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KENDRIYA VIDYALAYA SANGATHANAHMEDABAD REGION

FORMULAE OF CHEMISTRY AT GALANCE

Students must know at least the following formulae to solve numerical problems in chemistry specially low achiever who have fear for numerical problem. The given formulae will certainly help them out.

CLASS XIISUBJECT : CHEMISTRY

MAIN FORMULAE

SN CHAPTER/UNIT FORMULA

TERMS

1. UNIT ISOLID STATE

1.Calculation of density: z. M

d = a3. NA

z = no of atoms in a unit cellM = mass of unit cella = edge length NA= Avagadro’s number

2. Edge length of hcp structure : a = 2√2.r

a = Edge length r = radius

3. Edge length of bcc structure: 4.r

a = √3


4. Edge length for simple cubic: a = 2.r


2. UNIT IISOLUTION 1. Calculation of partial vapour

pressure p1 = p1

0 . xa

2. Colligative properties

(i) Relative lowering of vapour pressure p1

0 - p1 = w2

. M1

p10 M2 .w1

(ii) Elevation of boiling point ΔTb = kb. m

kb . 1000. w2

= M2 .w1

(iii)Depression of freezing point

ΔTf = kf. m

Kf . 1000. w2

= M2 .w1

(iv) Osmotic pressure

Π = C.R .T

w2. R.T =

M2 .V

p1 = vapour pressure of solution p1

0 = vapour pressure of pure component. xa = mole fraction

p1 = vapour pressure of solution p1

0 = vapour pressure of pure component co w1 = mass of the solvent

w2 = mass of the soluteM1 = molar mass of solventM2 = molar mass of solute

ΔTb = Increase in boiling point kb = Molal elevation constant m = Molarity of the solution w1 = mass of the solvent w2 = mass of the solute M2 = molar mass of solute

ΔTf = depression of freezing point kf = Molal depression constant m = Molarity of the solution w1 = mass of the solvent w2 = mass of the solute M2 = molar mass of solute

Π = Osmotic pressureC = concentration of solutionR = Gas constantT = TempV = Volume of the solution M2 = molar mass of solute

3 Unit IIIElectrochemistry 1. Nernst Equation

Ecell = emf of the cell E0

cell = Standard emf of the cell

4.

UNIT IVChemical kinetics

Ecell = E0cell -

R T ln [M]

n F [Mn+]

Ecell = E0cell -

0.059 log [Anodic con]

n [cathodic con]

2. Relation between Equilibrium constant and E0

cell

Ecell = E0cell -

2.303 R T log kc

n F

OR

Ecell = E0cell -

0.059 log kc n

3. Molar conductivity

Λm = k 1000

molarity

4. Cell constant

l

G* =

a

1st order Rate Reaction Equation

k = 2.303 log [R]

0 t [R]

2. Half life period 0.693

t1/2 = k

R = Gas constant T = Temp. [M] = Anodic concentration [Mn+] = Cathodic concentration n = no. of electrons in the balanced redox reaction. F = Faraday constant.

kc = Equilibrium constant Ecell = emf of the cell E0

cell = Standard emf of the cell R = Gas constant T = Temp. n = no. of electrons in the balanced redox reaction. F = Faraday constant.

k = conductivity

Λm = molar conductivity

G* = cell constantl = length of the conductora = cross sectional area of the conductor

k = Rate constant [R]

0 = Initial concentration of reactant [R] = final concentration of reactant t = time

t1/2 = Half life time

k = Rate constant

3. Arrehenius Equationlog k2 = Ea T

2 – T1

k1 2.303R T1 T2

k1 And

k2 are rate constants for the reaction at two different tempEa = Activation energyT1 & T2

are two different temp

R = Gas constant

CLASS XISUBJECT : CHEMISTRY

MAIN FORMULAE

SN CHAPTER/UNIT FORMULA

TERMS

1. UNIT ISTRUCTURE OF ATOM

1. de Broglie’s equation h

λ = p

λ = Wave lengthh= Planck’s constantp = momentum

2. Heisenberg’s Equation : hΔx.m. Δv= 4 Π

Δx = Change in position of particle m = mass of particle Δv = change in velocity h= Planck’s constant

3. Einstein Equation

E = hυ

or E = h c /λ

E = Energy of photonh= Planck’s constantυ = Frequency of radiation

c = Light velocity

λ = Wave length

4. Rydberg’s Equation:

υ = 109677 1 _ 1 cm-1

n21 n2

2

υ = wave number n2

1 = Initial energy level of electron n2

2 = Final energy level of electron

2. STATES OF MATTER

1 Boyle’s Law p = k. 1/V

2. Charle’s Law

V = k.T Or

V1/ V0 = T1/ T0

3. Ideal gas equationPV = n RT

Or

P1 V2 / T1 = P2 V2/ T2

4. van der Waal’s Equation

( p + an2/ v2 ) ( v - nb) = n RT

p = pressure of gask = constant of proportionalityv = Volume of the gas

V = Volume of the gask = constant of proportionality

V1 = Initial volume of the gas Vo = Final volume of the gasT1 = Initial Temp of the gasT2 = Final Temp of the gas

P = Pressure of gasV = Volume of the gasn = No. of moles of gasR = Gas constantT = Temp

P1 = Initial Pressure of the gasP2 = Final pressure of the gasV1 = Initial volume of the gas V2 = Final volume of the gasT1 = Initial Temp of the gasT2 = Final Temp of the gas

P = Pressure of gasV = Volume of the gasn = No. of moles of gasR = Gas constantT = Tempa & b are the van der Waal’s constant

3

4.

Unit VIThermodynamics

Unit VIIChemical Equilibrium

1. First law of thermodynamics ΔU = q + w

2. Enthalpy of system

ΔH = ΔU+ PΔV

3 Entropy of system

ΔS = q / T

3.Change in the enthalpy of a

reaction

ΔrH = Σ ( bond enthalpy of reactant) -

Σ( bond enthalpy of products)

4. Gibb’s Helmholtzs Equation

ΔG = ΔH - T ΔS

1. pH Calculation

pH = - log [H+]

2. Relation between kw , ka & kb

ΔU = Change in the internal energyq = Heat absorbed or given out w = Work done

ΔH = Change in enthalpy ΔU = Change in internal energy P = Pressure ΔV = Change in volume

ΔS = Change in entropy q = Heat absorbed or given out

T = Temp

F = Faraday constant.

ΔrH = Change in the enthalpy of a reaction

ΔG = Change in Gibb’s energy ΔH = Change in enthalpy T = Temp ΔS = Change in entropy

[H+] = Concentration of H+ ions

kw = Dissociation constant of water ka = Dissociation constant of weak acid kb = Dissociation constant of Weak base

kw = ka . kb

3. pH of aqueous solution of salt

of weak acid and strong base

pH = 7 + ½ [Pka – Pkb]

4. Solubility product :

Ksp = [M+] [X-]

Pka = - log ka Pkb = - log kb

Where

ka = Dissociation constant of weak acid kb = Dissociation constant of Weak base

Ksp = Solubility product [M+] = Concentration of Cations[X-] = Concentration of Anions

Thank you

CBSE QUESTION PAPER DESIGNING (PATTERN), BULE PRINT AND MARKING SCHEME

CBSE has it own and very specific way of design & pattern of question paper,blue print and marking scheme. Therefore, it is important to know about all these and to follow the same strictly, when we prapare que. Papers, Blue Print and marking scheme.

Designing of CBSE Question Paper for Chemistry : The whole designing of question paper of chemistry of Board Exam depends on the following details, which must be told to the students very categorically.

Before I start discussion about setting and designing of question papers let me discuss about the term Learning and the points to be kept in mind while setting/designing a question paper.

1. What is learning :

According to Hilgard and Hunter “ Learning is the process by which behaviour is organized or changed through practice or training” And to ensure whether learning has been successful or not there is a need of evaluation of students. And one of the best tools of evaluation is question papers. So, before we set Question papers some basics of its setting must be known. & hence lets have a glance over the point to be taken care of :

2.Points to be taken care of while framing a question paper :

Question paper must be balancedQuestion paper should not be too lengthyIt must neither be very easy nor very hardIt must contain questions for all category of students i.e. weak, bright and average

It is very clear from the above details that every aspect of the question paper is very well set and hence this must be brought to the knowledge of the students. Now second point is to discuss about blue print of question paper.Let us know what BLUE PRINT is all about :BLUE PRINT :*Blue print is a frame work that gives you the focus points that the examiner has decided for the examinee at a glance.*Blue print is schematic overview of the focus point that the examiner has to consider for evaluation of examinee.OBJECTIVES OF BLUE PRINT

• To overview whole syllabus for the purpose of evaluation• To make sure that total part of the syllabus is covered • Each unit has been given due weightage

• Question paper consists of different type of questions • To help students by letting them know about the pattern of question papers.

Unit VSA(1)

SA-I(2) SA-II (2)

LA(5) TOTAL

1. Solid State 2(2) - - 2(4)2. Solution - 1(2) 1(3) - 2(5)3. Electrochemistry - 1(2) 1(3) - 2(5)4. Chemical Kinetics - - - 1(5)- 1(5)5. Surface Chemistry 1(1) - 1(3) - 2(4)6. General Principles& Extraction - - 1(3) - 1(3)7. p-block elements -1(3) 1(5) 2(8)8. d&f block elements - 1(2)- 1(3) 2(5)9. Co-ordination compounds 1(1) 1(2) - - 2(3)10. Haloalkanes& Haloarenes - 2(2) - - 2(4)11. Alcohols& Phenols 2(1) 1(2) - 2(4)12. Aldehydes,Ketones& Carboxylic

acids1(1) - - 1(5) 2(6)

13. Organic compounds containing nitrogen

1(1) 1(3) - 2(4)

14. Biomolecules 1(1) - 1(3) - 2(4)15. Polymers 1(1) 1(2) - 2(3)16. Chemistry in everyday life - - 1(3) - 1(3)

8(1) 10(20) 9(27) 3(15) 30(70) A Blue Print for Chemistry Question Paper (Class XII )Let us discuss about the different type of a question as mentioned in a blue print

Types of questions Knowledge based Understanding based Application & skill based

What are KNOWLEDGE BASED questions ? These questions usually starts from “how, why, what, where, whom, when, which,

define, discuss, state, Name, write etc. and direct questions (eg. name reactions).e.g. Q1. Write Kohlrausch law. Q2. State Raoult’s law. Q3. Define mole fraction.What are Understanding based questions: These include * Reasoning Based questions * Direct numerical * Nomenclature * Reaction mechanism * To Differentiate beetween….. questions * Derivation Based Question *Questions starts from “to show that……”

What are Application based questions ? These include: * Questions based on graph * Numerical in which application of very concepts of chemistry are involved. * Organic conversions * Questions which are related to our day to day life.

What are Skill based questions ? These include *Balancing of equation questions *Draw structure of______ type questions *Plot the graph of……. *Question related to observations, practicals (eg. Smell, colour change, taste etc. related)

From single topic different type of questions can be framed :

Q. Write Kohlrausch’s law. (If asked like this it is a knowledge based question)

Q. If molar conductivities at infinite dilution for NH4Cl,NaOH and NaCl are 129.8,217.4 and108.9 Scm2 mol-1 respectively,calculate molar conductivity at infinite dilution for NH4OH. (If asked like this it is an understanding based question)

Q. If molar conductivities at infinite dilution for NH4Cl,NaOH and NaCl are 129.8,217.4 and108.9 Scm2 mol-1 respectively and the molar conductivity of a 10-2 M solution of NH4OH is 9.33 Scm2 mol-1 .Calculate degree of dissociation of NH4OH in its above

mentioned solution. (If asked like this it is an Application based question as degree of dissociation will be calculated only after calculation of molar conductivity at infinite dilution for NH4OH )

MARKING SCHEME :

*Marking scheme of question paper is prepared to make the evaluation procedure uniform*Step wise marking is ensured in a marking scheme.* All the answers must be mentioned ( in full) in the marking Scheme. Avoid writing like and give full details of answers :

1. For correct definition- 1Mark2. For correct reasoning - 1m3. For correct diagram - 2m4. For correct mechanism – 2m etc.

- Thanking you

HOW TO TAKE UP NUMERICAL PROBLEMS IN CHEMISTRY

In order to make students learn numerical problems in chemistry one must make it sure that students have been told about all the related concepts and formulae and a sufficient practice has been given in this regard, prior to solve any numerical problem. teacher must enlist all possible types of numericals of a particular unit and further sufficient practice for solution must be given, by giving different kinds question for the same formulae and concepts starting from simple and single concepts involving numerical to difficult and multiconceptual nemericals. In this context I am to put forward such an example by taking 1st Unit of Class XII Chemistry subject (Solid State Chemistry). In which we require the following 2 tables to make student confident and comfortable to solve problems:

TABLE 1Sn Type of unit

cellsEdge Length (l)

Radius (r)

No. Of Sphere (z)

Inter ionic distance

% of Packing

1 Simple/Primitive a=2r r = a/2 1 = a 52.4%

2 Body centred Cube/CsCl/CsBr

a=4.r/√3 r = a√3/4 2 = a√3/2 68%

3 FCC/hcp/ccp/NaCl/RockSalt /ZnS /Sphalerite/fluorite/Antifluorite

a= 4.r/√2 r = a√2/4 4 = a/√2(Note= for Metallic solid but a/2 (for ionic solids solid)

74%

Note : √2 = 1.414 √3 = 1.732

Table 2Radius Ratio Coordination

No.Structural Arrangement

Structure Type

Example

0.155-0.225 3 Trigonal Planar B2O3 B2O3

0.225-0.414 4 Tetrahedral ZnS /Sphalerite ZnS,CuCl,HgS,CuBr etc.

0.424-0.732 6 Octahedral NaCl/RockSalt NaCl, MgO, CaO etc.0.732-1.00 8 Cubic CsCl type CsCl, TlBr, NH4Br

etc.

Now, on the basis of the above table it is very easy for the students to find formulae/required data to solve any kind of problems.

Now let us have glance over all possible kind of numericals in the unit and give sufficient practice to each student for each of the following possible types of numericals:

1. To calculate density : Use the following formula - z. M

d = a3. NA

2. To calculate Cell dimensions (l, a, r, inter-ionic distance etc.) : Use the above formula according to need and see the table 1 to find out inter-ionic distance.

3. To calculate Molar mass or Molecular mass or Atomic mass etc : Use above equation given in the 1st point. (Note that molar mass must be reported in g/mol

4. To find NA or Number of particles/spheres/Atoms/ions/molecule etc.: Use above equation given in the 1st point.

5. To find fractions of ions in the given formula of oxides or other molecule: For such question use the oxidation concept taught in XI chemistry.

6. To Predict Molecular Formula of the given constituent particles/spheres/Atoms/ions: Use the following conceptual point to solve such problem

7. Doping Related problems : (see the problem related with doping of NaCl with SrCl2)

8. To find coordination no. of the given sphere.(Use table 2)

9. To predict nature of the unit cell on the basis of z (Use table no.1)

10. To find Radius Ratio on the basis of Co-ordination no. (Use table 2)

11.To find the range of the cation/anion on the basis of radius ration. (Use table 2)12. To find the nature of the unit cell on the basis of Radius ratio. (Use table 2) - Thank You

UNIT WISE DIFFICULT AREAS MENTIONED BY PARTICIPANTS

No. of Octahedral voids (N) = ccp arrangement of the sphere(N)=2*No. of tetrahedral voids(N)

THE SOLID STATE1. 3-D Representation of solids2. Numericals/calculations3. Magnetic properties4. Defects of solids

SOLUTIONS1. Numericals on Vant hoff factor2. Ideal&Non-ideal solutions

ELECTROCMEMISTRY1. Nerst equation

CHEMICAL KINETICS1. Fractional order of reaction2. Log table3. Numericals based on initial rate&activation energy

SURFACE CHEMISTRY1. Preferential adsorption of ions.2. Charge on colloids.

GENERAL PRINCIPLES OF EXTRACTION &ISOLATION OF ELEMENTSEllingham Diagram

p-BLOCK ELEMENTSDifficulty faced by exception cases

d&f BLOCK ELEMENTSElectrode potential of transition elementsM.Pt of transition elementsCu+/Cu+2 Stability

COORDINATION CHEMISTRY1.Valence bond theory2.Crystal field theory

ORGANIC CHEMISTRY(Unit 10-13)Basicity of amines ,

Note : All the above topic were discussed and solved by The Course Director and Resource Persons

UNIT WISE IMPORTANT TOPICS COMPILED BY THE PARTICIPANTS

UNIT-1 Marks-4

Solid stateGroup terms and Definitions:-

Anisotropy: - Property like refractive index, X-ray diffraction etc. are different through different axis. This property shown by crystalline solid.

Classification of crystalline solids:-

Types of solid Force of attraction Milting point ExamplesAtomic or molecular

Vander waal force, Hydrogen bonding etc.

Very less Dry ice, Solid NH3 etc.

Covalent network solid

Covalent force Very high Quartz,Diamand.

Ionic solid Ionic force Very high NaCl, KFMatallic solid Metallic force High Fe,Sn,Alloys etc.

Unit cell and cell dimension.

Unit cell Edge length Coordination no. % occupied space

% free space

Simple cubic cell

a=2r 06 52 48

B.C.C a=4r/√3 08 68 32F.C.C/HCC/CCP a=4r/√2 12 72 28

F- Centre: - It is the defect in which when an alkali metal halide is heated in the vapour of alkali metal the excess cation occupy the crystal lattice and anion vacancy filled with the valance electron. So that the crystal appear coloured.

P-type semiconductor: - The semiconductor formed by doping of elements of less valence electron with crystal. For example Si (VE=4) dopped with Ga(VE=3)

n – type semiconductor :- The semiconductor formed by dopping of element with a crystal. For example Si (VE=4) dopped with As (VE=5).

Photovoltaic Material: The material which converts sun light into electricity is called photo voltaic material. [ Amorphous silica ]Crystalline solid [1] In a crystalline solid the particles (atoms, molecules or ions) are arranged in a regular and repetitive three dimensional arrangements [2] These solids have sharp melting point.[3] These solids are anisotropic , i.e. their physical properties such as electrical conductivity, refractive index, thermal expansion etc. have different values in different directions.[4] These solids can under go a clean cleavage.[5] These solids are generally incompressible.Examples: All the metallic elements like iron, copper and silver; Non – metallic elements like sulphur, phosphorus and iodine and Compounds like sodium chloride, zinc sulphide and naphthalene Amorphous solid[1] In amorphous solid the particles (atoms, molecules or ions) are arranged in an irregular and non repetitive three dimensional arrangements.[2] Rapidly solidified liquids are amorphous substances, e.g. Glass, rubber etc.[3] These solids are generally isotropic , i.e. physical properties are same in all directions. [4] These solids on cleavage form smaller pieces with non-planar faces.[5] These solids do not have sharp melting point and boiling point i.e. they melt gradually over a temperature range.[6] These solids are compressible.What makes a glass different from a solid such as Quartz? Under what conditions quartz could be converted into glass?In glass, amorphous silica ( SiO2 ) is present. SiO4 tetrahedral have an irregular arrangement.In quartz, crystalline silica ( SiO2 ) is present. SiO4 tetrahedral have a regular arrangement.When quartz (SiO2 ) is melted and the melt is cooled very rapidly, quartz converted into glass.FluidLiquids and gases are called fluids because of their ability to flow. The fluidity is due to the fact that the molecules are free to move about.Why glass pans fixed to windows or doors of old building are invariably found to be slightly thicker at the bottom than at the top?Due to fluidity property, the glasses flows down very slowly and make the bottom portion slightly thicker.

Why solids are rigid in nature?The constituent particles in solid have fixed position and can only oscillate about their mean position, for which solids are rigid.

Name the factors which determine the stability of a substance.[1] Intermolecular forces tend to keep the molecules or constituent particles closer.[2] Thermal energy tends to keep them apart by making them move faster.

Characteristic properties of solid[1] They have definite mass, volume and shape.[2] The inter molecular distances are short.[3] The inter molecular forces are strong.

[4] Their constituent particles have fixed positions and can only oscillate about their mean positions.[5] They are incompressible and rigid.

Why glass is considered as a super cooled liquid?Amorphous solids have a tendency to flow. Since glass is an amorphous solid , so it is called super cooled liquid or pseudo solid.

Why some glass objects from ancient civilizations are found to become milky in appearance?Glass becomes crystalline at some temperature. For which glass objects from ancient civilizations become milky in appearance because of some crystallization.

Space lattice or crystal latticeA regular three dimensional arrangement of points in space is called space lattice or crystal lattice. The points represent the constituent particles of the crystal.Unit cellAn unit cell is the smallest portion of the crystal lattice. When it is moved repeatedly a distance equal to its own dimension along each direction, a three dimensional crystal lattice is generated. Types of unit cell

[2] Non-Primitive unit cell / centered unit cell- In this type of unit cells, particles as points are present not only at the corners but also at some other positions.

Packing in metallic crystal:-The identical solid spheres can be packed in a number of ways. [1] In the first layer the spheres are arranged in a hexagonal manner in which each sphere is in contact with six other spheres.[2] In the second layer the spheres will fit into the depression of the first layer.[3] For the third layer, there are two possibilities:- (a) The spheres can be placed in the depression of the second layer i.e. the third layer is directly above the first and the fourth layer is directly above the second. This leads to the arrangement ABABAB…… This type of arrangement is known as hexagonal closed packed (HCP) structure. e.g. Zinc, magnesium crystallizes in this type of structure. [b] Alternatively, the sphere can be placed in the depressions, of the second layer; do not lie directly above the atoms of the first layer i.e. the spheres in fourth layer lie exactly above the first, fifth above the second, sixth above the third and so on. This leads to the arrangement

ABCABCABC………. This type of arrangement is cubic closed packed or face centered cubic arrangement. e.g. Cu,Ag.Au crystallizes in this type of structure.Co-ordination numberIt is the number of atoms or spheres that surrounds the single sphere / atom in a crystal.C.Nof tetragonal arrangement = 3C.Nof tetrahedral arrangement = 4C.Nof octahedral arrangement = 6C.Nof body centered cubic arrangement = 8Any close ( tight ) packing having C.N = 12 i.e. hcp and ccp i.e. fcc having C.N = 12

Void / Hole/ Interstices The space which is left in between the closest pack arrangement is called void. In close packing two types of voids are created.

Relationship between edge length (a) and radius of the sphere ( r ) in unit cell.

Gold(At.radius= 0.144nm)crystallizes in aFCC unit cell. What is the length of the side oftheunitcell? Atomic radius= 0.144 nmSo, length of side a = 2 √ 2 r = 2 √ 2 x 0.144 = 0.406 nmAluminum crystallizes in a cubic closed packed structure. Its metallic radius is 125 pm.[1] What is the length of the side of the unit cell ?[2] How many unit cells are there in one c.c of aluminium?[1] For FCC , a = 2 √ 2 r = 2 √ 2 x 125 = 354 pm, So the edge length of the unit cell = 354 x 10 -10 cm[2] Volume of unit cell = a3 = ( 354 x 10 -10 ) 3 cm 3

Therefore, number of unit cells in 1 cc of aluminum = 1/ ( 354 x 10 -10 ) 3 cm 3 = 2.254 x 10 22

Unit cells

Ferric oxide crystallizes in a hexagonal closed pack array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Deduce the formula of the ferric oxide.No. of atoms in one unit cell of hcp structure = 6 No. of oxide ions per unit cell = 6No. of octahedral voids = 6Since ferric ions occupy only two out of every three octahedral voids, there fore, no. of octahedral voids occupied by ferric ions = ( 2/3 ) x 6 = 4 Stoichiometric ratio of Fe 3+ and O 2- is 4 : 6 = 2 : 3Hence the formula of ferric oxide is Fe2O3 Analysis shows that nickel oxide has the formula Ni 0.98 O 1.00. What fraction of nickel exist as Ni 2+ and Ni 3+ ions? Let amount of Ni3+ be x mol. Then amount of Ni2+ is (0.98 – x)Total oxidation number of Ni in the compound is 3x + 2 (0.98 – x) Oxidation number of oxygen is -2Since the sum of the oxidation number of all the constituents in a compound is zero.=> 3x + 2 (0.98 – x) – 2 = 0 => 3x + 1.96 – 2x – 2 = 0 => x = 0.04 Hence % of Ni3+ = ( 0.04 / 0.98 ) x 100 = 4.08 % % of Ni2+ = 100 – 4.08 = 95.92 %Relationship between density (d) and the dimension of unit cells.Let the edge length of unit cell be a There fore volume of unit cell = a 3

Let no. of atoms in unit cell = Z Gm. Atomic mass = M

There fore mass of one atom = M / NA Where NA = Avogadro’s number i.e. 6.023 x 1023

Silver crystallizes in FCC lattice. If edge length of the cell is 4.05 x 10 – 8 cm and density is 10.5 gm/cm 3 . Calculate atomic mass of silver.

Given data-Edge length ( a ) = 4.07 X 10 – 8 cm Density ( d ) = 10.5 gm/ cm3

Since silver crystallizes in fcc lattice, so rank , i.e. the no. of silver atoms per unit cell ( z ) = 4NA = 6.023 X 10 23

# Single crystals are formed when the process of crystallization occurs at extremely slow rate.# Point defects are the irregularities or deviations from ideal arrangement around a point or an atom in a crystalline substance.# Line defects are the irregularities or deviations from ideal arrangement in entire rows of lattice points.# Point defects do not disturb the stoichiometric of the solid is known as stoichiometric defect or intrinsic defect or thermodynamic defect.# Vacancy defect – When some of the lattice sides are vacant, the crystal is said to have vacancy defect. This results the decrease in density of the substance. This defect develops when a substance is heated.# Interstitial defect- When some constituent particles occupy an interstitial site, the crystal is said to have interstitial defect. This defect increases the density of the substance. Schottky defect [1] It occurs when a pair of ions of opposite charge are missing from the ideal lattice.[2] The presence of a large number of Schottky defect in a crystal lowers its density.[3] This defect occurs if cation and anion having similar size with high coordination number.[4] In sodium chloride, there is one schottky defect for 1016 ions. One c.c of sodium chloride contains 1022 ions. Therefore, one cc of sodium chloride possesses 106 schottky pair of ions.Frenkel defect[1] When an ion leaves its position in the lattice and occupy interstitial site leaving a gap in the crystal i.e. it creates a vacancy defect and interstitial defect.[2] This defect will occur if size of cation is smaller than anion, with low coordination number.[3] Frenkel defects are not found in pure alkali halide. Due to larger size of cations, ions can not accommodate in interstitial site.[4] Frenkel defect are found silver halide, because silver ions are smaller in size and can get into the interstitial site.[5] The Frenkel defect does not change the density of the solid.[6] In silver bromide, both Schottky and Frenkel defects are found.

Metal excess defect[a] F-Centers ( Farbe: means colour ) [1] When there is an excess of metal ions in non-stoichiometric compounds, the crystal lattice has vacant anion site. The anion sites occupied by electrons are called F-centre.[2] The F-centers are associated with the colour of the compounds. Excess of K in KCl makes the crystal violet. Excess of Li in LiCl makes the crystal pink.[3] Solid containing F-centre are paramagnetic, because the electrons occupying the F-centers are unpaired.[4] When the crystal having F-centers are exposed to light, they become photoconductor.[b] Metal excess defect due to presence of extra cation at the interstitial site Zinc oxide is white in colour at room temperature on heating, it looses oxygen and turns yellow.

