

(a) the formation of stable carbonimion

(b) resonance stabilization

(c) longer carbon-halogen bond

(d) sp2 hybridised carbon attached to halogen Q5. Which of the following state are correct (a)Benzyl halides are more reactive than vinyl and aryl halides (b)Aryl halides are less reactive than alkylhalide (c) Aryl halides are more reactive than benzyl halides

245

Q6.A narrow spectrum antibiotic is active against ............

(a) gram positive or gram negative bacteria (b) gram negative bacteria only (c) single organism or one desease (d) both gram positive and gram negative bacteria.

Q7. Which of the following statement is not true about glucose?

(e) it is an aldohexose (f) on heating with HI it forms n-hexane (g) it is present in furanose form (h) it does not give 2, 4-DNP test

Q8.The question given below consist of an Assertion and the Reason. Use the following key to choose

the appropriate answer. (a) Assertion and reason both are CORRECT and reason is the CORRECT explanation of the

assertion. (b) Assertion and reason both are wrong statements.

(c) Assertion is correct statement but reason is wrong statement.

(d) Assertion is wrong statement but reason is correct statement.

(e) Assertion and reason both are correct statements but reasson is not correct explanation of

assertion.

Q8 Assertion : n-Propylamine has higher boiling point than trimethylamine. Reason : Among n-propylamine molecules, there is hydrogen bonding but there is not hydrogen bonding in

trimethylamine.

Q 9. Assertion: Toxic metal ions are removed by the chelating ligands. Reason: Chelate complens tend to be move stable. Q10. Assertion: Hl cannot be prepared by the reaction of Kl with concentrated H2SO4. Reason : Hl has lowest H—X bond strength among halogen acids. Q11 Assertion : The α-hydrogen in carbonyl group is less acidic. Reason : The anion formed after loss of a—H atom is resonance stabilised. Q12 Assertion: O and p-nitrophenols can be separated by steam distillation. Reason: O-isomeris steamvolatile due to chelation and p-isomer is not steam volatile due to intermolecular hydrogen bonding Q13. Statement 1 : Galvanised iron does not rust. Statement 2 : Zn has more (–) ve electrode potential than

Fe.

Q14. Haematite is an ore of ...............

Q15. Starch is an example of ................... colloids.

246

Q16. Name the aldehyde which does not give Fehling’s solution test.

Q17 Collodion is a 4% solution of ................... in a alcohol or ether.

Q18. The reactions taking place in one step is called ............... reactions.

Q19. IUPAC name of [Pt(NH3)2Cl(NO2)] is:

a. Platinum diaminechloritrite

b. Chloronitrito-N-ammine platinum(II)

c. Diamminechloridonitrite-N-platinum (II)

d. Diamminechlornitrite-N-platinate(II) Q20. Name the antiseptic agents present in dettol.

Q21. Conductivity of 0.00241 M acetic acid is 7.896x10-5 S cm -1. Calculate its molar conductivity and

if ^0 for acetic acid is 390.5 S cm2 mol-1, what is its dissociation constant?

Q22. What is the effect of temperature on the rate constant of a reaction? How can this temperature

effect on the rate constant be represented quantitatively?

Q23. How does the acidified permanganate solution react with (a) iron(II) ions (b) oxalic acid? Write

the ionic equation for the reactions.

OR

What is meant by lanthanoid contraction? What is it due to and what consequences does it have on the

chemistry of element following lanthonoids in the periodic table?

Q24. Write the structure of the major organic product in each of the following reactions:

(CH3)3CBr + KOH --ethanol,heat-→

CH3CH2CH = CH2 + HBr –peroxide-→

OR

Write the reactions for obtaining benzaldehyde from benzal chloride and then benzal chloride from it.

Q25. Rearrange the compounds of each of the following sets in order of reactivity towards SN2

displacement:

247

(i) 2-bromo-2-methylbutane, 1-bromopentane, 2-bromopentane

(ii) 1-bromo-3-methylbutane, 2-bromo-2-methylbutane, 2-bromo-3-methylbutane

Q26. (a) How do antiseptics differ from disinfectants? Give one example of each.

(b)Why do soaps not work in hard water?

OR

(i)A child when dropped his soap in a bathing tub,the soap did not sink but was floating.Suggest the

method of manufacturing such soap.

(ii)Diabetic patients’ are advised to take natural sweeteners.Why? (iii)What is meant by a ‘broad

spectrum antibiotic? Give an example.

