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STUDENTS SUPPORT MATERIAL

CLASS-XII

CHEMISTRY

SESSION-2020-21

1

STUDENT SUPPORT MATERIAL

CHIEF PATRON

Dr. JAIDEEP DAS,

DEPUTY COMMISSIONER,

KVS R O AHMEDABAD

PATRON

SMT. SHRUTI BHARGAVA,

ASSISTANT COMMISSIONER,

KVS RO AHMEDABAD

SUPERVISION TEAM

(i) SMT RAJNI TANEJA (ii) SHRI JITENDRA DAROCH,

PRINCIPAL PRINCIPAL

KENDRIYA VIDYALAYA No. 1 KENDRIYA VIDYALAYA

BARODA V V NAGAR

(iii) SHRI MANISH JAIN (iv) SMT SUNITA DIWAKAR,

PRINCIPAL PRINCIPAL

KENDRIYA VIDYALAYA ONGC KENDRIYA VIDYALAYA

ANKLESHWAR EME BARODA

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CONTENT DEVELOPEMENT TEAM

Chapter 1 : The Solid State – Mr J L Agrawal

Chapter 2 : Solutions - Mr S K Soni

Chapter 3 : Electrochemistry – Ms Nageshwari

Chapter 4 : Chemical Kinetics - Mr Lokesh Suredia

Chapter 5 : Surface Chemistry – Mrs Uma Ummindini

Chapter 6 : The p-Block Elements- Mr Sher Singh

Chapter 7 : The d-and f-Block Elements- Mr Naupal S Rana

Chapter 8 : Coordination Compounds- Mrs Sabiha Shekh

Chapter 9 : Haloalkanes and Haloarenes- Mr Jitendra Singh

Chapter 10 : Alcohols, Phenols and Ethers- Mrs Anju Bala

Chapter 11 : Aldehydes, Ketones and Carboxylic Acids- Mr Bharat Pandiya

Chapter 12 : Amines- Mr Deepak Pandiya

Chapter 13 : Biomolecules- Ms Shilpi Negi

Compilation TEAM

(i) Mr Atul Tiwari, PGT CHEMISTRY KENDRIYA VIDYALAYA EME BARODA

(ii) MR M R CHOUDHARY, PGT CHEMISTRY KENDRIYA VIDYALAYA SAC AHMEDABAD

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CHEMISTRY Class-XII INDEX

S NO CONTENT PAGE NO

1 ALLOTED MARKS FOR THEORY & PRACTICALS 6

2 CH. 1 THE SOLID STATE 7-18

3 CH. 2 SOLUTION 19-29

4 CH. 3 ELECTROCHEMISTRY 30-40

5 CH 4 CHEMICAL KINETICS 41-50

6 CH 5 SURFACE CHEMISTRY 51-61

7 CH 6 THE P BLOCK ELEMENTS 62-73

8 CH 7 THE D & F- BLOCK ELEMENTS 74-82

9 CH 8 COORDINATION COMPOUNDS 83-89

10 CH 9 HALOALKANES & HALOARENES 90-102

11 CH 10 ALCOHOLS, PHENOLS & ETHERS 103-119

12 CH 11 ALDEHYDES, KETONES & CARBOXYLIC

ACIDS 120-134

13 CH 12 AMINES 135-144

14 CH 13 BIOMOLECULES 145-151

15 ORGANIC REVISION: IUPAC HELPING HINTS 152

ORGANIC REVISION: REASONING QUESTIONS 153-160

ORGANIC REVISION: DISTINGUISH TESTS 161-163

ORGANIC REVISION: MECHANISM 164-165

ORGANIC REVISION: KEY FOR CONVERSION 166-167

ORGANIC REVISION: NAMING REACTIONS 168-169

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CLASS XII (2020-21) (THEORY}

Unit No. Topic No. of Periods Marks

Unit I Solid State 8

23

Un‘rt II Solutions

Unit Ili Electrochemistry 7

Unit IV Chemical Kinetics 5

Unit V Surface Chemistry 5

Unit Vl p -elock Elements 7

19 Unit VII d -and f -Block Element 7

Un’it VIII Coordination Compounds

Unit IX Haloalkanes and Haloarenes 9

28

Unit X Alcohols, Phenols and Ethers 9

Unit XI Aldehydes, Ketones and Carboxylic Acids 10

Unit XII Amines 7

Unit XIII Biomolecules 8

Total 98 70

Votumetrlc Analysis

content Based Experiment

Project work

Ctass record arzl viva

Total 30

043

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CHAPTER-1 SOLID STATE

SOME IMPORTANT POINTS

1. 3 D close packing- CCP (ABC-ABC TYPE) and HCP (AB-AB TYPE). a) No. of Octahedral Voids = No. of spheres b) No. of Tetrahedral Voids = Twice the No. of spheres

2. Packing Efficiency: a) When unit cell is Simple Cubic – 52.4% (a=2r)

b) When unit cell is BCC – 68% , (√3a = 4r) co. no. = 8 c) When unit cell is FCC – 74 %, (√2a = 4r) co. no. = 12

3. Density of Crystal

𝑑 =𝑧𝑀

𝑎3 𝑁𝑜

4. Stoichiometric defects- a) Vacancy Defect, Interstitial Defect (in non-ionic crystals)

b) In ionic solids- Schottky and Frankel Defects c) Schottky Defects- Equal number of cations and anions are missing, so

decreases density of crystals, about same size of cations and anions.

examples, NaCl, KCl, CsCl and AgBr d) Frankel Defects- Large difference in size of cations and anions, smaller

ion (cation) leaves correct lattice site and occupy some interstitial site, no change in density. For example, ZnS, AgCl, AgBr and AgI

5. Impurity Defects- Doping of SrCl2 or CdCl2 in NaCl or AgCl. It creates cationic

vacancies. 6. Non-Stoichiometric Defects-

a) Metal excess defect due to anionic vacancies: On heating alkali metal halides in alkali metal vapours. It creates F- centres (colour bearing) and crystal possess particular colour. (LiCl- pink, NaCl- golden, KCl- violet)

b) Metal excess defect due to the presence of extra cations at interstitial sites: On heating of ZnO. Colour changes from white to yellow on heating due to

presence of electrons in interstitial sites. c) Metal Deficiency Defect- Mainly occurs in transition metal compounds like in

FeO, NiO. Transition metal ions found in different oxidation state in such

crystals.

Section A – Case Study Type Question (Type I)

Q. No.1 Questions Ans

Read the passage given below and answer the following questions:

Crystal Lattices and Unit Cells

A regular three dimensional arrangement of points in space is called a

crystal lattice.

Unit cell is the smallest portion of a crystal lattice which, when repeated

in different directions, generates the entire lattice. There are Seven Primitive Unit Cells and their Possible Variations as Centred Unit Cells. These seven types of Primitive Unit Cells are Cubic, tetragonal,

orthorhombic, monoclinic, hexagonal, rhombohedral and triclinic. There are some centred unit cell in addition to primitive unit cell. So there are

a total of 14 types of Bravais lattices of which unit cell is characterised by six parameters. The letters a, b, and c

have been used to denote the

dimensions of the unit cells whereas the

letters 𝛂, 𝞫, and 𝝲 denote the corresponding angles

in the unit cells.

In words, a Bravais lattice is an array of discrete points with an arrangement and orientation

that look exactly the same from any of the discrete points, that is the

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lattice points are indistinguishable from one another. These lattices are named after the French physicist Auguste Bravais.

The following questions (i to v) are multiple choice questions. Choose

the most appropriate answer: I If a unit cell has a = b ≠ c, α = β = γ = 90° : the crystal system will be

(a) Cubic (b) Tetragonal

(b) Rhombohedral (d) Triclinic

b

ii Orthorhombic lattices will have – (a) Only one type of unit cell (b) Two type of unit cells (c)Three type of unit cells (d) All the four type of unit cells

d

iii Which type of unit cell is possible in all the seven type of crystal lattice (a) Primitive

(b) Face centred (c) Body centred (d) End centred

A

iv The crystal system having different values of all the six parameters- (a) Cubic

(b) Tetragonal (c) Rhombohedral (d) Triclinic

D

v The crystal system in which a=b=c, α = β = 90°, γ = 120° will be (a) Cubic

(b) Tetragonal (c) Rhombohedral (d) Hexagonal

D

2 Read the passage given below and answer the following (I

to v) questions:

The formation of real lattices and structures take place through three-dimensional close packing. They are formed by stacking two-dimensional layers of spheres one above the other. This can be done

by two ways

Three-dimensional close packing from two dimensional square close

packed layers

Three-dimensional close packing from two dimensional hexagonal close packed layers

Suppose we take two hexagonal close packed layer ‘A’ and place it over the second layer B (as both layers have different alignment of

spheres) such that spheres of the second layer are placed in the depressions of the first layer. We observe that a tetrahedral void is formed when a sphere of the second layer is right above the void

(empty space) of the first layer. Adding further we notice octahedral voids at the points where the triangular voids of the second layer

are placed right triangular voids of the first one in such a way that triangular space doesn't overlap. Octahedral voids are surrounded

by six spheres.

I Coordination number of each sphere in ccp is (a) 4

(b) 12 (c) 6

(d) 8

B

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Ii Coordination number of each sphere in 2 D SCP and 2D HCP are respectively

(a) 6, 4

(b) 6, 12 (c) 4, 6

(d) 6, 6

C

Iii ABC – ABC type packing is also called as

(a) hcp (b) scp (c) Bcc

(d) ccp

D

Iv The ratio of radius of tetrahedral void to that of radius of sphere is

(a) 0.414 (b) 0.225 (c) 0.524

(d) 0.732

B

V The ratio of number of tetrahedral voids to number of octahedral voids

in a lattice is- (a) 1:2 (b) √2:1

(c) 2:1 (d) √3: √2

C

3 Case Study Type Question (Type 2)

Read the passage given below and answer the following

questions: Molecular Solids - Solids having molecules as their constituent particles are called Molecular solids. For, example, Hydrogen, Chlorine,

Water, HCl, solid carbon dioxide, sucrose, etc. Molecular solids are classified into three types on the basis of their bond:

a. Non-Polar Molecular solids

b. Polar Molecular Solids c. Hydrogen Bonded Molecular Solids

(a) Non Polar Molecular Solids Solids which are comprised of atoms only, such as helium and argon or molecules; formed because of the non polar covalent bonds are known as Non-Polar Molecular Solids. For

example – H2, Cl2, I2, etc. (b) Polar Molecular Solids The solids which are formed by polar

covalent bonds are known as Polar Molecular solids. For example – HCl, SO2, NH3, etc

(c) Hydrogen bonded Molecular Solids The molecules of hydrogen bonded molecular solids contain polar covalent bond between H and O, F or N. In solids such as H2O (ice) molecules are bound together strongly

with hydrogen bond. HF, H2O (ice), etc are the examples of hydrogen bond molecular solids.

Ionic Solids Solids, in which ions are the constituent particles, are called ionic solids. These solids are formed because of three dimensional arrangements of cations and anions bound together with strong

electrostatic forces (coulombic forces). For example NaCl. Metallic Solids All metals are referred as Metallic solids. Their

constituent particles are positive ions. These positive ions are surrounded by free moving electrons. For example – iron, aluminium, etc

Covalent Solids Crystalline solids are formed by non metals because of formation of covalent bonds between the adjacent molecules throughout

the crystal. These are also known as Network Solids. These are also called giant molecules. For example – diamond, graphite, silicon carbide, etc.

In these questions [Q. No (i) to (v)], a statement of

Assertion followed by a statement of Reason is given. Choose the correct answer out of the following choices. a) Assertion and Reason both are correct statements and Reason is

correct explanation for Assertion. b) Assertion and Reason both are correct statements but Reason is

not correct explanation for Assertion. c) Assertion is correct statement but Reason is wrong statement.

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d) Assertion is wrong statement but Reason is correct statement. I Assertion: Ice has higher melting point than that of iodine.

Reason: water and iodine are molecular solids. D

Ii Assertion: Covalent solids have a network of covalent bonds. Reason: In sodium chloride crystal, sodium ions are surrounded by a

number of chloride ions and vice versa.

B

Iii Assertion: NaCl has higher melting point than ice.

Reason: ionic bonds are stronger than H- bonds.

A

Iv Assertion: Diamond is hardest naturally known substance. Reason: Diamond has strong electrostatic force of attraction between

constituent particles.

C

V Assertion: Metallic solids are good conductor of electricity.

Reason: Valence electrons in metallic solids surrounds positively charged kernel and are mobile.

A

4 Read the passage given below and answer the following questions:

Solids are not perfect in structure. There are different types of imperfections or defects in them. Point defects and line defects are

common types of defects. Point defects are of three types - stoichiometric defects, impurity defects and non-stoichiometric defects. Vacancy defects and interstitial defects are the two basic types of

stoichiometric point defects. In ionic solids, these defects are present as Frenkel and Schottky defects. Impurity defects are caused by the

presence of an impurity in the crystal. In ionic solids, when the ionic impurity has a different valence than the main compound, some vacancies are created. Nonstoichiometric defects are of metal excess

type and metal deficient type.

In these questions [Q. No (i) to (v)], a statement of Assertion followed by a statement of Reason

is given. Choose the correct answer out of the following choices.

a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion. b) Assertion and Reason both are correct statements but Reason is

not correct explanation for Assertion. c) Assertion is correct statement but Reason is wrong statement.

d) Assertion is wrong statement but Reason is correct statement.

I Assertion: The Frenkel defect is not possible in crystal like NaCl. Reason: The size of sodium and chloride ions are about same.

A

Ii Assertion: Ni0.96O is an example of non-stoichiometric defects. Reason: Transition metal have different oxidation state.

B

Iii Assertion: Cation vacancy are created when SrCl2 is doped in NaCl crystal. Reason: Cation vacancy is due to creation of F- centres in NaCl

C

Iv Assertion: ZnO on heating turn into yellow colour. Reason: It is an example of metal excess defect.

B

V Assertion: LiCl turns to pink color when heated in Li vapors. Reason: It is due to creation of F- centres in the crystal

A

Multiple Choice Questions Note : In the following questions choose one correct option.

5 Which of the following is an amorphous solid? (i) Graphite (C) (ii) Silica gel

(iii) Chrome alum (iv) Silicon carbide (SiC)

ii

6 Which of the following statement is not true about amorphous solids?

(i) On heating they may become crystalline at certain temperature. (ii) They may become crystalline on keeping for long time.

(iii) Amorphous solids can be moulded by heating. (iv) They are anisotropic in nature.

iv

7 Iodine molecules are held in the crystals lattice by ____________.

(i) london forces (ii) dipole-dipole interactions (iii) covalent bonds (iv) coulombic forces

i

8 What type of solids are electrical conductors, malleable and ductile? (i) covalent

(ii) Metallic

(iii) Ionic (iv) Molecular

ii

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9 Which of the following is not the characteristic of ionic solids? (i) High value of electrical conductivity in the solid state. (ii) Brittle nature.

(iii) Very strong forces of interactions. (iv) Anisotropic nature.

i

10 Cations are present in the interstitial sites in __________. (i) Frenkel defect (ii) Schottky defect

(iii) Vacancy defect (iv) Metal deficiency defect

i

11 Schottky defect is observed in crystals when __________. (i) some cations move from their lattice site to interstitial sites.

(ii) equal number of cations and anions are missing from the lattice. (iii) some lattice sites are occupied by electrons.

(iv) some impurity is present in the lattice.

ii

12 Which of the following point defects are shown by AgBr(s) crystals? (A) Schottky defect (B) Frenkel defect

(C) Metal excess defect (D) Metal deficiency defect (i) (A) and (C)

(ii) (C) and (D) (iii) (A) and (B) (iv) (B) and (D)

iii

13 In which pair most efficient packing is present? (i) hcp and bcc

(ii) hcp and ccp (iii) bcc and ccp (iv) bcc and simple cubic cell

ii

14 F- centre may be possible in_. (i) FeO

(ii) LiCl (iii) Diamond (iv) SrCl2

ii

15 Correct order of number of lattice points in unit cell will be (i) simple< fcc < end centred < bcc

(ii) simple< fcc < end centred (iii) simple < bcc < end centred < fcc (iv) simple < end centred < bcc < fcc

iii

16 In which of the following structures coordination number for cations and anions in the packed structure will be same?

(i) Cl– ion form fcc lattice and Na+ ions occupy all octahedral voids of the unit cell. (ii) Ca2+ ions form fcc lattice and F– ions occupy all the eight tetrahedral

voids of the unit cell. (iii) O2– ions form fcc lattice and Na+ ions occupy all the eight tetrahedral

voids of the unit cell. (iv) S2– ions form fcc lattice and Zn2+ ions go into alternate tetrahedral

voids of the unit cell.

i

17 Graphite is a good conductor of electricity due to the presence of ___. (i) lone pair of electrons

(ii) free valence electrons (iii) cations

(iv) anions

ii

18 Which of the following statement is not true about the hexagonal close packing?

(i) The coordination number is 12. (ii) It has 74% packing efficiency.

(iii) Tetrahedral voids of the second layer are covered by the spheres of the third layer. (iv) In this arrangement spheres of the fourth layer are exactly aligned

with those of the first layer.

iv

19 The edge lengths of the unit cells in terms of the radius of spheres

constituting fcc, bcc and simple cubic unit cell are respectively_____.

i

12

Multiple Choice Questions

Note : In the following questions two or more options may be correct.

20 Which of the following statements are not true?

(i) Vacancy defect results in a decrease in the density of the substance. (ii) Interstitial defects results in an increase in the density of the

substance. (iii) Impurity defect has no effect on the density of the substance. (iv) Frankel defect results in an increase in the density of the substance.

iii

iv

21 An excess of potassium ions makes KCl crystals appear violet or lilac in colour since ________.

(i) some of the anionic sites are occupied by an unpaired electron. (ii) some of the anionic sites are occupied by a pair of electrons. (iii) there are vacancies at some anionic sites.

(iv) F-centres are created which impart colour to the crystals.

i iv

22 The number of tetrahedral voids per unit cell in NaCl crystal is ____.

(i) 4 (ii) 8 (iii) twice the number of octahedral voids.

(iv) four times the number of octahedral voids.

ii

iii

23 Which of the following defects decrease the density?

(i) Interstitial defect (ii) Vacancy defect

(iii) Frankel defect (iv) Schottky defect

ii

iv

24 Graphite cannot be classified as __________.

(i) hard solid (ii) network solid

(iii) covalent solid (iv) ionic solid

i

iv

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 choices.

(i) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion. (ii) Assertion and Reason both are correct statements but Reason

is not correct explanation for Assertion. (iii) Assertion is correct statement but Reason is wrong

statement. (iv) Assertion is wrong statement but Reason is correct statement.

25 Assertion: The total number of atoms present in a simple cubic unit cell is one.

Reason: Simple cubic unit cell has atoms at its corners, each of which is shared between eight adjacent unit cells.

i

26 Assertion: Graphite is a good conductor of electricity however diamond

belongs to the category of insulators. Reason: Graphite is soft in nature on the other hand diamond is very

hard and brittle.

ii

27 Assertion: Total number of octahedral voids present in unit cell of cubic close packing including the one that is present at the body centre, is

four. Reason: Besides the body centre there is one octahedral void present

at the centre of each of the six faces of the unit cell and each of which is shared between two adjacent unit cells.

iii

28 Assertion: The packing efficiency is maximum for the bcc structure.

Reason: The coordination number of each atom in crystal having bcc unit cell is 8.

iv

29 Assertion: The density of crystal decreases in Frenkel defect. Reason: In Frenkel defect smaller ion leave its correct lattice site and occupy some interstitial site.

iv

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Short answer type and carry 2 marks each. 1 If NaCl is doped with 10–3 mol % of SrCl2, what is the concentration of cation

vacancies? Ans Let moles of NaCl = 100

∴ Moles of SrCl2 doped = 10-3

Each Sr2+ will replace two Na+ ions. To maintain electrical neutrality it occupies one position and thus creates one cation vacancy.

∴ Moles of cation vacancy in 100 moles NaCl = 10-3

Moles of cation vacancy in one mole NaCI = 10-3 × 10-2 = 10-5

∴ Number of cation vacancies

= 10-5 × 6.022 × 1023 = 6.022 × 1018 mol-1

2 Silver crystallizes in fcc lattice. If edge length of the cell is 4.07 × 10-8 cm and density is 10.5 g cm-3, calculate the atomic mass of silver.

Ans It is given that the edge length, a = 4.07 × 10-8 cm and density is d = 10.5 g cm-3 As the lattice is fcc type, the number of atoms per unit cell, z = 4

3 A cubic solid is made of three elements P, Q & R. Atoms of Q are at the corners of

the cube and P are present at each face while atom R occupies half of the total

tetrahedral voids formed by atoms P. What is the formula of the compound? ANS Number of atoms of Q in one unit cell = 8 × (1/8) = 1

Number of atoms of P in one unit cell = 6x1/2 =3

Number of atoms of R in one unit cell = 1/2x (3x2) =3 So, the ratio of the number of atoms P, Q & R = 3:1:3

Hence, the formula of the compound is P3QR3 4 Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two

out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

ANS Number of oxide ions in one unit cell = 6x1/2 +8 × (1/8) = 4

Number of ferric ions in one unit cell = 4x2/3 =8/3 So, the ratio of the number Fe3+ and O2- = 8/3:4 =

= 2:3 Hence, the formula of the compound is Fe2O3

5 What types of stoichiometric defects are shown by

(i) ZnS (ii) AgBr? ANS (i) ZnS crystals may show Frenkel defects

(ii) AgBr crystals may show both Frenkel and Schottky defects 6 A compound forms hexagonal close-packed structure. What is the total number of

voids in 0.5 mol of it? How many of these are tetrahedral voids?

ANS No. of atoms in close packings 0.5 mol =0.5 x 6.022 x 1023 =3.011 x 1023 No. of octahedral voids = No. of atoms in packing =3.011 x 1023

No. of tetrahedral voids = 2 x No. of atoms in packing = 2 x 3.011 x 1023 = 6.022 x 1023 Total no. of voids = 3.011 x 1023 + 6.022 x 1023 = 9.033 x 1023

No. of tetrahedral voids = 2 x No. of atoms in packing 7 Why is glass considered as super cooled liquid?

ANS Glass is considered to be super cooled liquid because it shows some of the characteristics of liquids, though it is an amorphous solid. For example, it is slightly thicker at the bottom. This can be possible only if it has flown like liquid,

though very slowly. 8 Why ionic solids are conduct electricity in the molten state and not in the solid-

state? ANS In the ionic solids, the electrical conductivity is due to the movement of the ions.

Since the ionic mobility is negligible in the solid state, these are non-conducting

in this state. Upon melting, the ions present acquire some mobility. Therefore, the ionic solids become conducting.

9 A compound is formed by two elements M and N. The element N forms ccp and atoms of the element M occupy 1/3 of the tetrahedral voids. What is the formula of the compound?

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ANS Let us suppose that, the no. of atoms of N present in ccp = x Since 1/3rd of the tetrahedral voids are occupied by the atoms of M, therefore,

the no. of tetrahedral voids occupied = 2x/3 The ratio of atoms of N and M in the compound = x : 2x/3 or 3 : 2

∴ The formula of the compound = N3M2 or M2N3

10 If the radius of the octahedral void is r and radius of the atoms in close packing is

R, derive relation between r and R. ANS A sphere is fitted into the octahedral void as shown in the diagram.

Short answer type and carry 3 marks each. 1 How will you distinguish between the following pairs of terms :

(a) Hexagonal close packing and cubic close packing (b) Crystal lattice and unit cell (c) Tetrahedral void and octahedral void.

ANS (a) In hexagonal close packing (hcp), the spheres of the third layer are vertically

above the spheres of the first layer (ABABAB……. type). On the other hand, in cubic close packing (ccp), the spheres of the fourth layer are present above the

spheres of the first layer (ABCABC…..type). (b) Crystal lattice: It depicts the actual shape as well as size of the constituent

particles in the crystal. It is therefore, called space lattice or crystal lattice. Unit cell: Each brick represents the unit cell while the block is similar to the space or crystal lattice. Thus, a unit cell is the fundamental building block of the space

lattice. (c) Tetrahedral void: A tetrahedral void is formed when one sphere is placed over

triangular void made by three spheres of a particular layer. Octahedral void: An octahedral void or site is formed when three spheres arranged at the corners of an equilateral triangle are placed over another set of 3 spheres.

2 Analysis shows that nickel oxide has the formula Ni0.98 O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?

Ans 98 Ni-atoms are associated with 100 O – atoms. Out of 98 Ni-atoms, suppose Ni present as Ni2+ = x Then Ni present as Ni3+ = 98 – x

Total charge on x Ni2+ and (98 – x) Ni3+ should be equal to charge on 100 O2- ions. Hence, x × 2 + (98 – x) × 3 = 100 × 2 or 2x + 294 – 3x = 200 or x = 94

∴ Fraction of Ni present as Ni2+ = 94/98 × 100 = 95.92%

Fraction of Ni present as Ni3+ = 4/98 × 100 = 4.08%

3 How many lattice points and number of atoms are there in one unit cell of each of the following lattice? (i) Face-centred cubic (ii) Simple cubic unit cell (iii) Body-centred cubic

ANS (i) Face-centred cubic – lattice points 14, No. of atoms per unit cell= 4 (ii) Simple cubic unit cell - lattice points 8, No. of atoms per unit cell= 1

(iii) Body-centred cubic- lattice points 9, No. of atoms per unit cell= 2 4 Aluminium crystallizes in a cubic close-packed structure. Its metallic radius is 125

pm.

(i) What is the length of the side of the unit cell? (ii) How many unit cells are there in 1.00 cm3 of aluminium?

15

5 Explain the following terms with suitable examples:

(i) Schottky defect (ii) Frenkel defect (iii) F-centres. ANS (i) Schottky defect : In Schottky defect a pair of vacancies or holes exist in the

crystal lattice due to the absence of equal number of cations and anions from their

lattice points. It is a common defect in ionic compounds of high coordination number where both cations and anions are of the same size, e.g., KCl, NaCl, KBr,

etc. Due to this defect density of crystal decreases and it begins to conduct electricity to a smaller extent. (ii) Frenkel defect: This defect arises when some of the ions in the lattice occupy

interstitial sites leaving lattice sites vacant. This defect is generally found in ionic crystals where anion is much larger in size than the cation, e.g., AgBr, ZnS, etc.

Due to this defect density does not change, electrical conductivity increases to a small extent and there is no change in overall chemical composition of the crystal. (iii) F-Centres : These are the anionic sites occupied by unpaired electrons. F-

centres impart colour to crystals. The colour results by the excitation of electrons when they absorb energy from the visible light falling on the crystal.

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

ANS

Since there are four atoms per unit cell, the cubic unit cell must be face centred (fcc).

7 Explain how you can determine the atomic mass of an unknown metal if you know its mass density and the dimensions of unit cell of its crystal.

ANS

8 An element X crystallizes in f.c.c structure. 208 g of it has 4.2832 × 1024 atoms.

16

Calculate the edge of the unit cell, if density of X is 7.2 g cm-3. ANS

9 An element crystallizes in a structure having fee unit cell of an edge 200 pm.

Calculate the density if 200 g of this element contains 24 × 1023 atoms. ANS

10 Examine the given defective crystal

Answer the following questions with Reasons: (i) What type of stoichiometric defect is shown by the crystal?

(ii) How is the density of the crystal affected by this defect? (tii) What type of ionic substances show such defect?

ANS (i) Schottky defect, as same no of cations and anions are missing from lattice site. (ii) Density of the crystal decreases due to missing of ions from crystal. (iii) Ionic solids having same no of cations and anions and having approximate

equal size of cations and anions like NaCl.

Long answer type and carry 5 marks each. 1 (a) Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a

f.c.c. structure. (Atomic mass of Al = 27 g mol-1) (b) Give Reasons:

(i) In stoichiometric defects, NaCl exhibits Schottky defect and not Frenkel defect. (ii) Which stoichiometric defect increases the density of a solid?

(iii) Write a point of distinction between a metallic solid and an ionic solid regarding conductivity

ANS (a) Mass of Al = 8.1,

Atomic mass of Al = 27 g mol-1 No. of atoms = n × 6.022 × 1023

= (8.1/27) × 6.022 × 1023 = 0.3 × 6.022 × 1023 = 1.8066 × 1023 Since one f.c.c. unit cell has 4 atoms

∴ No. of unit cells =1.8066×1023/4 = 4.5 × 1022 unit cells

(b) (i) Schottky defect is shown by the ionic solids having very small difference in their cationic and anionic radius whereas Frenkel defect is shown by ionic solids having large difference in their cationic and anionic radius. NaCl

17

exhibits Schottky defect because radius of both Na+ and Cl– have very small difference.

(c) (ii) Interstitial defect increases the density of a solid.

Metallic solid conducts electricity in solid state but ionic solids do so only in molten state or in solution or metals conduct electricity through electrons and ionic

substances through ions. 2 (a) An element has an atomic mass 93 g mol-1 and density 11.5 g cm-3. If the

edge length of its unit cell is 300 pm, identify the type of unit cell. (b) Write any four differences between amorphous solids and crystalline solids.

ANS

(a)

As the number of atoms present in given unit cells are coming nearly equal

to 2, hence the given units cell is body centred cubic unit cell (BCC). 1. Crystalline – True solids, long range order, anisotropic, sharp M.P.

2. Amorphous – Pseudo solids / super cooled liquid, short range order, isotropic, No fixed M.P.

3 (a) Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm.

The density of iron is 7.874 g cm-3. Use this information to calculate Avogadro’s number. (Gram atomic mass of Fe = 55.84 g mol-1).

How will you distinguish between the following pairs of terms : (i) Tetrahedral and octahedral voids (ii) Crystal lattice and unit cell

ANS (a) Given : a = 286.65 pm = 286.65 × 10-10, d = 7.87 g cm-3, M = 56 g mol-1

Z = 2 NA = ?

= 6.023x1023

Tetrahedral void: A tetrahedral void is formed when one sphere is placed over triangular void made by three spheres of a particular layer. Octahedral void: An octahedral void or site is formed when three spheres

arranged at the corners of an equilateral triangle are placed over another set of 3 spheres.

A regular arrangement of the constituent particles of a crystal in a three dimensional space is called crystal lattice. The smallest three dimensional portion of a complete crystal lattice, which when

repeated over and again in different directions produces the complete crystal lattice is called the unit cell.

4 (a) Some of the glass objects recovered from ancient monuments look milky

instead of being transparent. Why? (b) Iron (II) oxide has a cubic structure and each side of the unit cell is 5Å. If density of the oxide is 4 g cm-3, calculate the number of Fe2+ and O2- ions present

in each unit cell. [Atomic mass: Fe = 56 u, O = 16 u; Avogadro’s number = 6.023 × 1023 mol-1]

ANS (a) Some of the glass objects found from ancient monuments look to be milky

in appearance because of conversion to crystalline nature of glass.

18

5 (i) Refractive index of a solid is observed to have the same value along all

directions. Comment on the nature of this solid. Would it show cleavage property?

Calculate packing efficiency in crystals having bcc unit cell. ANS (i) As the solid has same value of refractive index along all directions, it is

isotropic in nature and hence amorphous. Being amorphous solid, it will not show a clean cleavage and when cut, it will break into pieces with irregular surfaces.

19

CHAPTER 2: SOLUTION

Conc. of Solutions

(w/w)% :

(w/w)% = Mass of solute X100/Mass of solution

Molarity [M]: M= No. of Moles of solute / Volume of solution in litre

It is temperature dependent.

Molality [m]: m= No. of Moles of solute / Mass of Solvent in Kg It is temperature independent.

Normality[N]: N = No. of gm equivalent of solute / Volume of solution in litre

Parts Per Million: Ppm = Part of solute x 106 / Part of solution

Mole Fraction: Ratio of number of moles of a component in solution to the total no. of moles of all the components is called mole fraction (x) of the component.

RELATION BETWEEN MOLARITY (M) AND MASS %

Molarity = solute of massMolar

x10d x mass%

RELATION BETWEEN NORMALITY (N) AND MASS %

Normality = solute of mass Equivalent

x10d x mass%

Normality of a mixture (N) = 21

2211

VV

VNVN

+

+ ; Molarity of a mixture (M) =

21

2211

VV

VMVM

+

+

Relation between Normality (N) and Molarity (M): N = n M

RELATION BETWEEN MOLARITY (M) AND MOLALITY (m)

)]M(M-d[1000

1000 M

B

=m

‘M’ is the molarity and m is the molality, d is density of solution in g cm-3 MB is

molar mass of solute.

RELATION BETWEEN MOLE FRACTION )( B AND MOLALITY: ‘A’ stands for solvent

and ‘B’ stands for solute.

)( B : Mole fraction of component/solute B, m is the molality; MA is molar mass of

solvent.

A

B

Mm

m

1000+

= = 1000 M m

M

A

A

+

m also m =

AM.

1000

A

B

RELATION BETWEEN MOLE FRACTION )( B AND MOLARITY

d 1000)M - (M M

M M

BA

A

+=B Also M =

BBAA M M

X d X 1000

+

B

Henry’s law” P=KH

(Class XII)

m = KH P

(Class XI)

Raoult’s Law: “The V.P. of any volatile component in the solution is directly

proportional to its mole fraction”.

Raoult’s law for a solution

containing two miscible liquids:

Raoult’s law for a solution containing a

non-volatile solute:

20

PA A , PA = PA0 X A and

PB B , PB = PB0 X B

Therefore total vapour pressure of the solution: Ps = PA + PB

For a binary solution BA + = 1,

if BA -1 =

Ps = PA0 + (PB

0 - PA0) B and if

A -1 B =

Ps = PB0 + (PA

0 – PB0) A

0

0

B

A

AA

P

PP −=

Where PA0 is vapour pressure of pure

component ‘A’

PA is partial vapour pressure of component ‘A’

pressure vapour of lowering is P0

A AP−

pressure vapour of lowering relative is 0

0

A

AA

P

PP −

Ideal and Non ideal solution:

Ideal Solution

Non ideal solution Positive elevation

Negative Deviation

Colligative Properties:

CASE STUDY BASED QUESTIONS:

1. Read the passage given below and answer the following questions:

All those properties that depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the

solution are called colligative properties. Similar to lowering of vapour pressure, the elevation of boiling point also depends on the number of

solute molecules rather than their nature. Let Tb0 be the boiling point of pure solvent and Tb be the boiling point of solution. The increase in the

boiling point ΔTb = Tb – Tob is known as elevation of boiling point.

21

Experiments have shown that for dilute solutions the

elevation of boiling point (ΔTb) is

directly proportional to the

molal concentration of the solute in a solution.

Thus

ΔTb m [molality]

ΔTb = Kb m

Kb=Molal Elevation Constant (Ebullioscopic

Constant).

The following questions are multiple choice questions. Choose the most appropriate answer

(i) Which has the highest boiling point: (a) 1M Glucose (b) 1M NaCl (c) 1M CaCl2 (d) 1M AlF3 (ans: d) (ii) Molal elevation constant is also known as

(a) Cryoscopic consant (b) Ebullioscopic Constant (c) Molal depression constant (d) both a and b (Ans: b)

(iii) Unit of Molal elevation constant is - (a) K Kg mol-1 (b) K mol Kg-1 (c) Kg mol K-1 (d) mol Kg-1 K-1 (Ans: a)

OR Pressure cooker reduces cooking time for food because:

(a) Heat is more evenly distributed in the cooking space. (b) Boiling point of water involved in cooking is increased (c) The higher pressure inside the cooker crushes the food material

(d) Cooking involves chemical changes helped by a rise in temperature

(Ans: b)

(iv)The rise in the boiling point of a solution containing 1.8 gram of glucose in 100g of a solvent in 0.1o C. The molal elevation constant of the liquid is

(a) 0.01 K/m (b) 0.1 K/m (c) 1 K/m (d) 10 K/m (Ans: c)

2. Read the passage given below and answer the following questions:

Osmotic pressure results from a reduction in the chemical potential of a solven

t in the presence of a solute. The tendency of a system to have equal chemical potentials over its entire volume and to reach a state of lowest free energy gi

ves rise to the osmotic diffusion of matter. In ideal and dilute solutions, the osmotic pressure is independent of the nature of the solvent and solutes. At constant temperature it is determined only by the number of kinetically active par

ticles—ions, molecules, associated species, and colloidal particlesin a unit volume of t

he solution. For very dilute solutions of nondissociating compounds, osmotic pressure is d

escribed with sufficient accuracy by the equation πV = nRT, where n is t

he number of moles of solute, V is the volume of the solution, R is the universal gas constant, and T is the absolute temperature.

The following questions are multiple choice questions. Choose the most appropriate answer

(i) Which of the following process/method is used for purification of water?

(a) Osmosis (b) Reverse osmosis (c) Osmotic pressure (d) Elevation of boiling point

(ii) Which of the following colligative property is most suitable to measure molecular mass of proteins.

(a) Osmotic pressure (b) Elevation of boiling point

(c) Depression of freezing point (d) Lowering of vapours pressure (iii) What is the principle of reverse osmosis?

(a) The direction of solvent flow can be reversed by applying pressure

greater than osmotic pressure.

(b) The direction of solvent flow can be reversed by applying pressure less

than osmotic pressure

(c) The direction of solvent flow can be reversed by using a permeable

membrane

(d) The solvent flow can be reversed by concentrating the mixture

OR

Which of the following is not a use of reverse osmosis? a) Treatment of industrial waste water to remove heavy metal ions

b) Separation of sulfites and bisulfites from paper and processing industry

22

effluents c) Treatment of municipal water to remove inorganic salts d) Treatment of industrial effluents to decolorize i

(iv) Why is osmosis not a good separation process? a) Because it involves semipermeable membrane b) Because it transfers solvent from wrong direction c) Because it involves movement of solute particles d) Because it requires high temperatures

Ans: (i) b (ii) a (iii) a OR d (iv) b 3. Read the passage given below and answer the following questions:

The vapour pressure of the mixture (completely miscible binary mixture) is a function of the composition as well as the vapour. Pressure of the two pure

compounds in an ideal solution where the relation of vapour pressure and composition is given by Raoult’s law: the partial vapour pressure of each

volatile component is equal to the vapour pressure of the pure component multiplied by its mole fraction. Thus for a mixture of A and B:

PA = PA0

A

PB = PB0

B

Total vapour of the system is equal to sum of vapour pressure of A and B,

e.g., PA + PB Binary mixtures that follows Raoult’s law are those where the attraction

between A and B molecules is the same as those of pure components eg,

benzene/ toluene and paraffin mixtures. Thus for binary mixtures ΔHmix and ΔVmix are zero. When the interaction between A and B molecules is not same

as those for pure components, the mixture becomes non-ideal. Non ideal solutions show positive and negative deviations from Raoult’s law. Non ideal solutions showing positive deviation have A-B interactions weaker than A-A

interaction and B-B interactions and Non ideal solutions showing negative deviation have A-B interactions stronger than A-A interaction and B-B

interactions. In these questions (Q. No. i-iv), a statement of Assertion followed by a statement of Reason is given. Choose the correct answer out of the following

choices. a) Assertion and Reason both are correct statements and Reason is correct

explanation for Assertion. b) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion.

c) Assertion is correct statement but Reason is wrong statement. d) Assertion is wrong statement but Reason is correct statement.

(i) Assertion: Acetone and chloroform show negative deviations.

Reason: H-Bonding Acetone and chloroform is stronger than between

acetone – acetone and chloroform-chloroform. (Ans: a) (ii) Assertion: An ideal solution obeys Raoult’s law.

Reason: In an ideal solution solute-solute and solvent – solvent interactions are dissimilar to solute-solvent interactions. (Ans:c)

(iii) Assertion: ΔHmix and ΔVmix are zero for a binary mixture if it is an

ideal solution. Reason: Interactions between the particles of the components of a

solution are not identical as between particles in the liquids if it is a non ideal solution (Ans:b)

(iv) Assertion: A solution of Chlorobenzene and bromobenzene obeys

Raoult’s law. Reason: Solution of chlorobenzene and bromobenzene is nearly ideal

liquid solution. (Ans: a) OR

Assertion: Non ideal solutions showing positive deviation have A-B interactions stronger than A-A interaction

Reason: Non ideal solutions showing negative deviation have A-B

interactions stronger than A-A interaction and B-B interactions. (Ans: d) 4. Read the passage given below and answer the following questions:

An azeotropic mixture is basically a mixture that contains two or more liquids.

A azeotropic mixture basically has constant or same boiling points and the

mixtures’ vapour will also have the same composition as the liquid. Azeotropic mixtures are formed only by non - ideal solutions and they may have boiling

23

points either greater than both the components or lesser than both the components. There are two types of an azeotropic mixture.