Now there is excess of zinc in the crystal and its formula becomes Zn 1+ x OThe excess zinc ions move to interstitial site and the electrons to neighboring interstitial site.Metal deficiency defectFeO, mostly found with a composition of Fe 0.95 O i.e. range from Fe0.93O – Fe0.96O. In crystals of FeO, some Fe2+ ions are missing and the loss of positive charge is made up by the presence of required number of Fe3+ ions.Electron sea model of metallic bonding[1] A metal consists of a lattice of positive ions ( Kernel ) immersed in a sea of valence electrons ( mobile electrons )[2] The force of attraction between the mobile electrons and the the kernels is known as metallic bond.[3] The electrical and thermal conductivity of metals can be explained by the presence of mobile electrons in metal.The Band Model Of Metallic BondingThe band model of metal is based on molecular orbital theory. When a large no. of orbital overlap in metal, it results a continuous energy level produced by a large number of molecular orbital is called energy band.

Impurity defect ( Doping )The introduction of defects in a particular crystalline solid by the addition of other elements is known as doping.Doping increases the conductivity of crystal. For example, if we mix strontium chloride ( SrCl2 ) with sodium chloride, some strontium ( Sr2+ ) ions occupy the lattice sites of sodium ions ( Na +) and equal number of sodium ( Na+ ) sites remain vacant. Such vacancies in the crystal increase the electrical conductivity because certain ions from the neighboring sites can move into these vacant holes. In this defect the number of positive ions are less as compared to negative ions. Crystals with such defects also act as semiconductor. Since the conductivity is due to holes, these are known as P-type semiconductors.

If NaCl is doped with 10 -3 mol % of SrCl 2. What is the concentration of cation vacancies? The addition of SrCl2 to NaCl produces cation vacancies equal in number to that of Sr2+ ions.No. of moles of SrCl2 added to one mol of NaCl = 10-3 / 100 = 10-5 mol.No. of holes created in one mole of NaCl = 10-5 X 6.023 X 1023 = 6.023 x 1018

SemiconductorThese are solids whose conductivity lies in between those of conductors and insulators. The conductivity of semiconductors increases with increase of temperature.

Intrinsic semiconductor An insulator capable of conducting electric current at higher temperature or when irradiated with electromagnetic radiations, are known as intrinsic semiconductor.This happens because certain covalent bonds are broken and the released electrons are in a position to conduct electric current. e.g. Silicon , Germanium.Extrinsic semiconductorThese are formed when impurities of certain elements are added (doped) to insulator.N-type semiconductorsIt is obtained by doping group – 14 elements with group – 15 elements.Suppose Si is doped with P with 5 valence electrons, out of 5 valence electrons, only 4 valence electrons are involved in bond formation.The fifth electron is not bound any where and can be easily promoted to the conduction band. The conduction is thus mainly caused by the movement of electrons.

P-type semiconductorsIt is obtained by doping group – 14 elements with group – 13 elements.Suppose Si is doped with Ga which has 3 valence electrons, 3 valence electrons are involved in bond formation with neighboring Si atom.A vacancy is left which can be filled by the transfer of a valence electron from a neighboring Si atom.The movement of electron into the vacancy leaves behind a hole which carries positive charge.Another electron from a neighboring Si atom can move into the hole leaving behind another hole. It appears as if the hole has moved through the lattice.The movement of positively charged hole is responsible for the conduction of charge.Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor? [1] Since Cu2O is non-stoichiometric oxide, it contains Cu in two oxidation states, +1 and +2.[2] Cu2+ provides an excess of positive charge. As a result an electron from a neighboring Cu+ is transferred to Cu2+.

[3] The transfer of electron leaves behind a hole, which carries an extra positive charge and a negative hole is created.[4] It appears that the positive hole moves through the lattice, hence it appears as P-type semiconductor.12 – 16 and 13 – 15 compoundsCombination of elements of Gr – 13 and Gr – 15 or Gr – 12 and Gr – 16 produce compounds which stimulate average valence of four as in Ge or Si .12 – 16 compounds –> ZnS, CdS, CdSe, HgTe13 – 15 compounds –> InSn, AlP, GaAsMagnetic properties[1] Diamagnetic Diamagnetic substances are the substances which are weakly repelled by a magnetic field. The electrons in diamagnetic substances are all paired. They do not contain unpaired electrons. e.g. TiO2 , NaCl , C6H6 ,N2 , Zn[2] Paramagnetic Paramagnetic substances are those which are attracted by a magnetic field but they lose their magnetism in the absence of magnetic field. These substances have permanent magnetic dipole, due to presence of atoms, molecules or ions containing unpaired electrons. e.g. Cu2+, Fe3+, O2 , NO, CuO, etc. Substances containing unpaired electrons are further classified as:

(a) Ferromagnetic substances –> Ferromagnetic substances are those substances which are strongly attracted by a magnetic field and can be made into permanent magnets.

These substances show magnetism even in the absence of a magnetic field. The large magnetism in these substances is due to the spontaneous alignment of magnetic moment,i.e Unpaired electron in the same direction.

Example: Iron, cobalt, nickel, gadolinium and CrO2 CrO2 is used to make magnetic tapes for audio recording.

(b) Anti-ferromagnetic substances –> Anti-ferromagnetic substances are those substances in which equal number of magnetic moments are aligned in opposite directions so as to give zero net moment.

Example: MnO, MnO2 and Mn2O3(c) Ferrimagnetic substances Ferrimagnetism is observed when the magnetic moments of the domains in the substance are aligned in parallel and anti-parallel directions in unequal numbers They are weakly attracted by magnetic field as compared to ferromagnetic substances. Example: Fe3O4 (magnetite) and ferrites like MgFe2O4 and ZnFe2O4

These substances also lose ferrimagnetism on heating a

TERMS EXPLANATIONSAmorphous and Crystalline Solids

Amorphous- short range order, Irregular shape eg-glassCrystalline Solids- long range order, regular shape eg : NaCl

Molecular solids Ar, CCl4, H2O (ice)

Covalent or Network solid SiO2 diamondNo of lattice points per unit cell

Simple cubic -4, BCC- 9, FCC – 14 , End-Centred- 10

No of atoms per unit cell (z )

Simple cubic -1, BCC- 2, FCC – 4 , End-Centred- 2

Coordination Number FCC- 6:6 BCC- 8:8Calculation of number of voids

Let the number of close packed spheres be N, then:The number of octahedral voids generated = NThe number of tetrahedral voids generated = 2N

Relation between r and a Simple Cubic a = 2r , BCC 4r = a√3 FCC 4r = a√2

Packing Efficiency Simple Cubic52.4% , BCC 68% , FCC 74%

Calculations Involving Unit CellDimensions M=molar mass (g/mol) a = edge

length in cm ,NA = 6.023× 1023

Frenkel Defect: Ccation is dislocated to an interstitial site. It does not change the density of the solid. Frenkel defect is shown by ionic substance in which there is a large difference in the size of ions, for example, ZnS, AgCl, AgBr and AgI due to small size of Zn2+ and Ag+ ions.

Schottky Defect A vacancy defect. The number of missing cations and anions are equal. Density decreases. For example, NaCl, KCl, CsCl and AgBr.

Metal excess defect due to anionic vacancies (F-centres )

When NaCl heated in an atmosphere of Na vapour, the Na atoms deposited on the surface of the crystal. The Cl– ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl. Thereleased electrons diffuse into the crystal and occupy anionic sites The anionic sites occupied by unpaired electrons are called F-centres They impart yellow colour to the crystals of NaCl. Similarly, excess of lithium makes LiCl crystals pink and excess of potassiummakes KCl crystals violet (or lilac).

Doping The conductivity of intrinsic semiconductors is increased by adding an appropriate amount of suitable impurity. This process is called doping

n / p -type semiconductors n- type : Si + As or Sb or Bi p-type: Si + B or Ga or In or Tl

13 –15 compounds &12–16 compounds

13 – 15 compounds: InSb, AlP and GaAs. 12 – 16 compounds :ZnS, CdS, CdSe and HgTe

Paramagnetic substances Weakly attracted by a magnetic field. Examples: O2, Cu2+, Fe3+, Cr3+

Diamagnetic substances Weakly repelled by a magnetic field. Example: H2O, NaCl and C6H6

Ferromagnetism: A few substances like iron, cobalt, nickel, gadolinium and CrO2 are attracted very strongly by a magnetic

field. Antiferromagnetism MnO. Domains oppositely oriented and cancel out

magnetic momentFerrimagnetism: domains aligned in parallel and anti-parallel

directions in unequal numbers . Example: Fe3O4 (magnetite)

Impurity defect ( Doping )The introduction of defects in a particular crystalline solid by the addition of other elements is known as doping.Doping increases the conductivity of crystal. For example, if we mix strontium chloride ( SrCl2 ) with sodium chloride, some strontium ( Sr2+ ) ions occupy the lattice sites of sodium ions ( Na +) and equal number of sodium ( Na+ ) sites remain vacant. Such vacancies in the crystal increase the electrical conductivity because certain ions from the neighboring sites can move into these vacant holes. In this defect the number of positive ions are less as compared to negative ions. Crystals with such defects also act as semiconductor. Since the conductivity is due to holes, these are known as P-type semiconductors.

If NaCl is doped with 10 -3 mol % of SrCl 2. What is the concentration of cation vacancies? The addition of SrCl2 to NaCl produces cation vacancies equal in number to that of Sr2+ ions.No. of moles of SrCl2 added to one mol of NaCl = 10-3 / 100 = 10-5 mol.No. of holes created in one mole of NaCl = 10-5 X 6.023 X 1023 = 6.023 x 1018

SemiconductorThese are solids whose conductivity lies in between those of conductors and insulators. The conductivity of semiconductors increases with increase of temperature.

Intrinsic semiconductor An insulator capable of conducting electric current at higher temperature or when irradiated with electromagnetic radiations, are known as intrinsic semiconductor.This happens because certain covalent bonds are broken and the released electrons are in a position to conduct electric current. e.g. Silicon , Germanium.Extrinsic semiconductorThese are formed when impurities of certain elements are added (doped) to insulator.N-type semiconductorsIt is obtained by doping group – 14 elements with group – 15 elements.Suppose Si is doped with P with 5 valence electrons, out of 5 valence electrons, only 4 valence electrons are involved in bond formation.The fifth electron is not bound any where and can be easily promoted to the conduction band. The conduction is thus mainly caused by the movement of electrons.P-type semiconductorsIt is obtained by doping group – 14 elements with group – 13 elements.Suppose Si is doped with Ga which has 3 valence electrons, 3 valence electrons are involved in bond formation with neighboring Si atom.A vacancy is left which can be filled by the transfer of a valence electron from a neighboring Si atom.The movement of electron into the vacancy leaves behind a hole which carries positive charge.Another electron from a neighboring Si atom can move into the hole leaving behind another hole. It appears as if the hole has moved through the lattice.The movement of positively charged hole is responsible for the conduction of charge.

Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor? [1] Since Cu2O is non-stoichiometric oxide, it contains Cu in two oxidation states, +1 and +2.[2] Cu2+ provides an excess of positive charge. As a result an electron from a neighboring Cu+ is transferred to Cu2+.

[3] The transfer of electron leaves behind a hole, which carries an extra positive charge and a negative hole is created.[4] It appears that the positive hole moves through the lattice, hence it appears as P-type semiconductor.12 – 16 and 13 – 15 compoundsCombination of elements of Gr – 13 and Gr – 15 or Gr – 12 and Gr – 16 produce compounds which stimulate average valence of four as in Ge or Si .12 – 16 compounds –> ZnS, CdS, CdSe, HgTe13 – 15 compounds –> InSn, AlP, GaAsMagnetic properties[1] Diamagnetic Diamagnetic substances are the substances which are weakly repelled by a magnetic field. The electrons in diamagnetic substances are all paired. They do not contain unpaired electrons. e.g. TiO2 , NaCl , C6H6 ,N2 , Zn[2] Paramagnetic Paramagnetic substances are those which are attracted by a magnetic field but they lose their magnetism in the absence of magnetic field. These substances have permanent magnetic dipole, due to presence of atoms, molecules or ions containing unpaired electrons. e.g. Cu2+, Fe3+, O2 , NO, CuO, etc. Substances containing unpaired electrons are further classified as:

(d) Ferromagnetic substances –> Ferromagnetic substances are those substances which are strongly attracted by a magnetic field and can be made into permanent magnets.

These substances show magnetism even in the absence of a magnetic field. The large magnetism in these substances is due to the spontaneous alignment of magnetic moment,i.e Unpaired electron in the same direction.

Example: Iron, cobalt, nickel, gadolinium and CrO2 CrO2 is used to make magnetic tapes for audio recording.

(e) Anti-ferromagnetic substances –> Anti-ferromagnetic substances are those substances in which equal number of magnetic moments are aligned in opposite directions so as to give zero net moment.

Example: MnO, MnO2 and Mn2O3(f) Ferrimagnetic substances Ferrimagnetism is observed when the magnetic moments of the domains in the substance are aligned in parallel and anti-parallel directions in unequal numbers They are weakly attracted by magnetic field as compared to ferromagnetic substances. Example: Fe3O4 (magnetite) and ferrites like MgFe2O4 and ZnFe2O4 These substances also lose ferrimagnetism on heating and become paramagnetic.

(important questions with answer)

Q.1) What are the stoichiometric defects found in ionic crystals? (1 Mark)

(Ans) The point defects in which the ratio of cations to anions remains the same as shown by the molecular formula of the compound are known as stoichiometric defects. Q.2) Write two examples of amorphous solids. (1 Mark)(Ans) Rubber and quartz glass are examples of amorphous solids.(Q.3) Name two metals which have cubic closepacked structure. (1 Mark)(Ans) Silver (Ag) and copper (Cu) have cubic close packed structure.(Q.4) Name the type of solids which are malleable, ductile and electrical conductors. (1 Mark)(Ans) Metallic solids are malleable, ductile and electrical conductors.(Q.56) Sometimes,crystals of common salt (NaCl) are yellow instead of being pure white. Why?(1 Mark)(Ans) Sometimes,crystals of common salt (NaCl)are yellow instead of being pure white due to the presence of electrons in some lattice sites in place of anions.These sites actas F-centers and impart yellow colour to crystal of common salt.(Q.6) What are voids? (1 Mark)(Ans) Voids are the empty spaces present between atoms or ions, when they are packed within the crystal. (Q.7) How many atoms are present in the unit cell of bcc and fcc? (1 Mark)(Ans) Number of atoms in unit cell of bcc is2and infcc it is 4.(Q.8) Fe3O4 is ferrimagnetic at room temperature. What happens to its magnetic properties when it is heated to 850 K? (1 Mark)(Ans) When Fe3O4 is heated to 850 K it loses ferrimagnetism and becomes paramagnetic.(Q.9) Name the point defect which lowers the density of a crystal. (1 Mark)(Ans) Schottky defect lowers the density of a crystal. (Q.10) Write the unit in which the magnitude of magnetic moment is measured.

(1 Mark)(Ans) The magnetic moment is measured in Bohr magneton ( B).(Q.11) What do you mean by F-center? (1 Mark)(Ans) Theanionsites which are cccupied by unpairedelectron are called F-centres.Q.(12) Name the type of solids which have long range orders. (1 Mark)(Ans) crystalline solids have long range order.(Q.13) Name one solid which has both Frenkel and Schottky defects? (1 Mark)(Ans) Silver bromide (AgBr) has both Schottky and Frenkel defects.(Q.14) Why is the window glass of the old building thick at the bottom?

(1 Mark)(Ans) Glass is a pseudo solid, that is, it is a supercooled liquid of high viscosity. It flows down very slowly and makes the bottom portion of window glass of old building slightly thicker.(Q.15) What causes the conduction of electricity by semiconductors? (1 Mark)(Ans) Electrons and holes produced by defects cause the conduction of electricity by semiconductors.(Q.16) Name the salt which can be added to AgCl so as to produce cationic vacancies. (1 Mark)(Ans) SrCl2orCdCl2 is added to AgCl to produce cationic vacancies. (Q.17) What is a diode? (1 Mark)

(Ans) Diode is a combination of n- type and p- type semiconductors. It is used as a rectifier. (Q.18) Name a transition metal oxide which has appreance and conductivity like that of Cu? (1 Mark)(Ans) Rhenium oxide (ReO3)(Q.19) What is the co-ordination number of an atom present in octahedral void?

(1 Mark)(Ans) The co-ordination number of atom present in octahedral void is 6. (Q.20) Write one property which is caused due to the presence of F-center in a solid. (1 Mark)(Ans) The colours and paramagnetic behaviour of the solid is due to the presence of F-center in a solid.(Q.21) Draw body centred and face centred cubic unit cells. (2 Marks)(Ans)

(Q.22) Define: (i) Crystal lattice (ii) Packing efficiency (2 Marks) (Ans) (i)Crystal lattice is the well-defined regular arrangement of atoms, molecules or ions in three-dimensional space.(ii) Packing efficiency is the percentage of the total space filled by the particles.(Q.23) Explain the following with suitable example: (i) Molecular solids (ii) Metallic solids (2 Marks)(Ans) (i) In molecular solids, the constituents particles are molecules which areheld together by van der Waals forces of attraction. E.g. I2.(ii)Metallic solids consist of positive ions surrounded by andheld together bya sea of free electrons. These free electrons are mobile and are responsible for high thermal and electrical conductivity of metallic solids. E.g. Ag, Cu. (Q.24) What is rank? Find rank of face centered cubic unit cell. (2 Marks)(Ans) Rank is the number of atoms per unit cell of a crystal.

In f c c Contribution of atoms present at the corners=

Contribution of atoms present at faces = Rank = 1 + 3 = 4.

(Q.25) What is doping? How does n-type and p-type semiconductors formed? (2 Marks)(Ans) The process of introducing atoms of other elements as impurity into an insulator to make it semiconductor is called doping. Doping of silicon or germanium with electron rich impurities like P, As, Sb results in formation of n-type semiconductors whereas p-type semiconductors are formed by adding elements of group 13 like B, Al,Ga.(Q.26) Explain the following with one example each:

(i)Ferrimagnetism (ii)Antiferromagnetic substances (2 Marks)

(Ans) (i) When magnetic moments are aligned in parallel and anti parallel directions in unequal numbers itresults in net moment. It is called ferrimagnetism. These substances are weakly attracted by magnetic field as compared to ferromagnetic substances.E.g:Magnetite (Fe3O4)(ii) Antiferromagnetic substances are expected to posses paramagnetism or ferromagnetism but actually they possess zero net magnetic moment. It is due to the presence of equal number of domains in opposite direction. E.g: MnO.(Q.27) Write two differences between isotropy and anisotropy. (2 Marks)(Ans)

Isotropy Anisotropy(i)These substances show identical values of physical properties in all directions.

These substances show different values of physical properties in different directions.

(ii) Amorphous solids show isotropy. Crystalline solids show anisotropy.

(Q.28) Potassium metal crystallizes in bcc. The edge length of unit cell is 4.3 A0. Find the radius of potassium atom. (2 Marks)(Ans)

(Q.29) A solid is made of two elements A and B. Atoms of element A ocupy all the tetrahedral sites while atoms of element B are in ccp arrangement.From this data find the formula of the compound. (2 Marks)(Ans) There are 2 tetrahedral sites per atom of B because atoms of element B have ccp arrangement.There are 2 atoms of element A for each atom of element B because all tetrahedral sites are occupied by atoms of element A. Therefore, the formula of the compound is A2B.(Q.30) A solid A+ B– has NaCl closed packed structure. The radius of anion is 245 pm. Find radius of anion. (2 Marks)(Ans) The co-ordination number of A+ B– = 6 ( It has NaCl type structure.)

For this,

(Q.31) The edge length of unit cell of NaCl crystal lattice is 5.6A0. The density of NaCl is 2.2g cm–3. Find the number of formula units of NaCl per unit cell. (2 Marks)(Ans)

(Q.32) Find the number of NaCl molecules in a unit cell of its crystal. (2 Marks)(Ans)

No. ofNaCl molecules in aunit cellof NaCl= 4(Q.33) A compound is made of two atoms X and Y. Atom X is arranged in ccp and atom Y occupies tetrahedral site. Find the formula of compound.(2 Marks)(Ans) No. of atoms of X = 8No. of atoms of Y =(8/8 ) + (6/2) = 4

Ratio of X : Y is 2: 1Formula of compound is X2Y.

(Q.34) The unit cell of metallic silver is face-centred cubic. What is the mass of a silver unit cell? (Ans) (2Marks)

(Q.35) A metal crystallizes as face centered cubic lattice with edge length of 450pm. Molar mass of metal is 50g mol–1. Find the density of metal.(2 Marks)(Ans)

(Q.36) A solid has bcc structure. Distance of closest approach between two atoms is 1.73A0. Find edge length of cell. (2 Marks)(Ans) In bcc, distance of closest approach =[( 3)/2] edge length

Or Edge length =[1.73/ 3] 2A

= 200pm (Q.37) Classify the following solids as metallic; molecular,amorphous, covalent or ionic. (i) SO2 (ii) Diamond (iii) I2 (iv) MgO (iv) Ag (v) Quartz (vi) Ar (3 Marks)(Ans) Metallic solid - AgCovalent solid - QuartzMolecular solids - I2, Ar, SO2

Ionic solids - MgO(Q.38) (i) What are voids? (ii) How a tetrahedral void is different from octahedral void?(iii) Draw structure of tetrahedral and octahedral void. (3 Marks)(Ans) (i) Atomsand ions are spherical in shape. A crystal is formed by close packing of atomsor ions.Since,spheres touch each other only at points, some empty space is left between them.This space is called void or hole. (ii)A tetrahedral void is surrounded by four spheres(atoms), which lie at vertices of regular tetrahedron whereas an octahedral void is surrounded by six spheres(atoms).(iii)

(Q.) The density of an atom is 7.2g cm–3. It has bcc structure. The edge length is 288 pm. How many atoms of element does 208g of element has?

(3 Marks)(Ans)

(Q.39) Find the type of lattice for cube having edge length of 400 pm, atomic wt. = 60 and density = 6.25 g/cc. (3 Marks)(Ans) Let the no. of atoms in a unit cell = x

Mass of one unit cell =

=

Volume of unit cell = (edge length) 3 = = 64 x 10-24 cm3

Density =

or, Mass =

=

= 4The unit cell has4 atomsIt is face centered cubic lattice.

(Q.40) A mineral contains Ca, O and Ti. In its unit cell oxygen atoms are present at face centres, calcium atoms at corners and titanium atoms at centre of cube. Find the oxidation number of titanium in the mineral.(3 Marks)

(Ans) No. of Ca atoms =

No. of O atoms = No. of Ti atoms =

Formula of mineral is CaTiO3

Let oxidation number ofTi = x In CaTiO3

+2 + x + (-2 3) = 0x = +4Oxidation state of titanium is + 4 in this mineral.

(Q.41) A metal oxide crystallizes in a hexagonal close packed array of oxide ions with two out of every three octahedral holes occupied by metal. Find formula of metal oxide. (3 Marks)(Ans)

(Q.42) A metal has cubic lattice. Edge length of lattice cell is 2A0. The density of metal is 2.4g cm–3.How many units cell are present in 200g of metal.(3 Marks)(Ans)

(Q.43) The density of NaCl crystal is 2.155g cm–3 and distance between Na+ and Cl– is 280 pm. Find value of Avogadro’s number. (3 Marks)(Ans) NaCl has fcc structure.

In fcc, a (edge length) = = pm= 560 pm

For fcc , Z = 4, Na = Avogadro number =?

(Q.44) In a face centered cubic lattice atoms of A occupy corner of cell and that of B occupy face centers. One of the A atoms is missing from one corner of a unit cell. Find the simplest formula of compound. (3 Marks)(Ans)

********************************** Unit-2 mm - 5

SOLUTIONSSolid Solutions Gas in solid Solution of hydrogen in palladium

Liquid in solid Amalgam of mercury with sodium

Temp. Vs Conc.Mass %, ppm, mole fraction and molality are independent of temperature, whereas molarity depends on temperature. This is because volume depends on temperature.

Henry’s law.

At a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. p = KH . x KH = Henry’s law constant ( greater the KH value means lower the solubility.)

Application of Henry’s law.

1.To increase the solubility of CO2 in soft drinks, the bottle is sealed under high pressure.2. To avoid bends, the tanks used by scuba divers are filled with air diluted with helium

Temp and Solubilty of gas

Solubility of gas increases with decrease of temperature. It is due to this reason that aquatic species are more comfortable in cold waters rather than in warm waters.

Raoult’s law for volatile liquids

The partial vapour pressure of each component in the solutionis directly proportional to its mole fraction p1 α x1 p1 = p1

0 x1

Ideal Solutions

The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. ( For ideal solution ΔmixH = 0, ΔmixV = 0)Example : Solution of n-hexane and n-heptane,

Non-ideal Solutions Positive deviation : A-B interactions are weaker than those between A-A

or B-B, Example - Mixtures of ethanol and acetoneNegative deviations : Forces between A-A and B-B are weaker than those between A-B Example- mixture of phenol + aniline. a mixture of chloroform +acetone

Azeotropes

Mixtures have same composition in liquid and vapour phase and boil at a constant temp.minimum boiling azeotrope(positive deviation) eg- 95% aq ethanolmaximum boiling azeotrope(negative deviation) eg- 68% aq nitric acid

Colligative properties Depend on the number of solute particles not upon their nature.

Osmosis Solvent flows through the semi permeable membrane from pure solvent to the solution.

Osmotic pressure The extra pressure applied on the solution that just stops the flow of solvent is called osmotic pressure of the solution

Isotonic solutions Two solutions having same osmotic pressureHypertonic Higher osmotic pressure than a particular solnHypotonic Lower osmotic pressure than a particular soln

Reverse Osmosis

The direction of osmosis can be reversed if a pressure larger than the osmotic pressure is applied to the solution side. That is, now the pure solvent flows out of the solution Application : Desalination of sea water

van’t Hoff factor ( i )ratio of normal molar mass to experimentally determined molar mass or as the ratio of observed colligative property to the calculated colligative property.

(Q1) Define binary solution? (1 Mark)

(Ans)Binary solution is a solution containing only one solute dissolved in a solvent.(Q2) What is molarity?

(1 Mark)(Ans) The number of moles of solute dissolved in one litre or 1dm3 of solution is known as molarity.(Q3) What do you understand by saying that molality of a solution is 0.2?

(1 Mark)(Ans) This means that 0.2 mol of the solute is dissolved in 1Kg of the solvent.(Q4) Why is the vapour pressure of a liquid remains constant at constant temperature?

(1 Mark)(Ans) At equilibrium, the rate of evaporation = rate of condensation. Hence the vapour pressure of a liquid is constant at constant temperature.(Q5) Define Azeotropes?

(1 Mark)(Ans) Constant boiling mixtures are called Azeotropes.(Q6) Which substance is usually added into water in the car radiator to act as antifreeze?

(1 Mark)(Ans) Ethylene glycol is usually added into water in the car radiator to act as antifreeze.(Q7) Which liquids form ideal solution?

(1 Mark)(Ans) Liquids having similar structure and polarities.

(Q8) Which property of solution depend only upon the number of moles of solute dissolved and not on the nature of the solute?

(1 Mark)(Ans) Colligative properties.(Q9) Write one example each of solid in gas and liquid in gas solution?

(1 Mark)(Ans) Solid in gas e.g. Camphor in nitrogen gas. Liquid in gas – e.g. Chloroform mixed with N2 gas(Q10) What is molal elevation constant or ebullioscopic constant?

(1 Mark)(Ans) The elevation in boiling point which takes place when molality of the solution is unity, is known as ebullioscopic or molal elevation constant.(Q) Define van’t Hoff factor.

(1 Mark)(Ans) The ratio of the observed colligative property to the theoretical value is called van’t Hoff factor.(Q11) Two liquids A and B boil at 1200c and 1600c respectively. Which of them has higher vapour pressure at 700 c?