Q27. Arrange the following:

1.in decreasing order of pKb value C2H5NH2, C6H5NHCH3,(C2H5)2NH and C6H5NH2

2.in increasing order of basic strength C6H5NH2,C6H5N(CH3)2, (C2H5)2NH and

3.In increasing order of boiling point C2H5OH,(CH3)2NH,C2H5NH Q28. (a) Write the mechanism of

dehydration of ethanol to form ethene.

Q28. What is meant by hydroboration-oxidation reaction? Illustrate with an example.

OR

1.Benzoquinone from phenol

2.2-Methylpropan-2-ol from methylmagnesium bromide

3.Propan-2-ol from propene

Q29. Write the names and structure of the monomers of the following polymers:

(i) Buna-S (ii) Neoprene (iii) Dacron

Q30 (a) Give evidence to show that [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4] Cl are ionization isomers.

(b) What is crystal field splitting energy? How does the magnitude of ∆o decide the actual

configuration of d orbital in a coordination entity?

Q31. Assign appropriate reasons for each of the following statements:

1.The acid strength decreases in the order:HCl>H2S>PH3

2.SF6 is inert towards hydrolysis

3.NH3 form hydrogen bonds but PH3 does not ?

248

Q32..(a) Why do physisorption and chemisorption behave differently with rise in temperature?

(b)Give an example of heterogeneously catalysed reaction

(c) Which of the following electrolytes is the most effective for the coagulation of Fe(OH)3 sol which is a

positively charged sol ?

NaCl, Na2SO4, Na3PO4

Q33. Calculate the potential (emf )of the cell

Cd | Cd2+ (0.10M) || H+ (0.20M) | Pt ,H2 (0.5atm)

Given Eo for Cd2+/Cd= -0.403V, R=8.314JK- mol-, F=96,500C mol-

Q34. The rate constant for the first order decomposition of H2O2 is given by the following equation:

log k =14.34-1.25x104 K/T

(i) Fluorine forms only one oxo acid HOF.

(ii) Helium is used in diving apparatus.

(iii) SnCl4 is more covalent than SnCl2?

(b) Draw the structures of the following molecules

(i) XeF6 (ii) HOClO2

Q35. (a) Account for the following

OR

(a) Draw the following structures:

(i) H2S2O8 (ii) (HPO3)3 (iii) XeO3

(b) Complete the reaction:

(i) 3Cl2(g) + 6NaOH (cold and dil.)→

(ii) HgCl2(aq) + PH3(g) →

Q36. (a) An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The

molecular mass of the compound is 86. It does not reduce Tollen’s reagent butforms an addition

compound with sodium hydrogensulphite and gives positive iodoform test. On vigorous oxidation it

gives ethanoic and propanoic acids. Derivethe structure of the compound ‘A’.

(b) Give simple tests to distinguish between the following pairs of organic compounds:

249

i.Acetophenone and benzophenone

ii.Benzoic acid and phenol

OR

(a)Arrange the following compounds in increasing order of their property indicated:

i.Acetaldehyde,Acetone, Di tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN )

ii.CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength)

iii.Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-methoxybenzoicacid (acid strength)

(b) Write giving chemical equations a brief account of the following:

(i) Clemmensen reduction

(ii) Rosenmund reduction

(iii) H.V.Z

Q37. (a) State Kohlrausch law of independent migration of ions. Write an expression for the molar

conductivity of acetic acid at infinite dilution according to Kohlrausch law.

(b) A solution of urea in water has a boiling point of 100.128oC. Caculate the freezing point of same

solution. Molal constant for water Kf and Kb are 1.86oC and

0.512oC respectively.

OR

State the following:

(i) Raoult's law in its general form in reference to solutions.

(ii) Henry's law about partial pressure of a gas in a mixture.

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at

250C,assuming that it is completely dissociated.

250

Unsolved SamplePaper-2

Class: XII

Subject: Chemistry

Time Duration: 3 Hour Maximum Marks.70

General Instructions (i) All questions are compulsory. (ii) Questions numbers 1 to 20 are very short answer questions carrying 1 mark each. (iii) Questions numbers 21 to27 are short answer questions carrying 2 marks each. (iv) Questions numbers 28 to 34 are short answer questions carrying 3 marks each. (v) Questions numbers 35 to 37are short answer questions carrying 5 marks each. (vi) Use log tables if necessary. Calculators are not permitted.