Maximum boiling Azeotrope Minimum boiling Azeotrope

A non ideal solution with negative deviation. Eg:

The solutions that show large negative deviation from Raoult’s law form

maximum boiling azeotrope at a specific composition. Nitric acid and water is an example of this class of azeotrope. This

azeotrope has the approximate composition, 68% nitric acid and 32%

water by mass, with a boiling point of 393.5 K. The boiling point of the azeotropes is

greater than the boiling points of the either of the components.

A non ideal solution with positive deviation. Eg:

For example, ethanol-water mixture (obtained by fermentation of sugars)

on fractional distillation gives a solution containing approximately 95% by volume of ethanol. Once this

composition, known as azeotrope composition, has been achieved, the

liquid and vapour have the same composition, and no further separation occurs.

The boiling point of the azeotropes is lesser than the boiling points of the

either of the components.

(i) Assertion: Non ideal solutions form azeotropic mixture. Reason: The boiling point of an azeotropic mixture is only higher than boiling points of both components. (Ans: c)

(ii) Assertion: Azeotropic mixtures are formed only by non - ideal solutions and they may have boiling points either greater than both the components or

lesser than both the components. Reason: The composition of the vapour phase is the same as that of the

liquid phases of an azeotropic mixture. (Ans: b) (iii) Assertion: The solutions that show large negative deviation from Raoult’s law form minimum boiling azeotrope at a specific composition.

Reason: 95% by volume of ethanol is an example of positive azeotrope. (Ans: d)

(iv) Asserstion: Azeotrope, in chemistry, a mixture of liquids that has a constant boiling point.

Reason: The azeotropic mixture’s vapour has the same composition as the

liquid mixture. (Ans: a)

Multiple Choice questions: 1. A 5% solution of cane-sugar (molecular weight = 342) is isotonic with 1%

solution of substance X. The molecular weight of X is (a) 342 (b) 171.2 (c) 68.4 (d) 136.8

2. H2S is a toxic gas used in qualitative analysis. If solubility of H2S in water at STP is 0.195 m. what is the value of KH?

(a) 0.0263 bar (b) 69.16 bar (c) 192 bar (d) 282 bar

3. Henry’s law constant for molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. The mole fraction of methane is benzene at 298 K under 760 mm Hg is (a) 1.78 × 10-3 (b) 17.43 (c) 0.114 (d) 2.814

4. Among the following substances the lowest vapour pressure is exerted by (a) water (b) alcohol (c) ether (d) mercury

5. 3 moles of P and 2 moles of Q are mixed, what will be their total vapour pressure in the solution if their partial vapour pressures are 80 and 60 torr respectively?

(a) 80 torr (b) 140 torr (c) 72 torr (d) 70 torr 6. Which of the following solutions shows positive deviation from Raoult’s

law? (a) Acetone + Aniline (b) Acetone + Ethanol (c) Water + Nitric acid

(d) Chloroform + Benzene 7. The system that forms maximum boiling azetrope is

(a) Acetone-chloroform

(b) ethanol-acetone (c) n-hexane-n-heptane

(d) carbon disulphide-acetone

24

8. A plant cell shrinks when it is kept in a (a) hypotonic solution (b) hypertonic solution (c) isotonic solution (d) pure water

9. The relative lowering in vapour pressure is proportional to the ratio of number of

(a) solute molecules to solvent molecules (b) solvent molecules to solute molecules

(c) solute molecules to the total number of molecules in solution (d) solvent molecules to the total number of molecules in solution

10. What weight of glycerol should be added to 600 g of water in order to

lower its freezing point by 10°C? (Kf = 1.86 K Kg mol-1) (a) 496 g (b) 297 g (c) 310 g (d) 426 g

11.The osmotic pressure of a solution can be increased by (a) increasing the volume (b) increasing the number of solute molecules (c) decreasing the temperature (d) removing semipermeable membrane

12.Sprinkling of salt helps in clearing the snow covered roads in hills. The phenomenon involved in the process is

(a) lowering in vapour pressure of snow (b) depression in freezing point of snow (c) melting of ice due to increase in temperature by putting salt

(d) increase in freezing point of snow 13.For carrying reverse osmosis for desalination of water the material used for

making semipermeable membrane is (a) potassium nitrate (b) parchment membrane (c) cellulose acetate (d) cell membrane

14.Which of the following units is useful in relating concentration of solution with its vapour pressure?

(a) Mole fraction (b) Parts per million (c) Mass percentage (d) Molality 15.The atmospheric pollution is generally measured in the units of

(a) mass percentage (b) volume percentage (c) volume fraction (d) ppm

16. All form ideal solutions except (a) C2H5Br and C2H5I (b) C6H5Cl and C6H5Br (c) C6H6 and C6H5CH3 (d) C2H5I and C2H5OH

17. Which of the following is not correct for ideal solution? (a) ΔSmix = 0 (b) ΔHmix = 0 (c) It Obeys Raoult’s law (d) ΔVmix = 0

18. Which of the following solutions is water possesses the lowest vapour

pressure. (a) 0.1(M) NaCl (b) 0.1(M) BaCl2 (c) 0.1(M) KCl (d) None of these

19. The solubility of a gas in water depends on (a) Nature of the gas (b) Temperature (c) Pressure of the gas (d) All of the above.

20.The sum of the mole fraction of the components of a solution is (a) 0 (b) 1

(c) 2 (d) 4

(Ans: 1-c, 2-d,3-a,4-d,5-c,6-b, 7-a,8-b,9-c,10-b,11-b,12-b, 13-c,14-a,15-d,16-d,17-a, 18-b, 19-d, 20-b)

5. Choose the correct answer out of the following choices:

a) Assertion and Reason both are correct statements and Reason is correct

explanation for Assertion. b) Assertion and Reason both are correct statements but Reason is not

correct explanation for Assertion. c) Assertion is correct statement but Reason is wrong statement.

d) Assertion is wrong statement but Reason is correct statement.

(i) Assertion: When a solution is separated from the pure solvent by a semi-permeable membrane, the solvent molecules pass through it from pure

solvent side to the solution side. Reason: Diffusion of solvent occurs from a region of high concentration solution to a region of low concentration solution.

(Ans:c) (ii) Assertion: Molarity of a solution in liquid state changes with

temperature. Reason: The volume of a solution changes with change in temperature. (Ans: a)

(iii) Assertion: Molality of a solution in liquid state does not change with

temperature. Reason: The mass of solution does not change with change in temperature. (Ans: a)

25

(iv)Assertion: If on mixing the two liquids, the solution becomes hot, it implies that it shows negative deviation from Raoult’s law. Reason: Solution which shows negative deviation from Raoult’s law are

accompanied by decrease in volume. (Ans: b)

(v) Assertion: Greater the value of Henry’s constant of a gas in a particular solvent, greater is the solubility of the gas at the same pressure and

temperature. Reason: Solubility of a gas is inversely proportional to its Henry’s constant at the same pressure and temperature.

(Ans: d)

2 Marks question - answer

6. Calculate the mole fraction of benzene in a solution containing 30% by mass of

it in CCl4 Ans. Mass of benzene =30g Mass of CCl4 =70g

Molar mass of benzene= 78 g Molar mass of CCl4= 154g mol-1 nC6H6=30g/78g=0.385mol nCCl4=70g/154g=0.454mol X C6H6= nC6H6/n C6H6 + nCCl4 =0.385 mol /0.385mol + 0.454mol = 0.459

7. (i) Name any two compounds which can be used as semi permeable membrane .

(ii)When two liquids A and B are mixed the solution becomes warmer ,which deviation solution exhibits from Raoult’s law?

Ans. (i)Copper ferrocyanide and cellulose acetate.(ii)negative deviation 8. What do you mean by Raoult’slaw ? What are the limitations of Raoult’slaw ? Ans. Raoult’s law – partial vapour pressure of any volatile component of a solution

at Any tempreture is equal to the product of vapour pressure of the pure compenent

and mole fraction of the component in solution pA xA pA = pA0xA and pB xB

pB = pB0xB.

Limitations : (i)Applicable only to dilute sol.(ii)Applicable to solutions of only non

electrolyte. 9. Equimolal solutions of NaCl and BaCl2 are prepared in water. Freezing point of

NaCl is found to be -20C. What freezing point do you expect for BaCl2 solution? Ans: For NaCl, van’t Hoff factor is 2, whereas for BaCl2, the van’t Hoff factor is 3.

Hence the depression of freezing point for BaCl2 solution is 3/2than that of NaCl solution. Hence the freezing point of BaCl2 is -30C.

10. After removing the outer shell of two eggs in dil. HCl, one is placed in distilled water and the other in a saturated solution of NaCl. What will you observe and

why? Ans: Eggs in water will swell whereas in NaCl solution it will shrink. This is because

due to osmosis, the net flow of solvent is from less concentrated to more

concentrated solution. 11.What is the molality of 1.0 M solution of sodium nitrate (NaNO3) if its density is

1.25 g cm-3? Solution:

)]M(M-d[1000

1000 M

B

=m

‘M’ is the molarity and m is the molality, d is density of solution in g cm-3 MB is

molar mass of solute.

)]85(1-1.25[1000

1000 1

=m = 0.858 mol/kg (Ans)

12.Calculate molarity and molality of a 13% solution by weight of sulphuric acid? Its density is 1.02 gcm-3.

[Atomic wt. of H=1,S=32, O=16 amu]

WB = 13g, Mass of solution = 100 g , MB (H2SO4) = 98 g

Solution: Molarity (M) = 1-L mol 353.1

02.1

100

1000

98

13

solution ofdensity

solution of Mass

1000==

B

B

M

W

13. (i) Why is glycol and water mixture used in car radiators in cold countries?

(ii) Give Reason When 30 ml of ethyl alcohol and 30ml of water are mixed, the

26

volume of resulting solution is more than 60ml. Ans: (i) Ethylene Glycol reduces the freezing point of water. Due to this, the coolant

in radiators will not freeze.

(ii) Solution of water and ethyl alcohol shows positive deviation and hence 𝛥𝑉 > 0 14. (i) What will be the nature of the solution when ethyl alcohol and water are

mixed? Ans. Positive deviation from Raoult s law.

(ii) State the condition leading to reverse osmosis. Ans. Occurs when a pressure higher than osmotic pressure is applied. 15. The vapour pressure of water is 12.3 kPa at 300K; calculate the vapour

pressure of 1 molar solution of a solute in it. Ans: 1 molar solution = 1mol/1000 g of H2O (assuming solution to diluted)

Mole fraction of solute = 1/1+55.5 = 0.0177 P0-Ps / P0 = X2 so 12.3- Ps/12.3 = 0.0177 Therefore Ps = 12.08 kPa

3 Marks question - answer

16. A solution containing 18 g of non-volatile solute in 200g of water freezes at 272.07 K. calculate the molecular mass of solute (given Kf = 1.86 K/m)

Ans. W2 = 18 g W1= 200g, kf = 1.86k/m Tf= 273K – 272.07K = 0.93K

Tf =1000×kf×W2

M2×W1 M2 =

1000×kf×W2

∆Tf×W1

M2 = 1000×1.86×18

0.93×200 = 180 amu

17. Calculate the temperature at which a solution containing 54 g of glucose (C6H12O6) in250 g of water will freeze (kf for water =1.86 KKg/mol).

Ans. Given- Wsolute = 54 g ,Wsolvent = 250 g, Msolute = 180 Tf = kf x wsolute x 1000 Tf = 1.86 x 54 x 1000 Tf = 2.232 K

Msolutexwsolvent 180 x 250 Tf = T0

f - Tf Tf = 273- 2.232 = 270.768 K

18. Calculate the mass of urea required in making 2.5 kg of 0.25 molal aq.solution. Ans. Given- Molality (m) = 0.25 m Molar mass (MB) of urea (NH2CONH2)

= 60 g mol – 1

19. 200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The

osmotic pressure of such a solution at 300 K is found to be 2.57 × 10-3 bar. Calculate the molar mass of the protein.

Ans. The various quantities known to us are as follows: Π = 2.57 × 10–3 bar, V = 200 cm3 = 0.200 litre T = 300 K R = 0.083 L bar mol-1 K-1 Π = CRT = n x RT/V (n = W2/M2), Π = W2/M2 X RT/V Therefore M2 = W2/ Π

X RT/V Substituting values:

M2 = 1.26 g × 0.083 L bar K−1 mol−1 × 300 K / 2.57×10−3 bar × 0.200 L = 61,022 g mol-1

21. 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. Calculate-

(a) The freezing point depression (b) The freezing point of the solution.

Ans. Moles of ethylene glycol =45/62 g mol-1=0.73 mol

Mass of water in kg =600g/1000g kg-1 = 0.6 kg Hence molality of ethylene glycol = 0.73mol/0.60kg = 1.2 mol kg –1

∆ Tf = 1.86 K kg mol–1 × 1.2 mol kg –1 = 2.2 K Freezing point of the aqueous solution = 273.15 K – 2.2 K = 270.95 K

22. Miscible liquid pairs often show positive and negative deviation from Raoult’s

law. Explain the Reason for such deviation? Give one example of each of liquid pairs.

Answer: Miscible liquid pairs often show positive and negative deviation from Raoult’s law

are called

non ideal solutions. Explanation of the Reason of such deviations is as follows:

Negative deviation Positive deviation

27

(i) The vapour pressure of solution is lower than that of an ideal

solution of the same composition. (ii) A negative deviation is executed

by liquid pairs for which A-B

molecular interaction forces are stronger than the A-A and B-B

molecular interaction forces. This means that molecules of A and B will find it difficult to escape from

a solution than from the pure components. This will result in

decrease in vapour pressure which is known as –ve deviation.

Eg. Acetone + Chloroform Intermolecular force of interaction

between:

Acetone – Acetone, Diople dipole interaction

Chloroform – Chloroform, Diople dipole interaction

Acetone – Chloroform, H- Bonding

(Stronger than above two)

(i) The vapour pressure of solution is higher than that

of an ideal solution of the same composition.

(ii) A negative deviation is

executed by liquid pairs for which A-B molecular

interaction forces are weaker than the A-A and B-B molecular interaction

forces. This means that molecules of A and B will find

it easier to escape from a solution than from the pure

components. This will result in increase in vapour pressure which is known as +ve

deviation. Eg. Ethanol + Cyclohexane

Intermolecular force of

interaction between:

Ethanol – Ethanol, H- Bonding Cyclohexane – Cyclohexane,

London dispersions Ethanol – Cyclohexane (Weaker

than above two)

23. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at

the normal boiling point of the solvent. What is the molar mass of the solute? (Vapour pressure of pure water at the boiling point (P0) = 1 atm = 1.013 bar)

Ans:

Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar Vapour pressure of pure water at normal boiling

point Mass of solute, (w2) = 2 g

Mass of solvent (water), (w1) = 98 g

Molar mass of solvent (water), (M1) = 18 g mol−1 According to Raoult’s law,

= 41.35 g mol−1

Hence, the molar mass of the solute mol−1.

24. The element A and B formed purely covalent compounds having molecular

formula AB2 and AB4. When dissolved in 20 gram of benzene 1 gram of AB2 lowers the freezing point by 2.3 K and of AB4 by 1.3 K. Calculate atomic mass of A and B. The molar depression constant for freezing is 5.1 KKg mol-1

Ans:

Solution: Af

Bf B

WT

1000 WK M

=

Case 1: g/mol 87.11020 2.3

1000 1 1.5 M AB2

=

= Case 2: g/mol 196

20 1.3

1000 1 1.5 M AB4

=

=

Now 2B = AB4 – AB2 = 196 – 110.87 = 85.13 g/mol, Therefore B = 42.56 g/ mol Now A = AB2 – 85.13 = 25.74 g/ mol.

25. An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Determine (i) Molality of solution. (ii) Boiling point of solution (iii) Lowering of

vapour pressure of water at 298 K.

28

Ans: Given: Kf = 1.86 K Kg mol -1

. Kb = 0.512 K Kg mol -1,

V.P.of pure water at 298 K = 23.756 mm Hg.

Solution: (i) ΔTf = Kf m Therefore m = Kgmol /32.086.1

6.0

K

T

f

f ==

(ii) ΔTb = Kb m = 0.512 x 0.32 = 0.164 Therefore Tb = 373 + 0.164 = 373.16 K (iii) ∆P = P0

.m.MA= 23.756 x 0.32 x 0.018 = 0.137 mm Hg.

26.Calculate the osmotic pressure at 270C of a solution formed by mixing equal

volumes of two solutions, one containing 0.05 mole of glucose in 250 ml of solution

and the other containing 3.42 g of C12H22O11 in 250 ml of solution.[R = 0.082 L atm mol-1K-1]

Ans:

= CRT = (C1 + C2) RT = (Ans) atm 0.59112 300 .082 ]250

1000

342

3.42 1000

250

0.05[ =+

5 Marks questions

27. (i) State Henry’s law and mention some important applications. (ii) If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would

dissolve in 1 litre of water. Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N2 at 293 K is 76.48 kbar.

Ans. (i)The pressure of gas over a solution is directly proportional to the mole

fraction of the gas dissolved in the solution. Applications: (a) To increase the solubility of CO2 in soft drinks.

(b)To avoid painful effects during the decompression of scuba divers, oxygen diluted with less soluble helium gas is used by the sea divers.

Ans.(ii) The mole fraction of the gas in the solution is calculated by applying Henry’s law.

Thus:x(Nitrogen)=p(nitrogen)/KH=0.987bar/76,480bar=1.29×10-5

As 1 litre of water contains 55.5 mol of it, therefore if n represents no. of moles of N2in solutionx(Nitrogen) = n mol/ n mol + 55.5 mol =n/55.5= 1.29 x 10-5

Thus n = 1.29 × 10–5 × 55.5 mol = 7.16 × 10–4 mol =7.16×10−4 mol × 1000 millimoles /1 mol = 0.716 millimoles 28. (i) Two liquids A and B on mixing form an ideal solution. At 300C vapour pressure

of solution containing 3 mol of A and 1 mol of B is 550 mmHg. But when 4 mol of A and 1 mol of B are mixed. The vapour pressure of solution thus formed is 560

mm Hg. What would be the V.P of pure A and B? (ii) Explain the fact that Raoult’s Law is a special case of Henry’s Law Ans (i)

Components Case

A B Ps

I 3 mol 1 mol 550 mm Hg

II 4 mol 1 mol 560 mmHg

Formula: Ps = PA0 X A + PB

0 X B

Case I: A = 4

3 and B =

4

1 550 =

4

P 30

A + 4

P0

B or (550 = 4

P 30

A + 4

P0

B ) X 4,

2200 = 3 PA0 + PB

0 eq 1

Case II: A = 5

4 and B =

5

1 560 =

5

P 40

A + 5

P0

B or (560 = 5

P 40

A + 5

P0

B ) X 5,

2800 = 4 PA0 + PB

0 eq 2 Subtracting equation 1 from 2: 2800 = 4 PA

0 + PB0

2200 = 3 PA0 + PB

0 PA

0 = 600 mm Hg Substituting PA

0 in equation 1 or 2

PB0 = 400 mm Hg.

(ii) According to Raoult’s law, the vapour pressure of a volatile component in a given

solution is given by pi = xi pi0.

Ans (ii) In the solution of a gas in a liquid, one of the components is so volatile that

it exists as a gas. Its solubility is given by Henry’s law. p = KH x. If we compare the equations for Raoult’s law and Henry’s law, it can be seen that the

partial pressure of the volatile component or gas is directly proportional to its mole

fraction in solution. Only the proportionality constant KH differs from p10. Thus,

Raoult’s law becomes a special case of Henry’s law in which KH becomes equal to

p10.

29

29. (i) Which colligative property is most suitable to determine molecular mass of biomolecules. Also mention the Reason for its suitability.

Ans: Osmotic pressure method is most suitable to determine molecular mass of bio-

molecules. This is because in osmotic pressure method measurement is done at room temperature and biomolecules are stable at room temperature.

(ii) The osmotic pressure of blood is 8.21 atm at 370C. How much glucose should be used per liter for an intravenous system that is isotonic with blood?

Solution: (blood) = (glucose = solution) = 8.21 atm.

= nRT/V or RT

V n

= , V = 1.0 L, T = 273 + 37 = 310 K

(Molar mass of glucose =180)

31

10

310 0821.0

1.0 8.21

RT

V n =

==

, Weight of glucose = 180

31

10 = 58.06 g

30. (i) The molar freezing point depression of benzene is 4.9 Kg .mol-1. Selenium is known to exist as a polymer (Se)x with 3.26 gram of selenium dissolved in 226

gram of benzene. The depression of freezing point is 0.1120C. The freezing point is lower than the pure benzene. Deduce the molecular formula of Selenium.

(Atomic mass of SE is 78.96 gram)

Solution: Af

Bf B

WT

1000 WK M

= =

226 112.0

1000 3.26 9.4 M B

= = 631.09 g/mol

x = Ans. - (Se) 896.78

09.6318=

(ii) Draw a plot or a graph for vapour pressure vs temperature to explain the colligave property depression in freezing point.

Ans:

Depression of Freezing

Point: Difference in

freezing point of pure

solvent and freezing

point of

solution is called

‘Depression in Freezing

Point’. ΔTf = T0

f – Tf

ΔTf m [molality]

ΔTf = Kfm

Kf =Molal Depression Constant

(Cryoscopic Constant).

The unit of Kf is K kg mol-1

31. (i) The vapour pressure of pure liquids A and B are 450 and 700 mm Hg

respectively, at 350 K. Find out the composition of the liquid mixture if total

vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Ans.

It is given that: PA

0 = 450 mm of Hg PB0 = 700 mm of Hg ptotal = 600 mm of Hg

From Raoult’s law: Ptotal = PA0 X A + PB

0 X B

600 = 450 X A + 700 X ( A1 − )

On solving A = 0.4 Therefore, B = 1-0.4 = 0.6

In vapour phase, PA = 0.4 x 450 = 180 mm Hg and PB = 0.6 x 700 = 420 mm Hg

YA = 180/(180 + 420) = 0.3 and YB = 420/(180 + 420) = 0.7 (ii) a) State how does osmotic pressure vary with temperature?

Ans. Osmotic pressure increases with increase in temperature. b) What do you expect to happen when RBC’s are placed in 0.5% NaCl

solution?

Ans. The will swell and may even burst. This because RBC’s are isotonic with 0.9% NaCl solution.

30

CHAPTER 3 - ELECTRO CHEMISTRY

• ELECTRODE POTENTIAL OF A CELL E⁰cell = E⁰ cathode - E⁰anode

(in terms of reduction potential)

For a spontaneous cell Eocell must be positive

• NERNST EQUATION

• 𝐄⁰ 𝐜𝐞𝐥𝐥 =𝟎.𝟎𝟓𝟗

𝐧log

[𝐩𝐫𝐨𝐝𝐮𝐜𝐭]𝐩𝐨𝐰𝐞𝐫 𝐱

[𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕] 𝒑𝒐𝒘𝒆𝒓 𝒚 (Molar concentration of only ions will be used)

• 𝐄⁰ 𝐜𝐞𝐥𝐥 =𝟎.𝟎𝟓𝟗

𝐧log Kc (At 298 K) Kc is EQUILIBRIUM CONSTANT

• FREE ENERGY CHANGE FOR THE REACTION ∆G⁰ = -nF E⁰cell

• ∆G⁰ = -2.303 RT log Kc

• For an electrochemical cell if G⁰ < 0 then spontaneous process.

• E⁰cell is > 0 for spontaneous process.

• As per E⁰ values, a metal having more negative reduction potential is stronger

reducing agent than hydrogen while a metal having more positive reduction

potential is weaker reducing agent than hydrogen.

• Charge on one mole of electrons is equal to 96487 ≈ 96500 C/mol =1F.

• KOHLRAUSCH LAW

• SHE (Standard Hydrogen Electrode) is represented by

Pt/H2 (g, 1 bar)/ H+ (1M) and it is used as a reference half-cell.

2H+ (1M) + 2e– ⎯→ H2 (g, 1 bar), E⁰ H+/H2 = 0 V

• Conductance (G) is inverse of resistance (R). Unit : Ohm–1 or S. G=1/R

• Conductivity (specific conductance) denoted by κ, is reciprocal of resistivity and

is expressed in S cm–1 . ƿ =1/κ

• Conductivity (κ) = Conductance (G) x cell constant(l/A)

(Cell constant unit is cm–1 or m–1).

• Molar conductance (m) is conductance of solution containing 1 mol of solute

in a given volume. Unit Scm2mol-1

• MOLAR CONDUCTANCE M = κ x 1000/C

• DEGREE OF IONISATION α = Mc/M

O

• K = C α2/1- α

• Conductivity decreases but molar conductivity increases with decrease in

concentration.

CASE STUDY BASED QUESTIONS -ANSWERS

1 Redox reactions play a pivotal role in chemistry. The values of the standard reduction potential E⁰ of the two half cell reactions determine which way the reaction is expected

to proceed. A simple example is the Daniel cell in which zinc metal electrode dissolves in solution by losing two electrons and Cu2+ ion deposits as metallic copper by gaining

two electrons. E⁰ cell of Daniel cell is 1.1 V Answer the following questions. In these questions a statement of Assertion is

followed by a statement of Reason. Choose the correct option (a)Assertion and Reason both are correct and Reason is correct explanation

of Assertion. (b)Assertion and Reason both are correct but Reason is not correct

explanation of Reason (c)Assertion is correct but Reason is Incorrect. (d)Assertion is Incorrect but Reason is correct.

(e)Both Assertion and Reason is incorrect.

i) ASSERTION: It is not possible to measure the voltage of isolated half cell. REASON: Oxidation or reduction cannot take place alone.

a

ii) ASSERTION: Zinc can not displace copper from copper sulphate solution. REASON: The reduction potential of copper electrode is higher than that of zinc

d

31

metal. iii) ASSERTION: In the Daniel cell if the concentration of Zn2+ and Cu2+ is doubled

, the emf of the cell does not change.

REASON: If the concentration of the solution in metal is doubled the electrode potential is either doubled or halved depends upon the charge of the electrode.

c

iv) ASSERTION: Copper does not displace hydrogen from dilute acid. REASON: Metals do not evolve hydrogen gas with dilute nitric acid.

b

2 The concentration of potassium ions inside a biological cell is atleast 20 times higher than outside. The resulting potential difference across the cell plays a vital role in several processes such as the transmission of nerve impulses and maintenance of the

ionic balance, A simple model for such a concentration cell involving a metal M is M(s) /M+ (aq)(0.05M) // M+ (aq) (1M)/M (s) for which Ecell is 70mV

i) For the model cell a) Ecell<0, ∆G >0 b) Ecell>0, ∆G <0

c) Ecell<0, ∆G⁰ >0 d) Ecell>0, ∆G⁰ <0

b

ii) If the 0.05 m solution is replaced by a 0.0025 M solution the magnitude of the cell potential will be a) 35mV b) 70mV c) 140mV d) 700mV

c

iii)The value of ∆G in kJ/mol for the given cell is (1F is taken as 96500 C/mol ) is

a) - 6.7 b)-12.4 c) 12.4 d) 6.7

a

iv) If the concentration of ions in both the electrodes are same then ∆G value

a)increases b) decreases c) No change d) becomes zero

d

3 A non-spontaneous reaction brought about by means of electrical energy is called electrolysis. Electrical decomposition of compounds is carried out either in molten state or in aqueous solution because mobile ions are needed to carry charge through

cell. The products of electrolysis depend upon the reduction / oxidation potential of the

ions formed in the solution. Electrolysis play important role in metallurgy. Alkali metals are extracted by electrolysis of their molten salt. Answer the following questions. In these questions a statement of Assertion is

followed by a statement of Reason. Choose the correct option (a)Assertion and Reason both are correct and Reason is correct explanation

of Assertion. (b)Assertion and Reason both are correct but Reason is not correct explanation of Reason

(c)Assertion is correct but Reason is Incorrect. (d)Assertion is Incorrect but Reason is correct.

(e)Both Assertion and Reason is incorrect.

i) ASSERTION: sodium metal cannot be extracted by electrolysis aq NaCl. REASON: Reduction potential of sodium ion is less than Hydrogen ion.

a

ii) ASSERTION: Electrolysis reaction take place spontaneously.

REASON: During electrolysis electrical energy is converted in to chemical energy.

d

iii) ASSERTION: Electroplating of metal is also example of electrolysis.

REASON: In electrolytic cell oxidation take place at anode which is positive electrode.

b

iv) ASSERTION: The extra potential needed to speed up slow reaction is called over

voltage or over potential.. REASON: Over voltage is often large when gas is formed at electrode and small

when metal is deposited at electrode.

MCQ TYPE QUESTIONS

a

4 Tollens reagent (ammoniacal AgNO3) is used to test the presence of aldehyde group in carbohydrates. It is the reagent used to distinguish between reducing and non-

reducing sugar. Following reactions are involved, Ag+ + e- ---------→ Ag E⁰ = 0.80 V

C6H12O6 + H2O --------------→ C6H12O7 + 2H+ +2e- E⁰ =-0.05V [Ag(NH3)2]+ + e- -------→ Ag(s) + 2 NH3 E⁰ = 0.337 V

i) The value K for the reaction 2Ag+ + C6H12O6 + H2O ------------→ 2Ag + C6H12O7 + 2H+ is

a) 66.13 b) 58.38 c) 28.30 d) 46.29

b

32

ii) When ammonia is added to the solution, pH increases to 11, which half cell reaction

will be affected by pH ?

a)Eox will increase b)Eox will decrease

c)Ered will increase d)Ered will decrease

a

iii) In the given reaction [Ag(NH3)2]+ is stronger oxidising agent than Ag+ a) Correct

b) Not correct c) both have same oxidising power

d) partially correct

b

iv) The overall reaction will be spontaneous when

a) E⁰red> 0 b) E⁰ox>0 c) ∆G⁰>0 d) ∆G⁰<0

d

5. A gas X at 1atm is bubbled through a 1 M solution of Y as well as in 1M solution of Z at 25⁰ C. The reduction potential of Z>Y>X ,

a) Y will oxidise both X ans Z b)Y will reduce both X and Z

c) Y will oxidise X but not Z d) Y will oxidise Z but not X

d

6 Water cannot be electrolysed with electrode/electrodes of

a)Cu b)Ag c)Al d)Mg

c,d

7 The oxidation potential of Zn, Cu, Ag, H2 and Ni are 0.76, –0.34, – 0.80, 0, 0.55 volt

respectively. Which of the following reaction will provide maximum voltage ? (a) Zn + Cu2+--------→Cu + Zn2+

(b) Zn + 2Ag+--------→2Ag + Zn2+ (c) H2 + Cu2+---------→2H+ + Cu

(d) H2 + Ni2+---------→ 2H+ + Ni

b

8 The position of some metals in the electrochemical series in decreasing electropositive character is given as Mg > Al > Zn > Cu > Ag. What will happen if a copper spoon

is used to stir a solution of aluminium nitrate ? (a) The spoon will get coated with aluminium

b) An alloy of copper and aluminium is formed (c) The solution becomes blue (d) There is no reaction

d

9 The standard EMF of Daniel cell is 1.10 volt. The maximum electrical work obtained

from the Daniel cell is (a) 212.3 kJ (b) 175.4 kJ (c) 106.15 kJ (d) 53.07 kJ

a

10 A metal having negative reduction potential when dipped in the solution of its own

ions, has a tendency : (a) to pass into the solution (b) to be deposited from the solution (c) to become electrically positive (d) to remain neutral

a

11 Which of the following reactions is correct for a given electrochemical cell at 25oC ?

Pt / Br2 (g) / Br- (aq) // Cl- (aq) / Cl2 (g) / Pt. (a ) 2Br-+ Cl2 ------→ 2Cl- (aq)+ Br2 (g)

(b) Br2 + 2Cl- ------→ 2Br- (aq)+ Cl2 (g) (c)2 Cl- +2Br- ------→ Cl2 (aq)+ Br2 (g )

(d) Br2 +Cl2 (g)-----→2 Br-(aq)+ 2Cl- (g)

a

12 Standard solution of KNO3 is used to make salt bridge because

(a) velocity of K+ is greater than NO3-

(b) velocity of NO3- is greater than K+

(c) velocity of K+ and NO3- are same.

(d) KNO3 is highly soluble in water.

c

13 Na is used in reduction of Zn salt because. (a) E⁰ Zn(oxi)> E⁰ Na(oxi)

(b) E⁰ Zn(red)< E⁰ Na(red) (c) E⁰ Zn(oxi)< E⁰ Na(oxi)

c

33

(d) Both (a) and (b) 14 If hydrogen electrode dipped in two solutions of pH=3 and pH=6 and salt bridge is

connected, the emf of resulting cell is

(a) 0.177 V (b) 0.3 V (c) 0.052 V (d) 0.104 V

a

15 Which of the following solutions of NaCl will have the highest specific conductance?

(a) 0.001 N (b) 0.1 N (c) 0.01 N (d) 1.0 N

d

16 Equilibrium constant K is related to E0cell and not Ecell because

(a) E0cell is easier to measure than Ecell (b) Ecell becomes zero at equilibrium point but E0cell remains constant under all

conditions (c) at a given temperature, Ecell changes hence value of K can’t be measured (d) any of the terms Ecell or E0cell can be used

b

17 Electrical conductance through metals is called metallic or electronic conductance

and is due to the movement of electrons. The electronic conductance depends on (a) the nature and structure of the metal (b) the number of valence electrons per atom

(c) change in temperature (d) all of these

d

18 Faraday’s law of electrolysis is related to (a) Atomic number of cation

(b) Speed of cation (c) Speed of anion

(d) Equivalent weight of electrolyte

d

19 Fused NaCl on electrolysis gives ………….. on cathode.

(a) Chlroine (b) Sodium

(c) Sodium amalgam (d) Hydrogen

b

20 Units of the properties measured are given below. Which of the properties has been not matched correctly?

(a) Molar conductance = Sm2 mol-1 (b) Cell constant = m-1 (c) Specific conductance of = S m²

(d) Equivalence conductance = S m² (g eq)-1

c

21 The charge required for reducing 1 mole of MnO4 −to Mn2- is (a) 1.93 × 105 C (b) 2.895 × 105 C

(c) 4.28 × 105 C (d) 4.825 × 105 C

d

22

b

23 . The cell reaction of the galvanic cell.

d

24 The cell constant of a conductivity cell _____________. a) changes with change of electrolyte. b) changes with change of concentration of electrolyte. c) changes with temperature of electrolyte. d) remains constant for a cell.

d

34

25

Write the correct option for the following questions: (a) Both Assertion and Reason are true and Reason is correct explanation of the Assertion.

(b) Both Assertion and Reason are true and Reason is not the correct explanation of the Assertion.

(c) Assertion is true but Reason is false. (d) Both Assertion and Reason are false.

Assertion: Copper metal gets readily corroded in an acidic aqueous solution. Reason: Free energy change for this process is positive

c

26 Assertion: The cell potential of mercury cell is 1.35 V, which remains constant. Reason: In mercury cell, the electrolyte is a paste of KOH and ZnO.

b

27 Assertion: According to Kohlrausch’s law the molar conductivity of a strong electrolyte at infinite dilution is sum of molar conductivities of its ions. Reason: The current carried by cation and anion is always equal.

c

28 Assertion: Zn gives H2 gas with H2SO4 and HCl but not with HNO3 Reason: NO3

– is reduced in preference to hydronium ion. a

29 Assertion: Galvanised iron does not rust. Reason: Zinc has a more negative electrode potential than iron

2 MARKS QUESTIONS

a

30 Suggest a way to determine ᴧᵒm of water.

Ans: ᴧᵒm (H2O) = ʎᵒ(H+) + ʎᵒ(OH-)

= ʎᵒ(H+) + ʎᵒ(OH-) +ʎᵒ(Na+) - ʎᵒ(Na+) +ʎᵒ(Cl-) - ʎᵒ(Cl-)

= [ʎᵒ(H+) + ʎᵒ(Cl-) ]+ [ ʎᵒ(OH-) +ʎᵒ(Na+)] – [ʎᵒ(Na+) + - ʎᵒ(Cl-)]

ᴧᵒm (H2O) = ᴧᵒm (HCl) + ᴧᵒm (NaOH) -- ᴧᵒm (NaCl)

Thus the molar conductance of water can be determined.

2m

31 Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0X 104 A of

current is passed through molten Al2O3 for 6 hrs , how much Al is produced?

Ans: Aluminium is in +3 oxidation state so the production of 1 mol of Al requires 3 mol of electrons. i.e 3 F = 3X96500 C

Total charge passed = 4.0 X104 A X 6 X 60 X 60 = 8.64 X 108 C Mass of Al produced = 8.64 X 108 X27 / 3X96500

= 8.06 X104 g

2m

32 On the basis of following data given identify the strongest oxidising agent Fe2+ or Fe3+ or [Fe(CN)6]4- or [Fe(CN)6]3-

[Fe(CN)6]4-----------------→ [Fe(CN)6]3- +e- E⁰ = -0.35 V

Fe2+ ----------------------→ Fe3+ +e- E⁰= -0.77V

Ans: The species which have higher reduction potential can act as oxidising agent. Among these only Fe3+ (0.77V) and [Fe(CN)6]3- (0.35 V) have positive reduction

potential. Fe3+ is stronger oxidising agent as Reduction potential of Fe3+ is higher than[Fe(CN)6]3-

2m

33 ᴧᵒm for NaCl, HCl and CH3COONa are 126.0, 426, 1000 Scm2/mol respectively. If the conductance of 0.001 M CH3COOH is 5X10-5 S/cm, calculate the degree of dissociation of CH3COOH.

Ans:

ᴧᵒm CH3COOH = ᴧᵒm CH3COONa + ᴧᵒm HCl - ᴧᵒm NaCl ᴧᵒm CH3COOH = 100+426-126 = 400 Scm2/mol

ᴧcm = 1000к/M = 1000x5X10-5/0.001 = 50 Scm2/mol

α = ᴧcm /ᴧᵒm = 50/400 =0.125 or

α =0.125 x100 = 12.5%

2m

34 The molar conductance of an M/32 solution of weak monobasic acid is 8 mho cm 2 and that at infinite solution is 400 mho cm 2 . Find the dissociation constant of the acid?

Ans: α = M

c/Mꝏ = 8 mho cm 2 /400 mho cm 2 = 0.02

Ka = c α2/1- α =(1/32)x(0.02)2/ 1-0.02 Ka = 1.25 x10-5

2m

35

35 A galvanic cell has two hydrogen electrodes one is immersed in a solution with [H+] = 1M and other is immersed in 0.85 M KOH. Determine Ecell.

Ans: The galvanic cell is Pt/H2 (g, 1 bar)/ H+ (xM)// H+ (1M)/ H2 (g),Pt Where x= Kw/[OH-]

= 1X10-14/0.85 = 1.18 X10-14

Ecell = -0.0592log x = 0.0592 x1.18 X10-14

= 0.82 v

2m

36 Assess whether the following reaction is spontaneous or not? 2KMnO4 + 3H2SO4 + 5 H2C2O4 ---------------→ 2MnSO4 + K2SO4 + 10CO2 + 8H2O

Ans: Separate the reactions in to oxidation and reduction half reactions. +3 +4

Oxidation : H2C2O4-----------------→ 2CO2 + 2H+ +2e- E⁰ox = 0.49 V +7 +2

Reduction: MnO4 - + 8H+ + 5e----------------→ Mn2+ + 4 H2O E⁰red =1.47 V cell E⁰ = E⁰ox + E⁰red

E⁰ = 0.49 +1.47 = 1.98v E⁰ for the overall reaction is greater than zero so spontaneous.

2m

37 What happens if external potential applied is greater than Eᵒcell of the electrochemical cell?

Ans: If the external potential applied is greater than Eᵒcell , the reverse reaction will take place and the cell starts acting like as an electrolytic cell. In this case electrical energy is used to carry out non-spontaneous reaction.

2m

38 If a current of 0.5 amperes flow through a metallic wire for 2 hours, then how many electrons would flow through the wire?

Ans: Q (coulomb)= I (ampere) x t (second) Q = 0.5 x 2 x 60 x 60 = 3600 c

A flow of 96500 C is equivalent to the flow of 1 mole of electrons i.e 6.022x1023 electrons Therefore 3600 C is equivalent to = 6.022x1023x3600 /96500

= 2.246x1022 electrons.