(1 Mark)(Ans)Lower the boiling point, more volatile it is .So liquid A will have higher vapour pressure at 700c.(Q12) What happens when blood cells are placed in pure water?

(1 Mark)(Ans) Water molecules move into blood cells through the cell walls. So, blood cells swell and may even burst.(Q13) What is the effect of temperature on the molality of a solution?

(1 Mark)(Ans) No effect.(Q14) Write Henry’s law.

(1 Mark)(Ans) The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas at a given temperature.(Q15) What is the relation between normality and molarity of a given solution.

(1 Mark)(Ans) Normality = 2 x molarity.(Q16) What is an antifreeze?

(1 Mark)(Ans) An antifreeze is a substance which is added to water to lower its freezing point. e.g. Ethylene glycol(Q17) Why cutting onions taken from the fridge is more comfortable than cutting onions lying at room temperature?

(1 Mark)(Ans) The vapour pressure is low at lower temperature. So, less vapours of tear – producing chemicals are produced(Q18) What will be the van’t Hoff factor for O.1 M ideal solution?

(1 Mark)(Ans) Van't Hoff factor = 1, because ideal solution does not undergo dissociation or association.(Q19) Name the substances which are used by deep sea divers to neutralize the toxic effects of nitrogen dissolved in the blood.

(1 Mark)(Ans) Theyusemixture of helium andoxygen to neutralize the harmful effects of nitrogen.

(Q20) Why is Anoxia disease very common at higher altitudes? (1 Mark)

(Ans) Anoxia is very common at higher altitudes because of lower partial pressure of oxygen at higher altitudes.(Q21) What is the optimum concentration of fluoride ions for cleaning of tooth?

(1 Mark)(Ans) The optimum concentration of fluoride ions for the cleaning of tooth is 1.5 ppm. [If it is more than 1.5 ppm it can be poisonous and if less than 1.5 ppm it isineffective.](Q22) What happens to blood cells when they are placed in water containing less than 0.99% (mass/volume) salt?

(1 Mark)(Ans) The blood cells collapse due to loss of water by osmosis when placed in water containing less than 0.9% (mass/volume) salt. (Q23) Why is molality generally preferred over molarity as unit for expressing concentration of solutions?

(1 Mark)(Ans) Molality is preferred over molarity as unit for expressing concentration of solutions because it involves mass termwhich is not affected by change in temperature while molarity involves volume termwhichchanges with temperature.(Q24) Calculate the van't Hoff factor of K4[Fe(CN)6].

(1 Mark)(Ans) K4[Fe(CN)6] 4 K

+ +

[Fe(CN)6] 4-

Since one molecule of K4[Fe(CN)6] dissociates to produce 5 ions, the value of van't Hofff factor is 5.(Q25) Name the law which explains the relationship between solubility of a gas in a liquid and pressure above the liquid surface. Also write the name of the variable which is kept constant for this law.

(1 Mark)(Ans) This is Henry’s law and temperature is kept constant for this law.[According to Henry's law the partial pressure of gas in vapour phase is proportional to the mole fraction of gas in the solution at constant temperature.](Q26) How does Henry constant vary with the solubility of gas in the solution at a given pressure?(1 Mark)(Ans) Higher the value of “Kh” (Henry constant), lower is the solubility of gas in solution.(Q27) What care should be taken while preparing intravenous injection?

(1 Mark)(Ans) They should be prepared in aquatic medium and salt concentrations shouldbe same as blood plasma levels.(Q28) Why osmotic pressure is better technique for determination of molar mass of biomolecules? (1 Mark)(Ans) Osmotic pressure is better technique for the determination of molar mass of biomolecules because they are generally not stable at higher temperatures and have poor solubility. Therefore, other techniques like elevation in boiling point and depression in freezing point cannot be used to determine molecular masses of biomecules.(Q29) What will be the molecular mass of acetic acid when it is dissolved in benzene? (1 Mark) (Ans) Normal molecular mass of acetic acid (CH3COOH) is 60. It dimerises

when dissolved in benzene.Therefore, molecular mass of acetic acid will be 120 when it is dissolved in benzene.(Q30) How does increase in temperature affects the solubility in endothermic and exothermic dissolution of substances?

(1 Mark)(Ans) The solubility in endothermic dissolution increases while in exothermic it decreases.(Q31) What will be the shape of graph which is obtained by ploting partial pressure of a gas against mole fraction of gas in solution?

(1 Mark)(Ans) It is a straight line sincepartial pressure of a gasand mole fraction of gas are directly proportional to each other.(Q32) Aquatic animals are more comfortable in cold water than in warm water. Explain? (1 Mark)(Ans) This is because “Kh” (Henry constant) values for both N2 and O2 increase with increase in temperature indicating that the solubility of gases increases with decrease intemperature.(Q33) Deep sea divers are advised not to come to surface immediately from deep waters. Why? (1 Mark)(Ans) Deep sea divers are advised not to come to surface immidiately from deep waters because sudden change in outside pressure can be fatal for divers because N2 will bubble out of blood vessels causing severe pain and can be dangerous.(Q34) What are colligative properties? Name them.

(2 Marks)(Ans) The properties of solute which depend upon the number of particles present in definite amount of solvent but not on the chemical nature of solute are called colligative properties. They are-

(i) Relative lowering of vapour pressure (ii) Elevation in boiling point(iii) Depression in freezing point (iv) Osmotic pressure

(Q35) Which is better method for expressing concentration of solution – Molarity or Molality? (2 Marks)

(Ans)

Volume changes with temperature whereas mass does not change with temperature. So molality, which does not have volume term in it is better method for expressing concentration.

(Q36) Define- (i) Mole fraction (ii) Molality(2 Marks)

(Ans) ( i) The mole fraction of a particular component in a solution is the ratio of the number of moles of that component to the total number of moles of all the components present in the solution.(ii) Molality of a solution is defined as a number of moles of solute dissolved per 1000g of solvent.(Q37) The osmotic pressure of human blood is 7.65 atm at 37°C. For injecting glucose solution it is necessary the glucose solution has same osmotic pressure as of human blood. Find the molarity of glucose solution having same osmotic pressure as of human blood.

(2 Marks)

(Ans) =

Or 7.65 =

=

= Molarity = 0.30 (Q38) Vapour pressure of two liquid A & B are 120 and 180mm Hg at a given temp. If 2 mole of A and 3 mole of B are mixed to form an ideal soln, calculate the vapour pressure of solution at same temperature. (2 Marks)(Ans) Total moles = 2 + 3 = 5

P solution =

= = 48 + 108= 156mm.(Q39) Density of 1 M soln of glucose 1.18g/cm3. Kf for H2O is 1.86 Km–1. Find freezing point of solution. (2 Marks)(Ans) Mass of solution =volume x density =1000 x 1.18=1180gMass of water =1180 - 180=1000g

= 1 m=

=

= = 0 – 1.86 = –1.86°C(Q40) An aqueous solution freezes at –0.186°C. Kf = 1.86°, Kb = 0.512. Find elevation in boiling point.

(2 Marks)

(Ans) = 0 – (0.186) = 0.186°C =

Or m = =

= (Q41) Vapour pressure of dilute solution of glucose is 750 mm of Hg at 373K. Calculate the mole fraction of solute.

(2 Marks)(Ans) 373K = 100°C Vapour pressure of pure water = 760 mm Hg.P = 750 mm Hg (given)

Or (Q42) Ethylene glycol solution having molality 0.5 is used as coolant in a car. Calculate the freezing point of solution (given Kf=1.86 °C/mole)

(2 Marks)(Ans)

(Q46) Calculate the molality of solution prepared by dissolving 18g of glucose 500g of water. (2 Marks)

(Ans) Mol. Wt. of glucose =(12 x 6) + (1 x 12) + (16 x 6)

= 180

Molality = = (Q47) Find the vant Hoff factor for Al2(SO4)3

(2 Marks)(Ans)

Total ions produced = 2 + 3 = 5

(Q48) On a hill station pure water boils at 99.82°C. The Kb of water is 0.513°C Kg mol–1. Calculate the boiling point of 0.69m solution of urea.

(2 Marks)(Ans)

Boiling point of solution = Boiling point of water + = 99.82 + 0.3539= 100.17° C (Q49) A soln of ethanol in water is 1.6 molal. How many gms of ethanol is present in 500g of solution. (2 Marks)Ans) Mass of ethanol =Molality x Molecular weight= 1.6 x 46= 73.6 g

Total of mass of solution = 73.6 + 1000 = 1073.6g1073.6g of solution contain 7.6g of ethanol.

Mass of ethanol in 500g of solution = (73.6 / 1073.6)x 500

(Q50) List two conditions that ideal solutions must satisfy. (2 Marks)

(Ans) 1. Hmixing and Vmixing of ideal solutions should be zero.2.They should obey Raoult’s law over the entire range of concentrations.(Q51) Explain ideal and non-ideal solutions with respect to intermolecular interactions in a binary solution of A and B.

(2 Marks)(Ans) For the given binary solution of A and B, it would be ideal if A-B interactions are equal to A-A and B-B interactions and it would be non-ideal if they are different to each other.The deviation from ideal behaviour will be positive if A-B interactions are weaker as compared to A-A and B-B. The deviation will be negative if A-B interactions are stronger as compared to A-A and B-B.(Q52) (i) What are minimum boiling and maximum boiling azeotropes? (ii) Can azeotropes be separated by fractional distillation?

(2 Marks) (Ans) (i) Minimum boiling azeotropes are the non-ideal solutions showing positive deviation while maximum boiling azeotropes are those which show negative deviation. Because of positive deviation their vapour pressures are comparatively higher and so they boil at lower temperatures while in case of negative deviation, thevapour pressures are lesser and sohighertemperature arerequired for boiling them.(ii) No, azeotropes can’t be separated by fractional distillation. (Q53) (i) When a non-volatile solute is added to solvent,there is increase in boiling point of solution.Explain. (ii) Define ebullioscopic constant and give its units.

(2 Marks)(Ans) (i) When a non-volatile solute is added to a volatile solvent the vapour pressure of pure solvent decreases because a part of the surfaceis occupied by non-volatile solute which can’t volatilise. As a result, thevapour pressure of solution decreases andhence, the solution requires acomparatively higher temperature to boilcausing an elevation of boiling point.(ii) Ebullioscopic constant isdefined as the elevation in boiling point of a solution of a non-volatile solute when its molality is unity. Its units are K Kg mol-1

(Q54) One molal solution of a given solvent is always less concentrated than one molar solution. Explain.

(2 Marks)

(Ans) In one molar solution one gram mole of solute is dissolved in one litre of solution while in case of one molal solution same one gram mole of solute is dissolved in 1000 gm of solvent only, which on considering normal density parameters of water, can’t be lesser in amount than solvent part present in one litre solution.Therefore,more amount of solvent is present in one molal solution than in one molar solution.

(Q55) State Raoult’s law. Prove that it is a special case of Henry’ law? (2 Marks)

(Ans) Raoult’s law states that partial pressure of a volatile component of a solution is directly proportional to its mole fraction. It is a special case of Henry’ law because it becomes the same when “Kh” (Henry constant) is equal to pressure of pure solvent.(Q56) How did Van’t Hoff explain the abnormal molecular masses of electrolytes like KCl in water and non-electrolytes like benzoic acid in benzene.

(2 Marks)(Ans) The molecular mass of KCl in aqueous medium has been observed to be almost half than expected and it has been explained as dissociation of KCl into K+ ions and Cl- ions when actual no. of particles become double and so become the colligative properties but since molecular mass is always inversely proportional to colligative property it becomes almost half.In case of benzoic acid in benzene, association of molecules take place when they dimerise and their no. becomes almost half and so molecular mass doubles as a result.(Q57) Calculate molality of an aerated drink having 2.5 gm of carbonic acid dissolved in 150 gm of water.

(2 Marks)(Ans) Molality= moles of solute / Mass of solvent in Kg= [(2.5 / 62.)/ (150 / 1000)] = 0.268.(Q58) When 3.49 gm of a non-volatile solute was dissolved in 125 gm of benzene, its boiling raised by 1.23K. Calculate the molecular mass of non-volatile solute. (Kb for benzene is 2.53 K kg Mol-1)

(2 Marks)(Ans)

(Q59) When 5.29 g of a non-volatile solute was dissolved in 400 g of water, its freezing point decreased by 1.79 K. Calculate the molecular wt of solute if Kf for water is 1.86 KkgMol-1 (2 Marks)(Ans)

(Q60) Calculate the mass of ethanol which is present in 500g of 1.6m solution of ethanol in water.(2 Marks)(Ans)

(Q61) One litre seawater is found to contain 5.8 10-3g of dissolved oxygen. Calculated the ppm of dissolved oxygen in seawater. (Density of seawater = 1.03 g/cc).

(2 Marks)(Ans)

(Q62) When a non-volatile solute is added to pure water its vapour pressure decreases by 4 mm Hg. Calculate molality of solution. (Vapour pressure of pure water is 40mm Hg)(2 Marks)(Ans)

(Q63) What are azeotropes? Give an example of maximum boiling azeotrope.(2 Marks)

(Ans) Azeotropes are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. Mixture of nitric acid and water is an example of maximum boiling azeotrope.(Q64) A decimolar solution of NaCl exerts OP(∏) of 4.6 atm at 300K. Find the degree of dissolution. (3 Marks)

(Ans) NaCl → Na+ + Cl - Initial moles 1 0 0Moles at EquilibriumTotal moles at equilibrium =

= 0.868 = 86.8% (Q65) An aqueous solution of a non-volatile and non-electrolyte substance boils at 100.5°C. Calculate osmotic pressure of this solution at 27°C. Kb (for water) per 1000g = 0.50.

(3 Marks)(Ans) =

= 100.5 –100 = 0.5°C

= Molality of solution = 1Solvent is water. density of solution = 1

Volume of solution = volume of solvent = 1000/1 = 1000ml = 1 L

= nRT

= (Q66) The vapour pressure of benzene at certain temperature is 640mm Hg. To 39.08 of benzene, non-volatile and non-electrolyte solid-weighing 2.175g was added. The vapour pressure of solution was 600mm of Hg. Find the mass of the solute?

(3 Marks)(Ans)

or =

=

=

or m = (Q67) The vapour pressure of water at 296K is 19.8 mm of Hg, 0.1 mol of glucose dissolved in 178.2g of water. Calculate the vapour pressure of resultant solution. (3 Marks)(Ans) n glucose = 0.1 (given)

n H2O =

x (water) = (Q68) A solution is prepared by dissolving 30g of non-volatile non-electrolyte solute in 90g water. The vapour pressure of solution was 2.8 K Pa at 298K. When 18g of water was further added to it, the vapour pressure became 2.9 k Pa at 298K. Calculate molar mass of solute.

(3 Marks)

(Ans)

(When A is for H2O) n = moles of solute

2.8 = ________ (1)

2.9 = _______ (2)Dividing eq. (1) and (2), we get

=

n = Mass of solute = 30 and molecular mass = 23

(Q69) Vapour pressure of pure water is 40mm. If a non-volatile solute is added to it, vapour pressure falls by 4 mm. Calculate molality of solution.

(3 Marks)(Ans)

The solution has 0.1 moles of solute in 0.9 moles of water.Mass of water =

Molality of solution = = 6.17m

(Q70) Conc. H2SO4 has a density 1.9g/ml and is .99% H2SO4 by weight. Find molarity of solution.(3 Marks)

(Ans) Mass of 1000 ml of H2SO4 = density volume == 1900 gMass of H2SO4 present in 1900 g (1L) of H2SO4

= = 1881 g

Mole of H2SO4 present in 1L = Molarity =

= = 19.197 M (Q710) A solution contains 25% water, 25% ethanol and 50% acetic acid by mass. Find mole fraction of each of the component.

(3 Marks)(Ans) Moles of water = n1 =

Moles of = n2 =

Moles of = n3 = Total moles in solution =

Mole fraction of water =

Mole fraction of =

Mole fraction of = (Q72) Which will have more osmotic pressure and why? Solution prepared by dissolving 6g/L of CH3COOH or Solution prepared by dissolving 7.45g/L of KCl

(3 Marks)(Ans)

Moles of CH3COOH =

Moles of KCl = Molar concentration of both the solutions is same.

KCl ionizes into K+ and Cl– where as CH3COOH does not ionize:Osmotic pressure is colligative property.Its value depend on number of particles.

Since, KCl produces more ions so, osmotic pressure of KCl will be more than that of CH3COOH .(Q73) A solution is prepared by mixing 50ml of chloroform and 50ml of acetone. What will be the resulting volume of solution? 100 ml or >100 ml or <100 ml.

(3 Marks)(Ans)

When chloroform and acetone are mixed, they form intermolecular hydrogen bonds. The hydrogen bonds are strong forces of attraction. As a result, volume of the solution will be less than 100ml. (Q74) One litre sample of seawater is found to contain 5.8 x 10-3 g of dissolved oxygen. Calculate the concentration of dissolved oxygen in seawater in ppm. (Density of seawater = 1.03 g/cc).

(3 Marks)(Ans) Mass = Volume x density= = 1030gMass of oxygen present in 106(one million) gof seawater

=

= (Q75) A solution is prepared by mixing 50ml of chloroform and 50ml of acetone. (i) The volume of the resulting solution will be 100 ml or less than 100ml or more than 100 ml?(ii) What happens to the net vapour pressure of resultant solution- it increases or decreases or remains same?Support your answer with suitable explanation. (3 Marks)(Ans) (i) When chloroform and acetone are mixed they exhibit negative deviation from ideal behavior because intermolecular hydrogen bonds are formed between molecules of acetone and

chloroform.Since thehydrogen bonds are comparatively stronger forces of attraction, the volume of the solution decreases. Therefore, the volume of the resulting solution will be less than 100ml.(ii) The net vapour pressure of the solution decreases because newly formed hydrogen bonds are stronger forces of attraction than the original forces of attraction existing in separate solutions.(Q76) Find molarity of a conc. H2SO4 sample having ρ 1.9 g/ml & is 99% pure by weight.(3 Marks)(Ans) Mass of 1000 ml of H2SO4 = Density Volume= 1.9 g/ml 1000 ml = 1900 gMass of H2SO4 present in 1900 g (1Litre) of H2SO4

= = 1881 g

Molarity of sulphuric acid =

= = 19.197(Q77) What are colligative properties? Why are they also called democratic properties? Is osmotic pressure a colligative property? Prove it.

(3 Marks)(Ans) These are the properties of solution that depend on the number of solute particles in the solution not at all on the nature of solute particle. Above mentioned is also the reason as why are they sometime called as democratic properties. Yes, osmotic pressure is a colligative property because osmotic pressure of a solution is proportional to its molarity at a given temperature.(Q78) Why semi permeable membrane is so important in the phenomenon of osmosis? What are isotonic, hypo tonic and hyper tonic solutions? Does osmosis take place in all three types of solutions?

(3 Marks)(Ans) The semi permeable membrane is very important in the phenomenon of osmosis because they only permit the movement of solvent molecules through them. Two solutions having similar osmotic pressure at a given temperature are called isotonic solutions. If the given solution has less osmotic pressure it is called hypo tonic and it is hyper tonic if its osmotic pressure is higher than the the solution on the other side of semi permeable membrane. Osmosis takes place only in hypo tonic and hypertonic solutions.(Q79) List three general applications of phenomenon of osmosis?

(3 Marks)(Ans) 1. Shrinking and swelling of blood cells when put into aqueous NaCl having concentration more or less than 0.9%.2. Water movement from soil into plant body.3. Preservation of meat and fruits by salting against bacterial infection.(Q80) Why is reverse osmosis of great practical utility to meet potable water requirements these days? Can it make sea water drinkable? Do we use ordinary semi permeable membranes for this?

(3 Marks)

(Ans) Reverse osmosis is of great practical utility since it can result in absolute purification of any water sample because semi permeable membrane allows only solvent molecules to pass through.It can make sea water drinkable though ordinary semi permeable membrane may not be able to sustain high pressure observed in desalination of sea water i.e. reverse osmosis of sea water.Semi permeable membrane of cellulose acetate is generally used for this.(Q81) The following solutions which have higher osmotic pressure is 1. Solution having 6gL-1 of CH3COOH or 2. Solution having 7.45gL-1 of KCl Explain.

(3 Marks)(Ans)

(Q82) Calculate the boiling point of a solution of urea prepared by dissolving 10.36 g of urea in 250 g of water on a hill station where pure water boils at 99.82°C. (Kb

of water is 0.513°C Kg mol–1) (3 Marks)(Ans)

(Q83) Calculate the mole fraction of each of the component in a solution containing 25% water, 25% ethanol and 50% acetic acid by mass. (3 Marks)(Ans)

(Q84) A solution has been prepared by dissolving 30g of non-volatile solute in 90g of water. The vapour pressure of solution is 2.8 K Pa at 298K. When 18g of water was added to it, the vapour pressure rose to 2.9 kPa at same temperature. Calculate molar mass of solute. (3 Marks)(Ans) According to Raoult’s law the vapour pressure of a component of a solution is equal to the product of its vapour pressure in pure state and its mole fraction

So (When A is for H2O) n = moles of solute

Dividing equation (1) by (2), we get

= n = 0.882Molecular Mass of solute = 30 x 0.882 = 26.470

(Q85) Calculate van’t Hoff factor of the following solutions1. Al2(SO4)3 is dissolved in water2. MgCl2 is dissolved in water3. Benzoic acid is dissolved in benzene4. Glucose is dissolved in water5. 10 gm of KCl is dissolved in 1 Litre of water (5 Marks)(Ans) 1.When Al2(SO4)3 is dissolved in water, its one molecule dissociates into 2 Al3+ ions and 3 SO4

2- ions, so its van’t Hoff factor for this is 5.2. When MgCl2 is dissolved in water, its one molecule dissociates into 1 Mg2+ ions and 2 Cl- ions, so its van’t Hoff factoris 3.3.When benzoic acid is dissolved in benzene, its two molecule associate to form a dimer. Therefore,its van't Hoff factor is 1/2.4. Glucose does not undergo association or dissociation, hence van’t 'Hoff factor forglucose is 1.5.When KCl is dissolved in water, its every moleculedissociates intwo ions.So its van’t Hoff factor for this is 2.(Q86) The osmotic pressure of human blood is 7.65 atm. Find the molarity of glucose solution that can be injected into human blood. If adding some more amount of glucose into it disturbs its molarity and now it boils at 100.50C, find its

new molality at 270C. What will be the effect on osmotic pressure of solution? Kb (for water) per 1000g = 0.50 (5 Marks)(Ans) As we know for injecting glucose solution into human blood it is necessary that glucose solution has same osmotic pressure as of blood. So osmotic pressure of solution should be 7.65 atm.

=

7.65 = (Human body temperature is 370C)

=

= Molarity = 0.30Now since elevation in boiling point has taken place with the addition of more glucose, it can be calculated as:

= = 100.5 –100 = 0.5°C

= Molality of solution = 1

The osmotic pressure of solution will increase because no of solute particles will increase.(Q87) Calculate the vapour pressure of a solution prepared by mixing 349.5 gm of component A (Mol. Mass 134) and 416.8 gm of component B (Mol. Mass 169) at 298 K. Vapour pressures of pure components A and B are 421mm Hg and 562mm Hg respectively. Also calculate mole fraction of A and B in vapour phase.

(5 Marks)(Ans)

As shown above component B is more volatile than component A and vapour phase is richer in component B

(Q88) 3.9 gm of a protein was dissolved in 63 gm of benzene when its freezing point was observed to go down by 3.26 K. Kf for benzene is 4.9 K Kg Mol-1. Calculate the percentage association of protein if it forms dimers in the solution?( molar mass of protein is 925) (5 Marks)(Ans) Wt of protein= 3.9 gmKf = 4.9 K Kg Mol-1Wt of benzene= 63 gmDepression in freezing point= 3.26 K

Thus experimental molecular mass of protein in benzene is 930.470Now suppose the degree of association of protein be X and 1-X mol of protein will be left unassociated and corresponding X/2 as associated at equilibrium.Therefore total no. of moles of particles at equilibrium will be1-X+X/2 = 1-X/2So total no moles at equilibrium equals von’t hoff factor (i)But as we know i = Theoretical molar mass/ Experimental molar mass

Therefore, degree of association of protein in benzene is only 1.2%(Q89) A solution of 10 gm NaCl in 1000 gm water freezes at –0.604K. Calculate the degree of dissociation of NaCl. (Kf for water is 1.86Kkgmol-1) (5 Marks)(Ans)

Unit-3 mm -5

Electrochemistry

Cell Notation :Zn(s) | Zn2+ (aq) || Cu2+(aq)| Cu(s) ; LOAN : Left / Oxidation /Anode / Negative

Nernst Eqn

Store CuSO4 in Zn pot?

No. because the following reaction takes place Zn(s) + CuSO4(aq) → ZnSO4 (aq) + Cu(s) ;

Electrochemical Series

F2(g) + 2e– → 2F– ; E0 = 2.87 F2(g) is the strongest oxidizing elementLi+ + e– → Li(s) ; E0 = –3.05 Li is the strongest reducing elementA negative E0 means that the redox couple is a stronger reducing agent(Itself will be oxidized)

Relation betweenE0 / Kc & ΔG

E=E0 - 2.303RT log Kc ΔGo= - nFEocell

nF Molar Conductivity

(Λm) Λm=K x1000/M Scm2 mol-1

Kohlrauschlaw

Limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.

Faraday’s 1st LawsThe amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of electricity passed through the electrolyte (solution or melt).

Faraday’s 2nd Laws

The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal ÷ Number of electrons required to reduce the cation).

Product of Electrolysis NaCl (molten) Cathode : Na+(l) + e– → Na(s) Anode : Cl–→ ½Cl2+e–

NaCl (aq) Cathode : H2O (l ) + e– → ½H2(g) + OH– Anode : Cl–→ ½Cl2+e–

H2SO4(dil) Cathode : H+ + e- ½ H2 Anode: 2H2O(l )→ O2(g) + 4H+

(aq) + 4e–

H2SO4(conc) Cathode : H+ + e- ½ H2 Anode: 2SO4 2– (aq) → S2O8 2–

(aq) + 2e–

AgNO3(aq)-Ag electrodes

Cathode : Ag+(aq) + e- Ag(s) Anode: Ag(s) Ag+(aq) + e-

AgNO3(aq)- Pt electrodes

Cathode : Ag+(aq) + e- Ag(s) Anode: 2H2O(l )→ O2(g) + 4H+

(aq) + 4e–

CuCl(aq)- Pt electrodes Cathode : Cu+(aq) + e- Cu(s) Anode: 2H2O(l )→ O2(g) + 4H+

(aq) + 4e–

Pri & Sec Batteries In the primary batteries, the reaction occurs only once, and cannot be reused again

Dry Cell

Anode(Zn) : Zn(s) → Zn2+ + 2e– Cathode(Graphite) : MnO2 + NH4

+ + e– → MnO(OH) + NH3 The emf = 1.5 V

Mercury Cell

Anode(Zn-Hg) : Zn(Hg) + 2OH– → ZnO(s) + H2O + 2e–

Cathode(HgO-C) : HgO + H2O + 2e– → Hg(l ) + 2OH–

EMF= 1.35 V and remains constant during its life as the overall reaction does not involve any ion in solution whose concentration can change during its life time.