Q1 One of the characteristics of transition metals to form the complex ion is: (a) Having unpaired electron in d-subshell (b) Having paired electrons in d-subshells (c) Providing empty d-orbitals (d) Having small charge/size ratio 1

Q2. The colour of the coordination compounds depends on the crystals field splitting.

What will be the correct order of absorption of warleinth of light in the visible region. for

the complenes [CO(NH3)6]3+; [CO(CN)0]3–; [CO(H2O)6]3+.

a) [CO (CN)6]3– > [CO(NH3)6]3+ > [CO(H2O)6]3+

b) [CO (NH3)6]3+ > [CO(H2O)6]3+ > [CO(CN)6]3–

c) [CO (H2O)6]3+ > [CO(NH3)6]3+ > [CO(CN)6]3–

d) [CO (CN)6]3– > [CO(NH3)6]3+ > [CO(H2O)6]3+

Q3. The property which depends on number of particles of solute is called ............. Q4. Match the column and choose correct option: (A) Smoke P. Foam

(B) Butter Q. Emulsion

(C) Hair cream R. Aerosol

(D) Whipped cream S. Gel

(a) A–P, B–S, C–Q, D–R (b) A–R, B–Q, C–S, D–P

(c) A–R, B–S, C–Q, D–P (d) A–S, B–P, C–R, D–Q

251

Q5 Which can adsorb larger volume of hydrogen gas?

(a) Colloidal solution of platinum

(b) finely divided nickel

(c) finely divided platinum (d) colloidal Fe(OH)3

Q6 What is the composition of copper matte ?

Q7. Name the substance which is used as a flux in the extraction of copper.

Q8. The major product obtained on interactionof phenol with NaOH and CO2 is (a) Benzoic acid (b) Salicaldehyde

(c) Salicylic acid (d) Pthalic acid

Q9. Which of the following alcohol on dehydration with conc H2SO4 gives but-2-enc? (a) 2-methylpropan-2-ol (b) Butan-1-ol

(c) 2-methyl propan-1-ol (d) Butan-2-ol Q10What is nitrating mixture ? Q11In DNA the complementary bases are:

(a) a.adenine and thynine ; guanine and eytosine

(b) b.adenine and thymine; guanine and uracil

(c) c.adenine and guanine; thymine and cytosine

(d) d.uracil and adenine; cytosine and guanine 12Name polysaccharides which is stored in the liver of animals. 13Name the amino acid which is not optically active. In the following questions a statement of assertion followed by a statement of reason is

given. Choose the correct answer out of the following choice. Both assertion and reason are True, and reason is the correct explanation of the

assertion.Both assertion and reason are True, but reason is not the correct explanation

of the assertion.Assertion is not True, but reason is True Both assertion and reason are

False.

14Assertion: Cu2+ iodide is not known. Reason: Cu2+ oxidises I– to iodine. 15Assertion: Separation of Zr and Hf is difficult. Reason : Because Zr and Hf lie in the same graph of the periodic table.

252

16 Which of the following statements are correct?

Among halogens, radius ratio between iodine and fluorine is maximum

Leaving F–F bond, all halogens have weaker X—X bond than X—X’ bond in

interhalogens Among interhalogen compounds maximum number of atoms are present in

iodine fluoride. Interhalogen compounds are more reactive than halogen compounds.

17 Which of the following statements are correct for SO2 gas?

It act as bleaching agent in moist conditions

It’s molecule has linear geometry

It’s dilute solution is used as disinfectant.

It can be prepared by the reaction of dilute H2SO4 with metal sulphide. 18. Which of the following statements are correct?

All the three N—O bond lengths in HNO3 are equal.

All P–Cl bond lengths in PCl5 molecule in gaseous state are equal.

P4 molecule in white phosphorous have angular strain therefore white phosphours is very reactive.