2m

39 The electrical resistance of a column of 0.05 mol L–1 NaOH solution of diameter 1 cm

and length 50 cm is 5.55 × 103 ohm. Calculate its resistivity and conductivity. A = πr2 = 3.14 × 0.52 cm2 = 0.785 cm2 = 0.785 × 10–4 m2 , l = 50 cm

Ans: R = ƿ l/ A or Resistivity ƿ = R x A/l

= 5.55 × 103 x 0.785 / 50 = 87.135 ohm cm Conductivity = 1/ƿ = 1/ 87.135 = 0.01148 S cm

3 MARKS QUESTIONS

2m

40 Given two reactions

i) Cu2+ + 2 e- --------------→ Cu E⁰ Cu2+/Cu = 0.34 V ii) Cu2+ + e- --------------→ Cu+ E⁰ Cu2+/Cu+

=0.153 V

Calculate E⁰ Cu+/Cu for the reaction Cu+ + e- --------------→ Cu

Ans: The given reaction can be obtained by subtracting (ii) from (i) According to Thermodynamics ∆G⁰ is a state function, so ∆G⁰ Cu+/Cu = ∆G⁰ Cu2+/Cu - ∆G⁰ Cu2+/Cu +

-nF E⁰ Cu+/Cu = -(nF E⁰ Cu2+/Cu )- (-nF E⁰ Cu2+/Cu+)

= - (2F x 0. 34 V) + (1F x 0.153 V)

= -0.68F + 0.153 F = - F (0.68 – 0.153) V = -F x 0.527 V

So E⁰ Cu+/Cu = 0.527 V

3m

41 What is the reduction potential at pH =14 for the Cu2+/Cu couple ?

Given E⁰ Cu2+/Cu = 0.34 V , Ksp of [Cu(OH)2] = 1 X1019

Ans: Ksp = [Cu2+] [OH-]2 = 1 X1019

So [Cu2+] = 1 X1019/ [OH-]2 = 1 X1019

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36

From Nernst equation 𝐸cell = E0cell −0.059

2 log 1/Cu2+

E cell = 0.34-0.0295 log 1/1 X1019

E cell = 0.34-0.0295 x 19 = -0.22V 42 i) Calculate Eᵒcell for the following reaction at 298 K:

2Al(s) + 3 Cu2+(0.01M) -----------→ 2Al3+ (0.01M) + 3Cu (s) Given Ecell = 1.98V

ii) Can we store copper sulphate solution in aluminium container? Ans:

i) [Al(s) ----------→ Al3+(aq) + 3e-] x 2 [Cu2+(aq) + 2e- ----------→ Cu(s)] x3

___________________________________________ 2Al(s) + 3 Cu2+(aq) -----------→ 2Al3+ (aq) + 3Cu (s) n= 6

__________________________________________ Substituting in Nernst equation

Ecell = Eᵒ cell - 0.0591

𝑛 log

[Al3+]2

[𝐶𝑢2+]3

1.98 V = Eᵒ cell - 0.0591

6 log

[0.01]2

[𝐴𝑔+0.01]3

Eᵒ cell = 1.98 + 0.0197 = 1.9997 V

ii) No, we cannot store copper sulphate solution in aluminium container. Because as per above reaction Al will displace Cu2+ from its solution.

3m

43 One half cell in a voltaic cell is constructed from a silver nitrate solution of unknown concentration. The other electrode consists of a zinc electrode in a 0.10 M solution of

zinc nitrate. A voltage of 1.48 V is measured for this cell. Use this information to calculate the concentration of silver ions in the solution.

Given than Eᵒ Zn2+/Zn = - 0.763V and Eᵒ Ag+/Ag = +0.80V Ans: Electro chemical cell formed is Zn(s)/Zn2+ (0.010M) // Ag+ (x M) / Ag(s)

Eᵒcell = Eᵒ cathode - Eᵒ anode = 0.80-(-0.763) V = 1.563 V

Nernst equation for the reaction

E cell = Eᵒ cell - 0.0591

𝑛 log

[Zn2+]

[𝐴𝑔+]2

1.48 V= 1.563 v - 0.0591

2 log

[0.10]

[𝐴𝑔+]2

log [0.10]

[𝐴𝑔+]2 = 0.083/0.02955 = 2.8087

[0.10]

[𝐴𝑔+]2 = A.log(2.8087) = 643.7

[Ag+] =1.247 x10-2 M

3m

44 Calculate the cell potential of galvanic cells in which the following reactions take place. Also calculate equilibrium constant and ∆G⁰ of the reaction. 2Cr (s) + 3Cd2+ (aq) ----------------------------→ 2Cr 3+(aq) + 3 Cd(s)

Given E⁰ Cd2+/Cd = -0.40V, E⁰ Cr3+/Cr =-0.74 V

Ans: E⁰cell = E⁰ Cd2+/Cd - E⁰ Cr3+/Cr = -0.40 –(-0.74) V = 0.34 V ∆G⁰ = -nFE⁰Cell = - 6 x 96500 x 0.34 = -196.86kJ/mol

= 196.86 x 10 3 J/mol ∆G⁰ = -2.303 RT log K

log K = ∆G⁰/ 2.303 RT = 196.86 x 10 3 / 2.303 x 8.314 x 298 = 34.5014

K = A.log (34.5014) = 3.16 x1034

3m

45 In the Edison storage cell Fe(s)/KOH(aq) // Ni2O3(s) /Ni , the half cell reaction are as

follows. Ni2O3(s) + H2O(l) + 2e- ----------→ 2NiO(s) + 2OH- E⁰ = = +0.40V

FeO (s) + H2O(l) + 2e- ----------→ Fe(s) + 2OH- E⁰ = = -0.87V

i) What is the cell reaction? ii) What is the emf of the cell? How it depends upon the concentration of KOH? iii) What is the maximum amount of electrical energy obtained from the cell?

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37

Ans: i) The cell reaction is Fe(s) + Ni2O3(s) --------→ FeO (s)+ 2NiO(s)

ii) E⁰cell = E⁰NI2O3/NiO - E⁰ FeO/Fe = 0.40 –(-0.87) V = 1.27 V

iii) The consumption of 1mol of Ni2O3 will deliver 2F charge across the electrodes

under a potential difference of 1.27 V. Maximum amount of electrical energy available from the cell can be calculated from the relation ∆G⁰ = -w

or w = - ∆G⁰ = nF E⁰cell = 2 x 96500 x 1.27

= 245110 J =245.11 kJ Maximum amount of electrical energy available from the cell is 245.11 kJ

46 The emf of hydrogen-oxygen fuel cell is 1.23 V at 25ᵒC. What is the equilibrium

constant for the reaction 2H2 (g)+ O2(g) ↔ 2H2O(l)

Ans: Half cell reaction in hydrogen-oxygen fuel cell are as follows Oxidation: H2 (g) + 2H2O -------→ 2 H3O+ +2e-

Reduction: O2 (g) + 4H2O +4e- ------------→ 4OH-

Over all cell reaction is 2H2 (g)+ O2(g) ↔ 2H2O(l) which involves 4 F.

At 25ᵒC E⁰cell = 0.0591

𝑛 logK or log K = n E⁰cell / 0.0591

log K = 4 x 1.23 / 0.0591 = 83.1081 K = A.log (83.1081) = 1.26 x 1083

3m

47 Two students use the same stock of solution of ZnSO4 and a solution of CuSO4. The

emf of one cell is 0.03V higher than the other cell. The concentration of CuSO4 in the cell with higher emf is 0.5 M. Find the concentration of CuSO4 in the other cell.

Ans: Let the concentration of the stock of ZnSO4 solution be C and concentration of CuSO4 solution be C1 and C2 ( where C2>C1)

The emf of the two cells be E1 and E2

E1 = Eᵒcell - 0.0591

𝑛 log

[Zn2+]

[𝐶𝑢2+] so E1 = Eᵒcell -

0.0591

2 log

[C]

[𝐶1]

E2 = Eᵒcell - 0.0591

𝑛 log

[Zn2+]

[𝐶𝑢2+] so E2 = Eᵒcell -

0.0591

2 log

[C]

[𝐶2]

E2- E1 = 0.0591

2 log

[C2]

[𝐶1] But E2- E1 =0.03 V as given.

0.03 = 0.0296 log [C2]

[𝐶1] so 1= log

[C2]

[𝐶1] or

log [C2]

[𝐶1] = 𝐴. log (1) = 10

[C2]

[𝐶1] =10

C1 = C2/10 =0.05 M

3m

48 Calculate the emf of the following cell when 99.99% of Cu2+ ion is consumed.

Zn(s)/Zn2+ (1M) // Cu2+ (1M) / Cu(s) Eᵒcell = 1.10 V Ans: When 99.99% Cu2+ ion is consumed, its concentration decreases by 0.9999M

and Zn2+ ion concentration increases by 0.9999 M. Final concentration of Cu2+ ion = 0.0001M and Zn2+ ion =1.9999M

Ecell = Eᵒcell - 0.0591

𝑛 log

[Zn2+]

[𝐶𝑢2+] = 1.10 -

0.0591

2 log

1.9999

0.0001

Ecell = 1.10 - 0.0591

2 log (.9999x104)

Ecell = 0.97 V

3m

49 The emf of the cell corresponding to the reaction

Zn(s) + 2H+(aq) -----------→ Zn2+ (aq, 0.1M) + H2(g ,1 atm) Given Ecell = 0.28V

3m

38

Write the half cell reaction and calculate the pH of the solution. Given E⁰ Zn2+/Zn = -0.76V, E⁰ H+/H2 = 0 V

Ans: Anode reaction Zn(s) -----------→ Zn2+ (aq) + 2e- Cathode reaction 2H+(aq) +2e-----------→ H2(g)

Nernst equation for the above reaction

Ecell = Eᵒcell - 0.0591

𝑛 log

pH2 [Zn2+]

[𝐻+]2

Ecell = 0.76 V - 0.0591

2 log

1 atm [0.1[]

[𝐻+]2

Ecell = 0.76 V - { 0.0591

2 log (1 x0.1) -

0.0591

2 log [H+]2 }

0.28 = 0.76 V - { 0.0591

2 log (1 x0.1) -2 X

0.0591

2 log [H+] }

0.28 = 0.76 –(-0.03) -0.0591pH

0.28 = 0.79 -0.0591 pH

pH = (0.79-0.28) /0.0591 = 8.6

5 MARKS QUESTION

50 (a) What is limiting molar conductivity? Why there is steep rise in the molar conductivity of weak electrolyte on dilution? (b) Calculate the emf of the following cell at 298 K :

Mg (s) | Mg2+ (0.1 M) || Cu2+ (1.0 × 10-3M) | Cu (s) [Given = E0 Cell = 2.71 V].

Ans: (a) The molar conductivity of a solution at infinite dilution is called limiting molar

conductivity and is represented by the symbol Λ∘m

There is steep rise in the molar conductivity of weak electrolyte on dilution because as the concentration of the weak electrolyte is reduced, more of it ionizes and thus

rapid increase in the number of ions in the solution.

5m

51 Consider the figure given and answer the following questions :

(i) What is the direction of flow of electrons? (ii) Which is anode and which is cathode? (iii) What will happen if the salt bridge is removed?

5m

39

(iv) How will concentration of Zn2+ and Ag+ ions be affected when the cell functions? (v) How will concentration of these ions be affected when the cell becomes dead?

Ans: (i) Electrons flow from anode (Zinc plate) to cathode (Silver plate).

(ii) Zinc plate where oxidation occurs acts as anode and silver plate where reduction occurs acts as cathode.

(iii) If the salt bridge is removed then electrons from zinc electrode will flow to the silver electrode where they will neutralize some of Ag+ ions and the SO4

2− ions will be

left and the solution will acquire a negative charge. Secondly the Zn2+ ions from zinc plate will enter into ZnSO4 solution producing positive charge. Thus due to

accumulation of charges in two solutions, further flow of electrons will stop and hence the current stops flowing and the cell will stop functioning.

(iv) As silver from silver sulphate solution is deposited on the silver electrode and sulphate ions migrate to the other side, the concentration of Ag 2SO4 solution

decreases and of ZnSO4 solution increases as the cell operates. (v) When the cell becomes dead, the concentration of these ions become equal due

to attainment of equilibrium and zero EMF.

52 (a) Calculate E0cell for the following reaction at 298 K:

2Al(s) + 3Cu2+ (0.01M) → 2Al2+ (0.01M) + 3Cu(s)

Given: Ecell = 1.98 V

(b) 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 (All India 2016)

Ans: (a) For the reaction 2Al(s) + 3Cu2+ (0.01M) → 2Al3+ (0.01M) + 3Cu(s) Given: Ecell = 1.98 V E0

cell = ?

Nernst equation for the reaction is

(b) Element A will be better for coating the surface of iron than element B

because its E° value is more negative.

5m

53 One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration. Its other half-cell consists of a zinc electrode dipping in 1.0 M solution of Zn(NO3)2. A voltage of 1.48 V is measured for this cell.

Use this information to calculate the concentration of silver nitrate solution used. (E0

Zn2+/Zn=−0.76V,E0Ag+/Ag = + 0.80 V)

Ans:In this silver electrode act as cathode and zinc electrode act as anode. Half cell reactions take place at silver and zinc electrode are as follows

5m

40

41

CHAPTER-4 CHEMICAL KINETICS

SN CASE STUDY BASED Questions

1 Observe the following graphs and answer the questions based on these graphs.

(a) What is order of reaction shown in graph I?

(b) What is slope in graph II?

5

42

(c) How does t1/2 varies with initial concentration in zero order reaction. (d) If t1/2 of first order reaction is 40 minute, what will be t99.9% for first order

reaction? (e) What is t1/2 of zero order reaction in terms of‘k’?

2 Observe the table given showing volume of CO2 obtained by reaction of CaCO3 and dilute HCl after every minute. Answer the questions that follow:Table showing volume of CO2 at one minute interval by reaction of CaCO3 with dilute HCl.

(a) What happens to rate of reaction

with time? (b) Why does CaCO3 powder react

faster than marble chips? (c) What happens to rate of reaction if concentrated HCl is used?

(d) In manufacture of NH3, N2(g) + 3H2(g) →2NH3 + heat what is effect of

pressure on rate of reaction? (e) Why does rate of reaction becomes

almost double for energy 10° rise in temperature?

5

CASE STUDY BASED Questions type 2

3 Order of a reaction is an experimentally determined quantity. It corresponds to the

stoichiometric coefficients only for an elementary reaction (a reaction which occurs in just one step). A complex reaction occurs in a series of elementary reactions (multiple steps). Some steps are very fast as they do not require much energy. These

steps do not affect the overall rate of reaction. Hence the rate of reaction is mainly determined by the slowest step of the reaction. Order of a reaction, in case of a

complex reaction, corresponds to the stoichiometric coefficients of this rate determining step. (i) Both Assertion and Reason are correct and the Reason is correct explanation

of Assertion. (ii) Both Assertion and Reason are correct but Reason does not explain Assertion.

(iii) Assertion is correct but Reason is incorrect. (iv) Both Assertion and Reason are incorrect. (v) Assertion is incorrect but Reason is correct.

(a) Assertion : Order and molecularity are same. Reason : Order is determined experimentally and molecularity is the sum of the

stoichiometric coefficient of rate determining elementary step.

1

(b) Assertion : Order of the reaction can be zero or fractional.

Reason : We cannot determine order from balanced chemical equation.

1

(c)Assertion : The order of reaction may be negative.

Reason : In some cases, the rate of the reactions decreases as the concentration of the reactions increases.

1

(d) Assertion : Molecularity has no meaning for a complex reaction. Reason : The overall Molecularity of a complex reaction is equal to the

Molecularity of the slowest step.

1

4 The rate constant is defined as the proportionality constant which explains the relationship between the molar concentration of the reactants and the rate of a

chemical reaction. The rate constant is denoted by k and is also known as reaction rate constant or reaction rate coefficient. It is dependent on the temperature. There

are two possible ways to calculate rate constant and they are: Using the Arrhenius equation. Using the molar concentrations of the reactants and the order of the

reaction

(i) Both Assertion and Reason are correct and the Reason is correct explanation

of Assertion. (ii) Both Assertion and Reason are correct but Reason does not explain Assertion.

(iii) Assertion is correct but Reason is incorrect. (iv) Both Assertion and Reason are incorrect. (v) Assertion is incorrect but Reason is correct.

a Assertion : Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules.

Reason : Reactant molecules undergo chemical change irrespective of their orientation during collision.

1

b Assertion : The numerical value of specific rate constant is independent of the concentration of any species present in the reaction mixture

1

43

Reason : When a reaction is carried out in aqueous solution and some alcohol is added to the reaction mixture, the rate of reaction will not change

c Assertion : The numerical value of specific rate constant is independent of the

concentration of any species present in the reaction mixture Reason : When a reaction is carried out in aqueous solution and some alcohol is

added to the reaction mixture, the rate of reaction will not change

1

d Assertion: If the activation energy of a reaction is zero, temperature will have no effect on the rate constant. Reason: Lower the activation energy, faster is the reaction.

1

5 The rate law for a reaction between the substances A and B is given by Rate = k[A]n[B]m. On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

1

6 In respect of the equation k = Ae-Ea/RT in chemical kinetics, which one of the following

statements is correct? (1)A is adsorption factor (2)Ea is energy of activation

(3)R is Rydberg’s constant (4)k is equilibrium constant

1

7 For the equilibrium, A(g) ⇌ B(g), ΔH is –40 kJ/mol. If the ratio of the activation

energies of the forward (Ef) and reverse (Eb) reactions is 2/3 then: (1) Ef = 60kJ / mol; Eb = 100 kJ/mol

(2) Ef = 30kJ / mol; Eb = 70 kJ/mol (3) Ef = 80kJ / mol; Eb = 120 kJ/mol (4) Ef = 70kJ / mol; Eb = 30 kJ/mol

1

8 The rate of a chemical reaction doubles for every 10°C rise of temperature. If the

temperature is raised by 50°C, the rate of the reaction increases by about (1) 32 times (2) 64 times (3) 10 times (4) 24 times

1

9 At 5180 C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 s-1 when 5% had reacted and 0.5 Torr s-1 when

33% had reacted. The order of the reaction is : (1) 3 (2) 1 (3) 0 (4) 2

1

10 For the reaction A + 2B → C, rate is given by R = [A] [B]2 then the order of the reaction is

(1) 3 (2) 6 (3) 5 (4) 7

1

11 If 50% of a reaction occurs in 100 seconds and 75% of the reaction occurs in 200 seconds, the order of this reaction is

(1) 1 (2) 2 (3) zero (4) 3

1

12 The half-life period of a first-order chemical reaction is 6.93 minutes. The time

required for the completion of 99% of the chemical reaction will be (log 2 = 0.301): (1) 46.06 minutes (2) 460.6 minutes (3) 230.3 minutes (4) 23.03 minutes

1

13 The time for half-life period of a certain reaction A → Products is 1 hour when the initial concentration of the reactant ‘A’ is 2.0 mol L–1 , How much time does it take

for its concentration to come from 0.50 to 0.25 mol L–1 if it is a zero-order reaction? (1) 1 h (2) 4 h (3) 0.5 h (4) 0.25 h

1

14 Decomposition of X exhibits a rate constant of 0.05 mg/year. How many years are required for the decomposition of 5 mg of X into 2.5 mg?

(1) 25 (2) 50 (3) 20 (4) 40

1

15 In a first-order reaction, the concentration of the reactant, decreases from 0.8 M to

0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is (a) 30 minutes (b) 15 minutes (c) 7.5 minutes (d) 60 minutes.

1

16 The rate law for the reaction below is given by the expression k[A] [B] A + B → Product If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be (1) 3 k (2) 9 k (3) k/3 (4) k

1

17 For the reaction, 2A + B → products, when the concentrations of A and B both were doubled the rate of the reaction increased from 0.3 mol L–1 s –1 to 2.4 mol L–1 s –1 . When the concentration of A alone is doubled, the rate increased from 0.3 mol L–1 s –1 to 0.6 mol L–1 s –1 . Which one of the following statements is correct?

(1) Order of the reaction with respect to B is 2. (2) Order of the reaction with respect to B is 1.

(3) Order of the reaction with respect to A is 2. (4) Total order of the reaction is 4.

1

44

18 The formation of gas at the surface of tungsten due to adsorption is the reaction of order

(1) 0 (2) 1 (3) 2 (4) insufficient data.

1

19 A reaction was found to be second order with respect to the concentration of carbon

monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will be (1) remain unchanged (2) tripled (3) increased by a factor of 4 (4) doubled.

1

20 t1/4 can be taken as the time taken for the concentration of a reactant to drop to 3/4 of its initial value. If the rate constant for a first order reaction is K, the t1/4 can be

written a (1)10 / K (2) 0.29/K (3) 0.69 / K (4) 0.75 / K

1

21 A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else

kept the same, the rate of reaction will (2006) (1)remain unchanged (2) triple (3) increase by a factor of 4 (4) double

1

22 Consider the reaction, 2A+B —> Products. 5 When concentration of B alone was doubled, i. the half-life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction

is (1) L mol-1 s-1 (2) No unit (3) molL-1 sy1 (4) s-1

1

23 Consider an endothermic reaction, X -→Y with the activation energies Eb and Ef for the backward and forward reactions, respectively. In general

(1)Eb < Ef (2) Eb > Ef (3)Eb = Ef (4)There is no definite relation between Eb and Ef

1

24 A catalyst alters, which of the following in a chemical reaction? (1)Entropy (2)Enthalpy (3)Internal energy (4)Activation energy

1

25 A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room?

(1) 1000 days (2) 300 days (3) 10 days (4) 100 days

1

a) Assertion (A) is True, Reason (R)is True; Reason (R)is correct explanation

for Assertion (A) b) Assertion (A) is True, Reason (R) is True; Reason (R) is not correct

explanation for Assertion (A) c) Assertion (A) is True, Reason (R) is False d) Assertion (A) is False, Reason (R) is True

26 Assertion (A): In the reaction I2+2S2O32-→S4O6

2-+2I-, the two S2O32-ions are used

for every I2. Reason (R): The rate of disappearance of I2 is one half the rate of disappearance of 2S2O3

2-

1

27 Assertion (A): The enthapy of reaction remains constant in the presence of a catalyst.

Reason (R): A catalyst participating in the reaction, forms different activated complex and lowers down the activation energy but the difference in energy of

reactant and product remains the same.

1

28 Assertion (A): Hydrolysis of cane sugar is a first order reaction.

Reason (R): Water is present in large excess during hydrolysis.

1

29 Assertion : For a first order reaction the plots of rate vs. concentration is straight

line Reason (R): For a first order reaction, Rate=k[A]

1

30 Assertion (A): For the reaction 2N2O5 → 4NO2 + O2 Rate = k[N2O5] Reason (R): Rate of decomposition of N2O5 is determined by slow step.

1

2 MARKS QUESTIONS

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

2

32 (a)For a reaction, A + B →Product, the rate law is given by, Rate = k [A]1 [B]2 . What is the order of reaction? (b) Write the unit of rate constat ‘k’ for the first order reaction.

2

33 Define half-life of a reaction. Write the expression of half-life for (i) zero order reaction and (ii) first order reaction

2

34 Define the following terms: (i) Pseudo first order reaction (ii) Half life period of reaction (t1/2)

2

35 (а) For a reaction A + B →P, the rate law is given by r = k [A1/2 ] [A]2 What is the order of this reaction?

(b) A first order reaction is found to have a rate constant k = 5.5 X 10-14 s-1. Find the half life of the reaction.

2

45

36 The rate constant for a first order reaction is 60 s 1. How much time will it take to reduce the initial concentration of the reactant to 1/10 th of its initial value?

2

37 i) If half life period of first order reaction is x and 3/4 th life period of same reaction is ’y’, how are V and ’y’ related to each other?

(ii) In some cases it is found that a large number of colliding molecules have energy more than threshold energy but yet the reaction is slow. Why?

2

38 The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature. [R = 8.314 J K-1 mol-1, log 4 = 0.6021]

2

39 A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to

2

40 Distinguish between ‘rate expression’ and rate constant’ of a reaction. 2

41 For the given reaction A +B→ Product (a)Write the rate equation (b)Calculate the rate constant

Following data are given

2

42 Rate of reaction, A + B →Product is given below as a function of different initial

concentration of A and B

(a)Determine the order of the reaction with respect of A and B. (b)What is the half-life of A in the reaction

2

43 One of the hazards of nuclear explosion is the generation of Sr90 and its subsequent incorporation in bones. This nuclide has a half life of 28.1 year. Suppose one

microgram was absorbed by a new born-child, How much Sr90 will remain in his bones after 20 year

2

44 At 3800C,the half-life period for the first order decomposition of H2O2 is 360 min.The energy of activation of the reaction is 200 KJmol-1.Claculate the time required for 75% decomosition at 4500C

2

3 MARKS QUESTIONS

45 Rate constant is equal to the rate of reaction when molar cone, of reactants is unity. Its unit depends upon order of reaction. A reactant has a half-life of 10 minutes. (i) Calculate the rate constant for the first order reaction.

(ii) What fraction of the reactant will be left after an hour of the reaction has occurred?

3

46 A first order reaction takes 10 minutes for 25% decomposition. Calculate t1/2 for the

reaction. (Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021).

3

47 The rate constant of a first order reaction increases from 2 X 10-2 to 4×10-2 when the

temperature change from 300Kto310K. Calculate the energy of activation (Ea). (log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)log 3 = 0.4771, log 4 = 0.6021)

3

48 The following data were obtained during the first order thermal decomposition of S02Cl2 at a constant volume: Calculate the rate constant. (Given: log 4 = 0.6021,

log 2 = 0.3010)

3

49 A first order gas phase reaction: A2B2(g) → 2A(g) + 2B(g) at the temperature 400°C has the rate constant k = 2.0 × 10-4 sec-1. What percentage of A2B2 is decomposed on heating for 900 seconds? (Antilog 0.0781 = 1.197)

3

50 A first or der reaction is 20% complete in 10 min. Calculate (i) the specific rate constant of the reaction, and

(ii) Time taken for the reaction to go to 75% completion.

3

51 Radioactive de cay is a first or der process. Radioactive carbon in wood sample de cays with a half-life of 5770 year. What is the rate constant (in year-1) for the de cay?

What fraction would remain after 11540 year?

3

46

52 In Arrhenius equation for a certain reaction, the value of A and Ea(activation energy) are 4 X1013 s-1 and 98.6 kJ mol–1respectively. If the reaction is of first or der, at what

temperature will its half-life period be 10 min?

3

53 The rate constant for an isomerisation re action, A → B is 4.5X10-3 min. If the initial

concentration of A is 1M. Calculate the rate of the reaction after 1 h.

3

54 A first ordr reaction is 20% complete in 10 min.Calculate

(i)the specific rate constant of the reaction (ii)the time taken for the reaction to go to 75% completion,

3

5 MARKS QUESTIONS

55 (a)What do you mean by pseudo first order reaction? Explain with example (b)For a reaction: H2 + Cl2 (hv)⟶ 2HCl Rate = k

(i) Write the order and molecularity of this reaction. (ii) Write the unit of k.

5

56 The rate of first or der reaction is 0.04 mol L s–1 –1 at 10 min and 0.03 mol L s–1 –1 at 20 min after initiation. Find the half-life of the reaction.

5

57 For the hydrolysis of methyl acetate in aqueous solution, the following result were

obtained. (i) Show that it follows pseudo first order reaction, as the concentration of water

remains constant. (ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.

(Given: log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)

5

58 (a)For a reaction A + B → P, the rate is given by Rate = k[A]2 [B]

(i) How is the rate of reaction affected if the concentration of A is doubled? (ii) What is the overall order of reaction if B is present in large excess?

(b) A first order reaction takes 23.1 minutes for 50% completion. Calculate the time required for 75% completion of this reaction. (Given: log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)

5

59 The rate constant for the first or der de com position of a certain reaction is described by the equation

(i) What is the energy of activation for the reaction? (ii) At what temperature will its half-life period be 256 min?

5

ANSWER

SN VALUE POINT

1 (a) Zero order reaction. (b) k /2.303where ‘k’ is rate constant. (c) t1/2 is directly proportional to initial concentration. (d) t99.9% = 10 t1/2 = 10 × 40 = 400 minutes

(e) t1/2 = [R]0 /2k for zero order reaction.

2 (a) The rate of reaction first decreases with time then becomes

constant. (b) CaCO3 powder has more surface area than marble chips therefore, more rate of

reaction. (c) The rate of reaction will increase because rate of reaction increases with the increase in concentration.

(d) The rate of reaction increases with increase in pressure. (e) It is because number of molecules undergoing effective collisions become almost

double, hence rate of reaction almost doubled.

3 Question A B C D

Answer i ii i ii

4 Question A B C D

Answer iii ii i ii

Answer(MCQ)

Q A Q A Q A Q A Q A Q A

5 2 9 4 13 4 17 1 21 3 25 4

6 2 10 1 14 2 18 1 22 1

7 3 11 1 15 1 19 3 23 1

8 1 12 1 16 4 20 2 24 4

Answer(Assertion and Reason)

26 27 28 29 30

b a a a a

31 It is because the concentration of reactants goes on decreasing with time

32 The sum of powers to which concentration terms are raised in rate law or rate equation.

47

33 The extra energy which must be supplied to the reactants so that they can undergo effective collision to form products.

34 It is equal to the rate of reaction when molar concentration of reactant are taken as unity

35 (i)s-1is the unit of first order rate constant. (ii) L mol-1 s-1is the unit of second order rate

constant.

36 The change in the concentration of any one of the reactants or products per unit time is

called rate of a reaction

37

38 S-1 is the unit for rate constant of first order reaction.

39 Mol L-1 S-1

40 The t1/2 of a first order reaction is independent of initial concentration of reactants.

41

42

43

44

45

48

46

47

48

49 (a)Those reactions which are not truly of the first order but under certain conditions become first order reactions are called pseudo first order reaction.eg Hydrolysis of Ester (i)This reaction is zero order reaction and molecularity is two. (ii) Unit of k = mol L-1 s-1

50

51

49

52

53

54

55 (a)Those reactions which are not truly of the first order but under certain conditions become

first order reactions are called pseudo first order reaction.eg Hydrolysis of Ester (i)This reaction is zero order reaction and molecularity is two. (ii) Unit of k = mol L-1 s-1

56

57

50

58

59

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CHAPTER-5 SURFACE CHEMISTRY

52

ADSORPTION ISOTHERM

CASE STUDY BASED QUESTIONS:1

1. Read given passage and answer the questions that follow:

Glucose forms true solution in water given to patients. Suspension is

heterogeneous, particles can be separated by filtration. Medicines are more in

colloidal form. Lyophilic sols are more stable than lyophobic sols which need

which need protective like starch, gelatin. Sky is blue in colour due to Tyndall

effect. Brownian movement is responsible. For stability of colloids. Electric

chimney in kitchen and industries is based on electrophoresis. Milk is oil in water

emulsion whereas liquefied butter is water in oil emulsion. Emulsifier stabilise

emulsion whereas change in pH, heating leads to coagulation. Lyophobic sols

are coagulated by electrolyte. Greater the charge on ion, more effective it will

be for coagulation, is Hardy Schulze Rule.

(i)which is most effective to coagulate AgI/Ag+ sol.

(a)MgCl2 (b) K2SO4, (c)NaCl (d) K4[Fe(CN)

Ans. (d)K4[Fe(CN)6

(ii) Hardening of lather is done by tannin. What is process called?

(a) Mutual coagulation (b) Peptization (c) Dialysis (d) Sorption

Ans .(a) Mutual coagulation

(iii) What is charge on colloidal solution of haemoglobin?

(a)Negatively charged (b) Neutral (c) Positively charged (d)No charge

Ans. (c) Positively charged.

(iv) Name the protective colloid used in ice cream.

(a)Sugar (b)Glucose (c) Butter (d) Gelatin

Ans(d). Gelatin

II. Read the following passage and answer the questions that follow:

Surface chemistry deals with phenomenon that occur at the surfaces. Corrosion

electrode processes, heterogeneous catalysis, dissociation and crystallisation

occur at the surface. Evaporation is a surface phenomenon. Adsorption takes

place at the surface. Easily liquefiable gas are more easily adsorbed on the

surface of catalyst. Gas mask contain activated charcoal to adsorb poisonous

gases.

(i) Which gas will absorb to maximum extent

(a) H2 (b)N2, (c)CO (d) NH3 Ans (d)

2 .Which of the following interface cannot be obtained ?

(a) liquid-liquid (b)solid -liquid (c) liquid -gas (d)gas-gas Ans (d)

3. Which of the following process does not occur at the interface of phases ?

(a) crystallsation (b) corrosion

(c) Heterogenous catalysis (d) homogeneous catalysis Ans (c)

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4. In physorption ,adsorbent does not show specificity for any particular gas

because………

(a) involvedvander Waals forces are universal

(b)Gases involved behave like ideal gases

(c)enthalpy of adsorption is low

(d) it is areversible process. Ans (a)

CASE STUDY BASED QUESTIONS :2 Earlier the term “Colloid” was used for a category of substance..However later

the termcolloidal state of matte preferred.Colloidal dispersion have been

classified into different types depending upon the physical state of dispersion phase and dispersion medium orthe nature of the interactions between them or the nature of the colloidal particlesThey are prepared in the industry or in the

laboratory by number of methods and purified In these questions a statement of Assertion is followed by a statement of Reason. Choose the correct option.

(a) Assertion and Reason both are correct and Reason is correct explanation of Assertion.

(b) Assertion and Reason both are correct but Reason is not correct explanation of Reason

(c) Assertion is correct but Reason is Incorrect

(d) Assertion is Incorrect but Reason is correct

(e) Both Assertion and Reason are incorrect

1.Assertion :Lyophilic colloids are more stable than lyophobic colloids.

Reason : In lyophobic system,the dispersed particles are more solvated than in

lyophilic system. Ans : (c)

2 Assertion :Micelles are formed by surfactant molecules above the micellar

concentration (CMC)

Reason : Thte conductivity of a solution having sufficient molecules decreases sharply

at the CMC . Ans : (b)

3. Assertion : Small Quantity of soap is used to prepare a stable emulsion.

Reason :Soaps lowers the interfacial tension between oil and water . Ans (a)

4. Assertion : Gold sol may be positive ,negative or neutral depending upon its

method of preparation.

Reason :Gold sol may be obtained by electro-disintegration as well as by reduction of

AuCl3 with SnCl2 or hydrazine. Anc (d)

CASE STUDY BASED QUESTIONS :2

Colloidal solutions when prepared ,generally contains excessive amount of impurities.While the prescence of traces of electrolyte is essential for the stability of the colloidal solution ,larger quantities coagulate it .It is therefore

necessary to reduce the concentration of these soluble impurities to a requisite minium.The colloidal system show intresting optical,mechanical and electrical

properties. In these questions a statement of Assertion is followed by a statement of Reason. Choose the correct option.

(a) Assertion and Reason both are correct and Reason is correct explanation of Assertion.

(b) Assertion and Reason both are correct but Reason is not correct explanation of Reason

(c) Assertion is correct but Reason is Incorrect

(d) Assertion is Incorrect but Reason is correct

(e) Both Assertion and Reason are incorrect.

54

1 Assertion : An ordinary filter paper impregnated with colloidan solution stops the

flow ofcolloidal particles.

Reason :Pore size of the filter paper becomes more than the size of the colloidal

particle . Ans (c)

2. Assertion ;Colloidal solutions show colligative properties .

Reason :Colloidal particles are larger in size Ans (b)

3. Assertion : Colloidal solutions do not show Brownian motion .

Reason :Brownian motion is responsible for stability of sols Ans : (d)

4 Assertion:The conversion of fresh precipitate to colloidal state is called peptization

Reason :It is caused by addition of common ions Ans (b)

Multiple Choice Questions.

Question 1. 3 g of activated charcoal was added to 50 ml of acetic acid

solution(0.06N)in a flask. After an hour it was filtered and strength of the filtrate was found to be 0.042N.The amount of acetic acid absorbed(per gram of charcoal)is (a) 42 mg (b) 54 mg

(c) 18 mg (d) 36 mg Ans: (c)

Question 2. The term ‘sorption’ stands for (a) absorption (b) adsorption

(c) both absorption and adsorption (d) desorption Ans: (c)

Question 3. Extent of physisorption of a gas increases with-

(a) increase in temperature (b) decrease in temperature

(c) decrease in surface area of adsorbent (d) decrease in strength of van der Waals forces Ans: (b)

Question 4. Extent of adsorption of adsorbate from solution phase increases with (a) increase in amount of adsorbate in solution

(b) decrease in surface area of adsorbent (c) increase in temperature of solution

(d) decrease in amount of adsorbate in solution Ans: (a)

Question 5. Physical adsorption of a gaseous species may change to chemical

adsorption with (a) decrease in temperature

(b) increase in temperature (c) increase in surface area of adsorbent

(d) decrease in surface area of adsorbent Ans: (b) Question 6. The coagulation values in milimoles per litre of the electrolytes used for

the coagulation of As2O3 are given below: I (NaCl) = 52, II (BaCl2)=0.69, III (MgSO4) =0.22

The correct order of their coagulation power is (a) III>II>I (b)III>I>II (c)I>II>III (d) II>I>III Ans (a)

Question 7.. Which of the following is an example of absorption?

(a) Water on silica gel (b) Water on calcium chloride (c) Hydrogen on finely divided nickel

(d) Oxygen on metal surface Ans: (b) Calcium chloride absorbs water. Other examples show adsorption.

Question 8. At high concentration of soap in water, soap behaves as (a) molecular colloid (b) associated colloid

(c) macromolecular colloid (d) lyophilic colloid . Ans (b)

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Question 9. Which of the following will show Tyndall effect? (a) Aqueous solution of soap below critical micelle concentration. (b) Aqueous solution of soap above critical micelle concentration.

(c) Aqueous solution of sodium chloride. (d) Aqueous solution of sugar. Ans: (b)

Question 10. Method by which lyophobic sol can be protected?

(a) By addition of oppositely charged sol (b) By addition of an electrolyte (c) By addition of lyophilic sol

(d) By boiling Ans: (c)

Question 11. Which of the following options are correct? (a) Micelle formation by soap in aqueous solution is possible at all temperatures. (b) Micelle formation by soap in aqueous solution occurs above a particular

concentration. (c) On dilution of soap solution micelles may not revert to individual ions.

(d) Soap solution behaves as a normal strong electrolyte at all concentrations. Ans: (b ) .

Question 12. Which of the following process is not responsible for the presence of electric charge on the sol particles?

(a) Electron capture by sol particles (b) Adsorption of ionic species from solution (c) Formation of Helmholtz electrical double layer

(d) Absorption of ionic species from solution Ans: (d)

Question 13. Which of the following process is responsible for the formation of delta at a place where rivers meet the sea?

(a) Emulsification (b) Colloid formation (c) Coagulation (d) Peptisatiotion Ans: (c)

Question 14. Colloidion is 4% solution of which one of the following in alcohol-ether mixture.

(a) Nitroglycerin (b) Cellulose acetate (c) Glycol dinitrate (d) Nitrocellulose Ans: (d)

Question :15 Given: GAS : H2 CH4 CO2 SO2

Critical temperature /K 33 190 304 630 On the basis of the data given above predict which of the following gases shows

least adsorption on a definite amount of charcoal. (a)H2 (b)CH4 (c)CO2 (d) SO2 Ans(a)

Q16 In Freundlich adsorption isotherm ,the value of 1/n is (a)between 0 and 1 in all cases

(b)between 2 and 4 in all cases . (c) 1in case of physical adsorption

(d) 1 in case of chemisirption Ans(a) Q16 Which one of the following characterstic is associated with adsorption

(a) ΔG and ΔH are negative but ΔS is positive. (b) ΔG and ΔS are negative but ΔH is positive

(c) ΔG is negative but ΔS and ΔH are positive (d) ΔG , ΔS and ΔH all are negative. Ans (d)

Q 17.Fog is a colloidal solution of (a) solid in gas (b) gas in gas

(c) liquid in gas (d) gas in liquid Ans : (c) Q18.Which of the following is correctly matched

(a)Emulsion-soap (b)Gel-butter (c)Foam-mist (d)Aerosol-hair cream Ans (b)

Q19.When O2 is absorbed on a metallic surface,electron transfer occurs from the metal to O2 The true statement(s) regarding this adsorption is (are)π*

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(a) O2 is physisorbed (b)heat is released (c)occupancy of π*2p of O2 is decreased (d)bond length of O2 is decreased Ans (b)

Q20.The dispersed phase and dispersion medium in soap lather are respectively

(a) gas and liquid (b) liquid an gas (c) solid and gas (d)solid and liquid (e)gas and solid

Ans: (d)

ASSERTION -REASON TYPE A statement of Assertion is followed by a statement of Reason. Mark the correct choice from the options given below:

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion. (b) Both Assertion and Reason are true but Reason is not the correct explanation

of Assertion. (c) Assertion is true but Reason is false.