Lead StorageBattery

Anode : Pb Cathode : Pb packed with PbO2 Electrolyte : 38% H2SO4

Anode: Pb(s) + SO4 2–(aq) → PbSO4(s) + 2e–

Cathode: PbO2(s) + SO4 2–(aq) + 4H+(aq) + 2e– → PbSO4 (s) + 2H2O (l )On charging the battery the reaction is reversed

Fuel Cells

Anode&Cathode : Porous C Electrolyte : Aq NaOHCathode: O2(g) + 2H2O(l ) + 4e– 4OH–(aq)Anode: 2H2 (g) + 4OH–(aq) 4H2O(l) + 4e–

The cell runs continuously as long as the reactants are supplied. Efficiency of about 70 % Pollution free. The water vapours produced during the reaction were condensed and added to thedrinking water supply for the astronauts (Apollo space programme)

Corrosion of Iron(Rusting)

Oxidation: Fe (s)→ Fe2+ (aq) +2e–

Reduction: O2 (g) + 4H+(aq) +4e– → 2H2O(l)Atomospheric oxidation : 2Fe2+(aq) + 2H2O(l) + ½O2(g) → Fe2O3(s) + 4H+(aq)

Prevention of Corrosion

By covering the surface with paint or by some chemicals (e.g. bisphenol). / Cover the surface by other metals (Sn, Zn, etc.) that are inert or react to save the object. An electrochemical method (sacrificial electrode like Mg, Zn, etc.) which corrodes itself but saves the object.

Important questionsQ1. How much electricity in terms of Coulomb is required to reduce 1 mol of

Cr2O72- to Cr3+.

Ans:- 2Cr2O7-2-------------2Cr+3, 2Cr+6+6e----------------2Cr3+

Therefore the coulomb of electricity required =6F, =6x96500 C= 579000 C

Q2. What is Fuel Cell?Ans:- Fuel cell is a device which produce the energy during the combustion of fuels like

Hydrogen , Methane, Methanol. Q3. A solution of CuSO4 is electrolysed using a current of 1.5 amperes for 10

minutes. What mass of Cu is deposited at the cathode? (Atomic mass of Cu=63.7)

Ans:- The reaction is Cu+2(aq.) + 2e ------------ Cu (s)The mass of copper deposited=E Cu X I X t = 63.7*1.5*10*60/ 2*96500C =0.297 g.

96500C Q4. Calculate the equilibrium constant for the reaction Cu (s) + 2Ag+ Cu

+2+2Ag(s) Eo Cu2+/Cu = +0.34V, EoAg+/Ag = + 0.80V.Q5. Write the Nernst equation and emf of the following cells at 298K:

Sn/Sn2+(0.050M)//H+(0.020M)/H2(g)/Pt(s) EoSn2+/Sn= - 0.13VQ6. Calculate the standard free energy change for the following reaction

at250C,Au(s) +Ca+2 (1 M)Au3+(1M) + Ca (s) ,The electrode values are Ca2+ /Ca = –2.87V,Au3+ / Au = +1.50V. Predict whether the reaction will be spontaneous or not at 250C.

Q7. How do you account for conductivity of strong and weak electrolyte with concentration? Plot the graphs also.

Q8. State Kohlrausch law . Calculate Limiting molar conductivity of NaCl, HCl and NaAc are 126.4, 425.9 &91 SCm2 mol-1.Calculate Limiting molar conductivity of HAc.

Ans:- According to this law, Molar conductiviy of an electrolyte, at infinite dilution can be expressed as the sum of contributions form its ividual ions. It the molar conductivity of the cation is denoted by Λo

+and that of the anions by Λo

- then the law of independent migration of ions is Λo

m=v+ Λo+ + v- Λo

-.

Q9. Resistance of conductivity cell filled with 0.1molL-1 KCl solution is 100 ohm.If the resistance of the same cell when filled with 0.02molL-1 KCl solution is 520 ohm. Calculate the conductivity & molar conductivity of 0.02molL-1 KCl solution. The conductivity of 0.1 molL-1 solution of KCl is 1.29Sm-1.

Q10. A Copper –silver is set up.The copper ion concentration in its is 0.10M.The concentration of silver is not known.The cell potential measured 0.422V.Determine the concentration of silver ion in the cell. Eo (Ag+/Ag) = +0.80V, Eo (Cu2+/Cu)= +0.34V.

Q11. A voltaic cell is set up at 250C With the following half cells

:Al(s)/Al3+(0.001M) and Ni2+(0.50)/Ni(s) ,Write the equation for the cell reaction that occurs when the cell generates an electric current and determine the cell potential (givenEo Ni2+/Ni = -0.25V, EoAl(s)/Al3+=-1.66V)

Q12. Write the reaction involved in the following cells: (a) Fuel Cell (b) Lead Storage Battery.

Q13. Three electrolytic cells A,B,C containing solutions ZnSO4,AgNO3,and CuSO4 respectively are connected in series .a Steady current of 1.5 amperes was respectively are connected in series .A steady current of 1.5 amperes was passed though them until 1.45g of silver deposited at the cathode of cell B.How long did the current flow? What mass of copper and zinc were deposited?

Q14. Conductivity of 0.00241M acetic acid is 7.896 X 10-6 S cm-1. Calculate its molar conductivity. If Λ0 for acetic acid is 390.5 S cm2 mol-1. What is its dissociation constant?

Q15. (a) Two half cell reactions of an electrochemical cell are given below: MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O Eo =+1.51VSn2+ Sn4++ 2e- Eo = +0.15V.Construct the redox reaction from the two half cell reaction and predict if the reaction favours formation of reactants or product shown in the reaction(b). How much electricity in terms of Faraday is required to produce (i)20g of Ca from molten CaCl2 (ii) 40g of Al from molten AlCl3

Unit-4 mm -5

Chemical Kinetics

Rate of a Chemical Reaction For RP

Change in concentration of a reactant or product in unit time. Unit => mol L-1s–1

Rate law / Rate equation / Rate expression

Rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which mayor may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation.

For aA + bB → cC + dDThe rate expression for this reaction is :Rate α [A]x [B]y => Rate = k [A]x [B]y k is the Rate const. (x , y from expts)

Order of a ReactionThe sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction.If Rate = k [A]x [B]y Order = x + y

Elementary reactions The reactions taking place in one step are called Elementary Reactions

Unit of rate constant For n th order reaction : mol1-n Ln-1 s-1

Molecularity of a Reaction

The number of reacting species taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction (values are limited from 1 to 3)

Rate Determining Step The overall rate of the reaction is controlled by the slowest step in a reaction called the rate determining step

Half-life (t1/2)The half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is represented as t1/2.

Zero Order Reactions(Integrated Rate Equation)

K =[R0] – [R] t Mol L-1 Sec-1

First Order Reactions(Integrated Rate Equation) K = 2.303log [A0]

t [A]

Dependence of t1/2 on [R]0For zero order : t1/2 α [R]0. For first order reaction t1/2 is independent of [R]0

Example of Zero OrderReactions

The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at high pressure. Rate = k [NH3]0 = k

Radioactive decay All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics

Pseudo First OrderReaction

When one of two reactants is present in large excess. During the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water The concentration of water does not get altered much during the course of the reaction and the reaction behaves as first order reaction. Such reactions are called pseudo first order reactions.

Activation Energy Activation energy is given by the energy difference between activated complex and the reactant molecules

Arrhenius equation

k = A e -Ea /RT (exponential form) ln k = – Ea/RT + ln A (logarithmic form) A is the Arrhenius factor or the frequency factor or pre-exponential factor

e -Ea /RT The factor e -Ea /RT corresponds to the fraction of molecules that have kinetic energy greater than Ea

Action of catalystCatalyst provides an alternate pathway or reaction mechanismby reducing the activation energy between reactants and products and hence lowering the potential energy barrier

Collision frequency (Z)

The number of collisions per second per unit volume of the reaction mixture.For A + B → Products Rate = P.ZAB e−Ea/ RT P is probability or steric factor ZAB is collision frequency of reactants, A and B

Important questionsQ1. Define Pseudo order reaction?Ans:- Reaction showing higher order but actually follow lower order is known as Pseudo

order Q2. The decomposition reaction of ammonia gas on platinum surface has a rate

constant = 2.3 x 10-5 L mol-1s-1. What is the order of the reaction?Q3 Mention the factors that affect the rate of a chemical reaction.Q4 From the rate expression for the following reactions determine their order

of reaction and dimensions of the rate constants.a) H2O2 (aq) + 3 I - (aq) + 2H+ 2H2O (l) + 3I-1 Rate = k [H2O] [I-]b) CH3 CHO (g) CH4(g) + CO(g) Rate = k [CH3 CHO]3/2

Q5.. A reaction is first order in A and second order in B.i) Write differential rate equation.ii) How is the rate affected when concentration of B is tripled?iii) How is the rate affected when the concentration of both A and B isdoubled?

Q6. The decomposition of NH3 on platinum surface is zero order reaction.What are the rates of production of N2 and H2 if k = 2.5 x 10-4 mol L-1 S-1?

Q7 . Derive the Integrated rate equation for first order reaction. Also find halflife period and plot the graph associated to it.

Q8. For a first order reaction, show that time required for 99% completion istwice the time required for the completion of 90% of reaction.

Q9. A first order reaction has a rate constant 0.0051min-1 .If we begin with 0.10M concentration of the reactant, what concentration of the reactant will beleft after 3 hours.

Q10. The half-life for radioactive decay of 14C is 5730 years. An archaeologicalArtifact containing wood had only 80% of the 14C found in a living tree.Estimate the age of the sample

Q11. What is the effect of temperature on the rate constant of a reaction? HowCan this temperature effect on rate constant be represented quantitatively ?

Q12. The rate of a reaction quadruples when the temperature changes from 293K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Q13. A first order reaction takes 40 min for 30% decomposition. Calculate t1/2..Q14 (a) Distinguish between order of reaction & Molecularity.

(b) For a decomposition reaction the values of rate constant k at two different temperatures are given below:k1 =2.15 x 10-8 L mol-1s-1 at 650K, k2 =2.39 x 10-7 L mol-1s-1

at 700K Calculate the value of Activation Energy for this reaction.Q15. (i) Write short notes on the following:

(a) Activation energy of a reaction (b) Elementary step in a reaction (c)Rate of a reaction(ii) The following result has been obtained during the kinetic studies of thereaction2A + B C+D

Experiment [A] mol L-1 [B] mol L-1 Intial rate mol L-1min-1

I 0.1 0.1 6.0x10-3

II 0.3 0.2 7.2X10-2

III 0.3 0.4 2.88X10-1

IV 0.4 0.1 2.40X10-2

Determine the rate law and rate constant for the reaction.

Q16 From the concentrations of C4H9Cl (butyl chloride) at different times given

below, calculate the average rate of the reaction:C4H9Cl + H2O C4H9OH + HCl

during different intervals of time.t/s 0 50 100 150 200 300 400 700 800

[C4H9Cl]/mol L–1 0.100 0.0905 0.0820 0.0741 0.0671 0.0549 0.0439 0.0210 0.017Ans We can determine the difference in concentration over different intervals

of time and thus determine the average rate by dividing Δ [R] by Δt

Unit - 5SURFACE CHEMISTRY MARKS-4

(Q.) What is meant by critical temperature of gas? (1 Mark)(Ans) Critical temperature is the minimum temperature above which a gas cannot be liquefied

howsoever high the pressure may be applied. (Q.) Give the expression for Fruendlich adsorption isother (1 Mark)(Ans) x/m= kp1/n

(Q.) What do x and m represent in the expression x/m=kp1/n (1 Mark)(Ans) ‘m’ is the mass of the adsorbent and ‘x’ is the number of moles of the adsorbate when the

dynamic equilibrium has been achieved between the free gas and the adsorbed gas.(Q.) Why is heterogeneous catalysis also known as surface catalysis (1 Mark)(Ans) In heterogeneous catalysis the reaction always starts at the surface of the catalyst. So, it is

also known as surface catalysis. (Q.) What is a hydrosol? (1

Mark)(Ans) A colloid in which the dispersion medium is water is known as hydrosol.(Q.) Define peptization? (1

Mark)(Ans) Peptization is a process of converting a precipitate into colloidal particlesby adding

suitable electrolyte.(Q.) Define Brownian movement? (1 Mark)(Ans) Brownian movement can be defined as continuous zig- zag movement of the colloidal

particles in a colloidal sol.(Q.) Why is Brownian movement important? (1 Mark)(Ans) Brownian movement opposes the force of gravity and does not allow the colloidal

particles to settle down, thus making the colloidal solution stable.(Q.) Differentiate between adsorption and absorption. (1 Mark)(Ans) Adsorption Absorption

a)it occurs only at surface a) it is a bulk phenomenab)concentration on the surface b) concentration is sameis more than in the bulk through out the material

(Q.) What is the effect of temperature on adsorption? (1 Mark)(Ans) Adsorption processes, being exothermic, decreases with increase in temperature.(Q.) When a finely powdered active carbon is stirred into a solution of a dye, the intensity of

color in solution decreases. Why? (1 Mark)(Ans) The intensity of color in the solution decreases because of gas adsorbedon the surface of

carbon.(Q.) Why do finely divided substances have larger adsorption power? (1 Mark)(Ans) Finely divided substances have large surface area for adsorption and hence have larger

adsorption power. (Q.) What are zeolites? (1

Mark)(Ans) Zeolites are aluminosilicates i.e. three dimensional network silicates in which some

silicon atoms are replaced by aluminium atoms.

(Q.) Why are zeolites called shape selective catalysts? (1 Mark)(Ans) Zeolites are called shape selective catalysts because their catalytic action depends upon

the size and shape of the reactant and the product molecules as well as on their own pores and cavities.

(Q.) A small amount of silica gel and that of anhydrous CaCl2 are placed separately in two corners of a vessel containing water vapours. What phenomena will occur in the two corners? (1 Mark)

(Ans) Adsorption would occur where silica gel is kept in the vessel where as absorption will occur in the corner where CaCl2 is placed.

(Q.) Name the substance catalysed by Zymase. (1 Mark)(Ans) Glucose--Zymase->ethyl alcohol. (Q.) How can colloidal solution of ferric hydroxide be prepared by peptization?

(1 Mark)

(Ans) A colloidal sol. of ferric hydroxide can be prepared by adding small quantity of ferric chloride solution to freshly prepared precipitate of ferric hydroxide.

(Q.) What is the cause of Brownian movement? (1 Mark)(Ans) Brownian movement is caused by the striking of the colloidal particles with the molecules

of dispersion medium due to their kinetic energy. (Q.) Define Tyndall effect? (1 Mark)(Ans) It is defined as the scattering of light by the colloidal particles present in a colloidal

solution.(Q.) What happens to a gold sol. when gelatin is added to it? (1 Mark)(Ans) Gold sol. which is lyophobic starts behaving like lyophilic sol. (Q.) Write down the relation between pressure of the gas and the amount of it adsorbed?

(1 Mark)(Ans) x/m = K P1/n

(Q.) Which adsorption may be a multilayered formation phenomenon? (1 Mark)(Ans) Physisorption(Q.) Which is irreversible and why? Physisorption or chemisorption. (1 Mark)(Ans) chemisorption. Because of the formation of chemical bond.(Q.) Name the promoter used in Haber’s process? (1 Mark)(Ans) Molybdenum.(Q.) What is emulsion? What are their different types? (2 Marks)(Ans) An emulsion is the colloidal dispersion in which both the dispersed phase and the

dispersion mediums are liquids. They can be of two types:-i) Emulsion of oil in water.ii) Emulsion of water in oil.

(Q.) How are micelles formed in soap solution? (2 Marks)(Ans) Soap is sodium salt of fatty acids (RCOONa) which when dissolved in water dissociates

to give RCOO- and Na+. The RCOO- consists of polar group COO- which is hydrophilic and stays at the surface and the non polar group R which being hydrophobic stays away from the surface. At high concentrations RCOO- ions are pulled into the solution to form spherical aggregates with R pointing to the centre COO- part remaining outward. This aggregate is known as ionic micelle.

(Q.)How can lyophobic sols be prepared by mechanical disintegration?(2 Marks)

(Ans) The coarse suspension of the substance is introduced in the colloid mill that consists of two metal discs close together rotating at a high speed in the opposite directions. Here the suspension particles are broken to the colloidal size.

(Q.) Differentiate between chemisorption and physisorption? (2 Marks)(Ans) Physisorption:

a)The forces operating are weak vander Waal’s forcesb)The heat of adsorption is low 20-40 KJ Mol-1

c)Does not require any activation energyd)Forms multimoleculer layer

Chemisorption:a)Forces acting are similar to those of chemical bondsb) The heat of adsorption is high 80-240 KJ Mol-1

c) Requires activation energyd) Forms unimolecular layer

(Q.) Describe the mechanism of peptization? (2 Marks)(Ans) When electrolyte is added to the freshly precipitated substance, the particles of the

precipitate preferentially absorb one particular type of ions of the electrolyte and get dispersed due to electrostatic repulsions giving particles of colloidal size and hence cause peptization.

(Q.) Give any two reasons for the origin of electrical charge on the colloidal particles.(2 Marks)

(Ans) The two reasons are:i) Due to electron capture by sol particles during electro dispersion of metals, due to preferential adsorption of ions from solutionii) Dissociation of colloidal sols.

(Q.) How is the electrical charge of the colloidal particles responsible for the stability of colloidal sols? (2 Marks)

(Ans) The electrical charges of the particles prevent them from coming together due to electrostatic repulsion. All the dispersed particles in a colloidal solution carry the same charge while the dispersion medium has equal and opposite charge.

(Q.) What is demulsification? Name two demulsifiers. (2 Marks)(Ans) The process of separation of the constituent liquids of an emulsion is called

demulsification. Demulsification can be done by centrifugation or boiling. (Q.) Why lyophilic colloids are called reversible sols while lyophobic sols are called irreversible

sols? (3 Marks)(Ans) In the lyophilic colloids if the dispersed medium is separated from the dispersion medium

the sol can be made again by simply remixing with the dispersion medium. So they are called reversible sols. In lyophobic sols if small amount of electrolyte is added, the sols are readily precipitated and do not give back the colloid by simple addition of the dispersion medium. So they are called irreversible sols.

(Q.) Describe the preparation of the following colloidal solution.(a) Gold sol (b) Sulphur sol (3 Marks)

(Ans) (a) Preparation of Gold sol :- By the reduction of very dilute solution of silver salts with a suitable reducing agent

2AuCl3 + 3SnCl2 -------------> 2Au + 3SnCl4 Gold sol

(b) Preparation of Sulphur sol :- By the oxidation of H2S in the presence of suitable oxidizing agent like nitric acid, bromine water, etc.

H2S + Br2 --------------> S + 2HBr

H2S + 2HNO3 ------------> 2H2O + 2NO2 + S

(Q.) What are macromolecular and multimolecular colloids? How are they different from associated colloids? (3 Marks)

(Ans) Macromolecular colloids:-

i)They are molecules of large size. ii)They have lyophobic property.Multimolecular colloids:-i) They are formed by the aggregation of large number of atomsor molecules which have

diameter less than 1nm.ii) They have lyophilic property.Associated colloids:-i) They are formed by the aggregation of large number of ions in concentrated solutionii) They contain both lyophilic and lyophobic groups (Q.) What are lyophilic and lyophobic solutions? Give examples for each.

(3 Marks)

(Ans) Lyophilic solutions are those that can be prepared by directly mixing the dispersed phase with dispersion medium. For example starch dissolved in water.

Lyophobic solutions are those that can not be prepared directly but some special methods are used to prepare them. For example metal sulphides when mixed with a dispersion medium directly do not result in any colloid.

(Q.) "Action of soap is due to emulsification and micelle formation".Comment. (3 Marks)(Ans) Yes, action of soap is due to emulsification and micelle formation. Soaps are sodium salt

of higher fatty acids like sodium stearate, C17H35COO-Na+

The anionic head of stearate ion (-COO-) is hydrophobic in nature and has great affinity for water, while the hydrocarbon part (C17H35

-) is hydrophilic in nature and great affinity for oil,grease etc.When soap is used in water, the anions (C17H35COO-) form micelle and due emulsification encapsulate oil or grease inside. These micelle are removed by rinsing with water; while free dirt (from oil or grease) either settle down or are washed away by water. Thus the main function of a soap is to entrap oil or grease with the micelles through emulsification, thereby freeing dirt from grease and oil.

(Q.) Why the sun looks red at the time of setting? Explain on the basis of colloidal properties.(3 Marks)

(Ans) At the time of setting, the sun is at the horizon. The light emitted by the sun has to travel a longer distance through the atmosphere. As a result, blue part of the light is scattered away by the dust particles in the atmosphere. Hence, the red part is visible.

(Q.) Explain the reason for these: (3 Marks)(a) Sky looks blue in colour.(b) Delta is formed at the meeting place of river and sea water.(c) Blood coagulate on treatment of alum.

(Ans) (a) Sky looks blue in colour because colloidal particles suspended in environment scatter the light and blue light is scattered maximum.(b) The charged colloidal particles of river water neutralized by ions present in sea water so coagulation take place.(c) The charged colloidal particles present in blood are neutralized by ions of alum.

UNIT – 6 Marks-3General Principles and Processes of Isolation of elements

1. Give the name and composition of ore chosen for extraction of aluminium. (1 Mark)Ans: The ore chosen for the extraction of aluminium is bauxite and its composition is

Al2O3.xH2O.2. What is leaching? (1 Mark)Ans: Leaching is the process of extracting a substance from a solid by dissolving it in a liquid.

In metallurgy leaching is used for the ores that are soluble in a suitable solvent.3. Why cryolite & fluorospar added to alumina during electrolytic reduction? (1 Mark)Ans: Cryolite and fluorospar are added to alumina during electrolytic reduction to reduce the

melting point of alumina and to increase its conductivity4. Reduction with C for Cu2O can be done at lower temp. than ZnO. Why? (1 Mark)Ans: In the Ellingham diagram the curve for Cu2O lies higher than ZnO i.e. for the reduction of

Cu2O with C the negative value of Gibbs energy can be reached at a lower temperature than ZnO.

5. Although thermodynamically feasible in practice magnesium metal is not used for the reduction of alumina. Why? (1 Mark)

Ans: Magnesium can reduce alumina at the temperature above the intersection point of the curves for Al2O3 and MgO in the Gibbs Energy vs T plot (Ellingham diagram). But the temperature at which this is feasible is too high to be achieved economically and is also technologically difficult. So this reduction is not done.

6. What is the significance of leaching in extraction of aluminium? (1 Mark)Ans: In the extraction of aluminium leaching is used for the concentration of ore by removing

the impurities i.e. silica, iron oxides and titatinium oxides.7. Define Metallurgy. (1 Mark)Ans: Metallurgy is the process of extraction of metals from their ores that includes various

steps.8. Why is hydraulic washing a type of gravity separation? (1 Mark)Ans: The process of hydraulic washing is based on the differences in gravity of the ore and the

gangue particles and so is known as gravity separation.9. What is the use of van Arkel method? (1 Mark)Ans: Van Arkel method is used for removal of impurities like oxygen and nitrogen from the

metals like zirconium and titanium.10. How is distillation used for metal refining? (1 Mark)Ans: Distillation is used for the metals with boiling point lower then the impurities. So the

metals can be evaporated and separately obtained as distillate.11. Why do the anodes used in the electrolytic cell for the reduction of alumina need to be

replaced regularly? (1 Mark)

Ans: The oxygen liberated at the anode during the reduction of alumina, reacts with the carbon of the anode to form CO and CO burns away the anode and hence the anodes need to be replaced.

12. What is the role of depressant in froth floatation process? (1 Mark)

http://en.wikipedia.org/wiki/Liquid

http://en.wikipedia.org/wiki/Solid

Ans: In froth floatation process the depressant selectively prevents one of the ores from coming to the froth in a mixture of two ores hence enabling the separation of the other one with the froth.

13. State the role of silica in the metallurgy of copper. (1 Mark)Ans: Silica in the metallurgy of copper helps in removal of iron oxide as iron silicate (slag).14. What is the role of graphite rods in the electrometallurgy of aluminium?(1 Mark)Ans: In the electrometallurgy of aluminium graphite rods act as anodes in the electrolytic cell of

reduction and are the site for release of oxygen15. Give an example when an element is extracted by oxidation. (1 Mark)Ans: Extraction of chlorine from brine is based on oxidation.16. What will happen if aqueous solution of NaCl is subjected to electrolysis?(1 Mark)Ans: If aqueous Solution of NaCl is subjected to electrolysis, Cl2 will be obtained with NaOH

and H2 gas as the side products.17. What is refining of metals? (1

Mark)Ans: Refining of metal is the process of purification of a metal extracted from its ore.18. What is vapour phase refining? (1 Mark)Ans: Vapour phase refining is the method of metal refining by changing the metal into volatile

compound that can be collected separately leaving behind the impurities and can be decomposed to give the pure metal.

19. Give the principle underlying the process used for refining of gallium. (1 Mark)Ans: The process used for the refining of gallium is zone refining and the principle underlying

it is that the impurities are more soluble in the melt than in solid state of the metal.20. State the principle on which the chromatographic methods of metal refining are based?

(1 Mark)Ans: Chromatographic methods of metal refining are based on the principle that different

components of a mixture are differently adsorbed on an adsorbent21. Which is the purest form of iron and what are its uses? (1 Mark)Ans: The purest form of iron is wrought iron & is used in making anchors, wires, bolts etc.22. What are minerals and how are they different from ores? (2 Marks)Ans: Minerals are the naturally occurring chemical substances in the earth’s crust obtained by

mining. Its different from ores, as ores are the minerals that are used for the extraction of metals profitably.

23. Name one ore each for iron & copper & give their chemical compositions. (2 Marks)Ans: The ore of iron is hematite- Fe2O3 and the ore for copper is copper pyrites- CuFeS2.24. What is the purpose of adding collectors and froth stabilisers during froth floatation? Give an

example for each. (2 Marks)Ans: During froth floatation process collectors like pine oil and fatty acids are added to enhance

non wettability of the mineral particles and the froth stabilisers like cresol and aniline stabilise the froth.

25. How can the ores ZnS and PbS be separated from a mixture using froth floatation process? (2 Marks)

Ans: During the froth floatation process a depressant like NaCN is added to the tank. The depressant selectively prevents ZnS from coming to the froth but allows PbS to come to the froth and hence helps the separation of PbS with the froth.

26. Give the equations involved in the concentration of bauxite ore. (2 Marks)

Ans: i) Al2O3(s) + 2NaOH + 3H2O 2Na[Al(OH)4](aq)

ii) 2Na[Al(OH)4](aq) + CO2(g) Al2O3.xH2O(s) + 2NaHCO3

iii) Al2O3.xH2O(s) Al2O3(s) + xH2O27. Give one reaction each for roasting and Calcination. (2 Marks)Ans: Calcination: ZnCO3 ZnO(s) + CO2

Roasting: 2ZnS + 3O2 2ZnO + 2SO2

28. Why is coke preferred over CO for reducing FeO? (2 Marks)Ans: According to Ellingham diagram the point of intersection of the curves of C, CO and Fe,

FeO lies at temperature lower than that of the point of intersection of CO,CO2 and Fe, FeO curves. This means the reduction of FeO will occur at much lower temperature with C than with CO. So C is preferred to CO for reduction.

29. How is cast iron different from pig iron? (2 Marks)Ans: Pig iron has 4% carbon and can be easily cast into verity of shapes. Whereas cast iron has

lower carbon content and is extremely hard and brittle.30. Give the reactions that occur after the copper matte have been fed into silica lined converter.

(2 Marks)Ans: 2FeS+3O2 2FeO + 3O2

FeO + SiO2 FeSiO3

2Cu2S + 3O2 2Cu2O + 2SO2

2Cu2O + Cu2S 6Cu + SO2

31. Give the reactions taking place at the anode and the cathode during the electrolytic reduction of alumina. (2 Marks)

Ans: Anode: C(s) + O2-(melt) CO(g) + 2e- C(s) + 2O2-(melt) CO2(g) + 4e-

Cathode: Al3+(melt) + 3e- Al(l)32. Explain the process of magnetic separation for concentration of ores. (3 Marks)Ans: In magnetic separation ore is carried over a conveyer belt which passes over a magnetic

roller. If either the ore or the gangue is capable of being attracted by the magnetic field then it will collect near the roller and the particles showing non magnetic behaviour will be collected away from the roller.