PCl5 is ionic in solid state in which cation is tetrahedral and anion is octahedral. 19. The correct IUPAC name of [Pt(NH3)2 Cl2] is

Diamminedichloridoplatinum (II)

Diamminedichlorideplatinum (IV)

Diamminedichlorideplatinum (0)

Dimminedichlorideplatinum

20. Which of the following polymers can have strog intermolecular forces. (a) Nylon (b) Polystyrene (c) Rubber (d) Polyesters

253

Q21 (a) Write IUPAC name of the following complex: 2

[Co(NH3)6 ]Cl3 (b) Write the structure of the following complex:

Tris(ethane1,2-diamine)chromium (III)

Q22 Give reason for the following: 2 a) At higher altitiudes, people suffer from disease called anoxia. In this disease, they become weak and cannot think clearly. b) Why a person suffering from high blood pressure is advised to

take minimum quantity of common salt? OR

Calculate the mass of compound (molar mass=256g/mol) to be dissolved in 75g

of benzene to lower its freezing point by 0.48 K ( Kf =5.12 K kg/mol)

Q23 Complete the following reactions: 2

a) Ca3P2 + H2O → b) XeF4 + O2F2→

Q24 Calculate the emf of the cell in which the following reaction takes place: 2

Ni(s) + 2Ag+ (0.002M) → Ni2+ (0.160M) + 2Ag(s)

Given that E0 cell =1.05V

OR

Calculate the strength of the current required to deposit 1.2 g ofmagnesium

from molten MgCl2 in 1 hour.[1 F= 96500 C mol-1; Atomic mass ,Mg= 24u]

Q25 How do you prepare: 2

a) K2MnO4 from MnO2?

b) Na2Cr2O7 from Na2CrO4?

Q26. A first order reaction is 50 % complete in 50 minutes at 300 K and the same reaction is again 50 % complete in 25 minutes at 350 K. Calculate activation energy of the reaction.

3 Q27 Define the following terms: 3 a) Ideal solution b) Azeotrope c) Osmotic Pressure

OR

(i)Explain the solubility rule “like dissolves like” in terms of intermolecular forces that exist in solutions. (ii)Why is it not possible to obtained pure ethanol by fractional distillation? What general name is given to binary mixtures which show deviation from Raoult’s law and whose components cannot be separated by fractionsl distillation.How many types of such mixtures are there?

254

Q28 a) How many moles of mercury will be produced by electrolyzing 3

1.0M Hg(NO3) 2 solution with a current of 2.00A for 3 hours? b) Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of solution for a weak and strong electrolyte. Q29 Describe the following giving one example each: 3 a) Cationic detergent b) Food preservatives c) Analgesics Q30 a) Name a member of the lanthanoid series which is well known to exhibit +4 3 oxidation state. b) Complete the following equation:

MnO4 - + 8H+ + 5e- →

c) Out of Mn3+ and Cr3+, which is more paramagnetic and why? (Atomic nos: Mn= 25, Cr=24) Q31 Write the name and structure of the monomer of the following polymers: 3

a) Teflonb) Bakelitec) Buna-S

Q32 How are the following conversions carried out: 3 a) Benzene →Biphenyl

b) Benzyl chloride → Benzyl alcohol c) Methanol →Ethanoic acid

OR

Write equation involved in the following reactions:

a) Reimer- Tiemann reactionb) Carbylamine reactionc) Sandmeyer reaction Q33 Answer the following questions: 3

a) Which of the following compounds would undergo SN2 reaction faster and why? i) CH3CH2CH2 CH2Br and CH3CH2CHBr CH3

b) (±) 2 Butanol is optically inactive c) Grignard reagent should be prepared under anhydrous conditions.

Q34 a) Write the structures of main products when aniline reacts withthe following 3 reagents:

i) Br2 water ii) (CH3CO)2/pyridine b) Arrange the following in the increasing order of their boiling point:

C2H5NH2, C2H5OH, (CH3)3N

Q35 a) Account for the following: i) Oxygen is a gas but sulphur a solid. 5

ii) Flourine forms only one oxo acid, HOF. iii) Tendency to form pentahalide decreases down the group in group 15.

(b)Draw the structure of the following molecules:

i) H3PO3 ii) ClF3

OR

255

a) Answer the following: Why do noble gases have very low boiling point? H2S is less acidic than H2Te? Sulphur in vapour state exhibits paramagnetism b) What happens when: Chlorine gas is passed through a hot concentrated solution of NaOH? Sulphur di oxide gas is passed through an aqueous solution of Fe(III) salt? Q36 a) 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. The disappearance of N2O5 (g) is a first order reaction with a rate constant of 5 X 10-4 s-1 at 450C,

i.e, 2N2O5 (g) → 4NO2(g) + O2(g). If the initial concentration of N2O5 is 0.25M, calculate its

concentration after 2 minutes. Also, calculate the half life for decomposition of N2O5 (g).