(d) Both Assertion and Reason are false. 1. 1. Assertion : Alcohols are dehydrated to hydrocarbons in the presence of zeolite. Reason : Zeolite are porous catalyst ( Ans - b)

2. Assertion : Isoelectric point is the pH at which colloidal particles can move towards either of the electrode.

Reason : At the isoelectric point colloidal particles may carry positive or negative charge. (Ans - d)

3. Assertion : Hydrolysis of ester is an example of auto - catalytic reaction. Reason : A catalyst speeds up the process without participating in the mechanism. ( Ans - c)

4. Assertion : Hydrated ferric oxide can be easily coagulated by sodium phosphate in comparison to KCl.

Reason : Phosphate ions has higher negative charge than chloride ions. Hence, they are more effective for coagulation. ( Ans - a) 5.Assertion :In chemisorptions adsorption keeps on increasing on increasing with

temperature. Reason : Heat keeps on providing more and more actvation energy .

( Ans - d)

2 Marks Question – answer

Q1. Differentiate between lyophilic and lyophobic colloids. Ans. a) Lyophilic sols are easily prepared by directly mixing with the liquid dispersion

medium but lyophobic sols cannot be prepared directly by mixing with liquid. b)lyophilic sols are stable and are not easily coagulated but lyophobic sols can be

easily precipitated by the addition of suitable electrolyte. 2. Why is adsorption always exothermic in nature?

Ans :When a gas is adsorbed on the surface of a solid, its entropy decreases and S becomes negative. Now G = H- TΔ S and for the process to be spontaneous, free

energy change must be negative. As TΔ S is negative i.e. -TΔ S is positive and for free energy change to be negative enthalpy change should be negative hence reaction should be exothermic always

3. Why is it essential to wash the precipitate with water before estimating it

quantitatively? . Ans : Some amount of the electrolytes mixed to form the precipitate remains adsorbed on the surface of the particles of the precipitate. Hence, it is essential to wash the

precipitate with water to remove the sticking electrolytes or any other impurity before estimating it quantitatively.

4. Comment on the statement that “colloid is not a substance but state of a substance.” .

Ans: Some amount of the electrolytes mixed to form the precipitate remains adsorbed on the

surface of the particles of the precipitate. Hence, it is essential to wash the precipitate with water to remove the sticking electrolytes or any other impurity before estimating it quantitatively.

5. Why is the ester hydrolysis slow in the beginning and becomes faster after some

time?

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Ans : RCOOR’ + H2O ---------------------→RCOOH + R’OH

The acid produced in above reaction act as auto catalyst for the reaction. Hence the reaction becomes faster after some time.

6 .Write any four points of differences between phyisorption and chemisorptions Ans

Physical adsorption or physorption Chemical adsorption of chemisorption

1. It arises because of van der Waals’

forces. 2. It is not specific in nature.

3. It is reversible in nature. 4. Enthalpy of adsorption is low (20-40 kJ mol–1)

5. Low temperature is favourable for adsorption. It decreases with increase of

temperature. 6. It results into multimolecular layers

1. It is caused by chemical bondformation.

2. It is highly specific in nature. 3. It is irreversible.

4. Enthalpy of adsorption is high(80-240 kJ mol–1 5. High temperature is favourable for

adsorption. It increases with the increase of temperature.

6. It results into unimolecular layer.

7. Explain the term associated colloids (Micelles). Ans :There are some substances which at low concentrations behave as normal strong

electrolytes, but at higher concentrations exhibit colloidal behaviour due to the formation of aggregates. The aggregated particles thus formed are called micelles.

These are also known as associated colloids. Surface active agents such as soaps and synthetic detergents belong to this class.

These colloids have both lyophobic and lyophilic parts. Micelles may contain as many as 100 molecules or more

8.Explain Shape- Selective Catalysis by Zeolites. Ans:The catalytic reaction that depends upon the pore structure of the catalyst and

the size of the reactant and product molecules is called shape-selective catalysis. Zeolites are good shape-selective catalysts. ZSM-5. It converts alcohols directly into

gasoline (petrol) by dehydrating them to give a mixture of hydrocarbons. 9 Write two differences between multimolecular colloids and macromolecular

colloids Ans

Multimolecular Macromolecular

On dissolution, a large number of

atoms or smaller molecules of a substance aggregate together to

form species having size in the colloidal range (diameter<1nm). For example, a gold sol, a

Sulphur sol. Generally lyophilic.

Macromolecules in suitable solvents form

solutions in which the size of the macromolecules may be in the colloidal

range. These colloids are quite stable and resemble true solutions in many respects. Examples starch, cellulose, proteins and

enzymes; polythene, nylon, polystyrene, synthetic rubber, etc. Generally lyophobic.

10.State Hardy – Schulze Rule.

Ans :Hardy – Schulze Rule: coagulation of the sols is directly proportion to valency of the ions carrying opposite charge. In the coagulation of a negative sol (e.g. As2S3), the flocculating power is in the order:

Al3+> Mg2+/Ba2+> Na+ Similarly, in the coagulation of a positive sol (e.g. Fe(OH)3), the flocculating power is

in the order: [Fe(CN)6]4–> PO4

3-> SO42-> Cl-

3 Marks Question – answer

1. Explain what is observed when

(i) An electrolyte, NaCl is added to hydrated ferric oxide sol. (ii) Electric current is passed through a colloidal sol.

(iii) When a beam of light is passed through a colloidal sol. Ans :(i)Coagulation (ii) Electrophoresis resulting in to coagulation. (iii) Tyndal effect.

2.Explain :(i)Blue colour of the sky (ii) Formation of delta (iii) Electrical precipitation of smoke i) Blue colour of the sky: Dust particles along with water

suspended in air scatter blue light which reaches our eyes and the sky looks blue to us.

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(ii) Formation of delta: River water is a colloidal solution of clay. Sea water contains a number of electrolytes. When river water meets the sea water, the electrolytes present in sea water coagulate the colloidal solution of clay resulting

in its deposition with the formation of delta. (iii) Smoke particles are actually electrically charged colloidal particles of carbon

in the air peptization of the smoke particles is carried out by cotterell precipitator which is based upon the principle of electrophoresis.

3.Describe Electrical disintegration or Bredig’s Arc method for the preparation of colloids. Ans: This process involves dispersion as well as condensation. Colloidal sols of

metals such as gold, silver, platinum, etc., can be prepared by this method. In this method, electric arc is struck between electrodes of the metal immersed in

the dispersion medium. The intense heat produced vapourises the metal, which then condenses to form particles of colloidal size. 4.Discuss the effect of pressure and temperature on the adsorption of gases on

solids? Ans . ADSORPTION ISOTHERMS: The variation in the amount of gas adsorbed

(Adsorbate) by the adsorbent with pressure at constant temperature can be expressed by means of a curve termed as adsorption isotherm. Freundlich adsorption isotherm:Freundlich, in 1909, gave an empirical

relationship between the quantity of gas adsorbed by unit mass of solid adsorbent i.e. (extent of adsorption (x/m) and pressure (P) at a particular temperature

5.Explain the term: Coagulation. Discuss the ways by which coagulation of the lyophobic sols can be carried out.

Ans The process of aggregation of colloidal particles of lyophobic sol a in to an insoluble precipitate by the addition of some suitable electrolyte is called coagulation.

The coagulation of the lyophobic sols can be carried out in the following ways: (i) By electrophoresis (ii) By mixing two oppositely charged sols (iii) By boiling(iv)

By persistent dialysis (v) By addition of electrolytes

6) Account for the following:

(i) What is colloidion? (ii) Why do we add alum to purify water?

(iii) Of physisorption and chemisorption, which type of adsorption has a higher enthalpy of adsorption? A14. (i) It is a 4% sol. of nitrocellulose in a mixture of alcohol and ether.

(ii) Alum coagulates colloidal impurities present in water. (iii) Chemisorption has higher enthalpy of adsorption on account of formation of

chemical bond. 7. How are the following colloids different from each other in respect of

dispersion medium and dispersed phase? Give an example each. (i) An aerosol (ii) A hydrosol (iii) An emulsion

(i) An aerosol is a colloidal dispersion of liquid in a gas, eg, fog (ii) A hydrosol is a colloidal sol. of a solid in water as the dispersion medium, eg,

starch sol. (iii) An emulsion is a colloidal system with dispersed phase as well as dispersion medium as liquids, eg, oil in water

8.(i) How can colloidal sol. of sulphur in water be prepared? (ii) What is electrophoresis due to ?

(iii) Why is Fe(OH)3 colloid +vely charged when prepared by adding FeCl3 to hot water? (i) It is prepared by oxidation of H2S by dil.HNO3. H2S + [O] H2O + S

(ii) Colloidal particles carry a charge, either +ve or –ve. On passing electricity, they migrate towards the oppositely charged electrode.

(iii) The colloidal sol. o f hydrated ferric oxide adsorbs +vely charged Fe3+ ion and therefore the colloidal sol. becomes +vely charged.

59

9. i) Why is ferric chloride prefered over potassium chloride in the case of a cut leading to bleeding? (ii) Why is desorption important for a substance to act as a good catalyst

? (iii)Why are substances like platinum and palladium often used for

carrying out electrolysis of aqueous solutions

Ans. (i) i) Blood is +vely charged colloid.One molecule of ferric chloride produces 3 –ve chloride ions while one molecule of potassium chloride produces one –ve chloride ion. Greater the –ve charge, faster the coagulation

After the reaction is over between the adsorbed reactants, the process of desorption must take place to remove the product molecules and create space

for other reactant molecules to adsorb on the catalyst surface. (iii)Platinum and palladium are inert materials & are not attacked by the ions of the electrolyte or the

products of electrolysis therefore used as electrodes for carrying out the electrolysis.

10. Define : (i) Kraft temperature (ii) Zeta potential (iii) Brownian movement (i) Kraft temperature‐ a particular temperature only above which formation of

micelles takes place.

(ii) Zeta potential‐ it is the potential difference between the fixed and diffused

layer of opposite charges around the colloidal particles

(ii) Brownian movement‐ It is a continuous zig-zag motion of colloidal particles.

It is due the unbalanced bombardment / collision of the particles by the molecules

of dispersion medium. It depends upon the size of the particles and viscosity of the solution

5 MARKS QUESTION

Q1 What do you mean by colloidal solution and colloidal state? How do colloidal solutions differ from true solutions with respect to dispersed particle size, homogeneity ,Brownian movement and Tyndall effect.

Ans: Solutions which are formed by colloids are called colloidal solutions. Colloidal

state is a state in which the sizeof the particles such(1 to 1000 nm)that they can pass through filter paper but not through animal or vegetable membrane. The size of the solute particles in solution is 1 nm but the size of dispersed particles

in colloidal solutions is between 1nm to 1000nm.True solutions are homogeneous in character but colloidal solutions are heterogeneous. True solutions do not show

brownian movement but colloidal solution show zig-zag movement of colloidal particles in a colloidal sol.True solutions do not show Tyndal effect But colloidal solution show tyndall effect

Q 2Explain the terms -adsorption isotherm and adsorption isobars. Describe the

freundlich adsorption isotherm. Ans:Adsorption isotherm : Agraph between the amount of the gas adsorbed per

gram of the adsorbent(x/m) and the equilibrium pressure of the adsorbate at constant temperature is called the adsorption isotherm .

Adsorption isobars Agraph between the amount of gas adsorbed per gram of the adsorbent (x/m) And temperature (t) at constant pressure of adsorbate gas is known as adsorption isobars.

Freundlich adsorption isotherm :Freundlich gave an empirical relation

between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature .the

relationship can be expressed by the following equation ;

x/m=k.p 1/n (n>1) Where x is the

mass of the gas adsorbe on the mass of the adsorbent at pressure P,k and

n are the constants which depend on the nature of the adsorbent and the

gas at a particular temperature.

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Log x/m = logk + 1/n logP

The factor 1/n can have values between 0 & 1.

When 1/n=0, x/m = constant which shows that adsorption is independent of pressure.

When 1/n=1,x/m = kP, the adsorption varies directly with pressure

Q3.How are colloids are classified on the basis of (a) physical states of components. (b)nature of dispersion medium and

(c) association between dispersed phase and dispersion medium

Ans :Based on the physical state of the disperse phase and dispersion medium.Solid in solid(solid sol e.g.coloured glass)solid in liquid (sol e.g.muddy water),solid in gas (aerosol ,e.g.smoke),Liquid in solid(gel e.g. butter),liquid in

liquid (emulsion,e.g.milk)liquid in gas (aerosol e.g. clouds or fog)gas in solid (solid foam,e.g. foam,rubber)and gas in liquid (foam,e.g. lather).

Lyophobic colloids (solvent hating colloids ) : These colloids cannot be prepared

by simply mixing dispersed phase with dispersion medium, they need stabilizing

agent to preserve them , irreversible. Ex: colloidal solutions of gold, silver, Fe(OH)3,

As2S3, etc.

Lyophilic colloids ( solvent loving): Directly formed by mixing DP with a suitable

dispersion medium), self‐stabilizing , reversible sol, sol of starch, gum, gelatin, rubber.

Classification based on type of particles of the dispersed phase

Multimolecular colloids : Consists of aggregates of a large number of atoms or

whose diameter is less than 1 nm. Ex Au sol

Macromolecular colloids: In these colloids the molecules have sizes and

dimensions to colloidal particles. Ex: proteins, starch, cellulose.

Associated colloids: At low concentrations, behave as normal, strong electrolytes

and at higher concentrations exhibit colloidal state properties due to the formation of

aggregated particles (micelles). e.g Soaps and detergents.

Q4. A.How can you prepare the colloidal solutions of the following by using

chemical methods . B. Explain the process of purification of colloidal solution by the process of

Dialysis (i) Sol of arsenic sulphide (ii)Sol of gold (iii)Sol of sulphur (iv)sol of ferric hydroxide

Ans :A (i) Sol of arsenic sulphide can be prepared by Double decomposition : Arsenious sulphide colloidal solution is prepared by passing of H 2 S gas into asolution of As2O3. (ii)Sol of gold Can be prepared by reduction 2AuCl3 +3HCHO+ 3H2O ---------> 2AU (sol) +3HCOOH +6HCl (iii) : Sulphur colloids are prepared by oxidation of H2S by O2.

SO2 +2H2S → 3S (sol) +2H2O

(ivi)Sol of ferric hydroxide can be prepared by Hydrolysis : Dark brown Fe(OH)3

colloidal solution is prepared by adding FeCl3 into boiling water.

FeCl3 + 3H2O → Fe(OH)3 + 3HCl

Ans B :Purification of colloidal solutions from the impurities(electrolytes) by diffusion through a porous membrane such as parchment, collodion, etc

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Q5.Write a short note on adsorption.How does adsorption differ from absorption?

What are the different types of adsorption. Ans :

Adsorption:The phenomenon of attracting and retaining the molecules of a

substance at the surface of the solid or a liquid resulting into higher concentration of

the molecules on the surface than in the bulk.

Adsorbent: The substance where adsorption occurs.

Adsorbate: The substance that get adsorbed.

Mechanism of adsorption : arises due to unbalanced force of attraction on the surface of solid that are responsible for attracting the adsorbate particles on the surface.The extent of adsorption increases with increase in surface area.

Absorption: The phenomenon in which the particles of gas and liquid get uniformly distributed throughout the body of the solid.

Adsorption is essentially a surface phenomenon but absorption ia bulk phenomenon.

Types of Adsorption: physical and chemical adsorption

1. Physical adsorption : When the particles of adsorbate are held to the surface of

adsorbent by weak

Van der Waals forces,

Characteristics of physical adsorption: Lack of specificity, low enthalpy of

adsorption, reversible in nature, no

activation energy required,decrease with increase in temperature .

2. Chemical adsorption : When the molecules of adsorbate are held to the surface

of adsorbent by strong chemical forces

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CHAPTER-6 p-block elements

CASE STUDY QUESTIONS

Que.1) The noble gases have closed-shell electronic configuration and are monoatomic

gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.

The direct reaction of xenon with fluorine leads to a series of compounds with oxidation

numbers +2, +4 and +6. XeF4 reacts violently with water to given XeO3. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced

considering the total number of electron pairs in the valence shell.

(i) Argon is used in arc welding because of its

a) Low reactivity with metals b) Ability to lower the melting points

c) Flammability d) High calorific value

Ans. (a) Low reactivity with metals

(ii) The structure of XeO3

(a) Linear

(b) Planner (c) Pyramidal

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(d) T-Shaped

Ans. (c) Pyramidal

(iii) XeF4 and XeF6 are expected to be

a) Oxidizing b) Reducing

c) Unreactive d) Strongly Basic

Ans. (a) Oxidizing

(iv) The lowest boiling point of Helium is due to its…..

a) Inertness

b) Gaseous nature c) Weak Van der Waal force Between Atoms

d) Small Size

Ans. (c) Weak Van der Waal force Between Atoms

Que.2)There are some deposits of nitrates and phosphates in earth’s crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily.

Ammonia forms large number of complexes with transition metal ions. Hybridization easily explains the ease of sigma donation capability of NH3 and

PH3. Phosphine is a flammable gas and is prepared from white phosphorous.

(i) Among the following, the correct statement is

(a) Phosphates have no biological significance in humans (b) Between nitrates and phosphates, phosphates are less abundant in earth’s

crust (c) Between nitrates and phosphates, nitrates are less abundant in earth’s crust (d) Oxidation of nitrates is possible in soil

Ans. (C) Between nitrates and phosphates, nitrates are less abundant in earth’s crust

(ii) Among the following, the correct statement is

(a) Between NH3 and PH3, NH3 is a better electron donor because the lone pair of

electrons occupies spherical s-orbital and is less directional

(b) Between NH3 and PH3 , PH3 is a better electron donor because the lone pair of

electrons occupies sp3 orbital and is more directional

(C) Between NH3 and PH3 , NH3 is a better electron donor because the lone pair of

electrons occupies sp3 orbital and is more directional

(d) Between NH3 and PH3 , PH3 is a better electron donor because the lone pair of

electrons occupies spherical s-orbital and is less directional

Ans. (C) Between NH3 and PH3 , NH3 is a better electron donor because the lone

pair of electrons occupies sp3 orbital and is more directional

(iii) White phosphorus on reaction with NaOH gives PH3 as one of the

products. This is a

(a) Dimerization Reaction

(b) Dispraportination Reaction

(c) Condensation Reaction

(d) Precipitation Reaction

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Ans. (b) Dispraportination Reaction

(iv) Perfectly dry ammonia gas in the liquefied form is

(a) Acidic

(b) Basic

(c) Neutral

(d) Amphoteric

Ans. (c) Neutral

CASE STUDY (ASSERTION REASON) QUESTIONS

1. Read the passage carefully,

Choose the correct Option according to Question:

a. Both Assertion and Reason are correct and Reason is the correct

explaination of Assertion

b. Both Assertion and Reason are correct but Reason is not the correct

explaination of Assertion

c. Assertion is correct but Reason is Incorrect

d. Assertion is Incorrect but Reason is correct

(i) ASSERTION: The electronic structure of O3 is

REASON: The following structure is not allowed because octet

around O cannot be expanded.

ANS. (a) Both

Assertion and Reason are

correct and Reason is the

correct explanation of Assertion

(ii) ASSERTION: HNO3 is stronger acid than HNO2. REASON: In HNO3, there are two nitrogen to oxygen bonds whereas

in HNO2 there is only one.

ANS. (a) Both Assertion and Reason are correct and Reason is the

correct explanation of Assertion

(iii) ASSERTION: Although PF5, PCl5 and PBr5 are known, the pentahalides of

nitrogen have not been observed.

REASON: Phosphorus has lower electronegativity than nitrogen.

ANS. (b) Both Assertion and Reason are correct but Reason is not

the correct explanation of Assertion

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(iv) ASSERTION: N2 is less reactive than P4.

REASON: Nitrogen has more electron gain enthalpy than

Phosphorus.

ANS. (c) Assertion is correct but Reason is Incorrect

2. Read the passage carefully,

Choose the correct Option according to Question:

a. Both Assertion and Reason are correct and Reason is the correct explaination of Assertion

b. Both Assertion and Reason are correct but Reason is not the correct

explaination of Assertion c. Assertion is correct but Reason is Incorrect

d. Assertion is Incorrect but Reason is correct

(i). ASSERTION: HI cannot be prepared by the reaction of KI with

concentrated H2SO4

REASON: HI has lowest H–X bond strength among halogen acids.

Ans. (b)

(ii).ASSERTION: NaCl reacts with concentrated H2SO4 to give colourless

fumes with pungent smell. But on adding MnO2 the fumes become greenish

yellow.

REASON: MnO2 oxidises HCl to chlorine gas which is greenish yellow.

ANS. (a)

(iii). ASSETION: SF6 cannot be hydrolysed but SF4 can be.

REASON: Six F atoms in SF6 prevent the attack of H2O on sulphur atom

of SF6.

Ans. (a)

(iv).ASSERTION: Nitrogen and oxygen are the main components in the

atmosphere but these do not react to form oxides of nitrogen.

REASON: The reaction between nitrogen and oxygen requires high temperature.

ANS. (a)

Multiple Choice Questions 1) Ammonia can be dried over

a) CaCl2 b) conc.H2SO4 c)PCl5 d)Quicklime Answer: d) Quick lime

2) Hybridisation of chlorine atom in ClF5 is a) sp3 b) sp3d c) sp3d2 d) sp3d3

Answer: c) sp3d2 3) When Br2 is added to aqueous solution of NaF, NaCl, NaI separately it results into the liberation of

a) F2 , Cl2 and I2 b) F2 , Cl2 c) Cl2 only d) I2 only Answer: d) I2 only

4) Which of the polyhalide ions is not known? a)F3- b)Br2- c)Cl3- d)I2-

Answer: a) F3-

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5) Bleaching action of SO2 is due to a)reduction b)oxidation c)acidic nature d)hydrolysis

Answer: a) reduction

6)The geometry of XeOF2 is a)pyramidal b)T-shaped c)Octahedral d)Tetrahedral

Answer:b)T-shaped

7)The noble gas compound prepared by Bartlettn was a)XeO3 b)XePtF4 c)KrF2 d)XeF2

Answer: b) XePtF4

8) When Cl2 gas reacts with hot sodium hydroxide solution, oxidation number of chlorine changes from

a) 0 to +5 , 0 to +3 b) 0 to +5 , 0 to -1 c) 0 to +1 , 0 to +5

d) 0 to +3 , 0 to +1 Answer: b) 0 to +5 , 0 to -1

9) Nitrogen dioxide and sulphur dioxide have some properties in common . Which property is shown by one of these compounds , but not by other ? a) Is soluble in water

b) Is used as food preservatives c) Forms ‘acid-rain’

d) Is a reducing agent Answer: b) Is used as food preservatives (Nitrogen Dioxide)

10) Among the following , which one is a wrong statement ? a) PH5 and BiCl5 do not exist.

b) pπ-dπ bonds are present in SO2 . c) SeF4 and CH4 have same shape . d) I3+ has bent.

Answer: c) SeF4 and CH4 have same shape.

11) On addition of conc. H2SO4 to a chloride salt , colourless fumes are evolved but in case of iodide salt , violet fumes come out . This is because a) H2SO4 reduces HI to I2 b) HI is of violet colour

c) HI gets oxidised to I2 d) HI changes to HIO3 Answer: c) HI gets oxidised to I2 .

12) Which of the halogen acids have highest bond disassociation enthalpy ? a)HF b)HCL c)HBr d)HI

Answer: a) HF

13) On heating lead nitrate oxides of nitrogen and lead are formed. The oxides formed are _____ .

a)N2O, PbO b)NO2, PbO c)NO, PbO d)NO, PbO2 Answer: b)NO2, PbO

14 ) Which of the following is not tetrahedral in shape ? a) NH4

+ b) SiCl4 c) SF4 d) SO4 Answer: c) SF4

15) Which of the following element is oxidised by conc.H2SO4 into two gaseous products?

a) Cu b) S c) C d) Zn Answer: c) C

16) Which of the following order is correct for Acid Strength ?

a) As2O3 > SiO2 > P2O3 > SO2 b) As2O3 > P2O3 > SiO2 > SO2

c) As2O3 < SiO2 < P2O3 < SO2 d) SO2 > SiO2 > P2O3 > As2O3

Answer: c) As2O3 < SiO2 < P2O3 < SO2

17) Which of the following order is correct for Thermal Stability ? a) H2O > H2S > H2Se > H2Te

b) H2O > H2S > H2Te > H2Se c) H2O > H2Se > H2S > H2Te d) H2O < H2S < H2Se < H2Te

Answer: a) H2O > H2S > H2Se > H2Te 18) Which of the following is an electrophilic reagent ?

a) BF3 b) KH3 c) H2Od) None Of These Answer: a) BF3

19) Partial hydrolysis of XeF4 gives

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a) XeO4 b) XeFO4 c) XeFO2 d) XeF2 Answer: c) XeFO2

20) The brown ring test for nitrates depends on

a) Reduction of ferrous sulphate to iron b) Oxidation of nitric oxide to nitrogen dioxide

c) The reduction of nitrate to nitric oxide d) Oxidising action of sulphuric acid

Answer: c) The reduction of nitrate to nitric oxide.

ASSERTION REASON

Choose the correct Option according to Question: a Both Assertion and Reason are correct and Reason is the correct

explanation of Assertion b Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion

c Assertion is correct but Reason is Incorrect d Assertion is Incorrect but Reason is correct

1. ASSERTION: Ozone is a powerful oxidizing agent in comparison to 02. REASON: Ozone is diamagnetic but 02 is paramagnetic.

Ans. Correct option is (b).

2. ASSERTION: P4 is more reactive than N2. REASON: P-P single bond in P4 is much weaker than N triple bond N in N2

molecule. Ans. Correct option is (a).

3. ASSERTION: F2 has low reactivity. REASON: F-F bond has two bond dissociation Enthalpy.

Ans. Correct option is (d).

4. ASSERTION: Xenon forms fluorides. REASON: 5d orbitals are available for valence shell expansion.

Ans. Correct option is (a). 5. ASSERTION: The electron gain enthalpy of sulphur is greater than that of

Oxygen.

REASON: Due to much interelectronic repulsion in 2p orbitals of Oxygen Ans. Correct option is (a)

TWO MARKS QUESTIONS

Q1. (i) Account for following: HF is not stored in glass bottles but is

kept in wax coated bottles? (ii) Why does fluorine not play the role of a central atom in interhalogen compound ?

Ans. (i) HF forms fluorosilicate ion on reaction with glass.Hence, it is stored in wax coated in bottles.

(ii) Fluorine does not have d - orbitals and can not show higher oxidation state. Therefore, it does not play the role of a central atom in the interhalogen compounds.

Q2. (i) F2 is the most reactive of all the four common halogens. Explain.

(ii) Arrange the following in decreasing order of oxidising power: HClO, HClO2, HClO3, HClO4. Ans. (i) It is due to low bond dissociation energy and high hydration

energy and high electron affinity. (ii) HClO>HClO2>HClO3>HClO4

Q3. (i) All the bonds in SF4 are not equivalent. Explain, why? (ii) Sulphur has greater tendency for catenation than Oxygen? Ans. (i) As sulphur is sp3d hybridised. It has trigonal by-pyramidal

structure in which one of the equatorial position is occupied by an lone pair of electrons. That’s why, all bonds are not identical.

(ii) As, S-S bond energy is more than O-O bond energy and due to smaller size of O the lp-bp repulsion is high resulting in lower tendency for catenation.

Q4. (i) Complete the following reacHon: PbS + O3 ——>

(ii) OF6 compound is not known. Why? Ans. (i) PbS + 4 O3 ——> PbSO4 + 4 O2

(ii) Due to the absence of d orbital in O, it limits its covalency to 4,

thus OF6 is not formed. Q5. Give equations that involve during the brown ring test?

Ans. The equations involved in the brown ring test are:

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1. NO3- (aq) + 3Fe2+

(aq) + 4H+ (aq) NO (g) + 3Fe3+

(aq) +2H2O(l)

2.[Fe(H2O)6]2+ + NO (g) [Fe(H2O)5(NO)]2+ (aq) + H2O

Brown Complex

Q6. (i) Why are the two S-O bonds in SO2 molecule of equal strength? (ii) The +5 oxidation state of Bi is less stable than its +3 . Why ?

Ans (i) Due to the resonance in , the two bonds equally share the double bond character and thus are identical.

(ii) Due to inert pair effect . Q7. (i) All the bonds in PCl5 are not are not equivalent Why ? (ii) Why Chlorine Water on standing loses its colour ?

Ans (i) Because two axial P – Cl bonds are repelled more by three equatorial P – Cl bonds leading to longer axial P – Cl bonds.

(ii) Cl2+H2O → HCl+HOCl It forms HCl and HOCl , both are colourless . Q8. (i) Of all the noble gases only xenon forms known chemical

compounds . Why ? (ii) Sulphur disappears when boiled with Sodium Sulphite . Why ? Ans (i) Xe atom has large atomic size and low Ionization enthalpy.

(ii) Due to formation of soluble sodium thiosulphate . Q9. (i) Why NO2 readily forms dimer ?

(ii) The two O – O bond lengths in Ozone molecule are identical. Why? Ans (i) Due to presence of odd valence electron in NO2 it Dimerizes to form N2O4 .

(ii) It is due to Resonance. Q10. (i) How is O3 estimated quantitatively?

(ii) Why are Trihalides more stable than Pentahalides for Group 15? Ans (i) When Ozone reacts with excess of Potassium Iodide solution,

buffered with borate buffer, Iodine is liberates which can be titrated against a

standard solution of sodium thiosulphate. By estimating the iodine liberated in the reaction, O3 can also be estimated quantitatively.

2I- + H2O + O3 = 2OH- + I2 + O2 (ii) Trihalides are more stable than pentahalides because of the following Reasons:

a) The stability of +5 oxidation state decreases while that of +3 oxidation state increases down the group due to inert pair effect.

b) For Br and I, the size is large. In pentabromides and pentaiodide, Steric hindrance is high and Bond strength is low.

THREE MARKS QUESTIONS

1. Complete and balance :

(i) F2 + H2O Cold →

(ii) BrO3- + F2 + OH- →

(iii) Li + N2 (cold) →

ANS. (i) 2 F2 + H2O Cold → 4H+ + 4F- + O2

(ii) BrO3- + F2 + 2 OH- → BrO4 + 2F- + H2O

(iii) 6 Li + N2 (cold) → 2 Li3N 2. How will you account the following:

(i) NCl3 is an endothermic compound while NF3 is an exothermic one.

(ii) XeF2 is a linear molecule.

(iii) The electron gain enthalpy with negative sign for fluorine is less than that of chlorine, still fluorine is stronger oxidizing agent than chlorine.

ANS (i) NCl3 is endothermic compound because of its high enthalpy of formation

due to large difference in sizes of N and Cl atoms. This large difference makes N-Cl bond weak, while in NF3 the sizes of atoms are not so different, therefore

N-F bond is strong and thus it is an exothermic compound. (ii) In XeF2, there are 5 pairs (2 bond pairs and 3 lone pairs) of electrons around Xe and thus, its geometry is trigonal bipyramidal. The fluorine atoms

ocuppy linear positions while the three lone pairs occupy equatorial positions. Thus, XeF2 is a linear molecules.

(iii) The negative electron gain enthalpy of fluorine is less than that of chlorine due to the small size of F atom and the electron-electron repulsion in the small 2p orbital of fluorine. Due to that incoming electron

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is not accepted with the same ease. But the bond dissociation enthalpy of F2 is less than that of Cl2. Due to small size, lone pairs of F atoms repel the bond pair. Hence, bond dissociates easily and therefore fluorine is stronger

oxidizing agent than chlorine. 3. Give the Reasons for the following:

(i) (CH3)3P = O exists but (CH3)3N = O does not. (ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.

(iii) H3PO2 is a strong reducing agent than H3PO4. Ans. (i) Because N can’t form 5 covalent bonds as its maximum covalency

is three. The octet cannot be extended as it doesn’t have d

orbital, while P can extend its octet as it has empty d orbital. (ii) This is due to very small size of oxygen atom, repulsion between

electrons is large in relatively small 2p orbital. (iii) In H3PO2 there are 2 P-H bonds, whereas in H3PO3 there is 1 P-H bond.

4. Explain why the stability of oxoacids of chlorine increases in the order given

below HClO < HClO3 < HClO4.

Ans. Oxygen is more electronegative than chlorine, therefore, dispersal of negative charge present on chlorine increases from ClO- to ClO4

- ion because number of oxygen atoms attached to chlorine increases. Therefore, stability

of ions will increase in the order given below: ClO- < ClO2

- < ClO3- < ClO4

-

Thus due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the following order:

HClO < HClO2 < HClO3 < HClO4

5. (i) What happens when : (a) Chlorine gas reacts with cold and dilute solution of NaOH?

(b) XeF2 undergoes hydrolysis? (ii) Assign suitable Reason for the following:

Out of noble gases only Xenon is known to form established chemical

compounds.

Ans. (i)(a) 2 NaOH + Cl2 → NaCl + NaOCl + H2O (b) 2 XeF2 + 2H2O → 2Xe + 4HF + O2

(ii) Xe has least ionization energy among the noble gases and hence it forms

chemical compounds particularly with O2 and F2. 6. What happens when:

(i) conc. H2SO4 is added to Cu? (ii) SO3 is passed through water? (iii) conc. H2SO4 is added to calcium fluoride?

Ans. (i) Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O

(ii) SO3 + H2O → H2SO4

(iii) H2SO4(conc.) + CaF2 → 2HF + CaSO4 7. (i) Explain why fluorine forms only one oxoacid, HOF.

(ii) Why are halogens coloured? (iii) Write the reactions of F2 and Cl2 with water.

Ans. (i) Fluorine forms only one oxoacid, HOF, because of its high electronegativity and small size.

(ii) The halogens are coloured because their molecules absorb light in the visible region. As a result of which their electrons get excited to higher energy levels while the remaining light is transmitted. The color of halogens is

the color of this transmitted light.

(iii) 2 F2 + 2H2O → 4H+ + 4F- + O2 + 4HF Cl2 + H2O → 4HCl + HOCl

8. The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?

Ans. Hydrides : NH3 PH3 AsH3 SbH3 H-N-H bond angle : 107° 92° 91° 90°

The above trend in the H-M-H bond angle can be explained on the basis of the electronegativity of the central atom. Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes

greater repulsion between the electron pairs around nitrogen, resulting in maximum bond angle.

We know that electronegativity decreases on moving down the group. Consequently, the repulsive interactions between the electron pairs decrease, thereby decreasing the H-M-H bond angle.

9. (i) How is O2 estimated quantitatively? (ii) Why does O3 act as a powerful oxidizing agent?

(iii) Complete the following reactions:

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(a) PbS(s) + 4O3(g) →

(b) NO(g) + O3(g) → Ans. (i) When O3 is treated with an excess of KI (Potassium Iodide) solution buffered with a borate buffer (pH 9.2), I2 is liberated quantitatively.

2I- (aq) + H2O(l) + O3 → 2OH-(aq) + I2(s) + O2(g) I2 liberated can be titrated against a standard solution of sodium

thiosulphate using starch as an indicator.

2Na2S2O3 + I2 → Na2S2O6 + 2NaI (ii) O3 is an endothermic compound. On heating, it readily decomposes to

give nascent oxygen and dioxygen. O3 -heat→ O2 + O Since, nascent oxygen is very reactive, therefore, O3 acts as a powerful oxidizing agent.

(iii) (a) PbS(s) + 4O3(g) → PbSO4(s) + 4O2(g)

(b) NO(g) + O3(g) → NO2(g) + O2(g) 10. A compound X has a large scale preparation by Ostwald process, boiling point

as 355.6 K. When concentrated X is reacted with copper it gives copper nitrate and when dilute X is reacted with zinc it gives zinc nitrate. It is

also involved in a test to confirm [Fe(H2O)5(NO)]2+ and has major use in manufacture of ammonium nitrate for fertilizers. What is X and write the

chemical equations involved. Ans. X is nitric acid (HNO3). Preparation by Ostwald process:

Step 1 : 4NH3 + 5O2 → 4NO + 6H2O + Heat

Step 2 : 2NO + O2 → 2NO2

Step 3: 4NO2 + 2H2O + O2 → 4HNO3 Reaction with copper and zinc

HNO3(conc.) + Cu → Cu(NO3)2 + 2NO2 + 2H2O HNO3(dil.) + Zn → 4Zn(NO3)2 + 2H2O + 2NO2

Test to confirm [Fe(H2O)5(NO)]2+ is Ring Test

NO3- + 3Fe2+ + 4H+ → NO + 3Fe3+ + 2H2O

[Fe(H2O)6]2+ + NO → [Fe(H2O)5(NO)]2+ + H2O (brown)

FIVE MARKS QUESTIONS

Q.1. State the Reasons: (i) The N-O bond is shorter in NO2

- than the N-O bond in NO3-.

(ii) SF6 is kinetically an substance.

(iii) All the P-Cl bonds in PCl5 molecules are not equivalent. (iv) Sulphur has a greater tendency for catenation than oxygen.

(v) NF3 is an exothermic compound whereas NCl3 is not why ?

Ans. (i) The N-O bond in NO2- is shorter than the bond in NO3

- as in NO2- two

bonds are sharing double bond whereas in NO3- three bonds are sharing

double bond. (ii) SF6 is kinetically an inert substance as it shows steric hindrance due to 6

fluorine atoms, hence it is unable to react further with any other atom.

(iii) All the P-Cl bonds in PCl5 are not equivalent as it is sp3d hybridisation

and trigonal bi pyramidal geometry. Therefore, it has 3 equatorial P-Cl bonds and two axial P-Cl bonds are repelled by two bond pairs , so axial bonds are longer than equatorial bonds.

71

(iv) Sulphur has a greater tendency for catenation as the lone pair of electrons feel more repulsion in O-O bond than than in S-S bond. Also, when the size of an atom increases down the group from O-PO, the catenation

tendency increases due to the increase in the bond strength. (v) Because F – F bond energy is lower than that of N – F bond . Also Cl –

Cl bond energy is higher than that of the N – Cl bond. Therefore , NCl3 is an endothermic compound and is unstable .

Q.2. Explain:

(a) O2 and F2 both stabilize higher oxidation states of metals but O2 exceeds F2 in doing so.

(b) Structure of Xenon fluorides cannot be explained by valence bond approach. (C) NF3 is an exothermic compound whereas NCl3 is not. (d) Xenon is known to form real chemical compounds out of noble gases.

(e) NO2 readily forms a dimer. Ans. (a) O2 and F2 both stabilize higher oxidation states of metals but O2 exceeds

F2 in doing so because oxygen can form multiple bonds with metals. (b) Structures of Xenon fluorides cannot be explained by valence bond

approach because the promotion of electrons in Xenon requires very high

energy which doesn't favour valence bond approach. (c) NF3 is an exothermic compound whereas NCl3 is not because F-F bond

energy in lower than that of N-F bond. Also Cl-Cl bond energy is higher than that of the N-Cl bond. Therefore, NCl3 is an exothermic compound and is unstable.

(d) Xenon is known to form real chemical compounds out of all noble gases because Xenon atoms have large size and lower ionisation potential. The

force of nucleus over the electrons is weak and a very small amount of energy can excite the electrons.

(e) NO2 readily forms dimer as it contains 23 odd electrons. N has seven electrons in the valance shell and is less stable. It dimerizes to form N2O4 in order to acquire stability.

3. (a) Give Reasons:

(i) H3PO3 undergoes disproportionation reaction but H3PO4 does not. (ii) When Cl2 reacts with excess of F2, ClF3 is formed and not FCl3.

(iii) Dioxygen is a gas while sulphur is a solid at room temperature. (b) Draw the structures of the following:

(i) XeF4 (ii) HClO3 ANS. (a) (i) In H3PO3, phosphorous exhibits an oxidation state of +3 hence it tends

to undergo disproportionation reaction. However, in H3PO4 , phosphorous exhibits +5 oxidation state and cannot be oxidized further hence it does

not show the disproportionation reaction. (ii) In interhalogen compounds, the central atom is of larger size and more

electropositive than the surrounding atoms. As fluorine is more

electronegativity and smaller in size than chlorine, ClF3 is formed. (iii) Due to its small size and high electronegativity, oxygen forms pπ-pπ

bonds and exists as a diatomic molecule. The intermolecular forces are weak van der Waals forces which result in dioxygen to exist as gas at room temperature whereas, sulphur exists as S8 molecules held together

by strong covalent forces and forms puckered ring structures packed in solid crystals.

(b) (i) Structure of XeF4

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(ii) Structure of HClO3

4. (a) When concentrated sulphuric acid was added to an unknown salt present

in a test tube a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into

a colourless solid(B). (i) Identify (A) and (B). (ii) Write the structures of (A) and (B).

(iii) Why does gas (A) change to solid on cooling? (b) Arrange the following in the decreasing order of their reducing character:

HF, HCl, HBr, HI (c) Complete the following reaction :

XeF4 + SbF5 → ANS: (a) (i) (A) is NO2 and (B) is N2O4.