33. Differentiate between roasting and Calcination. (3 Marks)Ans: Calcination:

i) it involves heating of the ore in the absence of airii) it is generally used for carbonate ores Calcination: ZnCO3 ZnO(s) + CO2

Roasting: i) it involves the heating of the ore in the presence of airii) it is generally used for sulphide ores Roasting: 2ZnS + 3O2 2ZnO + 2SO2

34. Give the reactions involved in the reduction of iron oxide to give iron in a blast furnace. (3 Marks)

Ans: The reactions are as follows: C + O2 CO2

CaCO3 CaO + CO2

CO2+ C CO3Fe2O3 + CO Fe3O4 +CO2

Fe3O4 + CO 3FeO + CO2

FeO +CO Fe + CO2

FeO + C Fe + COCaO + SiO2 CaSiO3

35. How is copper extracted from low grade ores and scraps? (3 Marks)Ans: For extraction of copper from low grade ores and scraps the ore is first leached out using

acid or bacteria. The solution containing Cu2+ is treated with scarp iron or H2 and Cu is obtained.

Cu2+(aq) + H2(g) Cu(s) + 2H+(aq) Cu2+ + Fe Cu(s) + Fe2+

36. How is gold extracted from its ore? (3 Marks)

Ans: Extraction of gold involves leaching the metal with CN- giving metal complex. 4Au + 8CN-(aq) + 2H2O + O2(g) 4[Au(CN)2]-(aq) + 4OH-(aq)

the metal is later recovered by displacement method with zinc acting as reducing agent. 2[ Au(CN)2]-(aq) + Zn(s) 2Au(s) + [Zn(CN)4]- (aq)

37. Describe the method used for refining copper metal. (3 Marks)Ans: Copper metal is refined by using electrolytic method with impure copper metal as anode

and the pure copper metal strip as cathode. The electrolyte is acidified copper sulphate solution. Copper dissolves from the anode into the electrolyte and get reduced and deposited on the cathode as pure metal.

Anode: Cu Cu2+ + 2e- Cathode: Cu2+ + 2e- Cu

Impurities deposit as anode mud.38. How is nickel refined? (3

Marks)Ans: Nickel is refined by Mond’s process which is based upon vapour phase refining. In this process nickel is heated in stream of carbon monoxide giving a volatile complex, leaving

the impurities behind. The complex is further subjected to higher temperature so that it gets decomposed to giving pure metal. 330-350K

Ni + 4CO --------------> Ni(CO)4

450-470K Ni(CO)4 ------------------> Ni + 4CO

39. Describe briefly column chromatography. (3 Marks)

Ans: Column chromatography is the method of chromatographic refining of metals available in minute quantities and the impurities are not chemically very much different from the element. In this process the column of Al2O3 is prepared in glass tube that forms the stationary phase and the solution of the components to be separated are taken as solution that forms the mobile phase. The components would separate out based on their different solubilities in the mobile phase and the stationary phase.

40. What criterion is followed for the selection of the stationary phase in chromatography?(3 Marks)

Ans: Stationary phase is the immobile and immiscible phase in chromatographic method. Stationary phase is such chosen that the components to be separated present in the mobile phase have different solubilities in the mobile phase and the stationary phase.

41. How is zinc extracted from zinc blende? (3 Marks)Ans: Zinc blende is ZnS. For the extraction of zinc from zinc blende, the ore is first

concentrated by the method of froth floatation. The concentrated ore is then roasted by heating the ore in the presence of oxygen to give ZnO releasing SO2. The ZnS is further reduced using coke at temperature of 673k giving zinc metal.

2ZnS + 3O2 2ZnO + 2SO2

ZnO + C Zn + CO

UNIT-7 MARKS-8

p-Block ElementsQues: Why is N2 inert at room temperature? (1 Mark)Ans: N2 is inert at room temperature because of high bond enthalpy due tothe presence of triple

bond in N2.Ques: Why are the axial bonds in PCl5 longer than equatorial bonds? (1 Mark)Ans: Axial bonds are longer than equatorial bonds in PCl5 because of the higher repulsion

experienced by the axial bond pairs as three pair of electrons repel them whereas only two pair of electrons repel the equatorial bonds

Ques: What happens when SO2 passed through an aqueous solution of Fe3+ salt? (1 Mark)Ans: 2Fe3+ + SO2 + 2H2O 2Fe2+ + SO4

2- + 4H+

Ques: What is Oleum and How is it forms? (1 Mark)Ans: Oleum is H2S2O7. It forms during the preparation of sulphuric acid, as follows: SO3 + H2SO4 H2S2O7

sulphur trioxide OleumQues: Why is the bond angle in PH4

+ higher than in PH3? (1 Mark)Ans: Alone pair of electrons is not present in PH4

+. But in PH3, the presence oflone pair of electrons repel the bonds giving a smaller bond angle.

Ques: Why does PCl3 fume in atmosphere? (1 Mark)Ans: PCl3 fumes in atmosphere due to its hydrolysis in the presence of moisture in atmosphere.

PCl3 + 3H2O H3PO3 + 3HClQues: What happens when H3PO3 is heated? (1

Mark)Ans: H3PO3 when heated, disproportionate to give orthophosphoric acid and phosphine.

4H3PO3 3H3PO4 + PH3

Ques: Why are noble gases least reactive? (1 Mark)Ans: Noble gases are least reactive because their valence shells are completely filled or they

already have octet configuration, which gives them the maximum stability.Ques: Why is the ionization energy of group 15 elements higher than that of group 14 elements?

(1 Mark)

Ans: The ionization energy of group 15 elements is higher than that of group 14 elements because the elements of group 15 have extra stable half-filled p orbital configuration and their size is smaller due to the higher nuclear charge.

Ques: Why does the tendency to exhibit the -3 oxidation state decreases down the group for group 15 elements? (1 Mark)

Ans: The tendency to exhibit -3 oxidation state decreases down the group because of the increase in atomic radii and metallic character down the group.

Ques: Why does basicity decrease form NH3 to BiH3? (1 Mark)Ans: Basicity decreases from NH3 to BiH3 because as the size of atom increases the electron

density decreases and the tendency to donate electrons decreases.Thus, the basic character decreases from NH3 to BiH3.

Ques: Why does NH3 has higher boiling point than PH3? (1 Mark)Ans: NH3 has higher boiling point than PH3 because of the presence of inter molecular

hydrogen bonding in NH3, as the electronegativity difference is quite high in case of Nand H.

Ans: Thermal decomposition of ammonium dichromate: (NH4)2Cr2O7 N2 + 4H2O + Cr2O3

Ques: How does NH3 behaves as a Lewis base? (1 Mark)Ans: N in NH3 has a lone pair of electrons that can be donated to form linkage with the metal

ions and hence NH3 acts as Lewis base.Ques: Why is SF6 exceptionally stable? (1 Mark)Ans: In SF6, the six F atoms protect the central sulphur atom and does not allow any reagent to

attack the S atom, thus making the compound extra stable. It is due to the small size of F atom.

Ques: Which aerosols deplete ozone? (1 Mark)Ans: Freons in aerosol sprays and refrigerants deplete ozone.Ques: Fluorine has a lower electron gain enthalpy than chlorine. Give reason. (1 Mark)

Ans: Due to the small size of fluorine, thereis astrong inter electronic repulsions in the relativelycompact 2p-orbitals of flourine. So the electron gain enthalpy of fluorine is lower than that of chlorine.

Ques: Why are halogens coloured? (1 Mark)Ans: Halogens are coloured because they absorb radiations in visible region, resulting in the

excitation to higher energy level and transmit the remaining light.The colour of that transmitted light is the colour of Halogen.

Ques: All halogens except F exhibit +1, +3, +5 & +7 Oxidation states. Why? (1 Mark)Ans: F cannot show oxidation states other than -1 because of the lack of d orbitals and its higher

electronegativity.Other halogens can show higher oxidation states due the presence of vacant d orbitals.

Ques: HF is liquid at room temperature while other halides are gases. Why?(1 Mark)Ans: HF is liquid at room temperature due to the presence of hydrogen bonding between

Hydrogen and Flourine atoms But no H-bonding is present in other halides so they are gases.

Ques: Acidic strength increases in this order : HF<HCl<HBr<HI. Give reason. (1 Mark)Ans: Acidic strength increases from HF to HI because, down the group,as the size of the

halogen increases, the Bond dissociation enthalpy decreases and it becomes easier for thathalogen atom to lose its H+ ion.

Ques: Why are halogens Strong Oxidizing agent? (1 Mark)Ans: Halogens are strong oxidizing agents because they have high electron gain enthalpy as

they require only 1 electron tocompletetheir configuration.Ques: Explain: fluorine forms only one Oxyacid , HOF. (1 Mark)Ans: Fluorine forms only one oxyacid because ithas high electronegativity and a small size. It

does not show any oxidation state other than -1.Ques: What are the conditions, required to maximize the yield of sulphuric acid by Contact

Process? (1 Mark)

Ans: For the maximum yield of sulphuric acid by Contact Process, We require low temperature, high pressure, presence of a catalyst(V2O5),pure gases and excess of Oxygen.

Ques: Enthalpy of dissociation for F2 is smaller than that for Cl2. Why? (1 Mark)Ans: Enthalpy dissociation for F2 is smaller than thatfor Cl2 because of the large electron-

electron repulsion among the lone pairs in F2 molecule .They are very close to each other than in Cl2.

Ques: Why are Inter Halogen compounds more reactive than Halogens? (1 Mark)

Ans: Inter halogen compounds are more reactive than halogens because the X-X bond in inter halogen compounds is weaker than the X-X bonds in halogen compounds.It breaks easily and thus compounds react more.

Ques: Give equations that involve during the brown ring test. (2 Marks)Ans: The equations involved in the brown ring test are: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) + H2OBrown ComplexQues: H3PO3 is dibasic while H3PO4 is tribasic. Explain. (2 Marks)Ans: The P-H bonds in these compounds are not ionisable and do not play any role in basicity.

Only those atoms which are attached to O in P-OH form are ionisable and are responsible for the basicity. Thus H3PO3 is dibasic as it has two P-OH bonds and H3PO4 is tribasic as it has three P-OH bonds.

Ques: How is NH3 used for the detection of Cu2+ ions? (2 Marks)Ans: N in NH3 has a lone pair of electronwhich it can donates to Cu2+ ions and form linkage

with the metal.This leads to the formation of a deep blue colored complex which helps in the detection of Cu2+ ions.

Cu2+(aq) + 4NH3(aq) = [Cu(NH3)4]2+(aq)Ques: Why are Trihalides more stable than Pentahalides for Group 15? (2 Marks)Ans: Trihalides are more stable than pentahalides because of the following reasons: i)The stability of +5 oxidation state decreases while that of +3 oxidation state increases down the

group due to inert pair effect.ii) For Br and I, the size is large. In pentabromides and pentaiodide, Steric hindrance is high and

Bond strength is low.Ques: How is O3 estimated quantitatively? (2 Marks)Ans: 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

Ques: Why does the stability of +5 oxidation state decreases and that of +3 increases down the Group 15? (2 Marks)

Ans: In Group 15 while moving down, the stability of +5 oxidation state decreases due to inert pair effect i.e. the ns2 electrons tend to remain paired in p block and so they do not take part in chemical reactions for these elements. Hence, they show +3 oxidation state.

Ques: Give reactions for partial hydrolysis of XeF6. (2 Marks)

Ans: On partial hydrolysis XeF6 gives oxyfluorides, XeOF4 and XeO2F2. XeF6 + H2O = XeOF4 + 2HF XeF6 + 2H2O = XeO2F2 + 4HFQues: How does Cl2 react with

(i) cold and dilute NaOH (ii) hot and concentrated NaOH (2 Marks)Ans: When Chlorine reacts with cold sodium hydroxide, it forms sodium chloride. (i) 2NaOH + Cl2 = NaCl + NaOCl + H2O

When Chlorine reacts withhot sodium hydroxide, it forms sodium chloride and other products.(ii) 6NaOH + 3Cl2 = 5NaCl + NaClO3 + 3H2O

Ques: How can SO2 be tested using KMnO4 solution? (2 Marks)Ans: when SO2 ispassed through acidified potassium permanganate solution.Itact as a reducing

agent and decolorizes the solution . 5SO2 + 2MnO4

- + 2H2O = 5SO42- + 4H+ + 2Mn2+

Ques: How is reaction of Zn with dilute HNO3 different from its reaction with concentrated HNO3? (2 Marks)

Ans: With cold dilute Nitric acid Zn forms ammonium nitrate as follows:4Zn + 10HNO3 (dilute) 4 Zn(NO3) 2+ 5H2O + N2OWhere as when Zn react with conc. Nitric acid ,it forms Nitrogen dioxide:Zn + 4HNO3 Zn(NO3) + 2H2O 2NO2

Ques: Oxygen exists as gas at room temperature while sulphur is solid. Why?(2 Marks)Ans: Oxygen exists as O2 i.e. O=O with weak vander wall forces and low molecular mass.

Therefore, it is in gaseous form at room temperature. While sulphur exists as S8 at room temperature, it’s molecular mass is much higher and the forces of attraction are stronger. Therefore, it is solid at a room temperature.

Ques: How is chlorine manufactured by Deacon’s process? (2 Marks)Ans: Cl2 is produced in Deacon’s process by the oxidation of hydrogen chloride by atmospheric

oxygen in the presence of CuCl2 at 723 K. 4HCl + O2 --------> 2Cl2 + 2H2O 723 K,CuCl2

Ques: What is Aqua regia? How does it dissolve gold metal? (2 Marks)Ans: Aqua regia is formed by mixing 3 parts of concentrated HCl and 1 part of concentrated

HNO3. It can dissolve gold as per the following reaction: Au + 4H+ NO3- + 4Cl- = AuCl4

- + NO + 2H2O

Ques: Why Ammonia is a mild reducing agent while BiH3 is the strongest reducing agent among all the hydrides? (2 Marks)

Ans: From N to Bi, as the size of atom increases and tendency to form covalent bonds with H decreases. Therefore the Thermal stability decreases down the group and their tendency to liberate hydrogen increases. Hence, NH3 is a mild reducing agent while BiH3 is a very strong reducing agent.

Ques: Oxygen has lesser electron gain enthalpy than Sulphur. Why? (2 Marks)Ans: The size of oxygen atom is very small as it is the first member of Group 16. Its electron

cloud is distributed over a small region of space making the electron density high which repels the incoming electrons. Therefore, its electron gain enthalpy is lower than that of sulphur as sulphur is larger in size hence the electron cloud is distributed over larger area and the electron densityof the atom is low.

Ques: H2O is a liquid while H2S is a gas. Why? (2 Marks)Ans: Hydrogen bonding is present in H2O due to the small size.There is a high electronegativity

difference between oxygen and hydrogen leading to the polarizing of water molecule and making it easier to form hydrogen bonds. So it is liquid at room temperature. However, in H2S since the size of sulphur is large andthe electronegativity difference is not so much and hydrogen bonding is absent. Therefore, it is gas at a room temperature.

Ques: Give reasons for the following: a) Aqueous solution of ammonia is slightly basic. b) The bond angle in PH4

+ higher than in PH3.

c) Axial bonds in PCl5 longer than equatorial bonds. (3 Marks)

Ans: a) N in NH3 has a lone pair of electrons that can be donated to form linkage with the metal ions and hence NH3 acts as Lewis base.

b) Lone pair of electronsare not present in PH4+ but in PH3,theseare present and repel the bonds,

giving a smaller bond angle.c) Axial bonds are longer than equatorial bonds in PCl5 because of the higher repulsion

experienced by the axial bond pairs as three pairs of electrons repel them whereas only two pairs of electronsatrracts the equatorial bonds.

UNIT: 8 MARKS-5

d & f-Block Elements

1. Vanadium pentaoxide acts as a good catalyst. Why? (1 Mark)(Ans) Vanadium shows variable oxidation states because it has vacant d-orbitals so vanadium

pentaoxide acts as a good catalyst. 3 (Q.) What are the different oxidation states exhibited by lanthanoids? (1 Mark)(Ans) Lanthanoids exhibit +2, +3, and +4 oxidation state.4 (Q.) What are the transition elements? (1

Mark)(Ans) The elements which have the properties in between those of s-block and p-block elements

are called the transition elements.5 (Q.) Why are the transition elements called d-block elements? (1

Mark)(Ans) In transition elements, the last electron enters in the d-orbital. So they are called d- block

elements.6 (Q.) Copper, silver and gold have completely filled d-orbital but they are known as transition

elements. Why? (1 Mark)(Ans) Because their cations in +2, +3 and +3 oxidation state have only partially filled d-orbitals.7 (Q.) Why the chromates in acidic solution becomes orange and dichromates in alkaline

solution becomes yellow? (1 Mark)

(Ans) Because chromates in acidic medium are converted into dichromates and the dichromates in alkaline solution are converted into chromates.

9 (Q.) What is the lanthanide contraction? (1 Mark)

(Ans) In the lanthanide series as the atomic number increases there is a progressive decrease in the size of the atoms and trivalent ions which is known as lanthanide contraction.

12 (Q.) With the increase in atomic number the atomic radius does not change very much in the transition series. Why? (1 Mark)

(Ans) As atomic number increases, the nuclear charge increases but the addition of electron in d-subshell increases the screening effect which counterbalances the effect of increased nuclear charge.

14 (Q.) Among the first transition metals, which divalent metal ion has maximum paramagnetic character and why? (1 Mark)

(Ans) Mn2+, because of maximum number of unpaired electrons.18 (Q.) Give the chemical equation for the reaction between MnO4

− and C2O4−. (1 Mark)

(Ans) 2MnO4− + 16H+ + 5C2O4

− -------------------> 2Mn2+ + 8H2O + 10CO2.20 (Q.) Which is the stronger reducing agent Cr2+ or Fe2+ and why? (2

Marks)

(Ans) Cr2+ is stronger reducing agent than Fe2+ because E0 Cr3+/ Cr

2+ is -0.41V and E0 Fe3+/Fe

2+ is +0.77 V so Cr2+ is easily oxidized to Cr3+ but Fe2+ is not easily oxidized to Fe3+ .

21 (Q.) In the series Sc to Zn the enthalpy of atomization of zinc is lowest why? (2 Marks)(Ans) In this series all the elements have one or more unpaired electrons except zinc. Its outer

electronic configuration is 3d104s2.This shows that the atomic inter metallic bonding in zinc is weakest. so enthalpy of atomization is lowest.

22 (Q.) Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why? (2 Marks)

(Ans) Mn show largest number oxidation state. Its electronic configuration is 3d54s2. The energy of 3d and 4s are close. So, it has maximum number of electrons to loose or share and it shows maximum number of oxidation states from +2 to +7.

23 (Q.) Copper I compounds are white and diamagnetic but copper II compounds are coloured and paramagnetic. Why? (2 Marks)

(Ans) In copper I ion all orbitals are completely filled so its compounds are white and diamagnetic. The electronic configuration of copper II ion is 1s22s22p63s23p63d9. it has one unpaired electron so it is paramagnetic and forms blue coloured compounds.

24 (Q.) How does the ionic and covalent character of the compounds of a transitional metal vary with its oxidation states? (2 Marks)

(Ans) As the oxidation states increases more and more valence shell electrons are involved in bonding. The atomic core becomes less shielded and the force of attraction on the electrons increased because of this ionic character of bonds decreases with increase in oxidation state.

25 (Q.) The E0 (M2+/M) value for copper is positive (+ 0.34). What is the possible reason for this?

(2 Marks)(Ans) E0 (M2+ / M) for any metal is the sum of the enthalpy changes taking place in the

following steps: M(s) + Δa H -------------> M(g), ΔaH = enthalpy of atomization

M(g) + Δi H -------------> M2+(g), ΔiH = ionization enthalpy

M2+(g) + aq -------------> M2+

(aq) +Δhyd H,ΔhydH = hydration enthalpyCopper has high enthalpy of atomization and low enthalpy of hydration so E0 (Cu2+ / Cu) is positive.

26 (Q.) Calculate the “spin only” magnetic moment of M2+(aq) ion (Z= 27). (2 Marks)

(Ans) Electronic configuration of M atom is 1s22s22p63s23p63d74s2. It has three unpaired electrons in d orbitals. Magnetic moment = √ n(n+2) BM

= √3 (3+2) = √15 3.87 BM27 (Q.) How does +2 oxidation state becomes more and more stable in the first half of the first

row transition elements with increasing atomic number? (2 Marks)

(Ans) The sum of first and second ionization enthalpies increases with increasing atomic number so the standard reduction potentials become less and less negative. Hence the +2 oxidation state becomes more and more stable.

UNIT-9 COORDINATION COMPOUNDS 3 marks

1. Write the formula for the following: - (3)

(i) Tetrahydroxozincate (II)ion(ii) Hexaammineplatinum (IV) ion (iii) Hexaamminecobalt (III) sulphate

Ans: (i) [Zn(OH)4]2-

(ii) [Pt(NH3)6]4+

(iii)[Co(NH3)6]2(SO4)3

2. Write the IUPAC name of the following: - (2)

(i)[Pt(NH3)2Cl(NH2CH3)]Cl (ii) [Co(NH3)4Cl(NO2)]Cl

Ans (i) Diamminechloridomethyleamineplatinum (ii) chloride (ii) Tetraaminechloridonitrito-N-cobalt (iii) chloride

3. What is meant by ambidentate ligands? Give two examples. (1)Ans : Ligand which can ligate through two different atoms e.g. CN-, SCN-

4. Explain on the basis of VBT, the experimental findings that [Ni(CN)4]2- ion with a square-planar structure is diamagnetic and the [NiCl4]2- ion with tetrahedral geometry is paramagnetic. (2)

Ans: In [Ni(CN)4]2-

Ni2+ electronic configuration- 3d8

Hybridization - dSP2

Unpaired electron = 0 , therefore it is diamagnetic In [NiCl4]2-

Ni2+ Electronic configuration 3d8

Hybridization SP3

Unpaired electron = 2, therefore it is paramagnetic

5. Aqueous copper sulphate solution (blue in colour) gives( a) a green precipitate with aqueous potassium fluoride, and(b) a bright green solution with aqueous potassium chloride. Explain. (2)

Ans:(a) [Cu(H2O)4]2+ + 4F- → [CuF4]2- + 4H2O Blue green ppt(b) [Cu(H2O)4]2+ + 4Cl- → [CuCl4]2- + 4H2O Blue Bright green solution

6. Give a chemical test to distinguish between [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br. What kind of isomerism do they exhibit? (2)

Ans : (i) [Co(NH3)5Br]SO4 + Ba2+ → BaSO4 White ppt

[Co(NH3)5Br]SO4 + Ag+ → No ppt

(ii) [Co(NH3)5SO4]Br. + Ba2+ → No ppt

[Co(NH3)5SO4]Br. + Ag+ → AgBrYellow ppt

7. Name the metal present in (i) Chlorophyll (ii) Haemoglobin (iii) Vitamin B12 (iv) Cis-platin (2)

Ans: (i) Mg (ii) Fe (iii) Co (iv) Pt

8. What is the coordination number of central metal ion in (i) [Fe(C2O4)3]3- (ii) [Co(en)2Cl2]+. (1)

Ans: (i) 6 (ii) 6

9. Why are cyclic complexes more stable than open one? (1)

Ans : Cyclic complexes have more stability because of reduced strain in five or six member rings. e.g. Chlorophyll (cyclic) is more stable than [Co(NH3)5Br]SO4 .

10. CuSO4 on mixing with NH3 (1:4) does not give test for Cu2+ ions but gives test for SO42-

ions. Why? (2)

Ans: It is because when NH3 coordinates to Cu2 ions it forms the complex [Cu(NH3)4]SO4. Copper ions are present in coordination sphere, therefore, they are non- ionisable whereas SO4

2- ions are counter ions which are ionisable.

UNIT-10 MARKS-4

HALOALKANES AND HALOARENES

Q :1 Identify A, B in the following:-

Dry ether H2O ─Br + Mg ─────────→ A ───────→ B

2

Ans: A= ─MgBr B = Q : 2 Give the IUPAC names of the following:-

(a) Cl CH2 C ≡ C-CH2-Br (b) CH3CH2CH(CH3)CH(C2H5) Cl 2

Ans : A) 1-Bromo,4-Chloro But-2- yne B) 3-Chloro-4-Methyl-hexane

Q: 3 Explain why Aryl halides are extremely less reactive towards nucleophlic substitution 1 reactions.

Ans : due to resonance in aryl halide in C-X bond will acquire double bond character so difficult to substitute halogen.

Q: 4 Haloalkanes are only very slightly soluble in water explain.1

Ans: Unable to form Hydrogen bond with water.

Q : 5 Write short notes on:- (a) Wurtz reaction (b) Swarts reaction2

Dry Ether Ans :A) R-X + 2Na +R-X ------------ R-R + 2Na X

B) R-Br + AgF ----------- RF + AgBr

Q : 6 Arrange the following in increasing order of the property indicated 1

Bromomethane, Bromoform, Chloromethane, Dibromomethane ( boiling point)

Ans : Chloromethane < Bromomethane < Dibromomethane < Bromoform .

Q : 7 What do you mean by asymmetric carbon? Give one example. 1

Ans: If all the four atom or group of atoms attach to a carbon are different then carbon is called asymmetric carbon. e.g. CH3CHBrCl.

Q : 8 Explain why H2SO4 is not used during the reaction of alcohol with KI.1

Ans: Because HI produced get oxidize to Iodine in presence of Sulphuric acid.

Q : 9 Explain why Reaction of CH3Br with KCN yields CH3CN while with AgCN yields CH3NC? 1Ans: Because CN- is an ambidentate nucleophile.

Q : 10 Write short notes on:- (a) Sandmeyer’s reaction (b) Finkelstein reaction 2 CuCl /HClAns : (a) C6H5N2Cl --------------- C6H5Cl

Acetone (b) C6H5 Br + NaI ---------------- C6H5 I + Na Br

Q : 11Convert : -propane-1-ol to 2-iodopropane.1

Conc. H2SO4 HI , M.Rule

Ans : CH3 CH2 CH2 OH ---------------- CH3 CH=CH2 ---------------- CH3 CH I CH 3 443K

Q : 12 Arrange the following in increasing order of the property indicated 2

(a) 1-Bromobutane, 2-Bromobutane, 2-Bromo-2-methylpropane ( boiling point)

Ans : 2-Bromo-2-methylpropane <2-Bromobutane < 1-Bromobutane, 2-Bromobutane.

(b) ─CH2Cl and ─Cl (reactivity towards SN2 reaction)

Ans : ─CH2Cl

UNIT-11 MARKS-4

ALCOHOLS, PHENOLS AND ETHERS

Q : 1 Write the names of reagents and equations for the preparation of 2-Methoxy-2-methylpropane. 1

Ans: CH3Br + (CH3)3CO- Na+-----------------------------------> CH3-O-C(CH3)3 + NaBr

Q : 2 Give reasons:

2a. Ethanol has higher boiling point in comparison to methoxymethane .