OR Define the following terms: i. Rate constant ii. Molecularity of a reaction A first order reaction is 20% complete in 10 minutes. Calculate the time for 75% completion of reaction. Q37 a) Give chemical tests to distinguish between the following pairs of compounds:

Acetophenone and benzophenone Propanal and propanone How will you bring out the following conversions: Ethanal to 3- Hydoxy butanal Benzoic acid to benzaldehyde Propanone to propene

OR An organic compound with molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizaro reaction. On vigorous oxidation, it gives1,2-benzenedicarboxlyic acid. Identify the compound and write the reactions.

Which acid out of the following would you expect to be strongerand why? CH3CO2H or FCH2CO2H Arrange the following in increasing order of reactivity in nucleophilc addition reactions and give reason: Ethanol, propanal, propanone, butanone.

256

Solved SamplePaper-3

Class: XII

Subject: Chemistry

Time Duration: 3 Hour Maximum Marks.70

General Instructions I. All questions are compulsory.

II. Questions numbers 1 to 20 are very short answer questions carrying 1 mark each. III. Questions numbers 21 to27 are short answer questions carrying 2 marks each. IV. Questions numbers 28 to 34 are short answer questions carrying 3 marks each. V. Questions numbers 35 to 37are short answer questions carrying 5 marks each.

VI. Use log tables if necessary. Calculators are not permitted.

SECTION A

1.Which is the addition polymer. (a) Nylon-66 (b) Teflon

(c) Polyester

(d) PHBV

Or Polymer used for the insulation of electrical cables is: (a) PVC (b) Glyptal (c) Neoprene (d) All of these 2.Distillation is used for the refining of:

(a) Fe (b) Zn (c) Sn (d) Cu

3.Which of the following aqueous solutions should have the highest boiling point?

(a) 1 M NaOH (b) 1 M Na2SO4 (c) 1 M NH4NO3 (d) 1 M KNO3

Or

10% solution of urea is isotonic with 6% solution of a non-volatile solute X. What is the molecular

mass of solute X? (a) 6 g mol– (b) 60 g mol–1 (c) 36 g mol–1 (d) 32 g mol–1 4.The charge required for reducing 1 mole of MnO4– to Mn2+ is:

(a) 1.93 × 105 C (b) 2.8 × 105 C

(c) 4.3 × 105 C (d) 4.82 × 105 C 5.The unit of rate and rate constant are same for a: (a) Zero order reaction (b) First order reaction (c) Second order reaction (d) Third order reacti

257

6. Match the column I and column II and mark the appropriate choice.

(A) Diastase (i) Proteins → peptones

(B) Pepsin (ii) Glucose → ethyl alcohol

(C) Ptyalin (iii) Starch → Maltose

(D) Zymase (iv) Starch → Sugar

(a) A—(iv), B—(ii), C—(i), D—(iii) (b) A—(ii), B—(i), C—(iv), D—(iii)

(c) A—(i), B—(ii), C—(iii), D—(iv) (d) A—(iii), B—(i), C—(iv), D—(ii) 7.In XeF2, XeF4 and XeF6 the number of lone pairs on Xe is respectively. (a) 2, 3, 1 (b) 1, 2, 3 (c) 3, 2, 1 (d) 4, 1, 2

8.The number of moles of KMnO4 in acidic medium that will be needed to react with one mole of sulphide

ion is:

(a) (b) (c) (d)

9.Propanone on reaction with alkyl magnesium bromide followed by hydrolysis will not produce. (a) Primary alcohol (b) Secondary alcohol (c) Tertiary alcohol (d) Carboxylic acid

10.In the following questions, a statement of assertion is followed by a statement of reason. Mark the

Correct choice as:

(a) If both assertion and reason are true and reason in the correct explanation of assertion (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false.

Assertion: The order of basicity of amnines in the gaseous phase follows the order : 3° amines > 2° amines > 1° amines. Reason: Amines have an unshared pair of electrons on nitrogen atom due to which they behave as lewis

abse.

11.Why NaCl is used to clear snow from roads? Or

12.Give one example of pseudo first order reaction.

13.Thermal stability of hydrides of group-16 elements decreases down the group. Why?

Or Why ICl is more reactive than I2 14.Why do Zr and Hf exhibits similar properties?

Or Why do transition metal show variable oxidation states.

258

CuSO4 is colourless while CuSO4 5H2O is coloured. Why? Or [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless, why?