(ii) Sturcture of A (NO2):

Structure of B (N2O4):

(iii) NO2 is an odd electron species, so on cooling it dimerises to form N2O4 which is

a solid. (b) Decreasing order of their reducing character:

HI > HBr > HCl > HF

(c) XeF4 + SbF5 → [XeF3]+ [SbF6]- 5. Explain why.

(a) Halogens are highly reactive. (b) Flourine does not show variable valnency while other halogens exhibit variable

valency. (c) Flourine atom is more electronegative than iodine atom, yet HF has lower acid

strength than HI.

(d) ClF3 exists whereas FCl3 does not. (e) OF2 should be called oxygen diflouride and fluorine oxide.

ANS. (a) It is because the X - X bond dissociation enthalpy is very less in the case of

halogens making them highly reactive. Also high electronegativity is another

Reason for their high reactivity. (b) Flourine has only one half-filled orbital and there is no d-orbital available for

excitation of electrons. Further it is the most electronegative element. Hence it shows oxidation state of -1 only. In all other halogens d-orbitals are available for excitation for electrons. Moreover, they can combine with the more

electronegative element, oxygen. Hence they show variable oxidation states. (c) F – atom being small in size, the bond dissociation energy of H - F is very high

as compared to that of H-I bond because I - atom is very large in size. (d) Electronic configuration of chlorine is 1s2 2s2 2p6 3s2 3p5. An electron from 3p

can jump to 3d orbitals. So it can show an oxidation state of 3 and combine

with the more electronegative fluorine. Electronic configuration of fluorine is 1s2 2s2 2p5. No d-orbital is available for

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excitation of electron. Moreover it is most electronegative element. So it shows an oxidation state of -1 only. (e) It is because fluorine is more electronegative than oxygen.

6. (a) The tendency to show -2 oxidation state diminishes from sulphur to polonium. Why?

(b) Oxygen molecule has the formula O2 while sulphur has S8. Explain why. (c) Why H2S is acidic while H2O is neutral?

(d) SF6 is known but SH6 is not known. Explain. (e) Why H2O is a liquid while H2S is a gas?

ANS.

(a) This is because as we move down the group from sulphur to polonium, the electronegativity decreases.

(b) Oxygen atom, being small in size has the tendency to form multiple bonds while sulphur atom, being large in size, forms single bonds with other sulphur atoms. The puckered ring structure, S8 is most stable.

(c) The S – H bond is weaker than O – H bond because size of S – atom is greater than that of O – atom. Hence H2S can dissociate to give H+ ions in aqueous

solution. (d) It is because in the highest oxidation state, sulphur can combine only with highly

electronegative element element like fluorine.

(e) There is H – bonding in water molecules due to high electronegativity and small size of oxygen – atom but there is no H – bonding in H2S.

74

CHAPTER-7 The d-and f-Block Elements

CASE STUDT BASED QUESTIONS: Q.1. Read the passage given below and answer the following questions: (1x4=4)

The transition metals (with the exception of Zn, Cd and Hg) are very much hard and have low volatility. Their melting and

boiling points are high. Fig. depicts the melting points of the 3d, 4d and 5d transition metals. The high melting points of these metals are attributed to the involvement of greater

number of electrons from (n-1)d in addition to the ns electrons in the interatomic metallic bonding. In any row the

melting points of these metals rise to a maximum at d5 except for anomalous values of Mn and Tc and fall regularly as the atomic number increases.

The following questions are multiple choice questions. Choose the most appropriate answer: (i) Which series of d-element have high melting point – (a) 5d (b) 4d (c) 3d (d) None of these

(ii) What whould you expect , which series of d-element have high Enthalpy of Atomisation – (a) 3d (b) 4d (c) 5d (d) None of these

(iii) Which statement is not true for d-element (a) Manganese has low Melting Point than expected value

(b) Tungsten has higest Melting Point (c) Melting Point first increases to Cr than decrease (d) Melting Point of Silver is greater than Gold

(iv) Which statement is correct for d-block element (a) melting point of heavy transition element is more than lighter element

(b) metal – metal bonding of heavy transition element is more than lighter element (c) Enthalpy of atomisation of heavy transition element is more than lighter element (d) Electrode Potential of heavy transition element has higher negative value than

lighter element

Q.2. Read the passage given below and answer the following questions: (1x4=4)

Oxidation state of the first row transition metals

The elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example, exhibits all the oxidation states from +2 to +7. At the other end, the only oxidation state of zinc is +2 (no d electrons are

involved). The maximum oxidation states of Reasonable stability correspond in value to the sum of the s and d electrons upto manganese (TiIVO2, VV O2

+ , CrV1O4 2–, MnVIIO4 – ) followed by a rather abrupt decrease in stability of higher oxidation states, so that the typical species to follow are FeII,III, CoII,III, NiII, CuI,II, ZnII

2. In these questions, a statement of Assertion followed by a statement of Reason is given. Choose the correct answer out of the following choices. (a) Assertion and Reason both are correct statements and Reason is correct

explanation for Assertion. (b) Assertion and Reason both are correct statements but Reason is not correct

explanation for Assertion. (c) Assertion is correct statement but Reason is wrong statement. (d) Assertion is wrong statement but Reason is correct statement.

(i) Assertion: Scandium (II) is not known Reason: Scandium has d1 configuration

(ii) Assertion: Zinc only shows (+2) oxidation state Reason: d- electron are not involved (iii) Assertion: Manganese exhibits maximum of +5 oxidation state

Reason: It has d5s2 configuration (iv) Assertion:these elements shows variable oxidation state

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Reason: Transition elements have incomplete filled d-orbitals

MCQ TYPE OF QUESTIONS

Q.3. Electronic configuration of a transition element X in +3 oxidation state is [Ar]3 d 5. What is its atomic number?

(i) 25 (ii) 26 (iii) 27 (iv) 24

4.The electronic configuration of Cu(II) is 3 d 9 whereas that of Cu(I) is 3 d10. Which

of the following is correct? (i) Cu(II) is more stable (ii) Cu(II) is less stable

(iii) Cu(I) and Cu(II) are equally stable (iv) Stability of Cu(I) and Cu(II) depends on nature of copper salts

5.Metallic radii of some transition elements are given below. Which of these elements will have highest density?

Element Fe Co Ni Cu

Metallic radii/pm 126 125 125 128 (i) Fe (ii) Ni (iii) Co (iv)Cu

6. Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state?

(i)Ag2SO4 (ii)CuF2 (iii)ZnF2 (iv)Cu2Cl2

7. The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition element, which shows highest magnetic moment.

(i)3d7 (ii)3d5 (iii)3d8 (iv)3d2

8.Which of the following oxidation state is common for all lanthanoids?

(i) +2 (ii) +3 (iii) +4 (iv) +5 9.There are 14 elements in actinoid series. Which of the following elements does not

belong to this series?

(i)U (ii) Np (iii)Tm (iv)Fm

10.Which of the following is amphoteric oxide?

Mn2O7, CrO3, Cr2O3, CrO, V2O5, V2O4

(i)V2O5, Cr2O3 (ii)Mn2O7, CrO3 (iii)CrO, V2O5 (iv)V2O5, V2O4

11.Gadolinium belongs to 4f series. Its atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?

(i) [Xe] 4f 75d16s2 (ii) [Xe] 4f 65d26s2 (iii) [Xe] 4f 86d2 (iv) [Xe] 4f 95s1

12.Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following is not the characteristic property of interstitial

compounds?

(i)They have high melting points in comparison to pure metals. (ii)They are very hard.

(iii)They retain metallic conductivity.

(iv)They are chemically very reactive.

13.The magnetic moment is associated with its spin angular momentum and orbital

angular momentum. Value of Spin only magnetic moment value of Cr3+ ion is….?

(i)2.87 B.M. (ii)3.87 B.M. (iii)3.47 B.M. (iv)3.57 B.M.

14.Which of the following statements is not correct? (i) Copper liberates hydrogen from acids.

(ii) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine.

(iii) Mn3+ and Co3+ are oxidising agents in aqueous solution.

(iv) Ti2+ and Cr2+ are reducing agents in aqueous solution.

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15.Highest oxidation state of manganese in fluoride is +4 (MnF4) but highest

oxidation state in oxides is +7 (Mn2O7) because .

(i) fluorine is more electronegative than oxygen. (ii) fluorine does not possess d-orbitals.

(iii) fluorine stabilizes lower oxidation state. (iv) in covalent compounds fluorine can form single bond only while oxygen forms double

bond.

16.Although Zirconium belongs to 4 d transition series and Hafnium to 5 d transition series even then they show similar physical and chemical properties

because .

(i) both belong to d-block. (ii) both have same number of electrons.

(iii) both have similar atomic radius.

(iv) both belong to the same group of the periodic table. 17.Among the following pairs of ions, the lower oxidation state in aqueous solution is

more stable than other, in (i) Tl+,Tl3+ (ii) Cu+,Cu2+ (iii) Cr2+,Cr3+ (iv) V2+, VO2+

18.Which of the following ion has the electronic configuration 3d6 ? (i) Ni3+ (ii) Mn3+ (iii) Fe2+ (iv) Co3+ 19.Which of the following statements is not correct?

(i) La(OH)3 is less basic than Lu(OH)3

(ii) La is actually an element of transition series rather than Lanthanoids

(iii) Atomic radius of Zr and Hf is same (iv) In Lanthanoid series, the ionic radius of Lu3+ is smallest

20. Among the following actinide pairs, the maximum oxidation states is shown by (i)U and Np (ii)Np and Pu (iii)Pu and Th (iv)U and Pa 21. Which of the following element is used in treatment of cancer?

(i) Th (ii) U (iii) Pu (iv) Np 22. On addition of small amount of KMnO4 to concentrated H2SO4, a green oily

compound is obtained which is highly explosive in nature. Identify the compound from the following. (i) Mn2O7 (ii) MnO2 (iii) MnSO4 (iv) Mn2O3

Assertion and Reason Type Questions In the following questions (Q. No. 23 - 27) a statement of Assertion followed by a statement of Reason is given. Choose the correct answer out of the

following choices. 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.

(i) Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.

(ii) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (iii) Assertion is correct statement but Reason is wrong statement

(iv) Assertion is wrong statement but Reason correct statement. 23. Assertion : Cu2+ iodide is not known.

Reason : Cu2+ oxidises I– to iodine. 24. Assertion : Separation of Zr and Hf is difficult.

Reason : Because Zr and Hf lie in the same group of the periodic

table. 25. Assertion : Actinoids form relatively less stable complexes as compared to

lanthanoids. Reason : Actinoids can utilise their 5f orbitals along with 6d orbitals in

bonding but lanthanoids do not use their 4f orbital for bonding. 26. Assertion : Cu cannot liberate hydrogen from acids. Reason : Because it has positive electrode potential.

27. Assertion : The highest oxidation state of osmium is +8. Reason : Osmium is a 5d-block element.

2 Mark questions:

28. Why Eo values for Mn, Ni and Zn are more negative than expected?

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29. Why first ionisation enthalpy of Cr is lower than that of Zn ?

30.Transition elements show high melting points. Why?

31.Out of Cu2Cl2 and CuCl2, which is more stable and why?

32.Although fluorine is more electronegative than oxygen, but the ability of oxygen to

stabilise higher oxidation states exceeds that of fluorine. Why?

33.Although Cr3+ and Co2+ ions have same number of unpaired electrons but the

magnetic moment of Cr3+ is 3.87 B.M. and that of Co2+ is 4.87 B.M. Why?

34.Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U. Why?

35.Although Zr belongs to 4d and Hf belongs to 5d transition series but it is quite

difficult to separate them. Why?

36.A solution of KMnO4 on reduction yields either a colourless solution or a brown

precipitate or a green solution depending on pH of the solution. What different stages

of the reduction do these represent and how are they carried out?

37.The halides of transition elements become more covalent with increasing oxidation

state of the metal. Why?

THREE MARK QUESTIONS: 38. Explain briefly how +2 state becomes more and more stable in the first half

of the first row transition elements with increasing atomic number?

39. What may be the stable oxidation state of the transition element with the following

d electron configurations in the ground state of their atoms : 3d5 and 3d8 ?

40. What is lanthanoid contraction? What are the consequences of lanthanoid

contraction?

41. Explain giving Reasons:

(i) The enthalpies of atomisation of the transition metals are high.

(ii) The transition metals generally form coloured compounds.

(iii) Transition metals and their many compounds act as good catalyst.

42. Predict which of the following will be coloured in aqueous solution? Ti3+, V3+,

Cu+ , Sc3+, Mn2+, Fe3+ . Give Reasons for each.

43. How would you account for the following:

(i) Of the d4 species, Cr2+ is strongly reducing while manganese(III) is strongly

oxidising.

(ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents

it is easily oxidised.

44. Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate

its magnetic moment on the basis of ‘spin-only’ formula

45. What can be inferred from the magnetic moment values of the following complex

species ?

Compound Magnetic Moment (BM)

(i) K4[Mn(CN)6) 2.2

(ii) [Fe(H2O)6] 2+ 5.3

46. What are inner transition elements? Decide which of the following atomic numbers

are the atomic numbers of the inner transition elements : 29, 59, 74, 95, 102, 104.

47. Calculate the number of unpaired electrons in the following gaseous ions: Mn3+,

Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution?

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FIVE MARK QUESTIONS:

48. Compare the chemistry of the actinoids with that of lanthanoids with reference to:

(i) electronic configuration (ii) oxidation states

49.On the basis of Lanthanoid contraction, explain the following :

(i) Nature of bonding in La2O3 and Lu2O3.

(ii) Trends in the stability of oxo salts of lanthanoids from La to Lu.

(iii) Stability of the complexes of lanthanoids.

(iv) Radii of 4d and 5d block elements.

(v) Trends in acidic character of lanthanoid oxides.

50.(a) Answer the following questions :

(i) Which element of the first transition series has highest second ionisation enthalpy?

(ii) Which element of the first transition series has highest third ionisation enthalpy?

(iii) Which element of the first transition series has lowest enthalpy of atomisation?

(b)Identify the metal and justify your answer.

(i) Carbonyl M (CO)5 (ii) MO3F

51. Mention the type of compounds formed when small atoms like H, C and N get

trapped inside the crystal lattice of transition metals. Also give physical and chemical

characteristics of these compounds.

52. (a) Transition metals can act as catalysts because these can change their oxidation

state.

How does Fe(III) catalyse the reaction between iodide and persulphate ions?

(b)Mention any three processes where transition metals act as catalysts.

ANSWERS

1.(i) a (ii) c (iii)d (iv)d

2.(i)b (ii)a (iii)d (iv)a

3.(ii)-26 4. (i) Cu(II) is more stable 5. (iv)Cu 6. (ii)CuF2

7. (ii)3d5 8. (ii) +3 9. (iii)Tm 10. (i)V2O5, Cr2O3

11. (i) [Xe] 4f 75d16s2 12. (iv)They are chemically very reactive

13. (i)2.87 B.M. 14. (i) Copper liberates hydrogen from acids.

15. (iv)in covalent compounds fluorine can form single bond only while oxygen

forms double bond.

16. (iii)both have similar atomic radius. 17. (i) Tl+,Tl3+

18. (iv) Co3+ 19. (I) La(OH)3 is less basic than Lu(OH)3

20. (ii)Np and Pu 21. (i) Th 22. (i) Mn2O7

23.(i) 24.(ii) 25.(iv) 26.(i) 27.(ii)

28. Negative Eo values for Mn2+ and Zn2+ are related to stabilities of half filled and

fully filled configuration respectively. But for Ni2+ , Eo value is related to the highest

negative enthalpy of hydration.

29. Ionisation enthalpy of Cr is lower due to stability of d 5 and the value for Zn is

higher because its electron comes out from 4s orbital.

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30. The high melting points of transition metals are attributed to the involvement of

greater number of electrons in the interatomic metallic bonding from (n-1) d-orbitals

in addition to ns electrons

31. CuCl2 is more stable than Cu2Cl2. The stability of Cu2+ (aq.) rather

than Cu+(aq.) is due to the much more negative ΔhydH of Cu2+ (aq.) than Cu+(aq.).

32. It is due to the ability of oxygen to form multiple bonds to metals.

33. Due to symmetrical electronic configuration there is no orbital contribution in

Cr3+ ion. However appreciable orbital contribution takes place in Co2+ ion.

34. It is because in the beginning, when 5f orbitals begin to be occupied, they will

penetrate less into the inner core of electrons. The 5f electrons will therefore, be

more effectively shielded from the nuclear charge than 4f electrons of the

corresponding lanthanoids.

35. Due to lanthanoid contraction, they have almost same size (Zr, 160 pm) and (Hf,

159 pm).

36.

37. As the oxidation state increases, size of the ion of transition element decreases.

As per Fajan’s rule, as the size of metal ion decreases, covalent character of the bond

formed increases.

38. It can be easily observed that except Sc, all others metals display +2 oxidation

state. Also, on moving from Sc to Mn, the atomic number increases from 21 to 25.

This means the number of electrons in the 3d-orbital also increases from 1 to 5

+2 oxidation state is attained by the loss of the two 4s electrons by these metals.

Since the number of d electrons in (+2) state also increases from Ti(+2) to Mn(+ 2),

the stability of +2 state increases (as d-orbital is becoming more and more half-

filled). Mn (+2) has d5 electrons (that is half-filled d shell, which is highly stable).

39.

40. The 4f electrons have poor shielding effect. Therefore, the effective nuclear

80

charge experienced by the outer electrons increases. Consequently, the attraction of

the nucleus for the outermost electrons increases. This results in a steady decrease

in the size of lanthanoids with the increase in the atomic number.

Consequences of lanthanoid contraction

(i)There is similarity in the properties of second and third transition series.

(ii)It is due to lanthanide contraction that there is variation in the basic strength of

lanthanide hydroxides. (Basic strength decreases from La(OH)3 to Lu(OH)3.)

41. (i) Transition elements have high effective nuclear charge and a large number of

valence electrons. Therefore, they form very strong metallic bonds. As a result, the

enthalpy of atomization of transition metals is high.

(ii). This is because of the absorption of radiation from visible light region to promote

an electron from one of the d−orbitals to another. In the presence of ligands, the d-

orbitals split up into two sets of orbitals having different energies.

(iii) The catalytic activity of the transition elements can be explained by two basic

facts.

(a) Owing to their ability to show variable oxidation states

(b)Transition metals also provide a suitable surface for the reactions to occur

42. on the basis of electronic configuration that only Sc3+ has an empty d-orbital. All

other ions, except Sc3+, will be coloured in aqueous solution because of d−d

transitions.

43. (i)Cr2+ is strongly reducing in nature. It has a d4 configuration. While acting as a

reducing agent, it gets oxidized to Cr3+ (electronic configuration, d3). This d3

configuration can be written as configuration, which is a more stable configuration.

In the case of Mn3+ (d4), it acts as an oxidizing agent and gets reduced to Mn2+ (d5).

This has an exactly half-filled d-orbital and is highly stable.

(ii) Co(II) is stable in aqueous solutions. However, in the presence of strong field

complexing reagents, it is oxidized to Co (III). Although the 3rd ionization energy for

Co is high, but the higher amount of crystal field stabilization energy (CFSE) released

in the presence of strong field ligands overcomes this ionization energy.

44. n = number of unpaired electrons In Ce, n = 2 2.828 BM

45. (i) K4[Mn(CN)6]

For in transition metals, the magnetic moment is calculated from the spin-only

formula. Therefore,

We can see from the above calculation that the given value is closest to . Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons

in the d-orbital.

81

Hence, we can say that CN− is a strong field ligand that causes the pairing of electrons.

(ii) [Fe(H2O)6]2+

We can see from the above calculation that the given value is closest to . Also, in this complex, Fe is in the +2 oxidation state. This means that Fe has 6 electrons

in the d-orbital.

Hence, we can say that H2O is a weak field ligand and does not cause the pairing of

electrons.

46. Inner transition metals are those elements in which the last electron enters the f-orbital. The elements in which the 4f and the 5f orbitals are progressively filled are

called f-block elements. Among the given atomic numbers, the atomic numbers of the inner transition elements are 59, 95, and 102.

47. 4,3,2,1 Cr3+ is the most stable in aqueous solutions owing to a configuration.

48. Electronic configuration:

The general electronic configuration for lanthanoids is [Xe]54 4f0-14 actinoids is [Rn]86 5f1-14 6d0-1 7s2. Unlike 4f orbitals, 5f orbitals are participate in bonding to a greater

extent. 5d0-1 6s2 and that for not deeply buried and

Oxidation states:

This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids.

Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.

49. (i) As the size decreases covalent character increases. Therefore La2O3 is more

ionic and Lu2O3 is more covalent.

(ii)As the size decreases from La to Lu, stability of oxosalts also decreases.

(iii)Stability of complexes increases as the size of lanthanoids decreases.

(iv)Radii of 4d and 5d block elements will be almost same.

(v)Acidic character of oxides increases from La to Lu.

50. (a) (i) Cu, because the electronic configuration of Cu is 3d104s1. So second electron needs to be removed from completely filled d-orbital.

(ii) Zn

(iii) Zn [ No unpaired electron for metallic bonding]

(b) (i) Fe(CO)5 [ EAN rule]

(ii) MnO3F [ Mn shows +7 oxidation state; d-electrons are not involved in bonding.]

51. Interstitial compounds. Characteristic properties :

(i) High melting points, higher than those of pure metals.

(ii) Very hard.

(iii) Retain metallic conductivity.

(iv) Chemically inert.

52.

82

83

CHAPTER-8 COORDINATION COMPOUNDS

CASE-STUDY BASED QUESTIONS 1 Crystal Field Theory

CFT is based on interaction of the five (n − 1)d orbitals with ligands arranged

in a regular array around a transition-metal ion. According to CFT, an octahedral metal complex forms because of the electrostatic interaction of a positively charged metal ion with six negatively charged ligands or with the

negative ends of dipoles associated with the six ligands. The lowest-energy arrangement of six identical negative charges is an octahedron. The five d

orbitals split into two groups whose energies depend on their orientations. The energy of an electron in the orbitals dz

2 and dx2−y2 ( eg orbitals) will be

greater than it will be for a spherical distribution of negative charge because

of increased electrostatic repulsions. In contrast, the other three d orbitals (dxy, dxz, and dyz, collectively called the t2g orbitals) are all oriented at a 45°

angle to the coordinate axes, so they point between the six negative charges. The energy of an electron in any of these three orbitals is lower than the energy for a spherical distribution of negative charge. The difference in energy

between the two sets of d orbitals is called the crystal field splitting energy (Δo), where the subscript o stands for octahedral. It is important to note that

the splitting of the d orbitals in a crystal field does not change the total energy of the five d orbitals: the two eg orbitals increase in energy by 0.6Δo, whereas the three t2g orbitals decrease in energy by 0.4Δo. Thus, the total change in

energy is

2(0.6Δo)+3(−0.4Δo)=0. (Source of data: Duward F. Shriver, Peter W. Atkins, and Cooper H. Langford,

Inorganic Chemistry, 2nd ed) 1. Which of the following shall form an octahedral complex?

(a) d4 (low spin) (b) d8 (high spin) (c) d6 (low spin)

(d) All of these 2. For a high spin d4 octahedral complex the crystal field splitting energy will

be (a)(a) -1.6 Δo

(b) -0.8 Δo

(c) -0.6 Δo (d) -1.2 Δo

3. Which of the following factors affects the magnitude of CFSE ? (a) Charge on metal ion (b) Nature of ligand

(c) Nature of metal atom (d) All of these

4. The hybridisation involved in [Co(C2O4)3]3- is (a)sp3d2 (b)sp3d3

(c)dsp3 (d) d2sp3

5. Which of the following complexes shows zero crystal field stabilization energy?

(a) [Co(H2O)6]3+

(b) [Fe(H2O)6]3+

(c) [Co(H2O)6]2+

(d) [Mn(H2O)6]3+

2. A research was conducted for studying co-ordination No. & oxidation No. in complexes. Elements most often exhibit CN6.The next CN in decreasing order

are 4,5,7&8.CN 3,5,9 are less common, which are followed by10-12 & very uncommon is CN1. In control samples CN6&4 are dominant . most complexing agents have ON +2. ONs +3 & +1 are less common.

1.The CN & ON of Co in [Co(en)3] +3 is

84

a)3,6 b)6,6

c)3,3 d)6,3

Ans. (d)

2. In Ni(CO)4 the ON of Ni is a) 0

b) 1 c) 2 d) 3

Ans. 0

3. Give the oxidation number and coordination number of the central metal atom in

the complex compound K3[Fe(C2O4)3] a)3,6

b)0.0 c)6,0

d)2,3 Ans. (a)

4.Two common structures are observed for four-coordinate metal

complexes: a) tetrahedral and square planar.

b) Tetrahedral & square pyramidal c) Trigonal planar, tetrahedral d) Octahedral, square planar

Ans. (a)

CASE STUDT BASED QUESTIONS:TYPE 2

3. In the formation of metal complexes in an aqueous medium, equilibrium constant or stability constant is used to determine the strength of interaction

between reagents that make the final product after the formation of bonds. In general stability means how long a complex may be existing under suitable

conditions. Stability constant of the formation of metal complexes is used to measure interaction strength of reagents. The stability of a chelate complex depends on the size of the chelate rings

Study the two statements labelled as Assertion (A) and Reason (R). (a) Both, A and R, are true and R is the correct explanation of A

(b) Both, A and R, are true but R is not the correct explanation of A (c) If A is true but R is false (d) If A is false but R is true

1.Assertion: If formation constants denotes stability constants of complexes

, those for the reverse reaction will denote the instability of the complexes. Reason: The reverse reaction of formation of complex would be dissociation reaction, hence it will denote instability of the complex.

Ans : (a) 2. Assertion: Tetrahedral complexes are generally more stable than

Octahedral. Reason: Crystal field splitting energy is greater in octahedral complexes

Ans : (d)

3.. Assertion : Out of NH3 and CO, the ligand that forms a more stable complex with a transition metal is CO.

Reason : CO can form more stable complex than NH3 because it is involved in synergic bonding or back bonding.

Ans : (a)

4. Assertion : [CO(NH3)6]3+ is more stable since than [Co(en)3]3+ Reason : ‘en’ is didentate ligand NH3 is unidentate ligand.

Ans : (d)

4. The magnetic properties of a compound can be determined from its electron

configuration and the size of its atoms. Because magnetism is generated by electronic spin, the number of unpaired electrons in a specific compound indicates how magnetic the compound is. In this section, the magnetism of

the d-block elements (or transition metals) are evaluated. These compounds tend to have a large number of unpaired electrons. the diverse range of colors

seen in salts of transition-metal ions and their aqueous solutions. In general, the color of a complex depends on the identity of the metal ion, on its oxidation state, and on the ligands bound to it.

85

1. Assertion: Coloured complexes become colourless once the ligands are removed. Reason: The d – d transitions are not possible since d orbitals become degenerate. (A)

2. Assertion: The colour of a complex changes when one ligand is replaced by another. Reason: The magnitude of d-orbital splitting is

decide by the nature of the ligand. (D)

3. Assertion: ([Fe(CN)6]3- ion shows magnetic moment corresponding to two unpaired electrons.

Reason: Because it has d2sp3 type hybridisation. (D)

4. Assertion: [Fe(CN)6] 3- is more paramagnetic than[FeF6] 3- . Reason: [FeF6] 3- has more number of unpaired electrons. (D)

********************************************

ASSERTION-REASON Qs.

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.

(A) Assertion and Reason both are true, Reason is correct explanation of Assertion.

(B) Assertion and Reason both are true but Reason is not the correct explanation of Assertion. (C) Assertion is true, Reason is false.

(D) Assertion is false, Reason is true.

1. Assertion: EDTA is used to treat lead poisoning. Reason: EDTA forms very stable complexes with lead. (A)

2. Assertion: Carbon monoxide forms low spin complexes with metals. Reason: Carbon monoxide is neutral oxide. (B)

3. Assertion: Carbon monoxide is a deadly poison. Reason: CO can form strong complexes with haemoglobin. (A)

4. Assertion: [Co(NH3)6] 3+ is an inner orbital complex. Reason: [Co(NH3)6] 3+ forms an octahedral complex. (B)

5. Assertion. Tetrahedral complexes cannot exhibit geometrical isomerism. Reason: Tetrahedral complexes are chiral in nature. (C)

2MARKS Qs

1 Define Coordination entity of coordination compounds. ANS. A coordinate entity constitutes a central metal atom or ion bonded to

a fixed number of ions or molecules (ligands).e.g: [Fe(CN)6] 4-.

2 What are homoleptic complexes? Give an example.

Ans.Homoleptic complexes are the complexes in which central metal ion or atom is bound to only one type of donor groups.

e.g: K4[Fe(CN)6]

3 Square planer complexes with coordination number 4. Show geometrical

isomerism whereas tetrahedral complexes do not. Why? Ans: In tetrahedral complexes relative positions of the ligands with respect to each other is same.

4 Write the formula of following: a)Ferric hexacynoferrate(II)

b)Sodium tetrafluorido oxochromate(IV) Ans:a) Fe4 [Fe(CN)6]3

b)Na2[CrF4 O]

5 Write the IUPAC name of [Cr(NH3 )5 NCS] [ZnCl4 ]

Ans: Pentaammine isothiocyanato chromium(III) tetrachloridoZincate(II).

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6 CuSO4 on mixing with NH3 ratio(1:4) does not give test for Cu2+ but gives test for SO4

2- why?

Ans: Because complex formed is [Cu(NH3 )4 ]SO4 in which Cu2+ is non ionisable SO4

2- is ionisable.

7 What is meant by stability of coordination compounds state the factors which govern stability of complexes? Ans: The stability of complexes means the degree of association between

two species involved in the state of equilibrium. M+4L ML4

8 Which of two is more stable K4[Fe(CN)6]or K3[Fe(CN)6] and why? Ans) K3[Fe(CN)6] because Fe3+ has smaller size and higher charge than

Fe2+

9 NH3 has strong ligand but NH4+ is not why?

Ans) due to presence of loan pair of electron in NH3 but not in NH4+.

Moreover NH4+ is positively charged therefore repelled by central metal ion

.

10 [Ti(H2O)]3+ is coloured but [Sc(H2O)]3+ is colourless why? Ans.Ti3+ has one unpaired d-electron and can undergo d-d transition. But

Sc3+ does not have unpaired d electron.

3MARKS Qs

1 A 0.01m complex of Cocl3 and NH3 (molar ratio 1:4) is found to give freezing

point depression equal to 0.0372 . suggest the formula of the complex . Kf =1.86 K/m .

Ans: ∆Tf = Kf * m 0.0372 = 1.86 * m

m = 0.02 since molality becomes double (m = 0.02 ) it means complex will produce two ions . Hence the formula is

[ CO( NH3) Cl2 ] Cl .

2 [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

Also, NH3 is a weak field ligand that does not cause the pairing of electrons in 3d

orbital.

Therefore hybridisation

is d2sp3 & electrons are

unpaired, hence it is

paramagnetic in nature

2−

configuration.

In [Ni(CN)4] , Ni exists in the +2 oxidation state i.e., d8

Ni2+:

CN− is a strong field ligand. It causes the pairing of the 3d orbital electrons.

Then, Ni2+ undergoes dsp2 hybridization.

3 The central ion Co3+ with co-ordination number 6 is bonded to the ligands

NH3 and Br- to form a dipositive complex ion. Write the formula and IUAC name of the complex ion.

ANS.[Co(NH3)5 Br]+2 PentaamminebromoCobalt(III)

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4 [Fe(H2O)6] 3+ is strongly paramagnetic, whereas [Fe(CN)6] 3- is weakly paramagnetic. Write the Reason

Ans. H2O is a weak ligand, cannot cause pairing up of electrons whereas CN- is a strong ligand hence electrons are paired up so compound exhibits diamagnetism.

5 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 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.

6 NiCl4]2-is paramagnetic while [Ni(CO)4] is diamagnetic though both are

tetrahedral .Why? Ans: Ni is in +2 oxidation state in [NiCl4]2- with electronic configuration 3d84s0 . Cl- is a weak ligand . It cannot pair up the electrons .

In [Ni(CO)4] , Ni is in zero oxidation state and electronic configuration 3d84s2 . In the presence of CO ligand , The 4s electrons shift to 3d to pair

up with 3d electrons so no unpaired electron is left.

7 What is crystal field splitting? (1).

How does the magnitude of ∆o and P decide the actual configuration of d4 in the presence of octahedral field?(2) Ans. The splitting of otherwise degenerate d- orbitals into two sets in the

effect of asymmetrical negative field of ligands is called crystal field splitting. Ans7. When the crystal field splitting energy, ∆ > o pairing energy

,P, then the fourth electron of d 4 configuration gets paired and the actual configuration becomes t g4 eg0

8 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− < S2− < SCN− < Cl−< N3 < F− < OH− < C2O42− H2O < NCS−

H− < CN− < NH3< en SO32− < NO2− < phen < CO

9 State the postulates of Werner’s theory of coordination compounds.

Ans. Postulates: 1. Central metal ion in a complex shows two types of valences - primary valence and secondary valence. 2. The primary valence

is ionisable and satisfied by negative ions. 3. The secondary valence is non ionisable. It is equal to the coordination number of the central metal ion or atom. It is fixed for a metal. Secondary valences are satisfied by negative

ions or neural molecules (ligands). 4. The primary valence is non directional. The secondary valence is directional. Ions or molecules attached

to satisfy secondary valences have characteristic spatial arrangements. Secondary valence decides geometry of the complex compound.

10. What are the limitations of Werner’s theory of coordination compounds? Ans. This theory fails to explain why, a) a few elements have the property to form coordination compounds b) the bonds in coordination compounds

have directional properties c) coordination compounds have characteristic magnetic and optical properties.

5MARKS Qs 1 A metal ion Mn+ having d4 valence electronic configuration combines with

three bidentate ligands to form a complex. Assuming ∆0 is greater than P.

i) Draw diagram showing d-orbital splitting. ii) Write the electronic configuration of in terms of t2g and eg .

iii) What type of hybridization will MN+ have. iv) Name the type of isomerism it shows.

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Ans: i)

ii)t2g

4 eg

o

iii)sp3 d2

iv) Optical isomerism.

2 1. The central ion Ag+ with co-ordination number 2 forms a positive complex ion with NH3 ligand. Also, Ag+ forms a negative complex with CN- ligand.

a) Write the formulae of the above positive and negative complex ions. Give the IUPAC name of each.

b) Give the denticity of NH3 and CN- ligands. Ans. a) [Ag(NH3)2]+ & [Ag(CN)2]-

1.DiammineArgentum(I)

2.DicyanoArgentate(I) NH3 & CN- are unidentate ligands

3 1. Na2EDTA is used in the estimation of hardness of water.

a) Draw the structure of EDTA4- b) What is its denticity? c) What are the donor atoms in it?

d) Why is it called a chelating ligand? Ans.a) structure of EDTA

b) hexadentate c)N & O

d) Bcoz it forms ring like structures with central Metal ions.

4 What is crystal field splitting? Explain crystal field splitting in tetrahedral entities using energy level diagram.

Ans. In an isolated gaseous central metal atom or ion, all the five d orbitals are having same energy. i.e they are degenerated. In the presence of attacking ligands, it becomes asymmetric and the d orbitals lose

degeneracy resulting in splitting of d orbitals. This is called crystal field splitting. In an octahedral complex, six ligands surround the central metal

ion. dx2 – y2 and dz2 orbitals (called eg set) are directed along the direction of ligands and experience more repulsion. They have more energy. dxy, dyz and dzx orbitals (called t2g set) are directed between the axes of attacking

ligands and experience lesser repulsion by the ligands. They have lesser energy. The energy separation between two split sets is denoted as ∆o .

The energy of eg orbitals increase by 3/5 ∆o and that of t2g set decrease by 2/5 ∆

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5 a) Classify the following ligands into unidentate, didentate and polydentate ligands.

NH3, EDTA, oxalate. Ans. NH3-unidentate, EDTA-polydentate, Oxalate-bidentate b) Write one difference between double salts and complex salts with respect

to their ionisation. Give one example for each type of salt.

Double salt Complex salt Double salt is stable only in solid state, but dissociate into simple ions completely in solution state. Complex salt is stable both in solid and solution state and does not dissociate completely in

solution state. E.g: KCl.MgCl2.6H2O E.g: K4[Fe(CN)6]

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Chapter:9 Haloalkanes and Haloarenes

CASE STUDY BASED QUESTIONS

1 Read the passage given below and answer the following questions:

(1x4=4) Nucleophilic substitution reactions are characteristic reactions of alkyl halide.the polarity of the carbon halogen bond is responsible for this reaction where a nucleophile replaces the halogen substituent. These reactions occurs through

two main mechanism-SN1 andSN2.experiments have shown that the factors affecting the relative rates of SN1 and SN2reactions are structure of substrate, reactivity and

concentration of nucleophile, nature of solvent and nature of leaving group.SN1 reaction in which the leaving group departs from a stereogenic carbon, results in extensive and sometimes complete racemisation while reaction with SN2 mechanism,

results in completes inversion of configuration. (a) In the solvolysis of 3-Methyl-3 chlorohexane, which of the following statement is

not correct? (i)It involves carbocation intermediate (ii) the intermediate involves sp2hybridised carbon

(iii)it involve inversion of configuration (iv)the rate of reaction depends on 3-Methyl-3 chlorohexane concentration.

Ans

(iii)

(b)Optically active alkyl halides undergo SN1reaction with (i)Racemization

(ii) Inversion of configuration (iii) Retention of configuration

(iv)None of these

(i)

(c)An SN2reaction at an asymmetric carbon of a compound will result in formation of

(i) An enantiomer of substrate (ii) A product with opposite optical rotation (iii) A mixture of diastereomers

(iv)a single stereoisomer

(iv)

(d)Which of the following is an example of SN1 reaction

(i) CH3Br+OH- → CH3OH +Br-

(ii) (CH3)2CHBr+OH- → (CH3)2CHOH +Br- (iii) CH3CH2OH→ CH2= CH2

(iv) (CH3 )3CBr+OH- → (CH3)3COH +Br-

(IV)

2 Read the passage given below and answer the following questions: (1x4=4) Alcohols can be converted into alky halides by displacement of the

hydroxyl group by a halogen atom. This displacement can be carried out by treatment of the alcohol with a number of reagents as halogen acids, phosphorous halides and thionyl chlorides. Thionyl chloride method is however is preferred because it gives

expected products in almost pure state with- out the danger of rearrange product being formed.

(a)Thionyl chloride is preferred over phosphorous penta chloride method for the preparation of alkyl chloride because-

(i) the reaction goes to completion. (ii) the by products being gases escape into the atmosphere leaving behind almost

pure alky chloride. (iii) Thionyl chloride is cheap while phosphorus pentachloride is costly. (iv)None of these

(ii)

(b) Which reagent cannot be use to prepare an alky chloride from an alcohol: (i) HCl+ Anhydrous ZnCl2 (ii) NaCl

(iii) PCl5 (iv) SOCl2

(ii)

(c) Neo-pentyl alcohol on treatment with HBr gives (i) Neo-pentyl bromide (ii) 2-bromo-2-methylbutane

(iii) 2-methyl-2-butene (iv) 2-methyl-1-butene

(ii)

(d) Solvolysis Neo-pentyl bromide in ethanol mainly gives: (i) neo-pentyl alcohol (ii) t-butyl ethyl ether (iii) 2-methylbutan-2-ol (iv) 2-ethoxy-2-methylbutane

(iv)

3 Read the passage given below and answer the following questions: (1x4=4)

91

Elimination reactions of alkyl halides are important reactions that compete with substitution reactions. An alkyl halide with α-hydrogen atoms when

reacted with a base or a nucleophile has two competing routes: substitution (SN1 and SN2) and elimination. Which route will be taken up depends upon the nature of alkyl halide, strength and size of base/nucleophile and reaction conditions. Thus, a bulkier

nucleophile will prefer to act as a base and abstracts a proton rather than approach a tetravalent carbon atom (steric Reasons) and vice versa.

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.

(i) Assertion and Reason both are correct and Reason is correct explanation of Assertion.

(ii) Assertion and Reason both are correct statements but Reason is not correct explanation of Assertion.

(iii) Assertion is correct but Reason is wrong statement. (iv) Assertion and Reason both are wrong statements. (v) Assertion is wrong but Reason is correct statement

(a

)

Assertion : 2-Bromopropane on heating with CH3COONA predominantly gives propene.

Reason : In the presence of weak base elimination reaction predominates.

(iv)

(b)

Assertion : Iso propyl bromide on heating with concentrated solution of alcoholic KOH

predominantly gives propene. Reason :in the presence of strong base elimination reaction predominates.