Ans : Due to presence of intermolecular hydrogenbond in ethanol.

b. Phenols are more acidic than alcohols. Ans : Due to –R effect of phenoxide ion produced after the loss of proton from phenol, phenoxide ion is more stable than alkoxide ion

Q : 3 (a) o-nitrophenol is steam volatile while p-nitrophenol not. Ans : Due to presence of intramolecular hydrogen bond

(b)Cleavage of phenyl alkyl ether with HI always gives phenol and alkyl iodide Ans : Due to resonance C-O bond of phenyl will acquire double bond character hence difficult to break 2

Q : 4 How are the following conversions carried out: -

a. Methyl magnesium bromide to 2-methylpropane-2-ol

CH3COCH3 H+/H2OAns : CH3MgBr ---------------------- (CH3)3C-OMgBr -----------------(CH3)3C-OH Dry Ether

a. Propene to propan-2-ol

HBr,M.Rule aq.KOH, ∆Ans : CH3CH=CH2------------------- CH3CHBrCH3------------------ CH3CHOHCH3

b. Propan-2-one to 2-methyl-2-propanol

CH3MgBr H+/H2OAns : CH3COCH3 --------------------- (CH3)3C-OMgBr -----------------(CH3)3C-OH Dry Ether 2

Q : 5 Write IUPAC name- i. CH3-O-CH2-CH-CH3 ii. CH3-CH-CH-CH2-CH2OH

| | | CH3 Cl CH3

Ans: i) 2-methoxybutane ii) 4-Chloro,3-methyl butan-1-ol. 2

Q : 6 Arrange the following as property indicated: -

(i) pentan-1-ol, pentanal, ethoxyethane (increasing order of boiling point)

Ans : n-butane <ethoxyethan < pentanal < pentan-1-ol

(ii) pentan-1-ol, phenol, 4-methylphenol, 3-nitrophenol ( increasing order of acid strength)

Ans : pentan-1-ol < 4-methylphenol < phenol < 3-nitrophenol. 2

Q : 7 Give simple chemical tests to distinguish between the following pairs of compounds: (i) Phenol and Benzoic acid. (ii) Propan-1-ol and Propan-2-ol.

Ans: (i) C6H5OH + NaHCO3 -------------------- No effervescence

C6H5COOH + NaHCO3 ----------------- C6H5COONa + H2O + CO2

(effervescence )

Lucas Reasgent (ii) CH3-CH2 –CH2OH ------------------- No Turbidity CH3-CHOH–CH3 ----------‘’--------- Turbidity produced with in five minutes 2

Q:8 Write the mechanism of dehydration of ethanol. Ans : Conc.H2SO4

CH3CH2OH-------------------- CH2=CH2 + H2O Mechanism : STEP1 : Formation of a protonated alcohol CH3CH2OH + H+-------------- CH3CH2-O+-H

l H STEP: 2 formation of a carbocation

CH3CH2-O+-H -------------- CH3CH2+ + H2O

l H STEP : 3 Loss of proton : CH3CH2

+ --------------------- CH2=CH2 + H+

Q : 9 A compound ‘A’ with molecular formula C4H10O is a unreactive towards sodium metal . It does not add Bromine water and does not react with NaHSO3 solution .On refluxing ‘A’ with excess of HI gives ‘B’ which react with aq. NaOH to form ‘C’. ‘C’can be converted into ‘B’ by reacting with P and I2 . ‘C’ on heating with aqueous alkali to form ‘E’ which form ‘F’ on heating with conc. H2SO4. ‘F’ decolourises bromine water . Identify A to F and write the reactions involved. Ans : ‘A’is not alcohol therefore it does not react with Sodium metal . ‘A’ is not aldehyde and ketone as it does not react with NaHSO3 ‘A’ is not unsaturated hydrocarbon as it does not add Br2

(aq) . It is likely to be ether. Ans : CH3CH2OC2H5 + 2HI-------------------------2C2H5I + H2O ‘A’ excess ‘B’ (C4H10O)2C2H5I + NaOH (aq) --------------------------------C2H5OH + Na I ‘B’ ‘C’

P / I2

C2H5OH ------------------------------------- C2H5I ‘C’ ‘B’

Cu CH3CH2OH --------------------------- CH3CHO ‘C’ 573 K ‘D’

OH-

CH3CHO ----------------------------- CH3CHOHCH2CHO 3. ‘D’ (3-Hydroxybutanal)

UNIT-12 MARKS-6

ALDEHYDES, KETONES AND CARBOXYLIC ACIDS

1. Arrange the following compounds in increasing order of their property as indicated:(2)

(i) HCOOH, CH3COOH, CH2(F)COOH, CH2(Cl)COOH (acid strength)(ii) Acetaldehyde, Acetone, Formaldehyde, Ethylmethyl ketone (reactivity towards HCN)Ans: (i) CH3COOH< HCOOH< CH2(Cl)COOH < CH2(F)COOH

2. Give reasons (2)

(i) Carboxylic acids are stronger acids than phenols.Ans : Carboxylic acids are stronger acids than phenols because(ii) Aromatic carboxylic acids do not undergo Friedel – Crafts reaction

3. Give simple chemical test to distinguish between (2)

(i) Propanal and propane Ans: CH3 CH2CHO + Ag2O ────→ CH3 CH2COOH + Ag (silver mirror)

CH3 COCH3 + Ag2O ────→ no reaction

(ii) Phenol and benzoic acid Ans: Phenol does not give sodium carbonate test whereas benzoic acid give this test. Phenol give azodye test , benzoic acid does not.

4. Identify A, B and C (2)

KCN SnCl2-HCl Zn-Hg/ HClCH3CH2Br ──────→ A ──────→ B ───────→ C

Ans: A= CH3CH2CN , B= CH3CH2CHO , C= CH3CH2CH3

5. Name the following as IUPAC system (2)

(i) CH3 CH( CH3 )CH2 CH2CHO Ans: 4- methylpentanal

(ii) (CH3)3C CH2 COOH 3-methylbutanoic acid

6. Give reasons for the following: (2) (a). Ethanal is more reactive towards nucleophilic addition reactions than propanone. Ans ethanal (CH3CHO) , Propanone CH3 COCH3 Because in propanone presence of two constituent hindered the approach of nucleophile to carbonyl carbon . (b).Ethanoic acid is a stronger acid than ethanol. Ans Due to presence of resonance in ethanoate ion..

7. An organic compound A (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Identify A, B and C. also write equations for the reactions involved. (5)

Ans: A= CH3 CH2 CH2COO CH2 CH2 CH2 CH3

B= CH3 CH2 CH2COOH C= CH3 CH2 CH2CH2OH

Equation H2SO4 CH3 CH2 CH2COO CH2 CH2 CH2 CH3 ─────→ CH3 CH2 CH2COOH (B)

+ CH3 CH2 CH2CH2OH (C)

dehydration CH3 CH2 CH2CH2OH ─────→ CH3 CH2 CH=CH2

Chromic acid CH3 CH2 CH2CH2OH ─────→ CH3 CH2 CH2COOH

8. Complete the following reactions by identifying A, B and C. (2) Pd/BaSO4

(i) A + H2(g) ────────→ (CH3)2CH-CHO Pd/BaSO4

Ans: (CH3)2CH-CO Cl + H2(g) ────────→ (CH3)2CH-CHO

(ii) CH3 CH2 CH=CH2 + B ────────→ CH3 CH2 CHO +CH2 O ZnO/H2O Ans: CH3 CH2 CH=CH2 + O3───────→ CH3 CH2 CHO +CH2 O

9. Suggest a reason for the large difference in the boiling points of butanol and butanal, although they have the same solubility in water. (2)

Ans: The b.p. of butanol is higher than butanal because butanol has strong intermolecular hydrogen bonding while butanal has weak dipole-dipole attraction. However, both of them form H- bonding with water and hance are soluble.10. Name two methods which are used to convert a >C=O group into >CH2 group.Ans: Clemmensen reduction and Wolff-Kishner reduction

UNIT-13 MARKS-4

AminesTypes of Amines

Preparation of amines1. Reduction method:

2. Hoffman’s ammonolysis method:

3. Reduction of nitriles:

4. Reduction of amides:

5. Gabriel Phthalimide Synthesis:

6. Hoffman’s Degradation Reaction:

Physical PropertiesAmines are polar in nature and can form H-bonding therefore there boiling pt. is more than non-polar molecules but lower than alcohols and carboxylic acids having same molecular mass.Lower Amines are soluble in water but solubility decreases with increase in alkyle group.Methyl and ethyl amines have ammonical smell other have fishy order.

Amines are basic in nature due to presence of lone pair of electrons e.g

Amines(Q.) Why are benzenediazonium salts soluble in water?

(1 Mark)(Ans) Benzenediazonium salts are soluble in water because they are ionic in nature.(Q.) What are aromatic nitro compounds?

(1 Mark)(Ans) Nitro compounds in which one or more benzene rings are present are called aromatic nitro compounds.(Q.) Draw the structure of trimethyl phenyl ammonium bromide.

(1 Mark)(Ans) C6H5N(CH3)3Br(Q.) What is diazotization?

(1 Mark)(Ans) The reaction in which benzendiazonium chloride is prepared by the reaction of aniline with nitrous acid at 273-278K is called diazotization reaction.(Q.) Why do nitro compounds have higher boiling points than the hydrocarbons having almost same molecular mass?

(1 Mark)(Ans) Nitro compounds have large dipole moments and higher polarity. So, they have higher boiling point than hydrocarbons having almost same molecular mass.(Q.) How will you convert the acetonitrile to propionitrile?

(1 Mark)

(Ans) CH3CN CH3CONH2 CH3COOH + NH3

Acetonitrile Ethanamide Acetic acid

CH3COOH CH3CH2OH CH3CH2Br CH3CH2CNEthanol BromoethanePropionitrile(Q.) Why primary amines have higher boiling point than tertiary amines?

(1 Mark)

(Ans) Because the molecules of primary amines are joint by intermolecular H-bonding while molecules of tertiary amines are joint by weak van der Waals forces.(Q.) Why aniline does not undergo Friedel Crafts reaction?

(1 Mark)(Ans) In Friedel crafts reaction AlCl3 is used as the catalyst with which anilines give salts. so the reaction is not possible foraniline(Q.) How is the basic strength of aromatic amines affected by the presence of electron releasing group on the benzene ring?

(1 Mark)(Ans) An electron releasing group increases the electron density on the N-atom. So, its tendency to donate an electron pair to a proton increases and hence the basicity of the amine.(Q.) What for quaternary ammonium salts are widely used?

(1 Mark)(Ans) The quaternary ammonium salts are used as detergents.(Q.) Which is more basic CH3NH2 or (CH3)3N?

(1 Mark)(Ans) CH3NH2 is more basic than (CH3)3N due to greater stability of conjugate acid due to H-bonding.(Q.) Write the short note on reductive amination reactions?

(2 Marks)(Ans) When aldehydes or ketones react with ammonia imines is formed. These imines give the primary amine by catalytic reduction.

RCHO + NH3 RCH=NH RCH2NH2

Aldehyde imines Primary amine(Q.) What is the difference between the nitrite (-O-N-O) group and nitro (NO2) group?(2 Marks)(Ans)

Nitrite (-O-N-O) group Nitro (NO2) group1. Nitrite group attaches itself to an alkyl or

aryl group through oxygen atom.2. Nitrogen is trivalent in nitrite group.

1. Nitro group attaches itself to any alkyl or aryl group through N atom.

2. Nitrogen is pentavalent in nitro group.(Q.) Why is NO2 group called an ambident group?

(2 Marks)(Ans) Nitro group can attach itself through two different atoms, i.e., through O atom and N atom. Thus, nitro group is an ambident group(Q.) How will you differentiate in between primary, secondary and tertiary amines? (2 Marks)(Ans) We can differentiate primary, secondary and tertiary amines by ‘Hinsberg test’. When amines are treated with benzene sulphonyl chloride, the following changes occur: -(i) Primary amines form monoalkyl benzene sulphonamide, which is soluble in KOH.

RNH2 + C6H5SO2Cl C6H5SO2NHR C6H5SO2NKRPrimary Benzene sulphonyl Monoalkyl Potassium salt of MonoalkylAmine Chloride sulphonamide sulphonamide(ii) Secondary amines form dialkyl benzene sulphonamide, which is insoluble in KOH.

R2NH + C6H5SO2Cl C6H5SO2NR2 insoluble dialkylsulphonamide(iii) Tertiary amine does not react.Q.) How will you form quaternary ammonium salt from haloalkane?

(2 Marks)(Ans) When haloalkanes heated with an ethanolic solution of ammonia in a sealed tube at 1000C .RX + NH3 RNH2 R2NH R3N R4+NX-

Haloalkane P. amine S. amine T. amine QuaternaryAmmonium salt(Q.)What happens when diazoniumfluoroborate is heated with aqueous sodium nitrite solution in the presence of copper?Give reactions. (2 Marks)(Ans) In this reaction thediazonium group is replaced by nitro group.C6H5N2BF4 + NaNO2 + Cu C6H5NO2 + N2 + NaBF4

(Q.) Why is aniline a weaker base than methylamine? (2 Marks)

(Ans) Aniline and methylamine both have nitrogen with lone pair of electron. In aniline the phenyl group is electron attracting. It tend to decrease the electron density on nitrogen atom and hence decreases electron releasing tendency of nitrogen. While in methylamine CH3- group is electron repelling in nature. It tends to increase the electron density on the nitrogen atom and helps in electron releasing tendency of nitrogen. So, the aniline is a weaker base than methylamine.

(Q.) Arrange the following in increasing order of basic strength: - (i) Triethylamine, ethylamine and ammonia.(ii) p-nitroaniline, aniline and p-toludine.

(2 Marks)(Ans) (i) Ammonia <Triethylamine< Ethylamine(ii) Aniline < p-nitroaniline< p-toludine.(Q.) How will you distinguish between cyanides and isocyanides?

(2 Marks)(Ans)

Properties Cyanide Isocyanides1.Solubility in water2.Odour3.Hydrolysis4. Reduction

1. Soluble in water2. Pleasant smell3. Give acid on hydrolysis with acid and alkalies4. Give primary amine on complete reduction.

1. Sparingly soluble in water.2. Highly unpleasant smell3. Give primary amines and formic acid on hydrolysis with dil. Acids.4. Give secondary amines on complete reduction.

(Q.) Can we prepare aniline by Gabriel-phthalimide reaction?(2 Marks)

(Ans) We can not prepare aniline by Gabriel-phthalimide reaction because this preparation requires the reaction of pot. Phthalimide with chlorobenzene or bromobenzene. But aryl halides do not give nucleophilic substitution reaction at ordinary condition in laboratory.(Q.) Arrange the following in increasing order of their basic strength: C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH

(2 Marks)(Ans) C6H5NH2< NH3< C6H5CH2NH2< C2H5NH2< (C2H5)2NH(Q.) Complete the following reactions: (i) CH3NH2 + C6H5COCl ………… + …….. (ii) C6H5NH2 + (CH3CO)2O ………. + ………. (2 Marks)

(Ans) (i) CH3NH2 + C6H5COCl CH3NHCOC6H5 + HCl(ii) C6H5NH2 + (CH3CO)2O C6H5NH-COCH3 + CH3COOH (Q.) Write the physical properties of nitro compound.

(3 Marks)(Ans) Physical properties of nitro compound: -1.Nitroalkane arecolourless liquids with a pleasant odour. Aromatic nitro compounds are either pale-yellow liquids or solids with characteristic odour.2.Bothnitroalkane and nitroarenes have higher boiling point than the hydrocarbons of comparable mass due to higher polarity.3.Nitroalkane are sparingly soluble in water, while nitroarenes are insoluble in water. But both are soluble in organic solvents. (Q.) Describe the orbital structure of the nitro group in nitroalkane.

(3 Marks)(Ans) The orbital structure of the nitro group in nitroalkane is described as: - 1. The carbon atom in the alkyl group is sp3 hybridized.2. The nitrogen atom in the nitro group is sp2 hybridized.3. The three sp2 hybrid orbitals lie in a plane and inclined to each other at 1200.The unhybridized p-orbital of nitrogen atom is perpendicular to the plane of sp2orbitals.The three sp2 hybrid orbitals on nitrogen are occupied by one electron each, whereas the unhybridised p- orbital accommodates a lone pair of electrons.4.The oxygen atoms of the nitro group are sp2 hybridized. Thus each oxygen atom has three sp2 hybrid orbitals and one unhybridised p-orbital. One of the sp2 hybrid orbitals and the unhybridised p-orbital on each oxygen atom accommodate one lone pair of electrons each.

(Q.) Predict the products of and balance the following reactions:

(i) C2H5NO2 ………

(ii) C2H5-O-N=O ……… + …….. + ………..

(iii) C6H5 NO2 …….. ……… (3 Marks)

(Ans) (i) C2H5NO2 C2H5NH2 + 2H2O

(ii) C2H5-O-N=O C2H5OH + NH3 + H2O

(iii) C6H5 NO2 m-C6H4(NO2)2 H2N.C6H4.NO2

(Q.) Write the main uses of nitro compounds.(3 Marks)

(Ans) Uses of nitro compounds are:1. Lower nitroalkanes such as nitromethane, nitroethane and nitrobenzene are used as solvents for fats, oils, resins, dyes and cellulose because these are less inflammable and less toxic.2.Chloropicrin is used as an insecticide.3.Nitrobenzene is used for scenting cheap soaps, shoe and floor polishes under the name oil of mirbane.4. Nitroalkanes are used as intermediate for industrial preparation of dyes, drugs, propellants and explosives.(Q.) Explain the coupling reaction with example.

(3 Marks)(Ans) Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with the diazonium salt to form p-hydroxybenzene. This type of reaction is known as coupling reaction.

(Q.) Write the structure the following compounds: - (i) Azoxybenzene (ii) Sulphanilic acid (iii) Picric acid

(3 Marks)

(Ans)

(Q.) Describe the Gabriel-phthalimide reaction.(3 Marks)

(Ans) In this reaction potassium phthalimide is reacted with an alkyl halide to get N-alkyl phthalimde. N-alkyl phthalimide on hydrolysis with dil. HCl under pressure.

UNIT-14 MARKS-4

BIOMOLECULES

KEY POINTS EXPLANATIONSMonosaccharides Cannot be hydrolyzed further .eg- glucose, fructose, riboseDisaccharides Sucrose (α-D- glucose + β-D-fructose) , Maltose(α-D- glucose + α-D-

glucose) Lactose(β-D-galactose + β-D-glucose )

Polysaccharides Starch (two components—Amylose and Amylopectin) polymer of α-D- glucose

Amylose Water soluble , 15-20% of starch., unbranched chain , C1– C4 glycosidic linkage.

Amylopectin Water insoluble , 80-85% of starch., branched chain polymer, C1–C4 & C1–C6 glycosidic linkage

Cellulose Straight chain polysaccharide of β -D-glucose units/ joined by C1-C4glycosidic linkage (β-link), not digestible by human / constituent of cell wall of plant cells

Glycogen Highly branched polymer of α-D- glucose .found in liver, muscles and brain.

reducing sugars Aldehydic/ ketonic groups free so reduce Fehling’s/ Tollens solution and. Eg- maltose and lactose

Non reducing sugars

Aldehydic/ ketonicgroups are bonded so can not reduce Fehling’s solution and Tollens’ reagent. Eg- Sucrose

Anomers. The two cyclic hemiacetal forms of glucose differ only in the configuration of the hydroxyl group at C1, called anomeric carbon Such isomers, i.e., α –form and β -form, are called anomers.

Invert sugar Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory fructose. Since the laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°), the mixture is laevorotatory. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (–) and the product is named as invert sugar

Glycosidic linkage Linkage between two mono saccharide Importance ofCarbohydrates

Major portion of our food. / used as storage molecules as starch in plants and glycogen in animals/. Cell wall of bacteria and plants is made up of cellulose./wood and cloth are cellulose /provide raw materials for many important industries like textiles, paper, lacquers and breweries.

essential amino acids

which cannot be synthesised in the body and must be obtained through diet, eg- Valine, Leucine

Nonessential amino acids

which can be synthesised in the body, eg - Glycine, Alaninezwitter ion. In aqueous solution, amino acids exist as a dipolar ion known as zwitter

ion. peptide linkage peptide linkage is an amide formed between –COOH group and –NH2

group of two successive amino acids in peptide chain. 10- str. of proteins:

sequence of amino acids that is said to be the primary structure of protein20- str. of proteins:

secondary structure of protein refers to the shape in which a long polypeptide chain can exist. They are found to exist in two types of structures viz. α -helix and β -pleated sheet structure.

Tertiary structure of proteins:

further folding of the secondary structure. It gives rise to two major molecular shapes viz. fibrous and globular.

Fibrous proteins Polypeptide chains run parallel, held together by hydrogen and disulphide bonds, fibre– like structure. Water insoluble .Eg- are keratin(in hair, wool, silk) and myosin (present in muscles).

Globular proteins chains of polypeptides coil around to give a spherical shape. water soluble. Eg-Insulin and albumins

Stab.forces2°& 3° hydrogen bonds, disulphide linkages, van der Waals and electrostatic forces of attraction.

Denaturation ofProteins

When a protein is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein. (During denaturation 2° and 3° structures are destroyed but 1º structure remains intact.) eg- The coagulation of egg white on boiling, curdling of milk

Fat soluble vit These are vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues

Water soluble vit B , C . these vitamins must be supplied regularly in diet because they are readily excreted in urine

Vitamins –sources-Deficiency diseases

Vit- A (Fish liver oil, carrots)- Night blindness / Vitamin B1 (Yeast, milk,)- BeriberiVit-B2 (Milk, eggwhite)- Cheilosis / Vit- B6 (Yeast, milk,)- Convulsions / Vit- B12 (Meat, fish,)- anaemiaVit C(Citrus fruits)- Scurvy, / Vit D(Exposure to sunlight, fish and egg yolk)- Rickets, osteomalaciaVit E(wheat oil, sunflower oil)- fragility of RBCs / Vit K(leafy vegetables)- Increased blood clotting time

DNA pentose sugar (D-2-deoxyribose) + phosphoric acid + nitrogenious bases ( A , G , C, T )

RNA pentose sugar (ribose) + phosphoric acid + nitrogenious bases (A , G , C, U )

Nucleoside / tides Nucleoside sugar + base Nucleotides sugar + base + phosphate

Phosphodiester link

Linkage between two nucleotides in polynucleotidesFunctions of Nucleic Acids

DNA reserve genetic information, maintain the identity of different species e is capable of self duplication during cell division, synthesizes protein in the cell.

Biomolecules(Q.) Define the term bio molecules?

(1 Mark)(Ans) Bio molecules may be defined as the complex lifeless chemical substances which form the basis of life, i.e., they not only build up living systems (creatures) but are also responsible for their growth, maintenance and their ability to reproduce.(Q.) Define the term photosynthesis? Give its general chemical equation?

(1 Mark)(Ans) Photosynthesis may be defined as a chemical process through which plants make their own food by the reaction of carbon dioxide and water in the presence of sunlight with the help of plant chlorophyll.x CO2 + y H2O ----->Cx(H2O)y + x O2

(Q.) Define Monosaccharides?(1 Mark)

(Ans) These are the simplest carbohydrates which cannot be hydrolysed to smaller molecules. Their general formula is (CH2O)n, where n = 3-7.(Q.) Define the term Oligosaccharides? (1 Mark)(Ans) Those carbohydrates which give 2-10 molecules of monosaccharides in hydrolysis.(Q.) Define Disaccharides?

(1 Mark)(Ans) Carbohydrates which on hydrolysis give two molecules of the same or different monosaccharides are called disaccharides. e.g., C12H22O11 + H2O C6H12O6 + C6H12O6

sucrose glucose fructoes

(Q.) What is difference between Reducing and non-reducing sugars or carbohydrates?(1 Mark)(Ans) All those carbohydrates which contain aldehydic and ketonic group in the hemiacetal or hemiketal form and reduce Tollen’s reagent or Fehling’s solution are called reducing carbohydrates while others which do not reduce these reagents are called non-reducing reagents.(Q.) Explain the term mutarotation?

(1 Mark)(Ans) Mutarotation is the change in the specific rotation of an optically active compound with time, to an equilibrium mixture.(Q.) Define glycosidic linkage?

(1 Mark)(Ans) The two monosaccharide units are joined together through an ethereal or oxide linkage formed by the loss of a molecule of H2O. Such a linkage between two monosaccharide units through oxygen atoms is called glycosidic linkage.(Q.) Give a chemical equation for obtaining Maltose?

(1 Mark)(Ans) Maltose is obtained by partial hydrolysis of starch by the enzyme diastase present in malt i.e., sprouted barley seeds. 2(C6H10O5)n + n H2O n C6H12O6

(Q.) What are the main sources of vitamins?(1 Mark)

(Ans) The main sources of vitamins are milk, butter, cheese, fruits, green vegetables, meat, fish, eggs, etc.(Q.) Give two methods for the preparation of glucose?

(2 Marks)(Ans) The methods for the preparation of glucose are: (i) From sucrose (Cane Sugar).When sucrose is hydrolysed by boiling with dil. HCl or H2SO4 in alcoholic solution, an equimolar mixture of glucose or fructose is obtained.C12H22O11 + H2O C6H12O6 + C6H12O6

(ii) From starch.Commercially glucose is obtained by hydrolysis of starch by boiling it with dil. H2SO4 at 393 K under pressure.(C6H10O5)n + n H2O n C6H12O6

(Q.) Define Carbohydrates? Give their basic classification depending upon their behaviour towards hydrolysis.

(2 Marks)(Ans) Carbohydrates are defined as optically active polyhydroxy aldehydes or polyhydroxy ketonesubstances which give these on hydrolysis.These are broadly classified as:(i) Monosaccharides.(ii) Oligosaccharides.(iii) Polysaccharides.

(Q.) What is Milk sugar? Give its characteristics.(2 Marks)

(Ans) Lactose occurs in milk so, it is called milk sugar. Lactose on hydrolysis with dilute acids yields an equimolar mixture of D-glucose and D-galactose. It is a reducing sugar since it forms an osazone. It undergoes mutarotation and also reduces Tollen’s or Fehling’s solution.(Q.) Define the term vitamins? State its importance.

(2 Marks)(Ans) Vitamins may be defined as group of bio-molecules (other than fats, carbohydrates and proteins) which are required in small amounts for normal metabolic processes and for the life, growth and health of human beings and animal organisms.Vitamins neither supply energy nor help in building tissues of the cells. They play an important role in keeping good health of human beings and animals. Their deficiency causes serious disturbances and diseases in the body. (Q.) What do you understand by denaturation of proteins?

(2 Marks)(Ans) When a protein in its native form, is subjected to physical change like in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein.(Q.) Give the D and L configurations of Glyceraldehyde?

(2 Marks)(Ans)

(Q.) Give the chemical structure of sucrose & explain why sucrose is non reducing sugar.(2 Marks)(Ans)

The two monosaccharide are held together by a glycosidic linkage between C1 of -glucose and C2 of -fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non-reducing sugar.(Q.) Give a broad classification of vitamins?

(2 Marks)(Ans) Vitamins are complex organic molecules.They can be broadly classified as:

(i) Water soluble vitamins: These include vitamin B-complex and vitamin C.(ii) Fat soluble vitamins: These are oily substances that are not readily soluble in water. However, they are soluble in fat. These include vitamins A,D,E and K. Nucleic acids are bipolar (i.e. polymers present in the living system). They are also called polynucleotides since the repeating structural unit of nucleic acids is a nucleotide.General structure of a nucleotide can be given as:

(Q.) Write a short note on cellulose and give its chemical structure.(3 Marks)(Ans)

Cellulose occurs exclusively in plants and it is the most abundant organic substance in plant kingdom. It is a predominant constituent of cell wall of plant cells. Cellulose is a straight chain polysaccharide composed only of -D-glucose units which are joined by glycosidic linkage between C1 of one glucose unit and C4 of t he next glucose unit.(Q.) Give a short note on Zwitter ion?