16.Out of and Cl Cl, which undergo SN1 reaction faster and why? Or Why grignard reagent should be prepared under anhydrous conditions? 17.Why phenol is acidic in nature?

18.Arrange the following in order of their increasing reactivity toward HCN CH3CHO,

CH3COCH3., HCHO, C2H5 COCH3.

19.What happens when aniline is treated with bromine? 20Explain why aspirin finds use in prevention of heart attacks?

SECTION B

21.Show that time required for 99% completionis twice the time required for the completion of

90% of reactions for a first order reaction

OR

The decomposition of hydrocarbon follow the equation K = (4.5 × 1011 5–1) e–28000 K/T

Calculate Ea

22. Explain :

(i) Actinoid contraction is greater from element to element than canthanoid

contraction. Why? (ii) The enthalpies of atomisation of the transsition metals are high. Why? OR

Complete the following reaction:

(i) Fe+2 + MnO4- + H+ ------->

(ii) CrO4-2 + H+ ----------->

23. . When an oxide of Mn (A) is fused with KOH in the presence of an oxidising agent and

dissolved in water, it gives a dark solution of compound (B) Compound (B) disproportionate in

neutral or acidic solution to give purple compound (C). Identify A, B, C.

24. State the role of silica in the metallurgy of copper and cryolite in the metallurgy of aluminium.

OR

259

Differentiate between roasting and calcination.

25. Complete the following reactions: (a) C2H5OC(CH3)3 --HI---> (b) CH2COOCH3

------NaBH4--------->

26. Write the names of monomers of the following polymers:

(a) Nylon 6,6

(b) Buna-N 27. How do antiseptics differ from disinfectants? Give one example of each.

SECTION C 28. 2 g of benzoic acid (C6H5 COOH) dissolved in 25 g of benzene shows a depression in

freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol–1.

What is the percentage association of acid if it forms dinner in solution?

Or How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and

NaHCO3 containing equimolar amounts of both?

29. The following data were obtained during the first order thermal decomposition of N2O5(g) at constant volume S.No. Time/s 1. 0 2. 100 Calculate the rate constant. 30. Explain what is observed?

When a beam of light is passed through a colloidel sol. An electrolyte, NaCl is added to the hydrated ferric oxide sol. Electric current is passed through a colloidal sol.

Or

(a) Describe Freundlich adsorption isotherm. (b) What do you mean by activity and selectivity of catalysts?

31. Draw the figure to show the splitting of d-orbitals in an octahedral crystal field. How does

the magnitude of ∆0 decide the actual configuration of d-orbitals in a coordination entity? Or

Write IUPAC Name of the complen [Co(en)3]3+ Draw all the isomers (geometrical and optical) of

260

(a) [Pt Cl2 (en)2]2+ (b) [Cr Cl2(ox)2]3–

32. Carry out the following conversions: (i) Aniline to chlorobenzene (ii) 2-Bromopropane to 1-Bromopropane (iii) Benzene to diphenyl

33. Write short on the following : (i) Carbylamine reaction (ii) Hofmann's bromamide reaction (iii) Gabriel phthalimide synthesis.

34. (i) Write the important structural and functional differences between DNA and RNA. (ii) Write the hydrolysis products of sucrose.

SECTION D

35. (i) Calculate ∆G0 and log Kc for the following reaction at 298 K :

36. 2Al(s) + 3Cu2+ (aq) → 2Al3+ (aq) + 3Cu(s)

Given : E0 = 2.02 V cell

(ii) Using the E0 values of A and B, predict which is better for coating the surface of iron

[E0(Fe2+/Fe) = – 0.44 V] to prevent corrosion and why? Given : E0(A2+/A) = – 2.37 V : E0(B2+/B) = – 0.14 V

(i) The conductivity of 0.001 mol L–1 solution of CH3COOH is 3.905 × 10–5 S cm–1.

Calculate its molar conductivity and degree of dissociation (α). Given λ0(H+) = 349.6 S cm2

mol–1 and

λ0(CH3COO–) = 40.9 S cm2 mol–1 (ii) What type of battery is dry cell? Write the overall reaction occurring in dry cell.

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

CCH3 + Ni/H2 ?