(i)

(c

)

Assertion : Bromoethane on heating with aqueous NaOH gives ethanol. Reason : Primary alkyl halides prefers to undergo substitution reaction.

(i)

(d

)

Assertion : 2-bromopentane gives pent-2-ene as the major product when heated with alcoholic KOH.

Reason : In dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.

(i)

4 Read the passage given below and answer the following questions:

(1x4=4) Nucleophilic substitution reactions are characteristic reactions of haloalkanes. In this type of reaction, a nucleophile reacts with haloalkane (the

substrate) having a partial positive charge on the carbon atom bonded to halogen. A substitution reaction takes place and halogen atom, called leaving group departs as halide ion. Since the substitution reaction is initiated by a nucleophile, it is called

nucleophilic substitution reaction.

Assertion : Nucleophilic substitution is more difficult to carry out on vinyl chloride

than ethyl chloride. Reason : vinyl group is an electron donating group.

(iii)

Assertion : It is difficult to replace chlorine by –OH in chlorobenzene in comparison to that in chloroethane.

Reason : Chlorine-carbon (C—Cl) bond in chlorobenzene has a partial double bond character due to resonance.

(i)

Assertion : Presence of a nitro group at ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution. Reason : Nitro group, being an electron withdrawing group decreases the electron

density over the benzene ring.

(i)

Assertion : KCN reacts with methyl chloride to give methyl isocyanide

Reason : CN– is an ambident nucleophile. (v)

MCQ TYPE QUESTIONS Ans.

1 Out of the following ,the alkene that exhibits optical isomerism is:

(i) 2-Methyl-2-pentene (ii) 3-Methyle-2- pentene (iii) 4-Methyle-1- pentene (iv) 3-Methyle-1- pentene

(iv)

2 The organic chloro compound, which shows complete stereo chemical inversion during

a SN2 reaction ,is (i) (C2H5)2CHCl (ii) (CH3)3CCl

(iii) (CH3)2CHCl (iv) CH3Cl

(iv)

3 Which of the following reaction will yield 2,2-Dibromopropane:

(i) CH3-C≡CH +2HBr → (ii) CH3-CH=CH-Br +HBr →

(iii) CH≡CH +2HBr → (iv) CH3-CH=CH2 +HBr →

(i)

4 In the reaction (CH3)2CH-CH=CH2 +HBr → A ,The compound A (predominant) is (i) 2-Bromo-3 methylbutane (ii) 1-Bromo-3 methylbutane (iii) 2-Bromo-2 methylbutane (iv) 1-Bromo-2 methylbutane

(iii)

5 The reagent for following conversion is (are)

CH2Br-CH2Br → CH≡CH (ii)

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(i) Alcoholic KOH (ii) Alcoholic KOH followed byNaNH2 (iii) Aqueous KOH followed byNaNH2 (iv) Zn/methanol

6 Flurobengene can be synthesized in laboratory (i) by heating phenol with HF and KF

(ii) by heating the diazonium salt with HBF4 (iii) by direct fluorination of benzene with F2 gas (iv) by heating 1-Bromobenzene with NaF solution

(ii)

7 Toluene reacts with a halogen in the presence of iron (III) chloride gives X and

inpresence of light gives Y then X and Y are (i) X= m-Chlorotoluene ,Y= p-Chlorotoluene

(ii) X= o and p-Chlorotoluene ,Y= Trichloromethylbengene (iii) X= Benzyl chloride ,Y= m-Chlorotoluene (iv) X= Benzyl chloride ,Y= o-Chlorotoluene

(ii)

8 Chlorobenzene is formed by reaction of chlorine with benzene in the presence

of AlCl3. Which of the following species attacks the benzene ring in this reaction? (i) Cl– (ii) Cl+

(iii) AlCl3 (iv) [AlCl4]–

(ii)

9 Consider the following reaction and answer the questions:

Which of the following statements are correct about this reaction?

(i) In (c) carbon atom is sp3 hybridised. (ii) (b) and (d) have opposite configuration.

(iii) (b) and (d) have same configuration. (iv) The given reaction follows SN1 mechanism.

(ii)

10 Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho

and para halo compounds. The reaction is (i) Electrophilic elimination reaction (ii) Electrophilic substitution reaction

(iii) Free radical addition reaction (iv) Nucleophilic substitution reaction

(ii)

11 Which of the following compounds will give racemic mixture on nucleophilic substitution by OH– ion?

(i) (a) (ii) (a), (b), (c)

(iii) (b), (c) (iv) (a), (c)

(i)

12 Arrange the following compounds in increasing order of rate of reaction towards nucleophilic substitution:

(i) (a) < (b) < (c) (ii) (c) < (b) < (a)

(iii) (a) < (c) < (b) (iv) (c) < (a) < (b)

(iii)

13 Arrange the following compounds in increasing order of rate of reaction towards

nucleophilic substitution:

(i) (c) < (b) < (a) (ii) (b) < (c) < (a) (iii) (a) < (c) < (b) (iv) (a) < (b) < (c)

(iv)

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14 Arrange the following compounds in increasing order of rate of reaction towards nucleophilic substitution:

(i) (a) < (b) < (c) (ii) (a) < (c) < (b) (iii) (c) < (b) < (a)

(iv) (b) < (c) < (a)

(iv)

15

Which of the following statements are correct about the mechanism of above reaction?

(i) A carbocation will be formed as an intermediate in the reaction. (ii) OH– will attach the substrate (b) from one side and Cl– will leave it simultaneously

from other side. (iii) An unstable intermediate will be formed in which OH– and Cl– will be attached by weak bonds.

(iv) Reaction proceeds through SN2 mechanism.

(i)

16 Reaction of C6H5CH2Br with aqueous sodium hydroxide follows:

(i) SN1 mechanism (ii) SN2 mechanism

(iii) Any of the above two depending upon the temperature of reaction (iv) Saytzeff rule

(i)

17 Which of the carbon atoms present in the molecule given below are asymmetric?

(i) a, b, c, d (ii) b, c

(iii) a, d (iv) a, b, c

(ii)

18 Which of the following is not chiral?

(i) 2-Hydroxypropanoic acid (ii) 2-Butanol (iii) 2,3-Dibromobutane (iv) 3-Bromopentane

(iv)

19

What is name of following reaction: (i) Wurtz Reaction (ii) Fitting Reaction (iii) Wurtz- Fitting Reaction (iv) swartz reaction

(ii)

20 The position of –Br in the compound in CH3CH=CH-C(Br)(CH3)2 can be classified as

(i) Allyl (ii) Aryl (iii) Vinyl (iv) Secondary

(i)

Assertion and Reason

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. (i) Assertion and Reason both are correct and Reason is correct explanation of Assertion.

(ii) Assertion and Reason both are correct statements but Reason is not correct explanation of Assertion. (iii) Assertion is correct but Reason is wrong statement.

(iv) Assertion and Reason both are wrong statements.

1 Assertion : Phosphorus chlorides (tri and penta) are preferred over thionyl chloride

for the preparation of alkyl chlorides from alcohols. Reason : Phosphorus chlorides give pure alkyl halides.

(iv)

2 Assertion : Aryl iodides can be prepared by reaction of arenes with iodine in the presence of an oxidising agent.

Reason : Oxidising agent oxidises I2 into HI.

(iii)

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3 Assertion : No white precipitate is obtained on addition of silver nitrate to methyl chloride.

Reason : silver nitrate is an ionic compound.

(ii)

4 Assertion : tert-Butyl bromide undergoes Wurtz reaction to give 2, 2, 3, 3-

tetramethylbutane. Reason : In Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbon containing double the number of carbon atoms present in the halide.

(i)

5 Assertion : In reaction of 1-Butene with HBr, 1-Bromobutane is obtained in the presence of peroxide.

Reason :The reaction involved formation of primary radical

(iii)

2 Marks questions

1 Write the IUPAC Name of following compounds: (a)(CCl3)3CCl (b)CHF2CBrClF

Ans: (a) 2-(trichloromethyl)1,1,1,2,3,3,3heptachloropropane (b)1-Bromo-1-chloro-1,2,2-trifluroethane

2 Which one of the following has higher dipole moment: (a)CH2Cl2 (b)CHCl3 Ans : CH2Cl2 (1.60D) > CHCl3 (1.08D)

3 Differentiate between retention and inversion. Retention: Retention of configuration is the preservation of integrity of the spatial

arrangement of bonds to an asymmetric centre during a chemical reaction. Inversion: A process in which the relative configuration of the atoms/groups arounds a chiral center in the product is opposite to that in the reactant

B = Inversion reactant A= Retention

4 How many stereoisomers are possible for following compounds:? (a)2-Aminobutane (b)1,2 -Dichloropropane

Ans: (a)2-Aminobutane

There is only one chiral carbon atom in 2-Aminobutane so two stereoisomers are possible.

(b)1,2- Dichloropropane

There is only one chiral carbon atom in 1,2 -Dichloropropane so two stereoisomers are possible.

5 Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major

alkene: (i) 2-Chloro-2-methylbutane

(ii) 2,2,3-Trimethyl-3-bromopentane. Ans: (i)

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(ii) 2,2,3-Trimethyl-3-bromopentane

6 Out of o-and p-dibromobenzene which one has higher melting point and why?

Ans: p-Dibromobenzene has higher melting point than its o-isomer. It is due

to symmetry of p-isomer which fits in crystal lattice better than the o-isomer.

7 A hydrocarbon C5H10 does not react with chlorine in dark but gives a single

monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon. Ans: A hydrocarbon with the molecular formula, C5H10 belongs to the group with a general

molecular formula CnH2n. Therefore, it may either be an alkene or a cycloalkane. Since

hydrocarbon does not react with chlorine in the dark, it cannot be an alkene. Thus, it

should be a cycloalkane.

Further, the hydrocarbon gives a single monochloro compound, C5H9Cl by

reacting with chlorine in bright sunlight. Since a single monochloro compound is formed, the

hydrocarbon must contain H−atoms that are all equivalent. Also, as all H−atoms of a

cycloalkane are equivalent, the hydrocarbon must be a cycloalkane. Hence, the said

compound is cyclopentane

8 How will you bring about the following conversions? (i) Propene to propyne (ii) Ethanol to ethyl fluoride

Ans (i)

(ii)

9 Identify chiral molecules in each of the following pair of Compounds.

96

Ans:

(i) (ii)

10 A hydrocarbon of molecular mass 72 g mol–1 gives a single monochloro

derivative and two dichloro derivatives on photo chlorination. Give the structure of the hydrocarbon. Ans: C5H12, pentane has molecular mass 72 g mol–1, i.e. the isomer of pentane

which yields single monochloro derivative should have all the 12 hydrogens equivalent ie. neopentane.

3 Marks questions

1 Write the structures of the following organic halogen compounds.

(i) 2-(2-Chlorophenyl)-1-iodooctane (ii) 4-tert-Butyl-3-iodoheptane

(iii) Perfluorobenzene Ans: (i) 2-(2-Chlorophenyl)-1-iodooctane

(ii) 4-tert-Butyl-3-iodoheptane

3

97

(iii) Perfluorobenzene

2. How will you bring about the following conversions?

(i) Ethanol to ethyl fluoride (ii) Propene to 1-nitropropane

(iii) Toluene to benzyl alcohol Ans: (i) Ethanol to ethyl fluoride

(ii) Propene to 1-nitropropane

(iii) Toluene to benzyl alcohol

3. Give Reasons for the following: (i) Ethyl iodide undergoes SN2 reaction faster than ethyl bromide.

(ii) (±) 2–Butanol is optically inactive.

(iii) C–X bond length in halobenzene is smaller than C–X bond length in CH3–X. Ans:

(i) Since I-ion is a better leaving group than Br-ion, hence, CH3I reacts faster than CH3Br in SN2 reaction with OH– ion.

(ii) (±) 2-Butanol is a racemic mixture, i.e., there are two enantiomers in equal

proportions. The rotation by one enantiomer will be cancelled by the rotation due to the other isomer, making the mixture optically inactive.

98

(iii) In CH3–X the carbon atom is sp3 hybridised while in halobenzene the carbon atom

is sp2 hybridised. The sp2 hybridised carbon is more electronegative due to greater s-character and holds the electron pair of C–X bond more tightly than sp3 hybridised carbon with less s-character. Thus, C–X bond length in CH3–X is bigger than C–X in

halobenzene.

4 (a) Out of chlorocyclohexane and chlorobenzene, which one is more reactive

towards nucleophilic substitution reaction and why? (b) Predict all alkenes that would be formed by the dehydrohalogenation of

2-bromobutane. (c) Chloroform contains chlorine but it does not give white precipitate with silver nitrate solutionWhy ?

Ans: a) Cyclohexyl chloride ; Because of partial double bond character of C-Cl bond in

Chlorobenzene / Resonance effect / sp3 hybridised carbon in cyclohexyl chloride whereas sp2 carbon in chlorobenzene. b) 2-Butene and 1-Butene (or structures)

c) All chlorine atoms are bonded to carbon atom by covalent bonds.

5 Draw the structures of major monohalo products in each of the following

reactions: (a)

(b)

(c)

(a)

(b)

(c)

6. Explain why: (i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?

(ii) Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain.

Ans: (i) In cyclohexyl chloride, the carbon atom attached to halogen is sp3 hybridised while in case of chlorobenzene ( haloarene), the carbon atom attached to halogen is sp2-

hybridised.

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The sp2 hybridised carbon with a greater s-character is more electronegative so it has less positive charge density and can hold the electron pair of C—X bond more tightly

than sp3-hybridised carbon in haloalkane with less s-chararcter. So dipole moment of chlorobenzene is lower than that of cyclohexyl chloride

(ii) KCN is predominantly ionic and provides cyanide ions in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs, the attack takes

place mainly through carbon atom and not through nitrogen atom since C—C bond is more stable than C—N bond. However, AgCN is mainly covalent in nature and nitrogen

is free to donate electron pair forming isocyanide as the main product

7. (a) The presence of –NO2 group at ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution reactions (b)The following haloalkanes are hydrolysed in presence of aq KOH.

(i) 1- Chlorobutane (ii) 2-chloro-2-methylpropane Which of the above is most likely to give (i)an inverted product (ii) a racemic mixture: Justify your answer

Ans:(a) -NO2 is an electron withdrawing group / It stabilises carbanion through resonance.

(b) (i)inverted product will be given by 1 Chlorobutane as it undergoes SN2 reaction.

(ii)racemic mixture will be given by 2 chloro-2-methylpropane as it undergoes SN1 reaction

8 (i) Out of (CH3)3C–Br and (CH3)3C–I, which one is more reactive towards SN1 and why ?

(ii) Write the product formed when p-nitrochlorobenzene is heated with aqueous NaOH at 443 K followed by acidification.

(iii) Why dextro and laevo – rotatory isomers of Butan-2-ol are difficult to separate by fractional distillation ? Ans:

(i) (CH3)3C-I , Due to large size of iodine it is better leaving group / Due to lower electronegativity.

(ii)It form p-Nitrophenol as per following reaction.

(iii) It isBecause enantiomers have same boiling points / same physical properties.

9 Among all the isomers of molecular formula C4H9Br, identify (a) the one isomer which is optically active. (b) the one isomer which is highly reactive towards SN2.

(c) the two isomers which give same product on dehydrohalogenation with alcoholic KOH.

Ans:

10 (a) Write equation for preparation of 1-iodobutane from 1-chlorobutane. (b) Out of 2-bromopentane, 2-bromo-2-methylbutane and 1-bromopentane,

which compound is most reactive towards elimination reaction and why ? (c) Give IUPAC name of

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Ans: Dry Acetone

(a) CH3CH2CH2CH2-Cl + NaI ---------------→ CH3CH2CH2CH2-l + NaCI

(b) 2-Bromo-2-methylbutane is more reactive as it gives more substituted alkene on elimination / Gives more stable product according to Saytzeff Rule/ 30 carbocation is more stable.

(c) 4-Bromo-4-methylpent-2- ene.

5 Marks questions

1 Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound

(b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C8H18 which is

different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.

Ans:

(A)CH3-CH(CH3)-CH2Br Isobutylbromide

(B)CH3-C(CH3)=CH2 2-Methylpropene (C) (CH3)3C-Br t-Butyl Bromide

(D) CH3-CH(CH3)-CH2-CH2-CH(CH3)-CH3 2,5-Dimethylhexane There are two primary alkyl halides having the formula, C4H9Br. They are n − bulyl bromide

and isobutyl bromide

Therefore, compound (a) is either n−butyl bromide or isobutyl bromide.

Now, compound (a) reacts with Na metal to give compound (b) of molecular formula,

C8H18, which is different from the compound formed when n−butyl bromide reacts with Na

metal. Hence, compound (a) must be isobutyl bromide

Thus, compound (d) is 2, 5−dimethylhexane. It is given that compound (a) reacts with alcoholic

KOH to give compound (b). Hence, compound (b) is 2−methylpropene.

Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a).

Hence, compound (c) is 2−bromo−2−methylpropane.

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2 (i) Write the mechanism of the following reaction:

ii)Answer following questions:

Ans: (i)

The given reaction is an SN2 reaction. In this reaction, CN− acts as the nucleophile and

attacks the carbon atom to which Br is attached. CN− ion is an ambident nucleophile and

can attack through both C and N. In this case, it attacks through the C-atom.

(ii)

(a)

is more reactive Because it is a primary halide. (b)

It is more reactive due to the presence of electron withdrawing -NO2

group . (c)

It is optically active due to chiral carbon.

3 (i) Define the following terms : (a) Enantiomers (b) Racemic mixture

(ii) Write chemical equations for the following name reactions :

(a) Swarts reaction (b) Wurtz-Fittig reaction (c) Finkelstein reaction.

Ans:

(i) a) The stereoisomers related to each other as non superimposable mirror images

are called enantiomers.

b) Equimolar mixture of d- and l- form is known as racemic mixture (ii)

(a) Swarts reaction:

(b) ) Wurtz-Fittig reaction:

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(c) Finkelstein reaction:

4. (i) Write the mechanism of the reaction between t-Butylbromide and

KOH(aq.). (ii) Write the products of the following reactions: (a)C6H5ONa +C2H5Cl

(b)CH3CH2CH2OH +SOCl2 (c)CH3CH2CH=CH2 +HBr

Ans: (i) The reaction between tert-butyl bromide and hydroxide ion yields tert-butyl alcohol and follows the SN1mechanism.

It occurs in two steps. In step I, the polarised C—Br bond undergoes slow cleavage to produce a carbocation and a bromide ion. The carbocation thus formed is then attacked by nucleophile in step II to complete the substitution reaction

(ii) Write the products of the following reactions

(a)

(b)

(c)

5 (a)Name the alkene which will yield 1-chloro-1-methylcyclohexane by its reactionwith HCl. Write the reactions involved.

(b)Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution.The rate of this reaction depends upon the concentration of the

compound ‘A’only. When another optically active isomer ‘B’ of this compound was treatedwith aq. KOH solution, the rate of reaction was found to be dependent onconcentration of compound and KOH both.

(i) Write down the structural formula of both compounds ‘A’ and ‘B’. (ii) Out of these two compounds, which one will be converted to the product

with inverted configuration.

Ans: (a) 1-Methylcyclohex-1-ene

(b) (i)

(ii)Compound B

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UNIT -10 : ALCOHOLS , PHENOLS AND ETHERS CASE STUDY BASED QUESTIONS

Q 1 Read the passage given below and answer the following questions: (1x4=4)

Both symmetrical and unsymmetrical ethers can be prepared by Williamsons synthesis

which involves the reaction between an alkyl halide and an alkoxide ion. The reverse

process involving the cleavage of ethers to give back the original alkyl halide and the

alcohol can be carried oud by heating the ether with HI at 373K

Ans

(a)

What are the products of heating of tert-Butyl Ethyl Etherwith HI at 373K

(a) isopropyl bromide and sodium iso propoxide

(b) isopropyl bromide and sodium ethoxide

(c) Bromobenzene and sodium phenoxide

(d) sodium tert-butoxide and ethyl bromide

d

(b) which of the following ethers can be prepared by Williamson's synthesis?

(a) benzyl methyl ether (b) methyl vinyl ether

(c) Divinyl ether (d) Diphenyl ether

a

(c) Allyl phenyl ether can be prepared by heating –

(a) C6H5Br + CH2=CH-CH2-ONa (b) C6H5-CH=CH-Br + CH3-ONa

(c) CH2=CH-CH2-Br + C6H5ONa (d) CH2=CH-Br + C6H5-CH2-ONa

c

(d) Benzyl ethyl ether reacts with HI to form-

(a) p- Iodotoluene and ethyl alcohol (b) Benzyl alcohol and ethyl iodide (c) benzyl iodide and ethyl alcohol (d) iodobenzene and ethyl alcohol

c

2 Read the passage given below and answer the following questions: (1x4=4)

Both alcohols and phenols contain a hydroxyl group, but phenols are more acidic

than alcohols. The Reason being that the phenoxide ion left after the removal of

a proton is resonance stabilized while alkoxide is not. The relative strength of o-,

m- and p- substituted phenols, however, depends upon a combination of inductive

and resonance effects of the substituent.

(a) The order of acidic strength of alcohols decreases in the order -

(a) 1∘ alcohol >2∘ alcohol >3∘ alcohol (b) 1∘ alcohol >3∘ alcohol >2∘ alcohol

(b) 2∘ alcohol > 1∘ alcohol >3∘ alcohol (b) 3∘ alcohol >2∘ alcohol >1∘ alcohol

a

(b) The strongest acid among the following is-

(a) m-methoxy phenol (a) p-Methoxy phenol (c) p-Nitro phenol (d) Phenol

c

(c) which of the following does not give NaHCO3 test?

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(a) Phenol (b) alcohol (c) carboxylic acid (d)Both a and b

d

(d) correct order of reactivity of 1∘ alcohol ,2∘ alcohol and3∘ alcohol towards Na metals

is-

(a) 1∘ >2∘ >3∘ (b) 1∘ >3∘ >2∘

(b) 2∘ > 1∘ >3∘ (b) 3∘ >2∘ >1∘

a

CASE STUDT BASED QUESTIONS:TYPE 2

3

Read the passage given below and answer the following questions: (1x4=4)

Phenols are the organic compounds containing benzene ring bonded to a hydroxyl group. They are also known as carbolic acids. Phenols react with active

metals like sodium, potassium to form phenoxide. This reaction of phenol with

metals indicates its acidic nature.Phenols react with aqueous sodium hydroxide too to produce phenoxide ions. This indicates that the acidity of

phenols is higher in comparison to the alcohols and water molecules.Phenoxide ion has greater stability than phenols, as in the case of phenol charge separation takes place during resonance.Electron

withdrawing groups increase acidic nature of phenols whereas electron

donating groups decreases acidic nature of phenols

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.

(i) Assertion and Reason both are correct and Reason is correct explanation of Assertion. (ii) Assertion and Reason both are correct statements but Reason is not

correct explanation of Assertion. (iii) Assertion is correct but Reason is wrong statement.

(iv) Assertion and Reason both are wrong statements. (v) Assertion is wrong but Reason is correct statement

(a)

Assertion : Picric acid does nol turns FeCl3 solution purple Reason : picric acid does not have phenolic group

(iv)

(b) Assertion : Phenol is a strong acid than ethanol.

Reason: Groups with +M effect decreases acidity at p-position.

(ii)

(c) Assertion : : Benzyl bromide when kept in acetone water, it produces benzyl

alcohol. Reason : The reaction follows SN 2 mechanism.

(iii)

(d) Assertion : The major products formed by heating C6H5CH2OCH3 with HI are C6H5CH2 I and CH3OH: Reason : Benzyl cation is more stable than methyl

cation.

(i)

4. Read the passage given below and answer the following questions:

(1x4=4) Ethers are compounds that contain the functional group –O–.

In the IUPAC system, the oxygen atom and the smaller carbon branch are named as an alkoxy substituent and the remainder of the molecule as the base chain, as in alkanes. Ethers can be prepared by

(a) Dehydration of Alcohols with conc. H2SO4 at 413 K (b) Williamson Synthesis

Ethers generally undergo chemical reactions in two ways:

1. Cleavage of C-O bond 2. Electrophilic Substitution

Each of these questions contains an Assertion followed by Reason. Read them

carefully and answer the question on the basis of following options. You have to select the one that best describes the two statements. a) If both Assertion and Reason are correct and Reason is the correct

explanation of Assertion.

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(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(c) If Assertion is correct but Reason is incorrect. (d) If both the Assertion and Reason are incorrect.

(a) Assertion : tert-Butyl methyl ether is not prepared by the reaction of tert-butyl bromide with sodium methoxide

Reason : Sodium methoxide is a strong nucleophile.

b

(b)

Assertion : Ethers behave as bases in the presence of mineral acids.

Reason : Due to the presence of lone pairs of electrons on oxygen

a

(c) Assertion : Ethyl phenyl ether on reaction with HBr form phenol and ethyl bromide.

Reason : Cleavage of C–O bond takes place on ethyl-oxygen bond due to the more stable phenyl-oxygen bond.

a

(d) Assertion : Phenyl is used as a household germicide.

Reason : Phenyl is phenol derivative and phenol is an effective germicide.

a

MCQ TYPE QUESTIONS:

1 Monochlorination of toluene in sunlight followed by hydrolysis with aq. NaOH

yields.

(i) o-Cresol

(ii) m-Cresol

(iii) 2, 4-Dihydroxytoluene

(iv) Benzyl alcohol

iv

2 CH3CH2OH can be converted into CH3CHO by ______________.

(i) catalytic hydrogenation

(ii) treatment with LiAlH4

(iii) treatment with pyridinium chlorochromate

(iv) treatment with KMnO4

ii

3 The process of converting alkyl halides into alcohols involves_____________.

(i) addition reaction

(ii) substitution reaction

(iii) dehydrohalogenation reaction

(iv) rearrangement reaction

ii

4 Give IUPAC name of the compound given below.

(i) 2-Chloro-5-hydroxy hexane

(ii) 2-Hydroxy-5-chlorohexane

(iii) 5-Chlorohexan-2-ol

(iv) 2-Chlorohexan-5-ol

(iii)

106

5

IUPAC name of the compound is ______________.

(i) 1-methoxy-1-methylethane

(ii) 2-methoxy-2-methylethane

(iii) 2-methoxypropane

(iv) isopropyl methyl ether

(iii)

6 Mark the correct order of decreasing acid strength of the following compounds.

(i) e > d > b > a > c

(ii) b > d > a > c > e

(iii) d > e > c > b > a

(iv) e > d > c > b > a

(ii)

7 Arrange the following compounds in increasing order of boiling point.

Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol

(i) Propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol

(ii) Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol

(iii) Pentan-1-ol, butan-2-ol, butan-1-ol, propan-1-ol

(iv) Pentan-1-ol, butan-1-ol, butan-2-ol, propan-1-ol

(i)

8 What happens when tertiary butyl alcohol is passed over heated copper at 300°C? (a) Secondary butyl alcohol is formed

(b) 2-methylpropene is formed (c) 1-butene is formed (d) Butanol is formed

(b)

9 Acid catalysed dehydration of t-butanol is faster than that of n-butanol because (a) tertiary carbocation is more stable than primary carbocation

(b) primary carbocation is more stable than tertiary carbocation (c) t-butanol has a higher boiling point

(d) rearrangement takes place during dehydration of t- butanol

(a)

10 An unknown alcohol is treated with “Lucas reagent” to determine whether the

alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism? (a) Tertiary alcohol by SN

2

(b) Secondary alcohol by SN1

(c) Tertiary alcohol by SN1

(d) Secondary alcohol by SN2

(c)

11 Identify the final product of the reaction sequence.

(a) Benzophenone

(b) Acetophenone (c) Diphenyl (d) Methyl salicylate

b

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12 The reaction between phenol and chloroform in the presence of aqueous NaOH is (a) nucleophilic substitution reaction

(b) electrophilic addition reaction (c) electrophilic substitution reaction (d) nucleophilic addition reaction

c

13 The most suitable reagent for the conversion of RCH2OH → RCHO is (a) K2Cr2O7 (b) CrO3 (c) KMnO4

(d) PCC

d

14 In the following reaction sequence Z is

(a) butan-1-ol (b) butan-2-ol

(c) 2-methylpropan-2-ol (d) 1, 1-dimethylethanol

c

15 Tertiary butyl alcohol gives tertiary butyl chlorideon treatment with (a) Conc HCl/anhydrous ZnCl3 (b) KCN

(c) NaOCl (d) Cl2

a

16 Conversion of phenol to salicyclic acid and to salicyaldehyde are known as (respectively)

(a) Reimer-Tiemann reaction and Kolbe’s reaction (b) Williamson’s synthesis and Hydrobration-oxidation (c) Kolbe’s reaction and Williamson’s synthesis

(d) Kolbe’s reaction and Reimer-Tiemann reaction

d

17 Which of the following alcohols gives 2-butene on dehydration by conc. H2SO4?

(a) 2-methyl propene-2-ol (b) 2-methyl 1 -propanol

(c) Butane-2-ol (d) Butane 1-ol

c

18 A compound X with the molecular formula C3H8O can be oxidised to another compound Y whose molecular formulae is C3H6O2. The compound X may be (a) CH3CH2OCH3

(b) CH3CH2CHO (c) CH3CH2CH2OH

(d) CH3CHOHCH3

c

19 4. Phenol on reduction with H2 in the presence of Ni catalyst gives

a. benzene b. toluene c. cyclohexane d. Cyclohexanol

d

20 Picric acid is a yellow coloured compound. Its chemical name is (a) m-nitrobenzoic acid

(b) 2, 4, 6-trinitropheriol (c) 2, 4, 6-tribromophenol (d) p-nitrophenol

b

Assertion and Reason type questions:

Each of these questions contains an Assertion followed by Reason. Read them carefully and answer the question on the basis of following options. You have to

select the one that best describes the two statements.

a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. (b) If both Assertion and Reason are correct, but Reason is not the correct

explanation of Assertion. (c) If Assertion is correct but Reason is incorrect.

(d) If both the Assertion and Reason are incorrect.

108

1 Assertion: IUPAC name of the compound

is 2-Ethoxy-2-methylethane.

Reason: In IUPAC nomenclature, ether is regarded as hydrocarbon derivative in

which a hydrogen atom is replaced by —OR or —OAr group [where R = alkyl

group and Ar = aryl group]

c

2 Assertion: Bond angle in ethers is slightly less than the tetrahedral angle.

Reason: There is a repulsion between the two bulky (—R) groups.

d

3 Assertion: Phenol forms 2, 4, 6 – tribromophenol on treatment with Br2 in carbon

disulphide at 273K.

Reason: Bromine polarises in carbon disulphide.

b

4 Ethanol is a weaker acid than phenol.

Reason: Sodium ethoxide may be prepared by the reaction of ethanol with aqueous

NaOH.

c

5

Phenols give o– and p-nitrophenol on nitration with conc. HNO3 and

H2SO4 mixture.

Reason: —OH group in phenol is o–, p– directing.

d

2 MARKS QUESTIONS

1 1. Write the mechanism of acid catalysed hydration of an alkene to form corresponding alcohol.

Answer:

Acid catalysed hydration : Alkenes react with water in the presence of acid as catalyst to form alcohols

Mechanism : It involves three steps : (i) Protonation of alkene to form carbocation by electrophilic attack of H3O+

2

Explain the mechanism of dehydration steps of ethanol :

109

Answer:

3

Explain the mechanism of following reaction:

Answer:

4

Write the mechanism of the following reaction.

Answer:

110

5 Write the structures of the products when Butan-2-ol reacts with the following: (a) CrO3

(b) SOCl2 Answer:

6 Write the IUPAC name of following compounds:

(i)

: IUPAC name : 2-Bromo-3-methyl but-2-en-l-ol.

(ii)

1-phenylpropan-2-ol

7 Illustrate the following name reactions : (i) Reimer-Tiemann Reaction (ii) Williamson Synthesis.

Reimer-Tiemann Reaction:

Williamson Synthesis:

8 How are the following conversions carried out?

(i) Propene to Propan-2-ol (ii) Ethyl chloride to Ethanal

Answer: (i) Propene to propan-2-ol

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9 (i)Rearrange the following compounds in the increasing order of their boiling points:

CH3—CHO, CH3—CH2—OH, CH3—CH2—CH3 (ii)Arrange the following compounds in the increasing order of their acid

strengths: 4-Nitrophenol, Phenol, 2,4,6-Trinitrophenol.

Answer: (i)CH3CH2CH3 < CH3CHO < CH3CH2OH.

(ii)Phenol < 4-Nitrophenol < 2, 4, 6-Trinitrophenol.

10 Give a seperate chemical test to distinguish between the following pairs of compounds: (i) Ethanol and Phenol (ii) 2-Pentanol and 3-Pentanol

Ans: (i) Ethanol on reacting with I2 in NaOH gives yellow ppt of iodoform whereas

phenol does not respond to this test. (ii) 2-Pentanol on reacting with I2 in NaOH gives yellow ppt of iodoform whereas 3-pentanol does not respond to this test.

3 MARKS QUESTIONS

1

An organic compound ‘ A ‘ having molecular formula C3 H6 on treatment with aq. H2SO4

give ‘B’ which on treatment with Lucas reagent gives ‘C’. The compound ‘C’ on treatment

with ethanolic KOH gives back ‘ A’ .Identify A, B , C .

Ans:

2 Show how will you synthesise

(i) 1 -phenylethanol from a suitable alkene. (ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.

(iii) Pentan-l-ol using a suitable alkyl halide? Ans:

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3 Explain –

(a)why propanol has higher boiling point than that of the hydrocarbon, butane?

(b) Alcohols are comparatively more soluble in water than hydrocarbons of

comparable molecular masses. Explain this fact.

(c)Explain how does the -OH group attached to a carbon of benzene ring activate

it towards electrophilic substitution?

Answer

(i) Propanol undergoes intermolecular H-bonding because of the presence of -

OH group. On the other hand, butane does not

Therefore, extra energy is required to break hydrogen bonds. For this Reason,

propanol has a higher boiling point than hydrocarbon butane.

(ii) Alcohols form H-bonds with water due to the presence of -OH group.

However, hydrocarbons cannot form H-bonds with water.

As a result, alcohols are comparatively more soluble in water than hydrocarbons of

comparable molecular masses.

(iii) The -OH group is an electron-donating group. Thus, it increases the electron

density in the benzene ring as shown in the given resonance structure of phenol.

As a result, the benzene ring is activated towards electrophilic substitution.

4 (i)How will you convert phenol to benzoic acid?

(ii)An organic compound A having molecular formula C6H6O gives a violet colour with

neutral FeCl3 solution. A on treatment with CO2 and NaOH at 400 K under pressure

gives B which on acidification gives a compound C. The compound C reacts with acetyl

chloride to give D which is popular pain killer. Deduce the structure of A, B, C and D.

(i)

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(ii)

5 Give Reasons for the following :

(i) Phenol is more acidic than methanol. (ii) The C-O-H bond angle in alcohols is slightly less than the tetrahedral

angle (109∘28').(109∘28′).

(iii) (CH3)3C−O−CH3 on reaction with HI gives (CH3)3C−IandCH3−OH as the

main products and not (CH3)3C−OHandCH3−I Ans: (i) Phenol is more acidic than methanol because in phenol, phenoxide ion formed

is more stabilized by resonance than phenol. There is no resonance in methanol. (ii) The C—O—H bond angle in alcohols is slightly less than tetrahedral angle due

to repulsion between the lone pairs of electrons of oxygen. (iii) (CH3)3C+ is 3° carbo-cation which is more stable than CH+3 for SN1 reaction.

6 An aromatic compound 'A' on treatment with CHCl3/KOH gives two compounds 'B' and

'C'. Both B and C give the same product 'D' when distilled with Zinc dust. Oxidation of D

gives E having molecular formula C7H6O2. The salt of E on heating with soda lime gives

F which may be obtained by distilling A with zinc dust. Identify A to F.

7 The following is not an appropriate reaction for the preparation of tert-butyl methyl ether:

(i) What would be the major product of the given reaction? (ii) Write a suitable reaction for the preparation of tert-butyl methyl ether,

specifying the names of reagents used. Justify your answer in both cases.

Ans:((i) As the given alkyl halide is tertiary in nature, therefore on reaction with sodium methoxide, elimination reaction takes place in place of substitution and

hence, alkene is formed as a major product. Therefore, in the given reaction, 2-methyl propene is formed as a major product

114

(ii) For the preparation of tert-butyl methyl ether, i.e. an unsymmetrical ether, the alkyl halide should be primary because the reaction between alkyl halide and alkoxides follows mechanism and primary alkyl halides are most reactive towards

mechanism. Thus, for the preparation of terr-butyl methyl ether, methyl chloride should react with sodium tert-butoxide.

8

A compound 'A' is optically active. On

mild oxidation, it gives a compound 'B' but on

vigorous oxidation gives another compound

'C'. C along with D is also formed from B by

reaction with iodine and alkali. Deduce the

structures of A, B, C, and D.

9 Write structures of the products of the following reactions:

Ans:

115

10 Predict the products of the following reactions:

Ans:

5 MARKS QUESTIONS

1 Give Reasons:

(i) While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile.

(ii) Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?

(iii) Sodium metal can be used for drying diethyl ether but not ethanol

explain?

(iv) Phenol is an acid but does not react with sodium bicarbonate

solution. Why?

(v) phenol is more easily nitrated than benzene

Ans:

(i)Intramolecular H-bonding is present in o-nitrophenol. In p-nitrophenol, the

molecules are strongly associated due to the presence of intermolecular bonding.

Hence, o-nitrophenol is steam volatile.

(ii)

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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.

For this Reason, ortho-nitrophenol is more acidic than ortho-methoxyphenol.

(iii)Due to the presence of an active hydrogen atom, ethyl alcohol reacts with sodium metal.

2CH3-CH2-OH +2Na - > 2CH3-CH2-ONa +H2

Diethyl ether, on the other hand, does not have replaceable hydrogen atom therefore does not react with sodium metal. Hence Na can be used to dry ethers.

(iv Phenol is a weak acid compared to

p-nitrophenol. This is because in p-nitrophenol, there is no hydrogen bonding and hence react with bases like NaHCO3 which is a weaker base.

Phenol is less acidic due to the presence of polar O-H group. Since oxygen is more electronegative than hydrogen, the electron pair of O-H bond is pulled more towards O and H+ ion is easily released in aqueous solution.

(v) Nitration is an electrophilic substitution. The –OH group in phenol

increases the electron density at ortho and para position as follows Since phenol has higher electron density due to electron releasing nature of -OH

group , compared to benzene , therefore nitration is easy in phenol than benzene

2 (a) How are following obtained: (i) Toluene from phenol (ii) Phenol from Aniline.

(b) What happens when-

(i)ethanol is treated with Cu at 573 K,

(ii) phenol is treated with CH3COCl/anhydrous AlCl3, (iii) ethyl chloride is treated with NaOCH3?

Answer:

(b)

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3 (a)Draw the structure and name the product formed if the following alcohols are oxidized. Assume that an excess of oxidizing agent is used.

(i) CH3CH2CH2CH2OH (ii) 2-butenol (iii) 2-methyl-1-propanol

(b)Draw the structure of –

(i) 2-methylpropan-2-ol molecule

(ii) hex-l-en-3-ol compound.

Answer:

(b)

4 (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and

give their IUPAC names.

(ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary and

tertiary alcohols.

Answer

(i) The structures of all isomeric alcohols of molecular formula, C5H12O are shown

below:

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(ii) Primary alcohol: Pentan-1-ol; 2-Methylbutan-1-ol;

3-Methylbutan-1-ol; 2, 2 - Dimethylpropan-1-ol

Secondary alcohol: Pentan-2-ol; 3-Methylbutan-2-ol;

Pentan-3-ol

Tertiary alcohol: 2-methylbutan-2-ol

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5 (a) What is meant by hydroboration-oxidation reaction? Illustrate it with an example.

(b) Give the structures and IUPAC names of monohydric phenols of molecular formula,

C7H8O.

Ans:

(a) The addition of diborane to alkenes to form trialkyl boranes followed by their oxidation with alkaline hydrogen peroxide to form alcohols is called hydroboration-oxidation. For example,

(b) The three isomers are:

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CHAPTER: 11 Aldehydes, Ketones and Carboxylic

acids

Case study type-I (A) Aldehydes and ketones are quite succeptible towards attack of alcohols in the presence of dry HCl to form hemiacetal/acetal and hemiketal/ketal.

The reaction involves nucleophilic addition reaction catalyzed by acid.

Q1(i)

Product P (cyclic) will react with: (A) Tollen’s reagent (B) Fehling reagent (C) NaOH+I2 (D) saturated NaHSO3

C

Q1(ii) Which of the following compound is expected to react fastet with CH3OH/Con.HCl?