(3 Marks)(Ans) Amino acids are usually colourless, crystalline solids. These are water soluble , high melting solids and behave like salts rather than simple amines or carboxylic acids. This behaviour is due to the presence of both acidic (carboxylic group) and basic (amino group) groups in the same molecule. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion known as zwitter ion.

(Q.) How are peptides formed. Show the formation of peptide bond with diagram.(3 Marks)

(Ans) Peptides are amides formed by the condensation of amino group of one -amino acid with the carboxyl group of another molecule of the same or different -amino acid with the

elimination of awater molecule. They are classified as di-, tri-, tetra-, etc.E.g.

UNIT-15 MARKS-3

POLYMERS

POLYMER MONOMER(Name & Structure) USESAddition or Chain Growth Polymer

Polythene Ethene CH2=CH2 Insulator, ,Packing material,Teflon(Poly tetrfluoroethene)

Tetrfluoroethene CF2=CF2

Lubricant, Insulator and making non-stick cooking ware.

Poly acrylonitrile Acrylonitrile CH2=CH-CN For syntheticfibres and synthetic wool.

Buna S Buta-1,3-diene + StyreneCH2=CH-CH=CH2 C6H5CH=CH2

Automobile tyres and Foot wears

Buna N Buta-1,3-diene + AcrylonitrileCH2=CH-CH=CH2 CH2=CH-CN

Oil seals, Tank lining

Natural Rubber 2-Methylbuta-1,3-diene (Isoprene) Used for tyres after vulcanisationSynthetic Rubber(Neoprene) 2-Chlorobuta-1,3-diene (Chloroprene) Insulator, making conveyor belts and printing

rollersPolypropene Propene CH3-CH=CH2 Ropes, toys,pipes and fibres

Polystyrene Styrene C6H5CH=CH2Insulator, Wrapping material,toys, Radio and television cabinets.

Polyvinyl chloride(PVC) Vinyl Chloride CH2=CH-Cl Rain coats, Hand bags, Vinyl flooring and

water pipe.Condensation or Step Growth Polymers

Terylene(Dacron) Ethane-1,2-diol + Benzene-1,4-dicarboxylic acid

Used for making fibres, safety belts, tents

Nylon 66 Hexamethylenediamine + Adipic acidNH2(CH2)6NH2 HOOC(CH2)4COOH

For making brushes,paratutes and ropes

Nylon 6 Caprolactum Tyrescords,fabrics andropes

Bakelite Phenol + Methanal Combs,electricalswitches,handles of utensiles and computer discs

Melamine Melamine + Methanal Unbreakable crockeryPHBV (biodegradable)

3-Hydroxybutanoic acid + 3-Hydroxypentanoic acid

Specialty packaging, orthopedic devices, In controlled drug release

Nylon 2 – Nylon 6(biodegradable)

Glycine + Amino caproic acidH2N-CH2-COOH H2N (CH2)5-COOH

Urea-formaldehyde resin Urea + Formaldehyde Unbreakable cups , laminated sheet

Glyptal Ethane1,2-diol + Benzene-1,2-dicarboxylic acid

Binding material in preparation of mixed plastics and paints

******************************************************************************************Semi-synthetic poly Cellulose Acetate (Rayon)Thermoplastic polymers

Linear or slightly branched / capable of repeatedly softening on heating and hardening on cooling. Example : Polythene, Polystyrene, Polyvinyls, etc.

Thermosetting polymers

Cross linked or heavily branched molecules, / on heating undergo extensive cross linking in moulds and again become infusible. These cannot be reused. Examples : Bakelite, Urea-formaldelyde resins.

Homo-polymer &Co-polymer

Homo-polymer Polymer of a single monomeric species. Example : Polythene , PVCCo-polymer Polymer of more than one monomer .Example : Nylon66, Bakelite

Initiators Benzoyl Peroxide [ C6H5CO-O-O-CO-C6H5 ] (in freeradical addition polymerization)

Vulcanisation of Rubber.

Natural rubber is soft at high temp and brittle at low tempand absorbs water. To improve these physical properties, it is heated with sulphur and an appropriate additive at a temperature range between 373 K to 415 K. On vulcanisation, sulphur forms cross links at the reactive sites of double bonds and thus the rubber gets stiffened.

Polymers(Q.) Give the structure of natural rubber?

(1 Mark)(Ans)

(Q.) Define the term polymerization?(1 Mark)

(Ans) The repeating structural units formed from some simple and reactive molecules (monomers) are linked together by covalent bonds. This process of formation of polymers from their respective monomers is called polymerization.(Q.) Define Elastomers?

(1 Mark)(Ans) Elastomers are rubber-like solids with elastic properties. In elastomers, the polymer chain is heldtogether by weak intermolecular forces which allow the polymer to be stretched. e.g. buna-S, buna-N neoprene.(Q.) What are fibers?

(1 Mark)(Ans) Fibers are the thread forming solids which possess high tensile strength and high modulus. These polymers possess strong intermolecular forces.Thus leading to close packing of chains and imparting crystalline nature. e.g., polyamides( nylon 6,6) and polyesters( terylene), etc.(Q.) Define Thermoplastics polymers?

(1 Mark)(Ans) Thermoplastic polymers are the linear or slightly branched chain molecules capable of repeatedly softening on heating and hardening on cooling. Some examples of thermoplastics are polythene, polystyrene, polyvinyl, etc(Q.) Define Thermosetting plastics?

(1 Mark)(Ans) Thermosetting polymers are cross linked or heavily branched molecules which on heating undergo extensive cross linking in moulds and again become infusible. Some examples of thermosetting plastics are bakelite, urea-formaldehyde resins, etc.(Q.) Give the method of preparation of polyacrylonitrile?

(2 Marks)(Ans) The addition polymerization of acrylonitrile in presence of a peroxide catalyst leads to the formation of polyacrylonitrile. It is used as a substitute for wool in making fibers as orlon or acrilan.

(Q.) Define copolymerization? Give chemical reaction showing formation of copolymer.(2 Marks)(Ans) Copolymerisation is a polymerization reaction in which a mixture of more than one monomeric species is allowed to polymerise and form a copolymer. For example, a mixture of a 1,3 – butadiene and styrene form a copolymer.

(Q.) Describe the method for the preparation of neoprene? (2 Marks)

(Ans) Neoprene or poly chloroprene is formed by the free radical polymerization of chloroprene.

(Q.) Differentiate between Addition and Condensation polymers?(2 Marks)

(Ans) S.No Addition Polymers Condensation polymers

1. They are formed by the repeated addition of molecules possessing

double or triple bonds.

They are formed by repeated condensation reaction between two different bi-functional and tri-functional monomeric

units.2. E.g. polythene E.g. nylon 66

(Q.) Differentiate between Homopolymers and Copolymers with example.(2 Marks)

(Ans) Homopolymers: The addition polymers formed by the repeated addition of monomer molecules possessing double or triple bonds, are known as homopolymers. E.g., nCH2 = CH2----->---(CH2—CH2)---Ethene polytheneCopolymers: The polymers formed by addition polymerization of two different monomers are termed as copolymers.E.g., nCH2 = CH—CH = CH2 + nC6H5CH=CH2---->---[CH2—CH = CH—CH2—CH2—CH(C6H5)]n---

(Q.) Give the differences between homopolymers and copolymers?(2 Marks)

(Ans)S.No. Homopolymers Copolymers

1 The addition polymers formed by the polymerisation of a single monomeric unit are called Homopolymers.

The polymers formed by the addition polymerisation of two different monomers are

termed as copolymers.2 E.g. Polythene. E.g. Buna-S

(Q.) What is synthetic rubber?(2 Marks)

(Ans) Synthetic rubber is any vulcanisable rubber like polymer, which is capable of getting stretched twice its length. However, it returns to its original shape and size as soon as the external stretching force is released.(Q.) Give the method of preparation of Teflon and its uses.

(3 Marks)(Ans) Polytetrafluoroethene (Teflon) is manufactured by heating tetrafluroethene with a free radical or persulphate catalyst at high temperature. It is chemically inert in nature. It is used for making oil seals and gaskets and also used for non-stick surface coated utensils.

Tetraflouroethane Teflon(Q.) What are the different types of structural polymers?Give examples?

(3 Marks)(Ans) There are three different types of structural polymers:(1) Linear polymers: These polymers consist of long and straight chains.e.g. polythene,polyvinyl chloride, etc.(2) Branched chain polymers: These polymers contain linear chains having some branches.e.g. low density polythene.(3) Cross linked polymers (network polymers): These are usually formed from bi-functional and tri-functional monomers and contain strong covalent bonds between various linear polymer chains. e.g. bakelite, melamine etc.(Q.) Give the method for preparing Bakelite?

(3 Marks)(Ans) Bakelite is manufactured from Phenol-formaldehyde polymers. It is obtained by the condensation reaction of phenol with formaldehyde in thepresence of an acid or base catalyst. The initial product formed is a linear chain called – Novolac used in paints. Novolac on heating with formaldehyde undergoes cross linking to form infusible solid mass called bakelite.

(Q.) Give the method of preparation and uses of nylon 6?(3 Marks)

(Ans) Nylon 6 is obtained by heating caprolactum with water at high temperature.

Nylon 6 is used for the manufacture of tyre cords, and fabrics and ropes.(Q.) How are condensation polymers formed? Explain giving one example.

(3 Marks)(Ans) The condensation polymers are formed by repeated condensation reaction between two different bi-functional or tri-functional monomeric units. During this process, a small molecule of water, alcohol, hydrogen chloride, etc is eliminated. E.g. terylene, nylon 6,6, nylon 6, etc. Preparation of nylon 6,6:nH2N(CH2)6NH2 + nHOOC(CH2)4COOH ----->---(NH(CH2)6NHCO(CH2)4CO)n--- + n H2O(Q.) What are natural and synthetic polymers? Give 2 examples of each.

(3 Marks)(Ans) Natural polymers: Polymers which are found in plants and animals are called natural polymers. e.g. proteins, cellulose starch etc. Synthetic polymers: Man made polymers are called synthetic polymers. They consists of a number of smaller molecules to form large molecules. e.g. nylon 6,6 and Buna-S.(Q.) Define the term polyesters? How is it manufactured? (3 Marks)(Ans) Polyesters are the condensation products of dicarboxylic acids and diols. e.g. Dacron or Terylene.It is manufactured by heating a mixture of ethylene glycol and terephthalic acid at 420 to 460 K in the presence of zinc acetate-antimony trioxide catalyst.(Q.) Explainvulcanisation of rubber?

(3 Marks)(Ans) Natural rubber becomes soft at high temperatures and brittle at low temperatures. It has a high water absorption capacity. It is soluble in non-polar solvents and is non-resistant to attack by oxidizing agents. To improve upon these physical properties, the process of vulcanization is carried out. It consists of heating a mixture of raw rubber with sulphur with a temperature range between

373 K to 415 K. On vulcanisation, sulphur forms cross links at reactive sites of double bonds and thus the rubber gets stiffened.(Q.) Give three examples of biodegradable polymers?

(3 Marks)(Ans) (1) phbv(2) Nylon-2(3) nylon-6

UNIT-16 MARKS-3

CHEMISTRY IN EVERYDAY LIFE

THERAPEUTIC ACTION OF DIFFERENT DRUGSDrugs Action Example

Analgesics Relieving pain AspirinAnalgin, Seridon, Anacine, Analgesics(Narcotic )

Reduce tension and pain. produceunconsciouness.

Opium, Heroin , Pethidine , Codeine, Morphine

Antibiotics Produced by micro – organism, that can inhibit the growth or destroy other micro-organism.

Penicillin G(Narrow Spectrum)Streptomycin, Ampicillin ,Amoxycillin Chloramphenicol Vancomycin, ofloxacin,

Antiseptics Prevent the growth of micro-organism or kill them but not harmful to the living tissues.

Dettol, Bithional(in soap) Tincture iodine, 0.2% phenol,

Disinfectants Kills micro-organisms, not safe for living tissues. It is used for toilets, instruments.

1% phenol, chlorine (Cl2) , Sulphurdioxide ( SO2)

Antacids Reduce or neutralise the acidity.

Antihistamines Reduce release of acid.

It is also used to treat allergy

Cimetidine(Tegamet), Ranitidine (Zantac),

Brompheniramine ( Dimetapp)Terfenadine ( Seldane)

Tranquilizers Reduce the mental anxiety, stress, (sleeping pill)

Valium, Serotonin, Veronal, Equanil,Amytal,Nembutal,Luminal, Seconal

Antipyretics Reduce body temperature Aspirin, Paracetamol, Analgin, Phenacetin.Antifertility drugs

These are the steroids used to control the pregnancy

Norethindrone, Ethynylestradiol (novestrol )

CHEMICALS IN FOODSweetening Agent Saccharine,Aspartme(for cold foods) Alitame

Sucrolose(stable at cooking temp)Food Preservative

Salt, sugar, veg. oils, sodium benzoate

CLEANSING AGENTSSoap Na / K –salt of long chain fatty

acidsNot work in hard water becoz with Ca and Mg salt soap produce insoluble scum

Anaionicdetergen

Sodium laurylsulphate Used in household work / in tooth paste

Cationic detergent Cetyltrimethyl ammonium bromide Hair conditioner / germicidal propertiesNon ionic detergent

Ester of searic acid and polyethylene glycol

Liquid dishwashing

Detergents with highly branched hydrocarbon parts are non biodegradable and hence water pollutants so branching is minimized which are degradable and pollution is prevented.

Chemistry in Everyday Life(Q.) Why it is advised to consult a doctor before taking a sleeping drug?

(1 Mark)(Ans) Because itmay produce harmful effects and can act as a poison and cause even death.

(Q.) Give the name of the alkoloids which are used to cure Hypertension and Malaria.(1 Mark)(Ans) Alkoloid for hypertension is reserpine and for malaria is quinine.(Q.) Name the macromolecules which are chosen as drug targets.

(1 Mark)(Ans) Macromolecules like proteins, nucleic acid, carbohydrates and lipids are chosen as drug targets.(Q.) What do you understand by chemotherapy?

(1 Mark)(Ans) It is the branch of chemistry deals with the study of the treatment of the diseases using chemicals.(Q.) Name the forces which are involved in holding the drugs to the active site of the enzymes? (1 Mark)(Ans) Hydrogen bonding, ionic bonding, dipole-dipole interactions or van der Waals interactions.(Q.) Antacids and antiallergic drugs interfere with the function of histamines but why do these not interfere with the function of each other?

(1 Mark)(Ans) Because the antiallergic and antacids drugs work on different receptors, so antihistamines remove allergy and antacid remove acidity.(Q.) Give a name of the substance which can be used as antiseptic and disinfectant.(1 Mark)(Ans) 0.2% solution of phenol acts as antiseptic and 1% solution of it act as disinfectant.(Q.) What are the main constituents of dettol?

(1 Mark)(Ans) Chloroxylenol and terpineol in a suitable solvent.(Q.) What is tincture iodine? Give one of its use.

(1 Mark)(Ans) The 2-3% solution of iodine in alcohol and water is tincture iodine. It is used as an antiseptic for wounds.(Q.) Why is the aspartame used only for the cold foods and the drinks?

(1 Mark)(Ans) Aspartame decomposes at baking or cooking temperature.so it isusedonlyin cold drinks and cold foods. (Q.) What are artificial sweetening agents? Give examples.

(1 Mark)(Ans) The chemical substances which are sweet in taste but do not add any calorie to our body are called artificial sweetening agents. For example Saccharin and Aspartame.(Q.) Why alitame is not used as an artificial sweetening agent?

(1 Mark)(Ans) Because it is a high potency artificial sweetener.(Q.) Why detergents are called soapless soaps?

(1 Mark)

(Ans) Because the detergents are cleansing agents which have all the properties of a soap but do not contain any soap.(Q.) Why do soaps not work in hard water?

(1 Mark)(Ans) In hard water, there are calcium and magnesium salts. So the soaps get precipitated in hard water as calcium and magnesium salts(Q.) Name the antibiotic used for typhoid fever.

(1 Mark)(Ans) Chloramphenicol.(Q.) What are antipyretics? Give some examples.

(1 Mark)(Ans) These are the chemicals used to lower down the temperature during the fever. Example: paracetamol, aspirin etc.(Q.) Why are detergents preferred over soaps?

(1 Mark)(Ans) Detergents contain calcium and magnesium salts of sulphonic acids which are soluble even in hard water but soaps are not soluble in hard water.(Q.) Label the hydrophilic and hydrophobic parts in the given molecule.

(1 Mark)(Ans)

Hydrophobic Hyhrophilic(Q.) Why the medicines should not be taken without consulting the doctors?

(1 Mark)(Ans) When a drug binds to more than one receptor site it causes side effects. Secondly the dose of the drug is also decisive because higher doses of some drugs act as a poison. So a doctor must be consulted to choose the right drug.(Q.) Can soaps and detergents be used to check the hardness of water?

(1 Mark)(Ans) Detergents do not get precipitated in water but the soaps get precipitated in hard water as insoluble calcium and magnesium soaps. So soaps can be used to check the hardness of water but detergents not.(Q.) Give the name and structure of two common antihistamine drugs. (2 Marks)(Ans)1. Diphenylhydramine:

2. Promethazine:

(Q.) Explain the following: (i) Equanil (ii) Morphine(2 Marks)

(Ans) (i) Equanil: It possesses a good tranquilizing effect so it is used to cure depression and hypertension. (ii) Morphine: It is used as analgesic. It is also a habit forming drug.(Q.) Give structure of following antifertility drugs:a) Mifepristone b) Norethindrone(2 Marks)(Ans) a) Mifepristone:

b)Norethindrone:

(Q.) How do antiseptics differ from disinfectants?(2 Marks)

(Ans) Antiseptics are the chemicals which are used to prevent the growth of micro organisms or to kill them but these are not harmful to animal tissues. Examples- Furacin and Soframycin.Disinfectants are the chemicals which are used to kill micro organisms but these are harmful to the animal tissues. Examples- phenol and chlorine.

(Q.) What do you understand by “board spectrum antibiotics”?(2 Marks)

(Ans) These are effective against several types of bacteria. For example tetracycline, chloramphenicol, Ofloxacinwhichare used as antibiotics.(Q.) What refers to the statement “Ranitidine is an antacid”?

(2 Marks)(Ans) This statement refers to the classification of the drugs according to the pharmacological effect because any drug used to neutralize the excess acid present in the stomach will be called an antacid.(Q.) Define the terms Lead compounds and Target molecules.

(2 Marks)(Ans) Lead compounds: The knowledge of the physiological function of the drug target in the body helps us to choose a compound which can interact with the target and hence is expected to be therapeutically active .These compounds are called target compounds. Target molecules: The macromolecules like carbohydrates, proteins,lipidsetc.interacting with the drugs are called target molecules.(Q.) Why are cimetidine and ranitidine better antacid than sodium bicarbonate or magnesium or aluminium hydroxide.

(2 Marks)(Ans) If there is an excess use of sodium bicarbonate or aluminium hydroxide or magnesium hydroxide in the stomach,it becomes alkaline and the HCl release in more quantitymay cause ulcers in the stomach. On the other hand cimetidine and ranitidine prevent the interaction of histamine with the receptor cells in the stomach wall and thus release lesser amount of HCl.(Q.) Explain the term bactericidal and bacteriostatic drug.

(2 Marks)(Ans) A drug which kills the organisms in the body is called bactericidal drug and a drug which inhibits the growth of organisms is called bacteriostatic drug.(Q.) Give the name and structure of two antacids.

(2 Marks)(Ans) 1. Omeprazole:

2. Lansoprazole:

(Q.) What do you understand by (i) Tranquillizers (ii) Antifertility drugs (iii) Antihistamines

Explain with examples.(3 Marks)

(Ans) Tranquillizers: The drugs which act on central nervous system to cure mental diseases are called tranquillizers. Example-Chlordiazepoxide and Meprobamate etc.Antifertility drugs: The chemicals which are used to check pregnancy in women are called antifertility drugs. Examples- Norethindrone and Mifepristone.Antihistamines: The drugs which interfere with the natural action of histamine are called antihistamines. Examples- Brompheniramine and Terfenadine.(Q.) Define the following with examples.(i) Preservatives(ii) Biodegradable detergents(iii) Non-biodegradate detergents

(3 Marks)(Ans) i) The chemical substances which are used to protect the food against bacteria, yeast, and moulds are known as preservatives. Examples- sodium benzoate, sorbic acid and its salts etcii) Biodegradable detergents: The detergents having straight hydrocarbon chains are easily decomposed by micro organisms. Examples- sodium lauryl sulphate, 4-(1-dodyl) benzene sulphonateetciii) Non-biodegradable detergents: These contain branched hydrocarbon chains and are not easily decomposed by the micro organisms. Examples- sodium4-(1, 3, 5, 7-tetramethyloctyl) benzenesulphonate.(Q.) Explain the cleansing action of soaps.

(3 Marks)(Ans) Suppose some grease or oil is sticking on the surface of a cloth, When it comes incontact with the soap solution the stearate ions arrange themselves around it in such a way that hydrophobic parts of the stearate ions are in the oil and the hydrophilic parts project outside the oil droplets. As hydrophilic part is polar so it interacts with the water molecules present around the oil droplet. As a result, the oil droplet is pulled away from the surface of cloth into water to form ionic micelle which is then washed away with the excess of water.(Q.) Write the chemical equations for preparing sodium soap from glycerol oleate and glycerol palmitate.

(3 Marks)(Ans) Preparation of soap from glycerol palmitate(Saponification): Heat CH2OH

(C15H31COO)3C3H5 + 3NaOH -------------> | + 3C15H31COONa(glycerolpalmitate) sodium CHOH sodium palmitate

hydroxide | (soap)CH2OHglycerol

Preparation of soap from glycerol oleate (Saponification):heat CH2OH (C17H33COO)3C3H5 + 3NaOH -----------> l + 3C17H33COONa(glycerololeate) CHOH sodium oleate

| (soap)CH2OH

(Glycerol)(Q.) Describe the following terms with suitable examples. a) Anionic detergents b) Cationic detergents c) Neutral detergents.

(3 Marks)(Ans) a) Anionic detergents: The detergents in which a large part of their molecules are anions are called anionic detergents. These are of two types: i)Sodium alkyl sulphates: example- sodium lauryl sulphate (C11H23CH2OSO3Na).ii)Sodium alkylbenzenesulphonates: example- sodium 4- (1-dodecyl) benzenesulphonate.

b) Cationic detergents: These are quaternary ammonium salts. Example- cetyltrimethylammoniumchloride.c) Neutral detergents: These are esters of high molecular mass alcohols with fatty acids. Example- polyethylene glycol stearate

{CH3 (CH2)16COO(CH2CH2O)nCH2CH2OH}(Q.) Name the hydrophobic and hydrophilic parts in the following compounds.a) CH3(CH2)10CH2OSO3Na b) CH3(CH2)15─N(CH3)Br−

c) CH3(CH2)6COO(CH2CH2O)nCH2CH2OH (3 Marks)(Ans) (a)CH3(CH2)10CH2 — OSO3Na hydrophobic hydrophilic(b)CH3(CH2)15 — N(CH3)Br−

hydrophobic hydrophilic(c)CH3(CH2)6 — COO(CH2CH2O)nCH2CH2OHhydrophobic hydrophilic

Important tools / notes for Organic Chemistry (Class XII)Compiled by N Kar, PGT (Chem) , KV Kunjaban, Agartala

KEY FOR CONVERSIONS IN ORGANIC CHEMISTRYS.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 Dehydrogenation4 PCl5 or SOCl2 -OH -Cl5 Cl2 / Δ or Cl2 / UV -H -Cl Free radical substitution6 Aq NaOH / KOH -X -OH Nucleophilic substitution7 KCN -X -CN Step Up 8 AgCN -X -NC9 Alcoholic KOH -HX = Dehydrohalogenation (Stzf)10 Mg / dry ether Mg R-X R-MgX11 HBr >=< H, Br Merkovnikov12 H2 / Pd-BaSO4 -COCl -CHO Rosenmund Reduction13 Zn-Hg / HCl >C=O -CH2- Clemmension Reduction14 NH3 / Δ -COOH -CONH2 -COOH + NH3 -COONH4

15 Br2 / NaOH or NaOBr -CONH2 -NH2 Step Down ( Hoffmann)16 HNO2 or NaNO2/HCl -NH2 -OH HONO17 CHCl3 / alc KOH -NH2 -NC Carbyl amine18 P2O5 -CONH2 -CN Dehydration19 H3O+ -CN -COOH Hydrolysis20 OH- -CN -CONH2

21 LiAlH4 -CN -CH2NH2 Reduction22 Red P / Cl2 α-H of acid -Cl HVZ Reaction

In benzene ring23 Fe / X2 /dark -H -X Halogination24 CH3Cl / AlCl3(anhyd) -H -CH3 Friedel Craft alkylation25 CH3COCl / AlCl3(anhyd) -H -COCH3 Friedel Craft acylation26 Conc.HNO3/con.H2SO4 -H -NO2 Nitration27 Conc H2SO4 -H -SO3H Sulphonation28 KMnO4 / H+ -R -COOH Oxidation29 CrO2Cl2 / H+ -CH3 -CHO Mild oxidation(Etard Reaction)30 Sn / HCl or Fe/HCl -NO2 -NH2 Reduction31 NaOH / 623K / 300 atm -Cl -OH32 Zn dust / Δ -OH -H33 NaNO2 / dil HCl / 273-

278 K-NH2 -N2

+Cl- Diazo reaction

34 CuCl / HCl or Cu/HCl -N2+Cl- -Cl Sanmeyer or Gattermann

35 CuBr / HBr or Cu/HBr -N2+Cl- -Br Sanmeyer or Gattermann

36 CuCN / KCN -N2+Cl- -CN Sanmeyer

37 KI -N2+Cl- -I

38 HBF4 / Δ -N2+Cl- -F

39 H3PO2 or CH3CH2OH -N2+Cl- -H

40 H2O / 283 K -N2+Cl- -OH

41 HBF4/ NaNO2, Cu / Δ -N2+Cl- -NO2

42 C6H5-OH -N2+Cl- -N=N-C6H5-OH Coupling ( p-hydroxy)

43 C6H5-NH2 -N2+Cl- -N=N-C6H5-NH2 Coupling ( p-amino)

Reactions of Grignard Reagent

Grignard reagent + Any one below + H2O ProductR-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-COOHR-CN R-CO-R

HCOOR AldehydeRCOOR 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 , -NH3

+ ,

Example : See no 7 in the table

NAME REACTIONS (ORGANIC CHEMISTRY)

1. Finkelstein CH3Br + NaI CH3-I + NaBr 2. Swarts CH3Br + AgF CH3F + AgBr

3.

Friedel-Crafts Alkylation

4.

Friedel-Crafts Acylation

5.Wurtz

6.

Fittig

7.

Wurtz-Fittig

8.

Kolbe

9.

Reimer-Tiemann

10. Williamson CH3-Br + CH3-ONa CH3-O- CH3 + NaBr

11.Stephen

12.

Etard

13.

Gatterman – Koch

14.

Rosenmund reduction

15.

Clemmensenreduction

16.

Wolff-Kishner reduction

17. Tollens’ test R-CHO + 2 [Ag(NH3)2]+ + 3 OH- R-COO- + 2Ag + 2H2O + 4 NH3

18. Fehling’s test R-CHO + 2 Cu2+ + 5 OH- R-COO- + Cu2O + 3H2O

19.

Iodoform Reaction

20.

Aldol condensation

21.Cannizzaro

22.

Hell-Volhard-Zelinsky (HVZ)

23.

Hoffmann bromamide degradation

24.Carbylamine

R-NH2 + CHCl3 + 3 KOH R-NC + 3 KCl + 3 H2O

25.

Sandmeyer.

26.