COONa

+ NaOH CaO

(a) DIBAL-H (iii) CH3–CH=CH–CN ? (b) H2O

261

b. Give simple chemical tests to distinguish between the following pairs of compounds: i.Butanal and Butan-2-one ii.Benzoic acid and Phenol

Or a. Write the reactions involved in the following : i.Etard reaction ii.Stephen reduction b.How will you convert the following in not more than two steps: i.Benzoic acid to Benzaldehyde ii. ii.Acetophenone to Benzoic acid iiiEthanoic acid to 2-Hydroxyethanoic acid 37. (a) Account for the following: i.Ozone is thermodynamically unstable. ii.Solid PCl5 is ionic in nature. iii.Fluorine forms only one oxoacid HOF. (b)DRaw the structure of (i) BrF5 (ii) XeF4

Or 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.Arrange the following in the increasing order of property mentioned :

(a) H3PO3, H3PO4, H3PO2 (Reducing character) (b) NH3, PH3m AsH3, SbH3, BiH3 (Base strength)

1. b or a 2. b 3. b or c 4. d 5. a 6. d 7. c 8. a

9. a, b, d 10. 6

262

11. It lowers freezing point of water.

Or Negative deviation 12. Hydrolysis of an ester or any other correct example.

13. Because down the group E—H bond dissociation enthalpy decreases. Or Because I—Cl bond is weaker than I—I bond. 14. Due to lanthanide contraction. Or Due to presence of vacant d-orbitals. 15. CuSO4 does not have any ligand, so splitting of d-orbitals take place while

CuSO4, 5H2O has water as ligand.

Or Ti3+ has one unpaired electron while Sc3+ does not have any unpaired e–, hence no

d—d transition in SC3+.

Cl 16. (2° carbocation). or RMgX + H2O →

x R – H + Mg – – oH

17. Due to stability of phenoxide ion by resonance. 18. C2H5COCH3 < CH3COCH3 < CH3CHO M H—CHO

NH NH2 2 Br

Br + 3 HBr

+ 3 Br2

Br 20. Due to antiblood clotting activity. 21. For first order reaction, t =

∴ = = =

⇒ t99% = 2 × t90% Or k = Ae–Ea/RT

=

263

Ea = 28000 × R

= 28000 × 8.314 Ea = 232.79 KJ mol–1 22. (i) Due to poor shielding effect of 5f electrons. (ii)Because of large number of unpaired electrons in their atoms they have stronger

interatomic interaction resulting in higher enthalpies of atomisation.

Or (i) 5Fe2+ + MnO4

– + 8H+ → Mn2+ + 5Fe3+ + 4H2O (ii) 2CrO4

2– + 2H+ → Cr2O72– + H2O

23. A = MnO2 B = K2MnO4 C = KMnO4

24. Silica act as flux in the metallurgy of Cu SiO2 + FeO → FeSiO3 and Gyolite is going to

decrease the melting point of melt or mixture. Or

Calcination is heating are in limited supply of oxygen/air, while roasting is heating are in a

regular supply of air in a furnace at a temperature below the melting point of the metal. CH3 25. CH3 C I + CH3 CH3 OH CH3

oH CH2 C oCH3 o 26. (i) Hexamethylenediammine, Adipic acid (or their IUPAC names) (ii) Buta-1, 3-diene, Acrylonitide 27. Antiseptics and disinfectants are the chemicals which either kill or prevent the growth of

microorganisms, antiseptics are applied to the living tissues such as wounds, cuts, ulcers and

diseased spin surfaces, while disinfectants are applied on floors, drainage systems. Example of antiseptics – Bithionol 0.1% phenol example of disinfectant : 1% phenol

28. NCERT Part-1 Solved Example 2.12 Page 57 Let mass of Na2CO3 = xg ∴ mass of NaHCO3 = (1 – x) g ∴ moles of Na2CO3 = moles of NaHCO3

264

= x = 0.558 g Number of moles of Na2CO3 = = 0.00526 mol Number of moles of NaHCO3 = 0.00526 mol Na2CO3 + 2HCl → 2 NaCl + H2O + CO2 NaHCO3 + HCl → NaCl + H2O + CO2 Total number of moles of HCl required = (2 × 0.00526) + 0.00526 = 0.01578 mol ∴ Volume of 0.1 M HCl =

= = 0.1578 L = 157.8 mL 29. NCERT Part-1 solved example 4.6 page 107 30. (i) Scattering of light took place/Tyndal effect.

(ii) Coagulation took place

(iii) Electrophoresis/Coagulation took place. Or

a. = K.P1/n (n > 1) log = K + log p

Slope = 1/n log x/m log R

h log P

b. Activity means how many times catalyst is going to increase the rate of reaction and

selectivity is its ability to direct a reaction to yield a particular product.