A

Q1(iii) Which of the following reagent(s) do you expect to carry out given transformation?

A

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(A) Ethylene glycol in the presence of con. HCl followed by KMnO4 in acidic medium

(B) KMnO4 in acidic medium (C) KMnO4 in basic medium (D) H2O2 in acidic medium

Q1(iv) Which of the following compounds do you expect to react with dil HCl followed by

saturated solution of NaHSO3?

C

Case study type-I (B)

Those aldehydes that do not have alpha-H atoms undergo Cannizzaro’s reaction in the presence

of strong alkali to produce alcohol and salt of carboxylic acid. The reaction is base catalyzed nucleophilic addition reaction.

2(i) What is the product P of the following reaction?

B

2(ii) The nature of compound [X] is

(A) optically active either –d or –l (B) optically inactive

(C) Racemic mixture (D) Meso compound

A

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Case study type-2 (A) Assertion –Reason type Answer given below questions on the basis of given case study.

An aldol condensation is a condensation reaction in organic chemistry in which an enol or an enolate ion reacts with a carbonyl compound to form a β-hydroxyaldehyde or β-hydroxyketone, followed by dehydration to give a conjugated enone.

The name aldol condensation is also commonly used, especially in biochemistry, to refer to just the first (addition) stage of the process—the aldol reaction itself—as catalyzed by aldolases. However, the aldol reaction is not formally a condensation reaction because it does not involve

the loss of a small molecule.

The reaction between an aldehyde/ketone and an aromatic carbonyl compound lacking an alpha-hydrogen (cross aldol condensation) is called the Claisen-Schmidt condensation. This reaction

is named after two of its pioneering investigators Rainer Ludwig Claisen and J. G. Schmidt, who independently published on this topic in 1880 and 1881. An example is the synthesis of dibenzylideneacetone. Quantitative yields in Claisen-Schmidt reactions have been reported in the

absence of solvent using sodium hydroxide as the base and plus benzaldehydes.

(A) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.

(B) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion.

(C)Assertion is correct statement but Reason is wrong statement.

2(iii) In the reaction between methanal and Ph-CHO in the presence of KOH, it is methanal that gets oxidized because:

(A) Ph-CHO , the benzene ring does not favour nucleophic attack due to +M effect (B) Methanl is more reactive towards nucleophic attack that Ph-CHO (C) Benzene ring is more stable due to resonance

(D) A and B both

D

2(iv)

B

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(D)Assertion is wrong statement but Reason is correct statement.

2(i) Assertion: CH2(Cl)-CHO undergoes aldol condensation easier than CH3-CHO

Reason : The enolate ion of CH2(Cl)-CHO is more stable than that of CH3-CHO

A

2(ii) Assertion: Aldol condensation may be catalyzed acid as well as base.

Reason : Aldehydes are easily attacked by all nuclephiles.

C

2(iii) Assertion: Acetic acid has three α-H but does not give aldol condensation.

Reason : In basic/acidic medium deprotonation of α-H is difficult in acetic acid.

A

2(iv) Assertion:In aldol condensation enolate ion has more stability than conjugate

base of the carbonyl compound having α-H undergoing aldol condensation but it attacks from carbon site and not oxygen. Reason :Carbanion intermediate has more nucleophilicity then enolate ion

A

Case study type-2 (A) Assertion –Reason type Answer given below questions on the basis of given case study.

Haloform Reaction

This reaction has been used in qualitative analysis to indicate the presence of a methyl ketone.

The product iodoform is yellow and has a characteristic odour. The reaction has some synthetic utility in the oxidative demethylation of methyl ketones if the other substituent on the carbonyl groups bears no enolizable α-protons.

(A) Assertion and Reason both are correct statements and Reason is correct

explanation for Assertion.

(B) Assertion and Reason both are correct statements but Reason is not correct

explanation for Assertion.

(C)Assertion is correct statement but Reason is wrong statement.

(D)Assertion is wrong statement but Reason is correct statement.

2(i) Assertion: Haloform reaction can produce fluoroform. Reason : F2 is highly reactive and F does not have d-orbitals.

D

2(ii) Assertion: Acetaldehyde and acetone can not be distinguished with iodoform test. Reason : Both give yellow ppt with NaOH and I2 mixture.

A

2(iii) Assertion: Formaldehyde and acetaldehyde can be distinguished with Tollen’s test as well as Iodoform test.

Reason :Both give Tollen’s test but not Iodoforme test

D

2(iv) Assertion: Haloform reaction involves oxidation of carbonyl compound and reduction

of halogen. Reason : The carbon of enolate ion provides electron to halogen

A

MCQ-

Ans: (b)

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The product of the above reaction is? (a) Butanone (b) Propanone (c) Butanol (D) Butanal

Ans: (a)

Ans: (b)

Ans:(c)

Ans:(b)

Ans: (b)

Ans:(b)

Ans: (d)

Ans: (b)

Ans: (d)

125

Ans: (b)

Ans: (A)

Ans: (a)

Ans:( d)

Ans: (d)

Ans: (b) and (d)

Ans: (a) and (c) both

126

Ans: (b) and (d)

Ans: (a) and (c)

Ans: (a)

Assertion-Reason Type

(A) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.

(B) Assertion and Reason both are correct statements but Reason is not

correct explanation for Assertion. (C)Assertion is correct statement but Reason is wrong statement.

(D)Assertion is wrong statement but Reason is correct statement

1 Assertion: Benzoic acid is more acidic than acetic acid

Reason : In benzoic acid carbon of –COOH group is attached to sp2-carbon

A

2 Assertion: O- toluic acid is more acidic than p- isomer

Reason : In O- toluic acid the resonance is hindered due to steric effect.

A

3 Assertion: (CH3)2CH-CHO mainly undergoes cannizzaro’s reaction even though it has alpha-H

Reason : Due to +I effect of two methyl groups the acidic nature of alpha –H is not prounounced

A

4 Assertion: Oxidative ozonolysis of of cyclobutene followed by sodalime

decarboxilation produce ethane.

Reason : Cyclobutene is antiaromatic and unstable

C

5 Assertion: In stephen-Reduction cyanide act as lewis bases. Reason : R-CΞNH+1 formation initiates the reaction.

A

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2- Marks Questions

128

129

Three Marks-Questions:

130

Q2. (a)What happens when cyano benzene reacts with SnCl2 and conc.HCl and then

product is hydrolyzed?

(b) Which medium acidic or basic is better for addition of HCN on following

aldehyde?

(c) What happens when CH3-CO-CH2-CH2-CH2-CO-CH3 is treated with dil NaOH?

Ans: (a) Stephen reduction- C6H5-CHO will form

(b) Acidic medium as in acidic medium compound will readily go protonation

due to stable carbocation formation which is later attacked by CN-1

Q4. (a)Explain with Reason as to why 3-methyl cyclohexanone on reaction with

H-CN produce racemic mixture?

Ans: The C- of C=O group of 3-methyl cyclohexanone is sp2-hybride, so the

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nuclephile CN-1 can attack of the either side and produce a racemic mixture

(b) What is order of the nucleophilic addition of H-CN on ethanol?

Ans: rate= k[aldehyde][CN-1] Its second order reaction

(C) What is the colour of precipitate formed when I-Cl in the presence of NaOH

reacts with acetaldehyde?

Ans: Yellow of CHI3 as Cl-1 will be leaving group when enolate attacks I-Cl

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Five Mark- Questions:

133

134

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CHAPTER: 12 AMINES

IMPORTANT NAME REACTION OF AMINE

Hoffmann Ammonolysis reaction:

Hoffmann bromamide degradation reaction

Gabriel phthalimide synthesis:

HINSBERG’S TEST-

DIAZOTIZATION REACTION-

CASE STUDY BASED MCQ

Read the given passage and answer the questions that follow.

Amines are classified as primary, secondary and tertiary amines. Primary amines

cannot be obtained by ammonolysis of alkyl halide because we will get mixture of 1°, 2° and 3° amines. Cyanides, on reduction give primary amines where as isocyanides on reduction give secondary amines. Nitro compounds, on reduction

also give primary amines. Primary amines react with CHCl3 and KOH to form foul smelling isocyanide. They react with HNO2 and liberate N2 gas. They react

with Hinsberg’s reagent to form salt soluble in KOH. Secondary amine form yellow oily compounds with HNO2 and salt formed with C6H5SO2Cl, is insoluble in KOH. 3° amines form salt soluble in water with HNO2 but does not react with

C6H5SO2Cl.

Diazonium salts are prepared by reaction of Aniline with NaNO2 and conc. HCl at 0 –

5 °C. Aromatic diazonium salts are more stable because phenyl diazonium ion is stabilized by resonance. Benzene diazonium chloride can be used to prepare halo benzene, phenol, nitro benzene, benzene, p-hydroxy azo benzene (azo

dye) and large number of useful compounds.

(a) Write the isomer of C3H9N which does not react with Hinsberg reagent.

Ans. (CH3)3N, N, N–dimethyl methanamine

(b) CH2NH2, on heating with CHCl3 and KOH gives ‘X’. Identify ‘X’.

Ans. CH2NC, a foul smelling compound.

(c) C6H5N2+Cl– + CuCN → A H2O/H+ B, Identify ‘A’ and ‘B’.

Ans. ‘A’ is C6H5CN, ‘B’ is C6H5COOH

(d) Distinguish between Aniline and ethyl amine.

Ans. Add NaNO2 and conc. HCl. Cool it to 0 to 5° C. Then add alkaline solution of phenol. Aniline gives orange dye where as ethyl amino does not.

• Read the Assertion and Reason carefully to mark the correct option out of the options given below :

(a) If both Assertion and Reason are true and the

Reason is the correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is

136

not the correct explanation of the Assertion. (c) If Assertion is true but Reason is false. (d) If the Assertion and Reason both are false.

(e) If Assertion is false but Reason is true.

1. A- n Propylamnie has higher boiling point than trimethylamine. R- Among n propylamine there is H bonding but there is no H bond in

trimethylamine. 2. A- Tertiary amine undergoes acylation reaction. R- Tertiary amines have replaceable H atom.

3. A- Aniline hydrogen sulphate , on heating forms a mixture of ortho and para aminosulphonic acids.

R- The sulphonic acid group is electron withdrawing group. 4. A- Sulphanilic acid has high melting point and is practically insoluble in water. R- Sulphanilic acid exists as zwitter ion salt.

Ans:- 1.(a) 2. (d) .3 (b) . 4.(a)

MCQ TYPES QUESTIONS 1. Considering the basic strength of amines in aqueous solution,which one has the

smallest pkb value ?

(a) C6H5NH2 (b) (CH3)2NH (c) CH3NH2 (d) (CH3)3N

2. The strongest base in aq solution among the following amines is –

(a) N,N diethylethanamine (b) N- ethylethanamine

( c ) N- methylmethanamine ( d) Phenylmethanamine 3. A compound with molecular mass 180 is acylated with CH3COCl to get a

compound with molecular mass 390. The number of amino groups present per

molecule of the former compound is

(a) 6 (b) 2 (c) 5 (d) 4

4. The product formed by the reaction of an aldehyde with a primary amines is

(a) Carboxylic acid (b) aromatic acid (c) Schiff’s base (d) Ketone

5. On heating an aliphatic primary amine with chloroform and ethanolic KOH , the

organic compound formed is –

(a) Alkyl isocyanide (b) an alkanol (c) an alkanediol (d) an alkyl cyanide

6. Positive carbylamines test is shown by

(a) N,N dimethylamine (b) triethylamine (c) N-methylaniline

(d) p- methylbenzylamine 7. The compound which gives an oily nitrosoamine on reaction with

Nitrous acid at low temperature is

(a) CH3NH2 (b) (CH3)2CHNH2 (c) (CH3)2NH (d) (CH3)3N

8. Anilinium hydrogen sulphate on heating with sulphuric acid at 453-473 K

produces

(a) Benzene sulphonic acid (b) anthranilic acid

(c ) Aniline (d) Sulphanilic acid

9. The correct sequence of reactions to be performed to concert benzene into m-

bromoaniline is

(a) Nitration,reduction, bromination

(b) Bromination,nitration,reduction

(c) Nitration, bromination ,reduction

(d) Reduction, nitration ,bromination

10.Which of the following exists as zwitter ion

(a) P- aminophenol (b) Salicylic acid (c) sulphanilic acid (d)

Ethanolamine ANS- 1.(b) .2 (b).3 (c).4(c) 5.(a) .6 (d) 7.(c).8 (d) 9.(c) .10(c)

Q1. Benzonitrile can be described by the formula –

a)C6H5NC

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b) C6H5CN c) C6H5CH2CN d) C6H5CH2 – NC

Ans. (b) Q2. IUPAC name of NH C CH2 CH3 is-

a) N – Phenyl propanamide

b) Ethyl benzamide

c) N – Ethyl Methanamide

d) Phenyl – N – ethanamide

Ans (a) Q3. The degree of following amine (CH3)3 – C– NH2 is:

a) 10

b) 20

c) 30

d) 40 Ans. 10

Q4. Ethyl amine can be prepared by the action of bromine and caustic potash on which

Compound? a) Acetamide

b) Propanamide

c) Formamide

d) Methyl Cyanide Ans. (b)

Q.5 . The Hinsberg’s method is used for which of the following ? (a) Preparing of primary amines

(b) Preparing of secondary amines

(c) Separation of aliphatic and aromatic amines

(d) Separation of amines mixture

Ans. (d) Q.6. Organic compound of the type Ar N = N Cl are classified as --

(a) Amines (b) Amide (c) Dyes

(d) Diazonium Salts Ans. (d)

Q.7. Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine- (a) H2 (excess)/ Pt

(b) LiAlH4 in ether

(c) Fe and HCl

(d) Sn and HCl Ans. (b)

Q.8 In the diazotization of aryl amine the use of nitrous acid is-

(a) is suppresses hydrolysis of phenol (b) it is a source of electrophilic nitrosonium ion

(c ) it neutralizes the base liberated ( e) all the above

Ans. (b)

Q.9. The Sandmeyer’s reaction of diazonium salt is a replacement reaction of N2 by (a) halogen

(b) hydroxyl group (c) coupling (d) hydrogen

Ans. (a) Q.10. Amides can be converted into amines by the reaction named

(a) Carbyl amine (b) Sandmeyer

(c) Hoffmann degradation (d) Gabriel pthalimide

Ans. (c)

Q.11. The correct decreasing order of basic strength of the following species is

H2O , NH3 , OH , NH2 (where –OH and -NH2 are attached to alkyl group)

(a) –NH2 > -OH > NH3> H2O

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(b) OH > NH2 > H2O > NH3

(c) NH3 > H2O > NH2 > OH

(d) H2O > NH3 > OH > NH2 Ans. (a)

ASSERSION REASON TYPES QUESTIONS:-

In the following questions a statement of Assertion followed by a Reason is given. Choose the correct answer out of the following choices :-

(i) Both Assertion and Reason are wrong

(ii) Both Assertion and Reason are correct statements but Reason is not

correct explanation of Assertion.

(iii) Assertion is correct but Reason is wrong

(iv) Both Assertion and Reason are correct statements and Reason is correct

explanation of Assertion.

(v) Assertion is wrong but Reason is correct .

1. A- Acylationof amines gives a monosubstituted product whearas alkylation of

amines gives polysubstituted product.

R- Acyl group sterically hinders the approach of further acyl groups.

2. A-Hoffmann’s bromamide reaction is given by primary amines.

R- Primary amines are more basic than secondary amines. 3. A- N Ethylbenzene sulphonamide is soluble in alkali.

R- Hydrogen attached to nitrogen is strogly acidic.

4. A- Only a small amount of HCl is required in the reduction of nitro compound s

with iron scrap and HCl in the presence of steam.

R- FeCl2 formed gets hydrolyzed to release HCl during the reaction.

5. A- Aromatic 1˚ amines can be prepared by Gabriel phthalimide synthesis.

R- Aryl halides undergo nucleophilic substitution with anion formed by phthalimide.

Ans:- 1-(iii) , 2- (iii) , 3-(iv), 4- (iv) , 5- (i)

SHORT ANSWER TYPES QUESTIONS ( 2 MARKS)

1. Why aliphatic amines stronger are bases than aromatic amines? Give plausible

explanation.

Ans. In aryl amines have an electron withdrawing phenyl group on the nitrogen atom. It decreases electron density on nitrogen atom and makes it a weaker base than

ammonia. In aliphatic amines alkyl group is electron donating and increases electron density on nitrogen thus making it a stronger base than ammonia. This is why alkyl amines are stronger bases than aromatic amines.

2. Account for the following:

I. pKb of aniline is more than that of methylamine.

II. Gabriel phthalamide synthesis is preferred for synthesizing primary

amine.

Ans. I. Aniline has an electron withdrawing phenyl group on the nitrogen atom. It decreases electron density on nitrogen atom and makes it a weaker base. On the other hand, methyl group in methylamine is electron-donating group. It increases

electron density on the nitrogen and makes it a stronger base. Thus, methylamine is a stronger base than aniline and hence pKb of aniline is higher

than that of methylamine. III. In Gabriel phthalamide synthesis, no secondary tertiary amines are

formed and hence pure primary amines formed by this method. Hence

this method is preferred for synthesizing primary amines.

3. Why is carbon-nitrogen bond length in aromatic amines shorter rthan in

aliphatic amines?

Ans. Due to electron-donating resonance effect (+R effect) of NH2 group, the carbon-nitrogen bond in aromatic amines has some double bond character. In

contrast in aliphatic amines, the carbon-nitrogen bond has only single bond character. Therefore, carbon-nitrogen bond length in aromatic amines is shorter

than in aliphatic amines.

139

4. Suggest a convenient scheme for preparing aniline, N-methylaniline,

toluene and phenol, present together in a mixture. Distillation is not to be used.

Ans. Carry out following steps: Treat with HCl. Ti will separate aniline and N-methylanilne as their respective

hydrochlorides from toluene and phenols. Toluene, phenol can be separated by NaOH, which reacts with phenol to form

sodium phenoxide. Mixture of hydrochlorides of aniline and N-methylaniline is treated first with C6H5SO2Cl and then with KOH solution when benzene sulphonamides of aniline

dissolves leaving behind benzene sulphonamide of N-methylaniline. 5. Give one chemical test to distinguish between the Secondary and Tertiary

Amines.

Ans. SECONDARY AMINES-(with nitrous acid) secondary amines react with

nitrous acid to form a yellow oily layer of N-nitrosoamines, which are soluble in aqueous mineral acids.

R2NH + HONO R2N—NO + H2O TERTIARY AMINES- Tertiary aliphatic amines react with nitrous acid to form soluble nitrite salts with no visible change in reaction mixture.

R3N + HONO R3NH+ONO-

6. Account for the following –

(a) Aniline does not undergoes Friedel craft reaction.

(b) Gabriel phthalimide synthesis reaction is preferred fpr aliphatic primary

amines.

Ans. (a) Aniline being lewis base reacts with Lewis acid AlCl3 which is used as a catalyst and form a salt. C6H5NH2 + AlCl3 -----→ C6H5NH+

2AlCl-3

(b ) Aryl halide are less reactive for SN reaction so it is difficult to prepare Aromatic primary halides .

7. Identify A and B in each of the following.

a) C2H5Cl 𝑁𝑎𝐶𝑁→ A

𝑅𝐸𝑑𝑢𝑐𝑡𝑖𝑜𝑛

𝐻2/𝑁𝑖>B

b) C6H5NH2𝑁𝑎𝑁𝑂2

𝐻𝐶𝑙> A

C6H5NH2

𝑂𝐻−> B

Ans a) A → C2H5CN, B → C2H5CH2NH2

Prppanenitrile propanamine b) A → C6H5N2+ Cl- B → Diazoamino benzene Benzene diazonium

Chloride 8. An aromatic compound ‘A’ on treatment with ammonia followed by heating

forms compound ‘B’, which on heating with Br2 and KOH forms a compound ‘C’ (C6H7N). Give the structures of A, B and C and write the reactions involved.

Ans ‘A’ must be benzoic acid

A B C

9. Although amine group is O, P- directly in aromatic substitution reaction, aniline on nitroation gives some amount of m-nitro aniline.

Ans- Nitration is usually carried out with conc. HNO3+ con.H2SO4 In presence

of these acids, some amount of aniline undergoes protonation to form

anilinium ion, so the reaction mixture consists of aniline and anilinium ion, -NH2

group in aniline is o- and p- directing and activating while the –NH3 group in

anilinium ion is m-directing and deactivating. Nitration of aniline mainly gives p-

nitroaniline (due to steric hindrance at o- position) nitration of anilinium ion

gives m-nitroaniline.

10. How will you convert 4-Nitrotoluen 2

Bromobenzoic acid?

140

Ans

SHORT ANSWER TYPE QUESTIONS (3 MARKS)

1. In the following cases rearrange the compounds as directed:

• In an increasing order of basic strength:

C6H5NH2, C6H5N (CH3)2, (C2H5)2NH and CH3NH2

• In a decreasing order of basic strength:

Aniline, p-nitroaniline and p-toluidine

• In an increasing order of pKb values:

C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2

Ans. ✓ C6H5NH2 < C6H5N (CH3)2 < CH3NH2 < (C2H5)2NH

✓ p-toluidine > Aniline > p-nitroaniline

✓ (C2H5)2NH < C2H5NH2 < C6H5NHCH3 < C6H5NH2

2. Explain the observed Kb order for Et2NH > Et3N > EtNH2 in aqueous solution.

Ans. The value of Kb of an alkyl amine is determined by two factor i) induction and ii) solvation. Inductively, the alkyl groups, being electron releasing, increase the electron density on the nitrogen making the amine more basic (larger Kb).

Increasing the number of alkyl groups should increase the basic character. In terms of solvation the basic character of an amine depends on ease of formation of its

ammonium cation by accepting a proton. The more the number of hydrogen-atoms in the ammonium ion, the more it is stabilized by hydrogen-bonding. As alkyl groups replace hydrogen-atoms hydrogen-bonding decreases and basic character

of amine should decrease. Thus, induction and solvation operate in opposite directions and there is a

discontinuity in Kb values. Induction makes all the three alkyl amines stronger bases than ammonia. Thus, (C2H5)2NH is stronger than C2H5NH2. However, when the alkyl group is added, the basic character does not enhance and the opposing

solvation effect assumes more importance. Thus, (C2H5)2NH > (C2H5)3N > C2H5NH2

3. The C—N—C bond angle in Mg3N is 108o.

a) Describe the hybridization and shape of Me3N.

b) Why is an amine of the type R1R2R3N chiral?

c) Why is it not possible to separate the enantiomers?

Ans. a) N forms three sp3 hybridized s bonds to the C’s of the Me groups and has a

nonbonding electron pair in the fourth sp3 orbital. Me3N has a roughly pyramidal shape. b) Because of its pyramidal geometry, such an amine is chiral, the unshared pair being considered a fourth “different” group.

c) The enantiomers rapidly interconvert; having enough kinetic energy at room temperatures, by a process called nitrogen inversion. For this process H‡ » 6 kcal/ mol

(25kJ / mol) and the inversion thus does not involve bond – breaking and subsequent

141

formation. The TS for the inversion is planar, the N being sp2 hybridized, with the unshared pair in the pz orbital.

4. In terms of s character, explain why NH3 with bond angles of 107° is much more

basic than NF3 with bond angles of 103°. The bond angles of NH4+ are 109°.

Ans. The bond angle of 109° in NH4+ indicates that N uses pure sp3 HO’s. The bond

angle of 107° in NH3 shows that, although N essentially uses sp3 HO’s, it nevertheless has slightly more p character in its bonding HO’s and slightly more s character in its

lone pair HO. The bond angle of 103° in NF3 indicates that N has even more p character in its bonding HO’s and evens more s character in its lone pair HO. NF3 is less basic

than NH3 because its lone pair orbital has more s character.

5. Why is it not possible to separate the enantiomers?

Ans. The enantiomers rapidly interconvert, having enough kinetic energy at room temperature by a process called nitrogen inversion. For this process 4 kcal/mole (25

kJ/mole) and the inversion thus does not involve bond breaking and subsequent formation. The TS for the inversion is planar, the N being sp2 hybridized, with the

unshared pair in the 2pz orbital.

6. Write the structure of A,B and C in the following.

C6H5CONH2 𝐵𝑟2/𝐾𝑂𝐻→ A

𝑁𝑎𝑁𝑂2+𝐻𝐶𝑙

0−5˚𝐶>B

𝐾𝐼→ C

Ans- A= Aniline B= Benzenediazonium salt C= Iodobenzene

7. Suggest a convenient scheme for separating aniline, N methylaniline , toluene

and phenol. Distillation is not to be used .

Ans- Steps (i) Treat with HCl .It will separate aniline and N methylaniline as their

respective hydrochloride from Toluene and Phenol. (ii) Tolene ,phenol can be separated by Naoh which reacts with phenol to

form sodium phenoxide.

(iii) Mixture of hydrochlorides is treated first with C6H5SO2Cl and then with KOH solution when benzene sulphonamides of aniline dissolves leaving

behind benzene sulphonamide of N-methylaniline. 8. An organic compound A on boiling with alkali gives ammonia and sodium salt of

acid B , which is monobasic acid that can be obtained from propan 1 ol . When

reduced A form C , which on subsequent treatment with nitrous acid gives D . Give the sequence and structure of A,B,C and D.

Ans-

CH3CH2CN𝑁𝑎𝐶𝑁

𝐵𝑂𝐼𝐿> CH3CH2COONa + NH3

CH3CH2CH2OH 𝑂𝑋𝐼𝐷𝐴𝑇𝐼𝑂𝑁𝐾2𝐶𝑟2𝑂7

𝐻+

>CH3CH2COOH (B)

CH3CH2CN (A) 𝐿𝑖𝐴𝑙𝐻4→ CH3CH2CH2NH2 (C)

𝐻𝑁𝑂2→ CH3CH2CH2OH (D)

9. Give structures of A,B and C in the following reactions .

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(i) CH3COOH𝑁𝐻3

𝐻𝐸𝐴𝑇> A

𝑁𝑎𝑂𝐵𝑟→ B

𝑁𝑎𝑁𝑂2/𝐻𝐶𝑙→ C

(ii) C6H5NO2𝐹𝑒/𝐻𝐶𝑙→ A

𝐻𝑁𝑂3/273𝐾→ B

𝐶6𝐻5𝑂𝐻→ C

Ans- (i) A= Ethanamide B= Methanamine C= Methanol

( ii ) A= Aniline B = Benzene diazonium chloride C= p hydroxy azo benzene

10.Identify A to G in the following reaction.

CH3CH2NH2

𝐻𝑂𝑁𝑂→ A

𝑃𝐶𝑙5→ B

𝐾𝐶𝑁→ C

𝑁𝑎/𝐶2𝐻5𝑂𝐻→ D

↓ [𝑂]

E[𝑂]→ F

𝑁3𝐻

𝐻2𝑆𝑂4>G

Ans- A= CH3CH2OH B= CH3CH2Cl C= CH3CH2CN D= CH3CH2CH2NH2 E= CH3CHO

F= CH3COOH G= CH3NH2

LONG ANSWER TYPES QUESTIONS (5 MARKS)

1. Two isomeric compound A and B having molecular formula C15H11N , both lose N2 on treatment with HNO2 and gives compound C and D. C is resistant to oxidation but

immediately responds to oxidation to lucas reagent after 5 minutes and gives a positive Iodoform test. Identify A and B .

Ans:- CH3 – CH – CH2 – CH3 + HNO2 CH3 – CH – CH2 –CH3

NH2 (B) OH (D) + N2 CH3 CH3

CH3 – C – CH3 + HNO2 CH3 – C – CH3 NH2 OH (c) + N2

CH2CH3 C2H5

CH3 – C – CH3 CONC. HCL CH3 – C – CH3 OH ZNCl2 Cl (30)

But ‘D’ respond to lucas reagent in 5 minutes.

.. CH3 – CH – CH2 – CH3 + HCL CH3 – CH – CH2 – CH3

OH (D) Cl CH3 – CH(OH) – CH2 – CH3 + I2 + NaOH CHI3 + CH3 – CH2 COONa

2. Idomethane reacts with KCN to form a major product A .Compound ‛A’ on reduction in presence of LiAlH4 forms a higher amine 'B’. Compound B on treatment with CuCl2

forms a blue colur complex C . Identify A,B,C Ans .M CH3I + KCN CH3 – CN CH3CH2NH2 (B)

CuCl2

[Cu(CH3CH2NH2)4]Cl2

Bluecomplex 3. A compound ‛A’ of molecular formula C3H7O2N reaction with Fe and conc, HCl

gives a compound ‛B’ OF molecular formula C3H9N. Compound ‛B’ on treatment with NaNO2 and HCl gives another compound ‛C’ of molecular formula C3H8o.The compound ‛C’gives effervescences with Na on oxidation with CrO3.The compound ‛C’ gives a

saturated aldehyde containing three carbon atom deduce A,B,C. Ans:- CH3CH2CH2NO2 CH3CH2CH2NH2 (B)

NaN02/HCL CH3CH2CH2OH (C)

Na

CH3CH2CH2ONa Sodium propoxide

CH3CH2CH2OH CH3CH2CHO

(C) CrO3

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4. Give the structures of A, B and C in the following reactions:

(i) (ii)

(iii)

(iv)

(v)

ANS. (i) (ii)

(iii)

(iv)

(v)

144

5. C5H13N Aq.NaNo2/HCl Y + some other

products

X tertiary alcohol

(optically active)

(i)Identify (X) and (Y) (ii)Is (Y) optically active? (iii)Give structures of intermediate(s), if any, in the formation of (Y) from (X)?

ANS:

. Step 1. To determine the structure of compound (X)

(i)Since compound (X) on treatment with NaNO2/HCl evolves N2 gas, therefore, (X) must be a primary amine.

Further since compound (X) is optically active, therefore, it must contain a chiral carbon. Now NH2 group cannot

be directly attached to the chiral carbon since such amines readily undergo racemization due to nitrogen

inversion. Therefore, the structure of the compound (X) is

CH3

*│

CH3CH2 —CH— CH2NH2

2- Methylbutanamine (x) (optically active)

Step 2. To determine the structure of tertiary alcohol (X)

The formation of compound (Y) from compound (X) may be explained as follows:

CH3 CH3

*│ HNO2 *│ +

CH3CH2 —C— CH2NH2 CH3CH2 —CH— CH2 — N≡N H2NH2 Cl¯

│ -Na+Cl¯

H

(X)

CH3 CH3 CH3

*│ + │ H2O │

CH3CH2 —C— CH2 CH3CH2 —C— CH3 CH3CH2 —C— CH3

│ + -H │

H 3° carbocation OH

1° carbocation (more stable) 2- Methylbutan -2-ol(Y)

(less stable) (optically inactive)

Thus, the compound is 2-methylbutan-2-ol. It is optically inactive.

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CHAPTER: 13 BIOMOLECULES

IMPORTANT Reactions of Glucose HI/∆ CH3(CH2 )3CH3 n-haxane

CH=N–OH

NH2OH (CHOH)4

CH2OH CHO CH(CN)(OH)

(CHOH)4 HCN (CHOH)4

CH OH (CH2OH)

Br2water

COOH (CHOH)4

(CH2OH) (CH3CO)2O Acetic CHO

(CH–O–CO–CH3) 4

Anhydride CH2–O–CO–CH3

dil HNO3 Oxidation COOH

(CHOH)4 COOH

Biomolecules

CARBOHYDRATES

Optically active Polyhydroxy aldehydes or

ketones or compounds which produce such units

on hydrolysis.

PROTEINS : Polymer of

α- amino acid nn

NUCLEIC ACID : Long

chain polymer of Nucleotide

Classification: Monosaccharide : Cannot

be broken further into small units on hydrolysis. ❖ GLUCOSE (C6H12O6)

❖ FRUCTOSE (C6H12O6)

Classification : (I) on the basis of place

of synthesis ❖ Essential : not

synthesized in

body

❖ Non-essential :

synthesized in

body

(II) ON the basis of shape ❖ Fibrous : fibre like

❖ Globular : spherical

Structure : H2N-CH2-CO-NH-CH(CH3)-COOH PEPTIDE LINKAGE

Denaturation of Protein : When a protein in its native form is

subjected to physical change, globules unfold,

helix get uncoiled and protein loses its biological

activity.

Types: DNA and RNA

Composition : In DNA sugar is β-D-2-deoxyribose whereas

in RNA is β-D-ribose. Structure :

NUCLEOSIDE: Formed by attachment of a base to 1’ of sugar

NUCLEOTIDE: Formed by link of

nucleoside to Phosphoric acid at 5’of

sugar.

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CASE BASE MCQ QUESTIONS Monosaccharides containing an aldehyde group are called aldoses while those

containing a ketone group are called ketoses .The aldehyde group is always present at C1 while the keto group is usually present at C2. All monosaccharides containing 5 and 6 carbon atoms and have cyclic structure of furanose (five membered) and

pyranose (six membered) . During ring formation, C1 in aldoses and C2 in ketoses become chiral and hence all these monosaccharide exist in two stereoisomeric form

called the alpha anomer and the beta anomer while C1 and C2 are called glycosidic or anomeric carbon atoms. In contrast stereoisomer, which differ in configuration at

any other chiral carbon other than glycosidic carbon are called epimers. Two molecules of the same or different monosaccharide combine together through glycosidic linkage to form disaccharide.

1. In disaccharides the linkage connecting monosaccharides unit is called a) Glycoside linkage b) Nucleoside linkage

c) Glycogen linkage d) Peptide linkage Answers: a

2. Mutarotation does not occur in

a) Sucrose b) D- glucose c)L- glucose d) D- galactose

Answers: a 3. Which of the following pairs give positive tollens test a) Glucose ,sucrose

b) Glucose , fructose c) Hexanal, acetophenone

d) Fructose ,sucrose Answers: b

4. Two forms of D- glucopyranose are called as

a) Enantiomers b) Anomers

c ) Epimers d) Diasteomers

Answers: b

II. α- amino acids are the building blocks of proteins. About 20 α- amino acids have been isolated by the hydrolysis of proteins . All these amino acid except glycine are

chiral and have L- configuration . Ten amino acids which the body cannot synthesise are called essential amino acids . The remaining ten are called non essential amino acids. All . α- amino acid exist as zwitterion each of which has specific isoelectric point.

Above isoelectric point, a . α- amino acid exist as an anion. Two, three or many Alpha amino acids join together to form di-, tri aur polypeptides or proteins respectively.

Each polypeptide or protein has free amino group at one end called N-terminal end and free carboxyl group at the other end called the c-terminal end. 1. Which one of the following statements is correct?

(a) All amino acids except lysine are optically active (b) All amino acids are optically active

(c) All amino acids except glycine are optically active (d) All amino acids except glutamic acids are optically active

Answer: c 2. Observation of “Rhumann’s purple” is a confirmatory test for the presence of: (a) Starch

(b) Reducing sugar (c) Cupric ion

(d) Protein Answer: d

3. Which of the following statements most correctly defines the isoelectric point?

(a) The pH at which all molecular species are ionized and that carry the same charge. (b) The pH at which all molecular species are neutral and uncharge.

(c) The pH at which half of the molecular species are ionized and other half unionized. (d) The pH at which negative and positive charged molecular species are present in equal concentration.

Answer: d 4. α-amino acid behaves as crystalline ionic solid and have high melting point due to

the presence of; (a) -- NH2 group (b) --COOH group

(c) With both NH2 and --COOH group

147

(d) None of the above. Answer:c

5. Which compound does not exist in a dipolar (zwitter ion ) structure ?

(a) Glycine (b) Glutamic acid

( c) sulphanilic acid (d) p-aminobenzoic acid

ANS: d

CASE BASED ASSERTION REASONS

a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.

b) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion.

c) Assertion is correct statement but Reason is wrong statement. d) Assertion is wrong statement but Reason is correct statement I. All amino acids have the same basic structure

which is shown in following figure.

At the centre of each amino acid is a carbon called the α-carbon and attached to it are four groups a hydrogen and α- carboxyl group and α-

amino group and an R group, sometimes referred to as a side chain. The Alpha carbon, carboxyl and amino groups are common

to all common acids so the R group is only unique feature in each amino acid a minor exceptions to this structure is that of proline which the end of the R group is attached to α-amine with the exception of glycine which has R group consisting of a hydrogen

atom ,all of the amino acid in proteins have for different groups attached to them and consequently can exist in two mirror images form L & D. With only minor exceptions

every minor acid found in cell and protein is in the L-configuration. 1. Assertion: All naturaly occurring α-amino acid except glycine are optically active . Reason: Most naturaly occurring amino acid have L-configuration.

ANS:b 2. Assertion : Valine is a non essential amino acid.

Reason : The lack of essential amino acid in the diet causes Kwashiorkor. ANS: d

3.Assertion: Solubilities of proteins (which are made of α-amino acid) is minimum at the isoelectric point . Reason: At isoelectric point protein molecules stop behaving as Zwitter ion .

ANS: c 4. Assertion : Protein are made up of α-amino acid.

Reason: During denaturation , primary structure of protein is destroyed. ANS: c

II. When carboxyl group of glycine combines with the amino group of alanine we get

a dipeptide, glycylalanine. If a third amino acid combines to a dipeptide, the product is called a tripeptide. A tripeptide contains three amino acids linked by two peptide

linkages. Similarly when four, five or six amino acids are linked, the respective products are known as tetrapeptide, pentapeptide or hexapeptide, respectively. When the number of such amino acids is more than ten, then the products are called

polypeptides. A polypeptide with more than hundred amino acid residues, having molecular mass higher than 10,000u is called a protein. However, the distinction

between a polypeptide and a protein is not very sharp. Polypeptides with fewer amino acids are likely to be called proteins if they ordinarily have a well defined conformation of a protein such as insulin which contains 51 amino acids. Proteins can be classified

into two types on the basis of their molecular shape. (a) Fibrous proteins (b) Globular Protein.

1. Assertion: Keratin is a globular protein. Reason : Enzymes are globular proteins.

ANS: d

2. Assertion: All enzymes are proteins but all proteins are not enzymes. Reason: Enzymes are biocatalysts and posses a stable configuration having an active

site pocket. ANS: b

3. β- pleated sheet structure of protein show maximum extension.

Reason : Intermolecular Hydrogen bonding is present in them. ANS:b

4.Assertion: In proteins ,amino acid are linked through peptide bond.

148

Reason: Peptide bonds are glycosidic or oxygen bridges. ANS: c

MCQ TYPE OD QUESTIONS 1.During acetylation of glucose it needs ,v moles of.acetic anhydride. The value of x

would be (a)3 (b)5

(c)4 (d) 1

ANS: b

2. Identify ‘Z’

(a) 2-lodoheptane (b) Heptane-2-ol (c) 2-lodohexane

(d) Heptanoic acid ANS: d

3.Which of the following treatment will convert starch directly into glucose? (a)Heating with dilute H2SO4 (b)Fermentation by diastase

(c)Fermentation by zymase (d) Heating with dilute NaOH

ANS: a 4. Which one of the following is not correct? (a) D(-) Fructose exists in furanose structure

(b) D (+) Glucose exists in pyranose structure (c) In sucrose the two monosaccharides are held together by peptide linkage

(d) Maltose is a reducing sugar ANS: c

5. Denaturation of protein leads to loss of its biological activity by (a) formation of amino acids (b) loss of primary structure

(c) loss of both primary and secondary structure (d) loss of both secondary and tertiary structures

ANS: d 6.Which type of interactions are responsible for making the a-helix structure stable? (a) Peptide bonds between -NH2 and -CO groups of adjacent carbon chain

(b) Hydrogen bonds between -NH of amino acid in the one turn with -CO of amino acid to adjacent turn

(c) -OH group of one amino acid with -CO group of other amino acid on the turn (d) Hydrogen bonds between adjacent amino acids

ANS: b

7. The melting points of amino acids are higher than the corresponding halo acids because

(a) amino acids exist as zwitter ions resulting in strong dipole-dipole attraction (b) amino acids are optically active (c) due to higher molecular mass of-NH2 group molecular mass of amino acids is higher

(d) they interact with water more than halo-acids and have salt like structure ANS: a

8. Which of the following are not polymeric? (a)Nucleic acid

(b)Proteins (c)Polysaccharides

(d) Lipids ANS: d

9. Which one of the following statements is wrong?

(a) Cellulose is a polysaccharide (b)Uracil is a pyrimidine

(c)Glycine is a sulphur containing amino acid (d) Sucrose is a disaccharide

ANS: c

10. Which of the following biomolecules does have a phosphodiester bond? (a)Fatty acids in a diglyceride

(b)Monoscchrides in a polysaccharide

149

(c )Amino acids in a polypeptide (d)Nucleic acids in a nucleotide

ANS: d

11. Collagen is a (a) fibrous protein

(b) globular protein (c)lipid

(d) carbohydrate ANS: a

12. Nucleotides are building blocks of nucleic acids. Each nucleotide is a composite

molecule formed by (a) base-sugar-phosphate

(b) base-sugar-OH (c )base-sugar-phosphate)n (d) Sugar-phosphate

ANS: a 13. Which one of the following pairs of nitrogenous bases of nucleic acids, is wrongly

matched with the category mentioned against it?

(a)Thymine, Uracil – Pyrimidines (b) Uracil, Cytosine – Pyrimidines

(c) Guanine, Adenine – Purines (d) Adenine, Thymine – Purines

ANS: d 14. What structural feature distinguishes proline from other natural alpha amino acids? (a) It is optically inactive.

(b) It contains aromatic group. (c) It is dicarboxylic acid

(d) It has secondary amine ANS: d

15. Proteins give a white precipitate with Milon’s reagent which is

(a) Mercurous and mercuric nitrate in HNO3 (b) Mercurous and mercuric chloride in HCL

(c ) Mercurous and mercuric chloride in HNO3

(d) all the above ANS: a

ASSERTION AND REASON

In these questions a statement of Assertion followed by a statement of Reason is given. Choose the correct answer out of the following choices.

a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.

b) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion. c) Assertion is correct statement but Reason is wrong statement.

d) Assertion is wrong statement but Reason is correct statement. 1. Assertion: Glucose give a reddish brown precipitate with Fehling’s solution.

Reason: Reaction of glucose with Fehling's solution gives CuO and gluconic acid. ANS: c

2. Assertion: Deoxyribose, C5H10O4 is not a carbohydrate .

Reason : Carbohydrates are hydrates of carbon compounds which follow Cx(H2O)y formula are carbohydrates.

ANS: d 3. Assertion: α- amino acid exist as internal salt in solution as they have amino and carboxylic acid group in near vicinity.

Reason : H+ ion given by a carboxylic group is captured by amino group having lone pair of electrons.

ANS: a 4. Assertion: Thymine occurs in RNA. Reason : RNA controls the synthesis of proteins.

ANS: d 5. Assertion: A solution of sucrose in water is dextrorotatory but on hydrolysis in the

presence of little hydrochloric acid, it becomes levorotatory. Reason : Sucrose on hydrolysis gives unequal amount of glucose and fructose, as a result of which change in sign of rotation is observed.

150

ANS: c

2 MARKS QUESTION 1. Structure of a disaccharide formed by glucose and fructose is given below. Identify anomeric carbon atoms in monosaccharide units. Is the sugar reducing in nature ? Explain.

ANS: a carbon of glucose and b carbon of fructose. It’s non – reducing as –OH groups attached to anomeric carbon are involved in formation of glycosidic bond.

2. Write such reactions and facts about glucose which cannot be explained by open chain structure. ANS: The following facts and reactions cannot be explained by open chain structure of

glucose. (i) Despite having the aldehyde group, glucose does not give 2, 4-DNP test, Shiff’s

test and it does not form the hydrogen sulphite addition product with NaHSO3. (ii) The penta-acetate of glucose does not react with hydroxylamine indicating the absence of free aldehydic group.

3. Amino acids, may be acidic, alkaline or neutral. How does this happen? What are the products formed on complete hydrolysis of RNA?

ANS: This depends upon the relative number of amino and carboxyl group present in their molecule. Equal number of amino and carboxyl groups makes it neutral, more amino group means basic and more carboxylic group means acidic amino acid.

Products are ; Pentose sugar, phosphoric acid and nitrogen bases (G, C, U, A) 4. Ka and Kb values of α - amino acids are very low. Explain.

ANS: The Ka and Kb values of α – amino acids, the acidic group is -NH3 instead of -COOH in carboxylic acids and basic group is -COO instead of NH2 group in aliphatic

amines. 5. (i) During curdling of milk , what happens to sugar present in it? (ii) Coagulation of egg white on boiling is an example of denaturation of protein.

Explain in terms of structural changes. ANS: (i) Milk sugar lactose is converted into lactic acid.

(iii) Globular protein albumin is converted into insoluble fibrous protein.

3 MARKS QUESTION

1. Aspartame, an artificial sweetener, is a peptide and has the following structure

(i) Identify the four functional groups. (ii) Write the Zwitter ionic structure.

(iii) Write the structure of the amino acids obtained from the hydrolysis of aspartame. ANS: (i) Amine, acid, ester, sec-amide.

(ii) O CH2C6H5

H3N+-CH- C –NH- CH-COOCH3

CH2COO-

(iii) H2NCH(CH2COOH)COOH and H2NCH(CH2C6H5)COOH

2. Write the Zwitter ion structure of alanine at pH=2 and pH=10. ANS:

Basic group is ionized 3. An optically active compound having molecular formula C6 h12 o6 is found in two

151

isomeric forms A and B in nature when A and B are dissolved in water they show the following equilibrium (A ) < == > equilibrium mixture < == > ( B)

[α D ] =111o 52.2 o 19.2 o (i) What are such isomers called ?

(ii) Can they be called enantiomers? Justify your answer . (iv) Draw the cyclic structure of isomer.

ANS: (I) Anomers (ii) Not mirror images so they are not enantiomers.

(iii) 4. Explain the following : (i) Amino acids behave like salts rather than simple amines or carboxylic acids.

(ii) The two strands of DNA are complementary to each other. (iii) Reaction of glucose that indicates that the carbonyl group is present as an

aldehydic group in the open structure of glucose. ANS: (i) Due to the presence of both acidic (-COOH) and basic (-NH2) groups, exists as zwitter ion.

(ii) Two strands of DNA are complementary to each other because adenine forms H-binds with thymine and cytosine forms h-bond with guanine.

(iii) Reaction with bromine water.

152

ORGANIC CHEMISTRY REVISION FOR CBSE EXAM

1.IUPAC NOMENCLATURE : SOME BASIC RULES

1. The four components of IUPAC name are:

[Note : if the name of sec. suffix starts with a vowel then pri suffix will be :an / en / yn ]

2. locate the longest carbon chain by giving the preference as : functional group > double bond > triple bond > substituent / side chain

3. follow lowest SUM rule also

4. In a molecule the main group will be named as suffix. All the other groups will be written alphabetically as prefixes

5. Prefix and suffix names of the functional groups with decreasing

priority:

Sl

N

o

Class of Compound Functional Group

Structural formula

Prefix Suffix

1 Sulphonic Acid - SO3H sulphonic acid

2 Carboxylic Acid - COOH carboxy oic acid

3 Acid Anhydride - CO.O.CO- oic anhydride 4 Ester - COOR alkoxy

carbonyl oate

5 Acid Halides - COX haloformyl oyl halide

6 Amides -CONH2 carbamoyl amide 7 Nitriles(Cyanides) - CN cyano Nitrile

8 Isonitrile(Isonitrile) -NC isocyano isonitrile 9 Aldehyde - CHO formyl al

10 Ketone >C=O oxo / keto one

11 Alcohol - OH hydroxy ol 12 Thiol -SH mercepto thiol

13 Amines - NH2 amino amine 14 Ether -O- alkoxy Oxy alkane

Group Name(Prefix)

- Br Bromo - Cl Chloro

- OCH2CH3 Ethoxy - F Fluoro

- I Iodo

- OCH3 Methoxy - NO2 Nitro

- NO Nitroso -C6H5 Phenyl

Prefix(subst)→ Word root(meth, eth, etc)→ Pri. Suffix (ane/ene/yne)→ Sec Suffix(main group)

153

2. REASONING TYPE OF QUESTIONS

QUESTION-REASONING ANSWER- REASON

Arrange the following compounds in increasing order of their property as

indicated (i) CH3COCH3, C6H5 CO C6H5,

CH3CHO (reactivity towards nucleophilic addition reaction)

Give Reasons for the following : (i) Benzyl chloride is highly reactive

towards the SN1 reaction.

Due to the stability of benzyl carbocation/resonance/Diagram

Give Reasons for the following

2-bromobutane is optically active but 1-bromobutane is optically inactive

Because 2-Bromobutane has a chiral

centre

Give Reasons for the following Electrophilic reactions in haloarenes

occur slowly

Due to – I effect of halogen.

Give Reasons for the following :

(i) p-nitrophenol is more acidic than p-methylphenol

Due to –I / –R effect of –NO2 group &

+I / +R effect of –CH3 group or 4-nitrophenoxide ion is more stable than 4-methylphenoxide ion

(ii) Bond length of C – O bond in phenol is shorter than that in

methanol

Due to +R effect of – OH group in phenol

(iii) (CH3)3C – Br on reaction with

sodium methoxide (Na+ OCH3) gives alkene as the main product and

not an ether

(CH3)3C–Br being a 3o halide prefers

to undergo β – elimination on reacting with strong base like NaOCH3.

Arrange the following : (i) in

increasing order of basic strength C6H5NH2, CH3CH2NH2, C6H5NHCH3

C6H5-NH2 < C6H5-NH-CH3 < CH3-CH2-NH2

(ii) in increasing order of boiling point

C2H5 OH, CH3CH2NH2, CH3NHCH3

Give Reasons for the following:

(a) p-nitrophenol is more acidic than o-nitrophenol

Due to intramolecular H-bonding in o-

nitrophenol

(b) Bond angle C – O – C in ethers is slightly higher than the tetrahedral

angle (109o28’).

The mutual repulsion between bulky alkyl groups is stronger

than the l.p-l.p electronic repulsions

(c) (CH3)3C – Br on reaction with NaOCH3 gives an alkene instead of an ether

CH3ONa is not only nucleophile but also stronger base, thereby leads to elimination reaction of the alkyl

halide.

In increasing order of solubility in

water CH3NH2, (CH3)3N, CH3NHCH3

Aniline does not undergo Friedel Crafts reaction

Aniline being a base reacts with AlCl3(Lewis Acid) to form a salt

p-methylaniline is more basic than p-nitroaniline

―CH3 group shows +I – effect(electron releasing group)

whereas –NO2group shows –I- effect(electron withdrawing group)

Acetylation of – NH2 group is done in aniline before preparing its ortho and para compounds.

To reduce activating effect of –NH2.

154

QUESTION-REASONING ANSWER- REASON

Which would undergo SN1 reaction faster in the following pair and why ?

Because carbocation intermediate derived from (CH3)3CBr is more

stable than carbocation from CH3CH2Br.

o-nitrophenol is more acidic than o-

methoxyphenol

Because –NO2is an electron

withdrawing group

Butan-1-ol has a higher boiling point

than diethyl ether

Due to H-Bonding

Reaction occurs by SN1 mechanism ,

30-carbocation (CH3)3C+ is more stable than CH3+

Arrange the following in increasing order of their basic strength :

Give Reasons for the following :

(i) Phenol is more acidic than ethanol.

due to resonance in phenol, oxygen

acquires positive charge and releases H+ ion easily whereas there is no resonance in CH3CH2 OH

(ii) Boiling point of ethanol is higher

in comparison to methoxymethane.

Because of hydrogen bonding in

ethanol

(iii) (CH3)3C O CH3 on reaction with

HI gives CH3OH and (CH3)3C I as the main products and not (CH3)3COH

and CH3I

Because it follows S N1 path way

which results in the formation of stable (CH3)3C +.

Arrange the following compounds in

increasing order of their property as indicated : (i) CH3COCH3, C6H5COCH3, CH3CHO

(reactivity towards nucleophilic addition reaction)

(ii)ClCH2COOH, FCH2COOH, CH3COOH (acidic character)

CH3COOH<Cl-CH2-COOH < F-CH2-

COOH

Give Reasons for the following :

(i) Phenol is more acidic than methanol.

due to resonance in phenol, oxygen

acquires positive charge and releases H+ ion easily whereas there is no resonance in CH3 OH

(ii) The C – O – H bond angle in alcohols is slightly less than the

tetrahedral angle (109 28 ).

Due to lone pair- lone pair repulsion on oxygen

Arrange the following compounds in an increasing order of their acid

strengths: (CH3) 2 CHCOOH, CH3CH2CH(Br) COOH, CH3CH(Br) CH2COOH

why haloarenes are much less reactive than haloalkanes towards

nucleophilic substitution reactions

In haloarenes C—X bond acquires a partial double bond character due to

resonance

Which compound in each of the following pairs will react faster in SN2

reaction with —OH? Why? (i) CH3Br or CH3I (ii) (CH3 ) 3 CCl or CH3Cl

(i) CH3–I reacts faster than CH3–Br as iodine is a better leaving group

because of its larger size. (ii) CH3–Cl (1° halide) reacts faster than (CH3 ) 3 CCl (3° halide) since in

case tertiary butyl

155

chloride three bulky methyl group hinder the approaching nucleophile

QUESTION-REASONING ANSWER- REASON

pKb for aniline is more than that for

methylamine

In aniline due to resonance the lone

pair of electrons on the nitrogen atom are delocalized over the benzene ring.

As a result, the electron density on the nitrogen decreases. On the other hand, in methyl amine +I effect of

CH3 increases the electron density on the nitrogen atom. Therefore aniline is

a weaker base than methyl amine and hence its pKb value is higher than that of methyl amine

Methylamine solution in water reacts with ferric chloride solution to give a

precipitate of ferric hydroxide

Aniline does not undergo Friedel-Crafts reaction

Aniline being a Lewis base, reacts with lewis acid AlCl3 to form a salt.

Due to this N atom of aniline acquires positive charge and hence acts as a strong deactivation

group for further reaction.

Aldehydes are more reactive than ketones towards nucelophiles

due to steric and electronic Reasons

The boiling points of aldehydes and ketones are lower than of the

corresponding acids

due to intermolecular hydrogen bonding in carboxylic acids

The aldehydes and ketones undergo a

number of addition reactions

Due to greater electronegativity of

oxygen than carbon the C atom of the bond

acquires a partial positive charge in

aldehydes and ketones and hence readily undergo

nucleophilic addition reactions

Why is it that haloalkanes are more

reactive than haloarenes towards nucleophiles

In haloarenes C—X bond acquires a

partial double bond character due to resonance

Which one of the following reacts faster in an SN1 reaction and why?

Compound (I) reacts faster in SN1 reaction as it is a 2° alkyl halide

Monochloroethanoic acid has a higher pKa value than dichloroethanoic acid.

This is because dichloroethanoic acid is a stronger acid than

monochloroethanoic acid

Ethanoic acid is a weaker acid than

benzoic acid

because methyl group due to its

positive inductive effect destabilize the acetate anion by intensifying the

negative charge

Why are lower members of aldehydes easily miscible with water?

Lower members aldehydes are able to form intermolecular hydrogen bonds with water molecules.

Hence, they are easily miscible with water

Why do amines behave as nucleophiles?

Due to the presence of a lone pair of electrons on nitrogen atom

pKb of methylamine is less than that of aniline.

In aniline, due to resonance, the lone pair of electrons on the nitrogen atom

are delocalized overthe benzene ring. As a result, the electron density on the nitrogen decreases. On the other

hand, in methyl amine +ve I effect of CH3 increases the electrondensity on

156

the nitrogen atom. Therefore aniline is a weaker base

than methyl amine and hence its pKb value is higher than that of methyl amine

QUESTION-REASONING ANSWER- REASON

Ethylamine is freely soluble in water whereas aniline is only slightly soluble

Ethyl amine is freely soluble in water because it forms hydrogen bonds with water molecules

On the other hand in aniline due to large, hydrocarbon part, the extent of

hydrogen bonding decreases considerably and hence aniline is slightly soluble

Why is an alkylamine more basic than ammonia?

electron donating nature of alkyl group increases electron density on

‘N’ atom in alkyl amine.

Which ones in the following pairs of

substances undergoes SN2 substitution reaction faster and why?

Arrange the following compounds in

an increasing order of basic strengths in their aqueous solutions:

NH3, CH3NH2, (CH3)2NH, (CH3)3N

NH3 < (CH3)3N < CH3—NH2 <

(CH3)2 NH.

Which one in the following pairs undergoes SN1 substitution reaction

faster and why?

The boiling point of ethanol is higher

than that of methoxymethane

Ethanol undergoes intermolecular

hydrogen bonding due to the presence of a hydrogen

attached to oxygen atom

o-and p-nitrophenols are more acidic

than phenol

Due to –I effect or –R effect of –NO2

group

Arrange the following compounds in

an increasing order of their indicated property: (i) Benzoic acid, 4-Nitrobenzoic acid,

3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid

strength)

(i) Acid strength: 4-Methoxy benzoic

acid <Benzoic acid < 4-Nitrobenzoic acid < 3, 4-Dinitrobenzoic acid.

(ii) CH3CH2CH(Br)COOH, CH3CH(Br)

CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength)

(ii) Acid strength: (CH3)2CHCOOH <

CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH

Arrange the following compounds in an increasing order of their basic

strength in aqueous solutions: NH3 , RNH2 , R2NH, R3N

NH3 < R–NH2 < R3N < R2NH

A solution of KOH hydrolyses CH3CHClCH2CH3 and

CH3CH2CH2CH2Cl. Which one of these is more easily hydrolysed?

CH3CH2ClCHCH3 more easily hydrolysed as it forms secondary

carbocation which is more stable than primary carbocation

(i) In an increasing order of basic strength:

C6H5NH2 , C6H5N(CH3 ) 2 , (C2H5 ) 2 NH and CH3NH2

(iii) In an increasing order of pKb values:

C2H5NH2 , C6H5NHCH3 , ( C2H5 ) 2 NHand C6H5NH2

(ii) In a decreasing order of basic strength:

p-Toluidine > Aniline > tr-nitroaniline

157

Aniline, p-nitroaniline and p-toluidine

(iii) In an increasing order of pKb

values: C2H5NH2 , C6H5NHCH3 , ( C2H5 ) 2

NH and C6H5NH2

(C2H5)2 NH2 < C2H5NH2 <

C6H5NHCH3 < C6H5NH2

QUESTION-REASONING ANSWER- REASON

Which compound in the following couples will react faster in SN2

displacement and why? (a) 1-Bromopentane or 2-bromopentane

(b) 1-Bromo-2-methylbutane or 2-bromo-2-methylbutane

(a)1-Bromopentane, as it is a primary alkyl halide. and

(b) 1-Bromo-2-methyl butane, as it is a primary alkylhalide

State one use each of DDT and iodoform

DDT: It is used as insecticide to control flies, mosquitoes, etc.

Iodoform: Iodoform is used as an antiseptic.

What is Tollen’s reagent? Write one usefulness of this reagent

A solution of AgNO3 dissolved in NH4OH is known as Tollen’s reagent. This is

used to detect the presence of –CHO group in an organic compound. For

example: RCHO + 2Ag (NH3)2 OH 3/43/4® RCOONH4 + 2Ag¯ + H2O + 3NH3.

Why is an alkylamine more basic than ammonia?

Due to electron releasing nature, the alkyl group (R) pushes electrons

towards nitrogen in alkyl amine and thus makes the unshared

electron pair more available for sharing with the proton of the acid. Therefore alkyl amine are

more basic than ammonia

Why do primary amines have higher

boiling points than the tertiary amines?

due to ntermolecular hydrogen

bonding

Identify chiral in CH3CHOHCH2CH3 and CH3CHOHCH3

Arrange the following compounds in an increasing order of their solubility

in water: C6H5NH2, (C2H5)2NH, C2H5NH2

Increasing order of solubility of amines in water

C6H5NH2 < (C2H5)2NH < C2H5NH2

Haloalkanes easily dissolve in organic solvents, why?

Haloalkanes dissolve in organic solvents because the new

intermolecular attractions between haloalkanes and organic solvent molecules have much the same

strength as ones being broken in the separate haloalkanes and

solvent molecules

What is known as a racemic mixture?

Give an example.

An equimolar mixture of a pair of

enantiomers is called racemic mixture. For example, butan-2-ol. A racemic mixture is

optically inactive due to external compensation

Of the two bromoderivatives, C6H5CH(CH3)Br and

C6H5CH(C6H5)Br, which one is more reactive in SN1 substitution reaction and why?

Of the two bromo derivatives, C6H5CH(CH3)Br and

C6H5CH(C6H5)Br, the intermediate obtained from C6H5CH (C6H5) Br is more stable than obtained from

C6H5CH(CH3) Br because it is stabilised by two phenyl groups

due to resonance. Therefore, C6H5CH (C6H5)Br is more

158

reactive than C6H5(CH3) Br.

Rearrange the following in an

increasing order of their basic strengths:

C6H5NH2, C6H5N(CH3)2, (C6H5)2 NH and CH3NH2

(C6H5)2 NH < C6H5NH2 <

C6H5N(CH3)2 < CH3NH2

QUESTION-REASONING ANSWER- REASON

Rearrange the compounds of each of

the following sets in order of reactivity towards SN2

displacement:

(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

(ii) 1-Bromo-3-methylbutane, 2-Bromo-2methylbutane, 3-Bromo-2-methylbutane

(iii) 1-Bromobutane, 1-Bromo-2, 2-dimethylpropane,

1-Bromo-2-methylbutane

(i) 1-Bromopentane > 2-

Bromopentane > 2-Bromo-2methyl butane. (ii) 1-Bromo-2-methyl butane > 3-

Bromo-2-methyl butane > 2-Bromo-2-methyl butane

(iii) 1-Bromobutane > 1-Bromo-2-methylbutane > 1-Bromo-2,2-dimethyl butane

pKb value for aniline is more than

that for methylamine

In aniline, the lone pair of electrons

on N-atom are delocalised over benzene ring due to resonance. As a result, electron

density on the nitrogen atom decreases. In contrast, in

methylamine, +I-effect of CH3 – group increases electron density on the nitrogen atom.

Therefore, aniline is a weaker base than methylamine, hence its pKb

value is more than that for methylamine

Ethylamine is soluble in water whereas aniline is not soluble in water.

Ethylamine is soluble in water due to formation of inter-molecular hydrogen bonds with water

molecules. However, in aniline due to large hydrophobic part, i.e.,

hydrocarbon part, the extent of hydrogen bonding decreases considerably and hence aniline is

insoluble in water

Primary amines have higher boiling

points than tertiary amines.

Due to the presence of two H-atoms

on N-atom, primary amines undergo extensive

intermolecular hydrogen bonding whereas tertiary amines have no H-atoms on the nitrogen

atom, do not undergo H-bonding. As a result, primary amines have higher

boiling points than tertiary amines.

Arrange the following compounds in increasing order of their reactivity in nucleophilic addition

reactions: ethanal, propanal, propanone, butanone

Butanone < Propanone < Propanal < Ethanal

Arrange the following in decreasing order of their basic strength in

aqueous solutions: CH3NH2, (CH3)2NH, (CH3)3N and NH3

(CH3)2NH > CH3NH2 > (CH3)3N > NH3

Although chlorine is an electron withdrawing group, yet it is ortho-,

para-directing in electrophilic aromatic substitution reactions.

Explain why it is so?

Through resonance effect, chlorine tends to stabilize the carbocation and

the effect is more pronounced at ortho and para-positions

159

There are two —NH2 groups in

semicarbazide. However, only one such group is involved

in the formation of semicarbazones

Although semicarbazide has two —

NH2 group but one which is directly attached to C=O

is involved in the resonance. Consequently, electron density on this —NH2 group decreases and hence it

does not act as a nucleophile

QUESTION-REASONING ANSWER- REASON

Cyclohexanone forms cyanohydrin in

good yield but 2, 4, 6-trimethylcyclohexanone does not.

Due to presence of three methyl

groups, the nucleophilic attack by CN– ion does not occur due to steric hindrance in 2, 4, 6-

trimethyl cylcohexanone. As there is no such steric hindrance in

cyclohexanone so nucleophilic attack by the CN– ion occurs readily and hence cyclohexanone cyanohydrin is

obtained in

good yield

Alcohols are more soluble in water

than the hydrocarbons of comparable molecular masses

due to their ability to form hydrogen

bonds with water molecules

Ortho-nitrophenol is more acidic than ortho-methoxyphenol

Due to —R and —I effect of —NO2 group, the electron density in the O—H bond decreases in

ortho-nitrophenol and hence the release of H+ ion becomes easy. On

the other hand, due to +R effect of the —OCH3 group, the electron density in the O—H bond in ortho-

methoxyphenol increases, thereby making the release

of H+ ion difficult

Acetaldehyde, Acetone, Methyl tert-

butyl ketone (reactivity towards HCN)

Methyl tert-butyl ketone < Acetone <

Acetaldehyde

Benzoic acid, 3,4-Dinitrobenzoic acid,

4-Methoxybenzoic acid (acid strength)

4-Methoxy benzoic acid < Benzoic

acid < 3,4-Dinitrobenzoic acid

CH3CH2CH(Br)COOH,

CH3CH(Br)CH2COOH, (CH3)2CHCOOH (acid strength)

(CH3)2CHCOOH <

CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH

Alkyl halides, though polar, are immiscible with water.

due to inability of alkyl halides to form intermolecular hydrogen bonds with water

molecules.

Grignard’s reagents should be

prepared under anhydrous conditions

This is because Grignard reagent

forms alkanes by reacting with moisture.

RMgX +H2O RH +Mg(OH)X

Of the two alcohols; (a) CH2=CH—

CH2OH and (b) CH2=CH—CH2—CH2OH, which one will react more easily with conc. HCl in the presence

of ZnCl2?

the dipole moment of chlorobenzene

is lower than that of cyclohexyl chloride

Since chlorobenzene has lower

magnitude of negative charge on Cl atom and

shorter C—Cl bond than cyclohexyl chloride due to resonance therefore chlorobenzene

has lower dipole moment than cyclohexyl chloride

Although phenoxide ion has more Resonating structures of carboxylate

160

number of resonating structures than carboxylate ion,

carboxylic acid is a stronger acid than phenol. Give two Reasons.

ion are more stable than phenoxide ion

QUESTION-REASONING ANSWER- REASON

The C—Cl bond length in chlorobenzene is shorter than that in CH3—Cl

Due to resonance C—Cl bond acquires a partial double bond character which is difficult to

cleave.

Chloroform is stored in closed dark

brown bottles

Because chloroform is slowly oxidised

by air in the presence of light to an extremely poisonous

gas phosgene.

Chlorobenzene is extremely less

reactive towards a nucleophilic substitution reaction

Due to repulsion between nucleophile

and electron-rich arenes

Ethanal is more reactive than acetone towards nucleophilic addition reaction

due to steric and electronic Reasons

(CH3)3C—CHO does not undergo aldol condensation

∝ −hydrogen is not present in

(CH3)3C—CHO .

Carboxylic acids are higher boiling liquids than alcohols

due to more extensive association of carboxylic acid molecules through

intermolecular hydrogen bonding

Ethylamine is soluble in water whereas aniline is almost insoluble.

Why?

Due to hydrogen bonding ability of ethylamine

Why are diazonium salts of aromatic

amines more stable than those of aliphatic amines?

due to dispersal of the positive charge

on the benzene ring .

Alkyl halides on treatment with aq KOHforms alcohols but with alc KOH alkenes are produced,Why?

An AqKOH provides OH- where as alc KOH provides RO- which is a stronger base than OH-,so it abstracts H+ from 𝛽 − 𝐶 𝑜𝑓 𝑎𝑙𝑘𝑦𝑙 ℎ𝑎𝑙𝑖𝑑𝑒

161

3. DISTINGUISH TEST

1) ETHYL BROMIDE AND BROM BENZENE:- AgNO3 TEST

C2H5Br + KOH(aq) C2H5OH + K+Br- K+ Br- + AgNO3 AgBr (Yellow PPT) + KNO3

C6H5Br + KOH No REACTION

2) p-ClC6H4CH3 AND C6H5CH2 Cl :- AgNO3 TEST

C6H5CH2 Cl + KOH(aq) C6H5CH2 OH + KCl

KCl +AgNO3 AgCl (WHITE ppt) + KNO3

3) CCl4 AND CHCl3 :- CARBYLAMINE TEST

CHCl3 +C6H5NH2 + KOH C6H5NC +KCl +H2O

CCl4 + C6H5NH2 + KOH NO REACTION

4) CH3OH AND CH3CH2OH OR CH3CH2OH AND CH3CH2CH2OH :- IODOFORM TEST

CH3CH2OH +NaOI CH3CHO + NaI +H2O

CH3CHO +NaOI CHI3 (YELLOW ppt) +HCOONa +NaOH

CH3OH +NaOI NO REACTION

OR CH3CH2CH2OH +NaOI NO REACTION

5) CH3CH2OH AND CH3CH(OH)CH3 :- LUCAS TEST

CH3CH(OH)CH3 + Anhy. ZnCl2 + Con. HCl CH3CHClCH3(WHITE TURBIDITY AFTER 5

MINS.)

CH3CH2OH+ Anhy. ZnCl2 +Con. HCl NO REACTION

6) CH3CH2CH2OH AND CH3CHOHCH3 :- IODOFORM TEST

CH3CHOHCH3 +NaOI CH3COCH3 + NaI +H2O

CH3COCH3 +NaOI CHI3 (YELLOW ppt) +CH3COONa +NaOH CH3CH2CH2OH + NaOI NO REACTION

7) CH3CHOHCH2CH3 AND (CH3)3COH :- LUCAS TEST OR IODOFORM TEST CH3CHOHCH2CH3 + Anhy. ZnCl2 +Con. HCl CH3CHClCH3(WHITE TURBIDITY AFTER 5

MINS.) (CH3)3COH + Anhy. ZnCl2 +Con. HCl (CH3)3CCl(WHITETURBIDITY

IMMEDIATELLY)

8) CH3CH2OH AND C6H5OH :- FeCl3 TEST C6H5OH + FeCl3 (C6H5O)3Fe (VIOLET COLOURATION) + HCl

CH3CH2OH +FeCl3 NO REACTION

9) (R)2NH AND RNH2 :- (CARBYLAMINE TEST)

R-NH2 + CHCl3 +3KOH R-NC +3KCl +3H2O

(R)2NH2+ CHCl3 +3KOH NO REACTION

10) C2H5OH AND CH3CH2NH2 :- ( IODOFORM TEST)

CH3CH2OH +NaOI CH3CHO + NaI +H2O CH3CHO +NaOI CHI3 (YELLOW ppt) +HCOONa +NaOH

11) CH3COCH2CH2CH3 AND CH3CH3COCH2CH3 :- ( IODOFORM TEST)

CH3COCH2CH2CH3 +NaOI CHI3 (YELLOW ppt) +CH3CH2CH2COONa

+NaOH

CH3CH3COCH2CH3 NO REACTION

12) C6H5COCH3 AND C6H5CHO :- (IODOFORM TEST)

C6H5COCH3 + NaOI CHI3 (YELLOW ppt) +C6H5COONa +NaOH

C6H5CHO + NaOI NO REACTION

162

13) CH3CHO AND CH3CH2CHO :- (IODOFORM TEST)

CH3CHO +NaOI CHI3(YELLOW ppt)+HCOONa +NaOH

CH3CH2CHO NO REACTION

14) CH3CHO AND C6H5CHO :- ( IODOFORM TEST)

CH3CHO +NaOI CHI3 (YELLOW ppt) +HCOONa +NaOH

C6H5CHO NO REACTION

15) C6H5CHO AND CH3COCH3 :- ( IODOFORM TEST) CH3COCH3 +NaOI CHI3(YELLOW ppt)+CH3COONa +NaOH

C6H5CHO NO REACTION

16) CH3COC6H5 AND C6H5COC6H5 :- ( IODOFORM TEST)

C6H5COCH3+ NaOI CHI3 (YELLOW ppt) +C6H5COONa

+NaOH

C6H5COC6H5 NO REACTION

17) CH3COCH3 AND CH3CH2CHO :- ( IODOFORM TEST)

CH3COCH3 +NaOI CHI3 (YELLOW ppt) +CH3COONa +NaOH

CH3CH2CHO NO REACTION (TOLLENS TEST IS ALSO

POSSIBLE)

18) CH3CHO AND HCHO :- ( IODOFORM TEST)

CH3CHO +NaOI CHI3(YELLOWppt)+HCOONa +NaOH

HCHO NO REACTION

19) C2H5Cl AND CH2=CHCl :- ( AgNO3 TEST)

C2H5 Cl + KOH (aq)+ AgNO3 C2H5 OH + AgCl (WHITE ppt) +

KNO3

CH2=CHCl NO REACTION

20) C6H5Cl AND C6H11Cl :- ( AgNO3 TEST)

C6H11 Cl + KOH (aq)+ AgNO3 C6H11 OH + AgCl (WHITE ppt)

+KNO3

C6H5Cl NO REACTION

21) C6H5Cl AND C6H5CH2Cl :- ( AgNO3 TEST)

C6H5CH2 Cl + KOH (aq)+ AgNO3 C6H5CH2 OH + AgCl (WHITE ppt)

+KNO3

C6H5Cl NO REACTION

22) CH3CH2CH2OH AND CH3COCH3 :- ( IODOFORM TEST) CH3COCH3 +NaOI CHI3 (YELLOW ppt) +CH3COONa +NaOH

CH3CH2CH2OH NO REACTION

23) C6H5CH2OH AND CH3CHOHCH3 :- ( IODOFORM TEST)

CH3CHOHCH3 +NaOI CH3COCH3 + NaI +H2O

CH3COCH3 +NaOI CHI3 +CH3COONa +NaOH

C6H5CH2OH NO REACTION

24) CH3OH AND C6H5OH :- (FeCl3 TEST)

C6H5OH + FeCl3 (C6H5O)3Fe (VIOLET COLOURATION) +

HCl

CH3OH NO REACTION

25) C6H5OH AND C6H5NH2 :- (AZO DYE TEST)

C6H5N2+ Cl- + C6H5OH+OH- C6H5N=NC6H4OH (ORANGE DYE) + Cl-

+H2O

163

C6H5N2+ Cl- + C6H5NH2+H+ C6H5N=NC6H4NH2 (YELLOW DYE) + Cl- +H2O

26) C6H5NH2 AND C6H5NH CH3(CARBYLAMINE TEST)

C6H5-NH2 + CHCl3 +3KOH R-NC (FOUL SMELL) +3KCl +3H2O

C6H5NHCH3+ CHCl3 +3KOH NO REACTION

27) C2H5NH2 AND (C2H5)2NH(HINSBERG TEST) C6H5SO2Cl + NH2C2H5 C6H5SO2NH-C2H5 (SOLUBLE IN ALKALI) +

HCl

C6H5SO2Cl + NH(C2H5)2 C6H5SO2N-(C2H5)2 (INSOLUBLE IN ALKALI)+ HCl

164

2 3

4. Mechanism

1. HYDRATION OF ETHENE TO YIELD ETHANOL.

Step (i) : — Protonation of alkene to form carbocation by electrophilic attack : H H

— C = C + H — O+ — H ———

Step (ii) : — Nucleophilic attack of water on carbocation : H H

— C — C+ + H2O ———

H

Step (iii) : — Deprotonation to form an alcohol :

H

— C — C — O+ — H + H O ——— — C — C — + H O

+

H

2.ETHANOL TO ETHENE Con H2SO4

a) CH3CH2OH CH2 = CH2 +

H2O 443 K Mechanism:

(i) H2SO4 H+ + HSO+

+ (ii) CH3CH2 – O – H + H+ CH3 – CH2 – O – H

|

H +

(iii) CH3CH2 – O – H + H+ CH3 CH2+ + H2O

| H

(iv) CH3 CH2 + CH2 = CH2 + H+

(v) H+ + HSO4- H2SO4

3. ETHANOL TO ETHER Con H2SO4

2CH3 CH2 OH CH3 CH2 O CH2 CH3 + H2O

413 K Mechanism:-

i) H2SO4 H+ + HSO4-

+ ii) CH3 CH2 OH + H+ CH3 – CH2 – O – H

|

H

+ (iii) CH3CH2 – O – H CH3 CH2

+ + H2O

|

H

+

iv) CH3 CH2 – O – H + CH3 CH+ CH3 – CH2 –O – H

|

CH2CH3

v) CH3CH2 – O –H CH3CH2 – O – CH2CH3 + H+

|

CH2CH3

vi) HSO4-+ H+ H2SO4

+

165

4 . Substitution nucleophilic unimolecular reactions (SN1) : Two Steps

5. Substitution nucleophilic Bimolecular reactions (SN2) : One Step

6.

166

5. KEY FOR CONVERSIONS

Sl

No

Reagent Group Out Group In Remark

1 KMnO4 / H+ -CH2OH -COOH Strong Oxidation (20 alc→ ketone)

2 LiAlH4 -COOH -CH2OH Strong Reduction (ketone → 20 alc)

3 Cu /573 K or CrO3 -CH2OH -CHO Dehydrogenation

4 PCl5 or SOCl2 -OH -Cl

5 Cl2 / Δ or Cl2 / UV -H -Cl Free radical substitution

6 Aq NaOH / KOH -X -OH Nucleophilic substitution

7 KCN -X -CN Step Up

8 AgCN -X -NC

9 Alcoholic KOH -HX = Dehydrohalogenation (Stzf)

10 Mg / dry ether Mg R-X → R-MgX

11 HBr >=< H, Br Merkovnikov

12 H2 / Pd-BaSO4 -COCl -CHO Rosenmund Reduction 13 Zn-Hg / HCl >C=O -CH2- Clemmension Reduction

14 NH3 / Δ -COOH -CONH2 -COOH + NH3 → -COONH4

15 Br2/NaOH or NaOBr -CONH2 -NH2 Step Down ( Hoffmann) 16 HNO2 orNaNO2/HCl -NH2 -OH HONO

17 CHCl3 / alc KOH -NH2 -NC Carbyl amine

18 P2O5 -CONH2 -CN Dehydration

19 H3O+ -CN -COOH Hydrolysis 20 OH- -CN -CONH2

21 LiAlH4 -CN -CH2NH2 Reduction

22 Red P / Cl2 α-H of acid -Cl HVZ Reaction In benzene ring

23 Fe / X2 /dark -H -X Halogination

24 CH3Cl / AlCl3(anhyd) -H -CH3 Friedel Craft alkylation

25 CH3COCl / AlCl3(anhyd)

-H -COCH3 Friedel Craft acylation

26 Conc.HNO3/con.H2SO4 -H -NO2 Nitration

27 Conc H2SO4 -H -SO3H Sulphonation 28 KMnO4 / H+ -R -COOH Oxidation

29 CrO2Cl2 / H+ -CH3 -CHO Mild oxidation(Etard Reaction)

30 Sn / HCl or Fe/HCl -NO2 -NH2 Reduction

31 NaOH / 623K / 300 atm

-Cl -OH

32 Zn dust / Δ -OH -H

33 NaNO2 / dil HCl / 273-

278 K

-NH2 -N +Cl- 2 Diazo reaction

34 CuCl/HCl or Cu/HCl -N +Cl- 2 -Cl Sandmeyer or Gattermann

35 CuBr/HBr orCu/HBr -N +Cl- 2 -Br Sandmeyer or Gattermann

36 CuCN / KCN -N +Cl- 2 -CN Sandmeyer

37 KI -N +Cl- 2 -I

38 HBF4 / Δ -N +Cl- 2 -F

39 H3PO2 or CH3CH2OH -N +Cl- 2 -H

40 H2O / 283 K -N +Cl- 2 -OH

41 HBF4/ NaNO2 ,Cu/ Δ -N +Cl- 2 -NO2

42 C6H5-OH -N +Cl- 2 -N=N-

C6H5-OH

Coupling ( p-hydroxy)

43 C6H5-NH2 -N +Cl- 2 -N=N- C6H5-NH2

Coupling ( p-amino)

3

Reactions of Grignard Reagent Grignard reagent + Any one below + H2O → Product

R-MgX

H2O or ROH or RNH2 R-H

H-CHO R-CH2-OH (10 alc) R-CHO R-CH(OH)-R (20 alc)

R-CO-R R2C(OH)-R (30 alc)

CO2 R-COOH R-CN R-CO-R

HCOOR Aldehyde

RCOOR Ketone

NB: i) During reaction generally changes take place in the functional group only so see the functional

group very carefully. ii) Remember structural formula of all the common organic

compounds ( with their IUPAC and common

names) iii) Wurtz Reaction and Aldol Condensation are not included in the table

although they are very important for conversions so study them . iv) By taking examples practice all the above cases ( from 1 to 43 and Grignard)

v) Practice only from NCERT book. vi) Start practicing NOW !

How to use the table? See below.

Directional Properties of groups in benzene ring for electrophilic substitution

Ortho-para directing group: -R , -OH, -NH2, -X, -OR, -NHR, -NR2, -NHCOCH3, -CH2Cl, -SH,

- Ph

Meta-directing group: -NO2 , -CHO , -COOH , COOR , -CN , -SO3H , -

COCH3 , -CCl3 , - NH + ,

Example : See no 7 in the table