265

31. NCERT Page 251, Fig. 9.8. If ∆0 < P, the fourth electron enters one of the eg orbitals giving the configuration t2g

2 eg1. (High spin complex) If ∆0 > P, the fourth electron enters t2g orbital with configuration t2g

4 eg0. (Low spin complex)

Or (i) trisethylenediamminecobalt (III) ion. (ii) Fig 9.7 NCERT Page-245 (b) Example 9.6 NCERT 246

+ –

(i)

NH2 N2Cl Cl

0–5C

(ii) CH3 CH CH3 alc.KoH CH3 CH3 CH2 HBr CH3 CH2 CH2 Br

Peroxide

Br

Cl

Cl2

anhy.Alcl3 Dry ether

33. (i) R—NH2 + CHCl3 + alc. 3KOH CH3—CH==CH2 CH3—CH2—CH2—Br

(ii) R—CONH2 + Br2 + 4NaOH → R—NH2 + Na2CO3 + 2NaBr + 2H2O O o o

C C C

NH NK N – R C C C

O o o O

C N – R 2

o

266

35. (a) DF0 = – nF E0

cell DG0 = – 6 × 96500 × 2.02

E0cell =

log Kc = = 205.42 (b) A because is E0 value is more negative. Or (a) ^°

m = κ × 1000/C

= 3.905 × 10–5 × 1000/0.001

= 39.05 S cm2/mole CH3 COOH → CH3COO– + H+ ^0 CH3COOH = λ0 CH3COO– + λ0H+ = 349.6 + 40.9 ^0 CH3COOH = 390.5 S cm2/mol

= 0.1 (b) Primary cell Zn + 2NH+

4 + 2MnO2 → Zn++ + 2NH3 + 2MnO(OH) 36. (a) (i)

OH

CN

(ii) (iii) CH3—CH==CH—CHO

b. (i) Tollen's reagent test : Add ammoniacal solution of silver nitrate (Tollen's Reagent) in both

the solutions. Butanal gives silver mirror whereas Butan-2-one does not. (ii) Add neutral FeCl3 in both the solutions, phenol forms violet colour but benzoic acid does not.

Or (a) (i) Etard reaction

CH3 CS2 CH(OCrOHCl2)2 H2O CHO

+ CrO2Cl2

Chromium complex

Benzaldehyde

or

CH3 (i) Cro2cl2. CS2 CHO

(ii) H30+

267

(ii) Stephen reaction RCN + SnCl2 + HCl → RCH== NH RCHO or (i) SnCl2 + HCl RCN RCHO (ii) H3 O+ (i) SnCl2 + HCl (b) (i) RCN RCHO (ii) H3O+

COOH COCl CHO SOCl2 Rosenmund's (ii) reduction Pd/BaSO4 Benzoic Benzoyl Benzaldehyde acid chloride 37. (a) (i) Endothermic compound/decomposition of ozone is exothermic is

nature and ∆G is negative / decomposition of ozone is spontaneous.

(ii) Exists as [PCl4]+ [PCl6]+ (iii) Shows only-1 oxidation state/most electronegative element/

absence of d-orbitals F F (b) (i) F Br F F

F F (ii)

Xe

F F

Or

(i) F2 is the stronger oxidising agent than chlorine (a) low enthalpy of dissociation of F—F bond (b) less negative electron gain enthalpy of F (c) high hydration enthalpyof F+ ion (ii) Low temperature, high pressure and presence of catalyst (iii) (a) H3PO4 < H3PO3 < H3PO2 (c) BiH3 < SbH3 < AsH3 < PH3 < NH3

268

BIBLIOGRAPHY

1. Ncert Text Books

2. New course of chemistr by Pradeep

3. Companion chemistry by Dinesh

4. www.excellup.com

5. mycbseguide.com

6. www.studiestoday.com

7. www.www.ncerthelp.com

269

SALWAN EDUCATION TRUST

REGD. OFFICE: Salwan Education Trust, Secretariat, C/o

Salwan Schools Complex,

Pandit Girdhari Lal Salwan Marg,

Rajendra Nagar, New Delhi-110060.

Phone : +91-11-49254500

www.salwaneducationtrust.org

salwan public school mayur vihar chemistry.pdf · chemistry, inorganic chemistry and physical...

Documents