Gatterman

27.

Coupling

ELECTRON DISPLACEMENT EFFECTS

+ I : O- , COO- , (CH3)3C , (CH3)2CH , CH3CH2 , CH3 (electron

donating)

- I : NR3+ , SR2

+ , NH3+, NO2 , SO2R , CN , COOH , F , Cl , Br , I , OR , OH, NH2

(withdrawing)

+ R ( + M ) : OH , NH2 , OR , NHR , X (electron

donating)

- R ( - M ) : NO2 , CN , CHO , COOH , COCH3 (electron

withdrawing)

DIRECTIVE INFLUENCE OF SUBSTITUENTS IN BENZENE RING

(for electrophilic substitution reactions)

EFFECT OF THE GROUP DIRECTING ACTIVATING / DEACTIVATING

+ I Ortho / Para Activating

+ I , + R Ortho / Para Activating

- I < + R Ortho / Para Activating

- I > + R Ortho / Para Deactivating

- I Meta Deactivating

- I , - R Meta Deactivating

Example: - I > + R : - X , - CH=CH2 , -CH=CH-COOH , -CH2-Cl

These groups are deactivating but exceptionally o / p directing due to +E effect by the attacking

reagents electron density increases at ortho and para position .

If two groups are present initially

1. When both the groups present in benzene ring are o/p directing than the order of

influence :

O- > NH2 > NR2 > OH > OCH3 > NHCOCH3 > CH3 > X

2. When both the groups present in benzene ring are meta directing than the order of

influence :

(CH3)3N+ > NO2 > CN > SO3H > CHO > COCH3 > COOH

3. When one group is o/p and another is m – directing than o/p directing group takes priority

IUPAC & Common Names of Some Common Organic Compounds (Aliphatic)

Sl IUPAC Name Formula / Structure Common Name

1. Methane CH4 Methane2. Ethane CH3-CH3 Ethane3. Propane CH3-CH2-CH3 Propane

4. Ethene CH2=CH2 Ethylene5. Ethyne CH≡CH Acetylene

6. Chloromethane CH3Cl Methyl chloride7. Chloroethane CH3CH2-Cl Ethyl chloride8. Trichloromethane CHCl3 Chloroform9. Triiodomethane CHI3 Iodoform

10. Methanol CH3-OH Methyl alcohol11. Ethanol CH3CH2-OH Ethyl alcohol12. Propanol CH3CH2CH2-OH Propyl alcohol13. Propan-2-ol CH3CH(OH)CH3 Isopropyl alcohol14. Ethane-1,2-diol CH2(OH)-CH2(OH) Ethylene glycol

15. Methanal H-CHO Formaldehyde16. Ethanal CH3-CHO Acetaldehyde17. Propanal CH3CH2-CHO Propionaldehyde

18. Propanone CH3-CO-CH3 Acetone

19. Methanoic acid H-COOH Formic acid20. Ethanoic acid CH3-COOH Acetic acid21. Propanoic acid CH3CH2-COOH Propionic acid

22. Ethanoylchloride CH3-COCl Acetyl chloride

23. Ethanamide CH3-CONH2 Acetamide24. Ethanoicanhydride (CH3CO)2O Acetic anhydride25. Methylethanoate CH3-COOCH3 Methyl acetate

26. Methanamine CH3-NH2 Methyl amine27. Ethanamine CH3CH2-NH2 Ethyl amine28. N-methylmethanamine CH3-NH-CH3 Dimethyl amine29. Ethanenitrile CH3-CN Methyl cyanide30. Methyl magnesium bromide CH3-MgBr Methyl magnesium bromide

Distinguish By a Single Chemical Test

1. All aldehydes ( R- CHO ) give Tollens’ Test and produce silver mirror.

RCHO + 2 [Ag(NH3)2]+ + 3 OH- RCOO- + 2 Ag + 2H2O + 4 NH3

Tollens’ Reagent silver pptNote: HCOOH(methanoic acid ) also gives this test, ketones(RCOR) do not give this test

2. All aldehydes (R- CHO ) and ketones(R CO R) give 2,4-DNP test RCOR + 2,4-DNP Orange pptR-CHO + 2,4-DNP Orange ppt

3. Aldehydes and ketones having CH 3CO - (keto methyl) group give Iodoform Test . Alcohols having CH3CH- group also give Iodoform Test.

| OH CH3CHO + 3I2 + 4 NaOH CHI3 + HCOONa + 3 NaI + 3H2O Yellow ppt The following compounds give Iodoform Test: ethanol (C2H5OH), propan-2-ol (CH3CH(OH)CH3), ethanal(CH3CHO), propanone(CH3COCH3), butanone ( CH3COCH2CH3) , pentan-2-one (CH3COCH2 CH2CH3) , acetophenone ( PhCOCH3 )

4. All carboxylic acids ( R- COOH ) give Bicarbonate Test RCOOH + NaHCO3 RCOONa + CO2 + H2O effervescence

5. Phenol gives FeCl 3 Test C6H5OH + FeCl3 (C6H5O)3Fe + 3 HCl (neutral) (violet color) 6. All primary amines (R/Ar - NH 2) give Carbyl Amine Test R-NH2 + CHCl3 + KOH(alc) R-NC + KCl + H2O offensive smell 7. Aniline gives Azo Dye Test ( Only for aromatic amines) C6H5NH2 + NaNO2 + HCl C6H5N2

+Cl- ; then add β-naphthol orange dye 8. All alcohols (R OH ) give Na-metal test R-OH + Na R-ONa + H2 bubbles 9. For esters (RCOOR) : Hydrolyses first. Then see the products ( acid & alcohol) and give a test to identify them 10. All alkenes (C=C) and alkynes (C≡C) decolorizes Br2 – water from red to colorless

11. Lucas Test to distinguish primary, secondary and tertiary alcohols Lucas reagent: ZnCl2/HCl 30-alcohol + Lucas reagent immediate turbidity 20-alcohol + Lucas reagent turbidity after sometime

10-alcohol + Lucas reagent no turbidity

SAMPLE QUESTION PAPERS WITH MARKING SCHEME AND BLUE PRINT

SAMPLE PAPER –I (PREPARED BY GROUP –Platinum)

BLUE PRINTS.NO.

NAME OF UNIT VSA SA(I) SA(II) LONG TOTAL

1 SOLID STATE 2X2 4

2 SOLUTION 2 3 53 ELECTROCHEMISTR

Y2 3 5

4 CHEMICAL KINETICS

5 5

5 SURFACE CHEMISTRY

1 3 4

6 GENERAL PRINCIPLE OF EXTRACTION

3 3

7 P BLOCK 3 5 88 THE d-& f-BLOCK

ELEMENTS2 3 5

9 CO-ORDINATION COMPOUNDS

1 2 3

10 HALOALKANES& HALO ARENES

2X2 4

11 ALCOHALS, PHENOLS & ETHERS.

1 3 4

12 ALDEHYDE, KETONES & CARBOXYLIC ACIDS

1 5 6

13 AMINES 1 3 414 BIOMOLECULES 1 3 415 POLYMERS 1 2 316 CHEMISTRY IN

EVERYDAY LIFE1 2 3

TOTAL 8 10 9 3 70

SAMPLE QUESTION PAPER -ICLASS XII

SUBJECT CHEMISTRYTIME: 3Hrs

M.M. 70General instruction:

1. Adsorption of a gas on the surface of solid is generally accompanied by decrease in entropy still it is spontaneous process. Explain.

2. Square planer complexes with co-ordination number of four exhibit geometric al isomerism where as tetrahedral complexes do not. Why?

3. How will you prepare methanol from formaldehyde without using a reducing agent?

4. Give the IUPAC name of CH3 CH3

CH3 CH2 C C CH2 CH 2 OH

Br Br5. Write steps involved in the conversion of 1-propaneamine to acetone.6. Why can not vitamin C be stored in our body.7. Define Elastomers?

8. Define non-biodegradable detergent with one example.9. Analysis shows that a metal oxide has empirical formula of M0.96O . calculate the

percentage of M2+ and M3+ ion in the crystal.10. Account for the following: a) some of glass object recovered from ancient monuments look like milky

instead of being transparentb)Zinc oxide is white but turn yellow on heating. Explain

11. What do you understand by “board spectrum antibiotics 12 Differentiate between Homopolymers and Copolymers with example. OR Describe the method for the preparation of neoprene? 13 The osmotic pressure of human blood is 7.65 atm at 37°C. For injecting Glucose solution it is necessary the glucose solution has same osmotic pressure as of human blood. Find the molarity of glucose solution having same osmotic pressure as of human blood.

scribe the method for the preparation of neoprene?14 . Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.15 What is lanthanoid contraction? Why lanthanides are known as f-block elements?

.16) Copper I compounds are white and diamagnetic but copper II compounds are coloured and paramagnetic. Why? 17.) Complete the following reactions: (a) C6H5ONa + C2H5Cl --------> (b) CH3CH2CH2OH + SOCl2 ------->18.) Why is aniline a weaker base than methylamine? 19 a) Define glycosidic linkage

b)What are the main sources of vitamins? c )Define the term Oligosaccharides? OR a) Explain the term mutarotation? b)What is difference between Reducing and non-reducing sugars or carbohydrates?(1 20 How are the following preparations carried out?

(a) Salicylic acid from phenol.(b) n-propyl alcohol from ethene.

21 a) explain the coupling reaction with example. b)Describe the Gabriel-phthalimide reaction. 22 a) Explain ideal and non-ideal solutions with respect to intermolecular interactions in a binary solution of A and B.b) Aquatic animals are more comfortable in cold water than in warm water. Explain?23 The standard electrode potential for Daniell cellis 1.1V. Calculate the standard Gibbs energy for the reaction:Zn(s) + Cu2+(aq) ⎯→Zn2+(aq) + Cu(s)24(a)What is the effect of temperature on adsorption?

.b) Why are zeolites called shape selective catalysts?

.c) Define tyndall effect25(a) What is leaching?(b)How can the ores ZnS and PbS be separated from a mixture using froth floatation process?26 (a)Why does NH3 has higher boiling point than PH3?(b)Why is the ionization energy of group 15 elements higher than that of group 14 elements(C)Acidic strength increases in thisorder : HF<HCl<HBr<HI. Give reason. 27 (a)How does +2 oxidation state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

(b)Vanadium pentaoxide acts as a good catalyst. Why28 a) The initial concentration of N2O5 in the following first order reaction

N2O5(g) →2 NO2(g) + 1/2O2 (g) was 1.24 × 10–2 mol L–1 at 318 K. Theconcentration of N2O5 after 60 minutes was 0.20 × 10–2 mol L–1.Calculate the rate constant of the reaction at 318 K.b)A reaction is second order with respect to a reactant. How is the rate ofreaction affected if the concentration of the reactant is

(i) doubled (ii) reduced to half OR

a)The rate constants of a reaction at 500K and 700K are 0.02s–1 and

0.07s–1 respectively. Calculate the values of Ea and A.b)For a reaction, A + B →Product; the rate law is given by, r = k [ A]1/2 [B]2.

What is the order of the reaction?c)Identify the reaction order from the following rate constants.

(i) k = 2.3 × 10–5 L mol–1 s–1

29. a)

ORa)

b. Describe the following (i) Aldol condensation (ii)cannizaro reaction

(iii)Friedel-Crafts reaction. Write reactions only30 a) Why is ammonia a mild reducing agent while BiH3 the strongest reducing agent among all the hydrides?

b) Explain the brown ring test for nitrates. Give the equations involved for the reactions occurring during the brown ring test. Ora)Give reasons:

(i) size of oxygen atom exceptionally small (ii) SF6 exceptionally stable (iii) PCl3 fumes in atmosphereb)How does Cl2 react with

(i) cold and dilute NaOH(ii) hot and concentrated NaOH

Answer key1. because adsorption is an exothermic process 1marks

2. because relative position of ligands attached to central atom are same with respect to one another 1marks

3. give cannizarro reaction. 1marks

4. 3,4-dibromo-3,4-dimethylhexan-1-ol. 1marks

5. CH3CH2CH2NH2 HONO CH3CH2CH2OH CONC sulphuric acid,443K CH3CH=CH2 mark.rule H+/H

2o CH3CHOHCH3

Cu powder 573k CH3COCH3 1marks

6. Because it is soluble in water so excreted in urine 1marks

7. Elastomers are rubber-like solids with elastic properties. In elastomers, the polymer chain is heldtogether by weak intermolecular forces which allow the polymer to be stretched. e.g. buna-S, buna-N neoprene 1marks8. Detergent containing contaioning branchned hydrocarbon chain are not easily degraded by microorganism and hence are called non-biodegradable detergent e.g. sodium-4-(1,3,5,7-tetramethyloctyl)benzenesulphonate. 1marks9. Ratio of M2+ : M3+ =96:100Total charge on M2+and M3+

(+2)X +3(96-X) =200X=88% of M2+ and M3+ =91.7 and 8.3 respectively 1+1marks10 a) Due to annealing over a number of years glass acquires some crystalline character b) because it loose oxygenZnO Zn2+ + 1/2O2 + 2e-

Metal excess defect electron in interstitial voids the colour is yellow as the remaining colour of white light are absorbed by these electron

1+1marks

11 These are effective against several types of bacteria. For example tetracycline, chloramphenicol, Ofloxacin whichare used as antibiotics. 1+1marks

12) Homopolymers: The addition polymers formed by the repeated addition of monomer molecules possessing double or triple bonds, are known as homopolymers. E.g., nCH2 = CH2 -----> ---(CH2—CH2)---

Ethene polytheneCopolymers: The polymers formed by addition polymerization of two different monomers are termed as copolymers.E.g., nCH2 = CH—CH = CH2 + nC6H5CH=CH2 ----> ---[CH2—CH = CH—CH2—CH2—CH(C6H5)]n--- 1+1marks

ORNeoprene or poly chloroprene is formed by the free radical polymerization of chloroprene.

1+1marks

13. (Ans) =

Or 7.65 =

=

= Molarity = 0.30 1+1marks14. Conductivity : conductance of a one cm3 of a solution is known as conductivity.Conductivity decrease with dilution i.e. decrease with decrease of concentrationMolar conductance : Molar conductance is the conductance of a solution having one mole of electrolyte when whole of the solution is present between the two electrode.Molar conductance increase with dilution i.e. increase with decrease of concentration. 1+1mark15. ) In the lanthanide series as the atomic number increases there is a progressive decrease in the size of the atoms and trivalent ions which is known as lanthanide contraction.The last electron enters in the f-orbital, so lanthanides are knowns as f- block elements 1+1marks16. ) In copper I ion all orbitals are completely filled so its compounds are white and diamagnetic. The electronic configuration of copper II ion is 1s22s22p63s23p63d9. it has one unpaired electron so it is paramagnetic and forms blue coloured compounds. 1+1marks17. ) (a) C6H5ONa + C2H5Cl C6H5-O-C2H5 + NaCl Sodium Ethyl chloride PhenetolePhenoxide(b) CH3CH2CH2OH + SOCl2 CH3CH2CH2Cl + HCl + SO2Propanol-1 Thionyl 1-chloropropaneChloride 1+1marks

18. Aniline and methylamine both have nitrogen with lone pair of electron. In aniline the phenyl group is electron attracting. It tend to decrease the electron density on nitrogen atom and hence decreases electron releasing tendency of nitrogen. While in methylamine CH3- group is electron repelling in nature. It tends to increase the electron density on the nitrogen atom and helps in electron releasing tendency of nitrogen. So, the aniline is a weaker base than methylamine.

1+1marks19. a)The two monosaccharide units are joined together through an ethereal or oxide linkage formed by the loss of a molecule of H2O. Such a linkage between two monosaccharide units through oxygen atoms is called glycosidic linkageb)The main sources of vitamins are milk, butter, cheese, fruits, green vegetables, meat, fish, eggs, etc.c) Those carbohydrates which give 2-10 molecules of monosaccharides in hydrolysis. 1+1+1marks ORa)Mutarotation is the change in the specific rotation of an optically active compound with time, to an equilibrium mixtureb)All those carbohydrates which contain aldehydic and ketonic group in the solution are called reducing carbohydrates while others which do not reduce these reagents are called non-reducing reagents

1+1+1marks20) (a) Salicylic acid from phenol: -

(b) n-propyl alcohol from ethane: - By oxo processCH2=CH2 + (CO + H2) CH3-CH2-CHO CH3-CH2-CH2-OH 3/2+3/2 MARKS

21.a) Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with the diazonium salt to form p-hydroxybenzene. This type of reaction is known as coupling reaction.

b) In this reaction potassium phthalimide is reacted with an alkyl halide to get N-alkyl phthalimde. N-alkyl phthalimide on hydrolysis with dil. HCl under pressure.

3/2+3/2 MARKS22. a) ) For the given binary solution of A and B, it would be ideal if A-B interactions are equal to A-A and B-B interactions and it would be non-ideal if they are different to each other.The deviation from ideal behaviour will be positive if A-B interactions are weaker as compared to A-A and B-B. The deviation will be negative if A-B interactions are stronger as compared to A-A and B-B.b) This is because “Kh” (Henry constant) values for both N2 and O2 increase with increase in temperature indicating that the solubility of gases increases with decrease intemperature 2+1 marks23. rG_ = – nF (cell) EVn in the above equation is 2, F = 96487 C mol–1 andcellEV = 1.1 VTherefore, rG_ = – 2 × 1.1V × 96487 C mol–1= –21227 J mol–1= –21.227 kJ mol–1 1+1+1 marks24. a) Adsorption processes, being exothermic, decreases with increase in temperature.b) Zeolites are called shape selective catalysts because their catalytic action depends upon the size and shape of the reactant and the product molecules as well as ontheir own pores and cavities.c) ) Itis defined as the scattering of light by the colloidal particles present in a colloidal solution. 1+1+1 marks25.a) Leaching is the process of extracting a substance from a solid by dissolving it in a liquid. In metallurgy leaching is used for the ores that are soluble in a suitable solvent.b) During the froth floatation process a depressent like NaCN is added to the tank. The depressent selectively prevents ZnS from coming to the froth but allows PbS to come to the froth and hence helps the separation of PbS with the froth.3/2+3/2 marks26.a) NH3 has higher boiling point than PH3 because of the presence of inter molecular hydrogen bonding in NH3, as the electronegativity difference is quite high in case of Nand H.b) The ionization energy of group 15 elements is higher than that of group 14 elements because the elements of group 15 have extra stable half-filled p orbital configuration and their size is smaller due to the higher nuclear charge

http://en.wikipedia.org/wiki/Liquid

http://en.wikipedia.org/wiki/Solid

c) Acidic strength increases from HF to HI because, down the group,as the size of the halogen increases, the Bond dissociation enthalpy decreases and it becomes easier for thathalogen atom to lose its H+ ion 1+1+1 marks27. a) The sum of first and second ionization enthalpies increases with increasing atomic number so the standard reduction potentials become less and less negative. Hence the +2 oxidation state becomes more and more stable.b) Vanadium shows different-different oxidation states because it has vacant d-orbitals.so vanadium pentaoxide acts as a good catalyst.2+1 marks28 a)

b) Reaction is of second order w.r.t. reactant i)therefore when concentration is double rate of reaction become four times rate =[2R]2 =4[R]2

ii) when concentration is half rate of reaction become ¼ of the initial rate. rate =[1/2R]2 =1/4[R]2 3+1+1 marks OR

a)

b)2+1/2 =2.5c)

3+1+1 marks29 a)

1+ 1+1+1+1 marks OR

b) give reaction onlyc)give reaction only 3+1+1 marks30 a) ) a) From N to Bi the size of atom increases and tendency to form covalent bonds with H decreases. Therefore, the thermal stability also decreases from the hydrides, down the group. As a result, their tendency to liberate hydrogen increases for hydrides down the group. Hence, NH3 is a mild reducing agent while BiH3 is a very strong reducing agent. b) The brown ring test is carried out by adding dilute ferrous sulphate solution to an aqueous solution containing nitrate ions and then carefully adding conc. H2SO4 along the sides of the test tube. The Fe2+ ions reduce

nitrates to nitric oxide, which reacts with Fe2+ ions to form a brown colored complex that forms a brown ring at the interface between the solution and sulphuric acid layers.The equations involved in the brown ring test are:NO3- + 3Fe2+ + 4H+ = NO + 3Fe3+ +2H2O[Fe(H2O)6]2+ +NO = [Fe(H2O)5(NO)]2+ + H2O 2+3marks ORi) The size of oxygen is exceptionally small because of its high electron density and hence higher attraction between the electron cloud and the nucleus. ii) In SF6 the six F atoms protect the central sulphur atomfrom the reagents making the compound extra stable.iii) PCl3fumes in atmosphere due to its hydrolysis in atmosphere.

PCl3 + 3H2O = H3PO3 + 3HClb) ) When Chlorine reacts with cold sodium hydroxide, it forms sodium chloride. (i) 2NaOH + Cl2 = NaCl + NaOCl + H2OWhen Chlorine reacts withhot sodium hydroxide, it forms sodium chloride and other products.(ii) 6NaOH + 3Cl2 = 5NaCl + NaClO3 + 3H2O 1+1+1+1+1 marks

SAMPLE QUESTION PAPER-IIBLUE PRINT

Time : 3 Hours: Class-XIISubject :Chemistry

Max Time: 3Hrs

Unit Types of questions --> VSA SA(I) SA(II) LA Total

no. Name of the unit (1mark) (2Marks) (3Marks) (5Marks)

1 Solid State 1(1) 3(1) 4(2)

2 Solutions 5(1) 5(1)

3 Electrochemistry 2(1) 3(1) 5(2)

4 Chemical Kinetics 2(1) 3(1) 5(2)

5 Surface Chemistry 1(1) 3(1) 4(2)

6General Principles and Processes of Isolation of Elements 3(1) 3(1)

7 p-Block Elements 1(1) 2(1) 5(1) 8(3)

8 d and f - Block Elements 2(1) 3(1) 5(2)

9 Co-ordination Compounds 1(1) 2(1) 3(2)

10 Haloalkanes and haloarenes 4(2) 4(2)

11 Alcohols, Penols and ethers 4(2) 4(2)

12Aldehydes, Ketones and Carboxylic acid 1(1) 5(1) 6(2)

13 Organic compds containing nitrogen 1(1) 3(1) 4(2)

14 Biomolecules 1(1) 3(1) 4(2)

15 Polymers 3(1) 3(1)

16 Chemistry in Everday life 1(1) 2(1) 3(2)

Total 8(8) 20(10) 27(9) 15(3) 70(30)

SAMPLE QUESTION PAPER-II Class-XII Max Marks:70 Subject- Chemistry Max. Time: 3 hrs

Total no. of questions: 30 Total no. of printed pages:03 General Instructions

1. All questions are compulsory.2. Question nos. 1 to 8 are very short answer questions and carry 1 mark

each.3. Question nos. 9 to 18 are short answer questions and carry 2 marks each.4. Question nos. 19 to 27 are also short answer questions and carry 3 marks

each5. Question nos. 28 to 30 are long answer questions and carry 5 marks each6. Use log tables if necessary, use of calculators is not allowed.

1.Primary amines have higher boiling point than tertiary amines. Why? (1)

2.In liquid state HCl is stronger acid than HF. Why? (1)

3.What is coordination no. of ions in a rock salt type crystal structure. (1)

4.What is meant by the term broad spectrum antibiotic? (1)

5.Why are carbohydrates generally optically active? (1)

6.Write the IUPAC name of the following: (1) CH3

CO-N C2H5

7.Write the IUPAC name of the complex compound [Co(NH3)6][Cr(CN)6] (1)

8.How does adsorption of a gas on a solid surface vary with (1)increase in pressure?

9.Give reasons for the following: (2)(a)Oxygen is diatomic while sulphure is octatomic.(b)H2S is less acidic than H2Te

10.Using valence bond approach, deduce the shape and magnetic character of [CoF6]3- ion. (2)

11.Account for the following: (2)(a)Dry cell become dead after a long time even it has not been used.(b)Iron does not rust even if zinc coating is scratched in galvanized iron.

12.Discuss the two ways in which drugs prevent attachment of natural substrate active site of an enzyme.

(2)Or

Explain the following terms with suitable examples:(i) Cationic detergents (ii) Food preservatives

13.Give reasons for the following:(2)

(i)Alcohols act as weak base.(ii)m-aminophenol is stronger acid than o-aminophenol.

14.Describe a chemical test to distinguish between the following pairs:(2)

(i)Acetaldehyde and acetone(ii)Benzoic acid and phenol

15.Distinguish order and molecularity of a reaction. When could order and molecularity of a reaction be the same?

(2)

16.Account for the following:(2)

(i)There are irregularities in the electronic configurations of actinoids(ii)Compounds of transition elements are often coloured.

17.State the IUPAC name of the compound (2)

(ii)Complete the equation: CH3CH2CH=CH2 HBr/Peroxide

18.Account for the following:(2)

(i) Cl* reacts faster than Cl

(ii)The treatment of an alkyl halide with aqueous KOH leads to the formation of an alcohol whereas in the presence of alcoholic KOH, alkene is the major product.

19.The decomposition of phosphine: 4PH3(g) P4(g) + 6H2(g) has the rate law, Rate=k[PH3]. The rate constant is 6.0 x 10-4 S-1 at 300K and activation energy is 3.05 x 105 J/mole. Calculate the value of rate constant at 310K.

(3)

20(a)What is the difference between a colloidal solution and an emulsion? Give one example of each.(b)What are emulsifiers?

(3)

21.Describe the role of following:(3)

(i)Depressant in froth floatation process.(ii)Cryolite in metallurgy of aluminium.(iii)Silica in the extraction of copper from copper pyrites ore.

22(a)Describe the commercial preparation of potassium permanganate from pyrolusite ore.(b)Write ionic equation to represent the reaction of acidified KMnO4 solution with oxalic acid.

23(a)Distinguish between homo polymers and copolymers. Give one example of each. (3)(b)Is –(-CH2-CH(C6H5)-)-n a homo polymer or a copolymer? Why?

24.What do you understand from:(3)

(a)Denaturation of protein (b)Primary structure of protein (c) Replication

25.Illustrate the following reaction with suitable example(3)

(i)Coupling reaction(ii)Hell Volhard Zeilnsky reaction

26.Calculate the standard cell potential of a galvanic cell in which the following reaction takes place:

(3)2 Cr(s) + 3 Cd2+(aq) 2 Cr3+(aq) + 3 Cd(s)

Calculate ∆G0 and equilibrium constant, K of the above reaction at 250C.Given that E0 Cr3+/Cr = - 0.74V; E0 Cd2+/Cd = - 0.40V

27.Explain the following terms with examples:(3)

(i)Schottky defect (ii)FerromagnetismOr

(a)What is the coordination number of an octahedral void? (b)Name one defect which causes colour in ionic solids. Describe that defect.

28.(a)Arrange HClO3, HClO2, HOCl and HClO4 in order of increasing acid strength. Give reasons for your answer.(b)Write the balanced chemical equation for the reaction of Cl2 with hot and conc NaOH solution. Justify that this reaction is a disproportionation reaction.(c)Give one use of SF6.

(5)Or

(a)Why do some noble gases form compounds with fluorine and oxygen only?(b)Draw the structure of IF7

(c)How are the following compounds prepared from XeOF6?(i)XeOF4 (ii)XeO3

29(a)An organic compound A, molecular formula C8H16O2 was hydrolysed with dilute sulphuric acid to give a carboxylic acid B and an alcohol C. Oxidation of C with chromic acid produced B. C on dehydration gives but-1-ene. Write chemical equations for the reactions involved.(b)How will you carryout following conversion

(a) Propanone to Propene (b)Methane to ethyl amineOr

An organic compound A(C7H6Cl2) on treatment with NaOH solution gives another compound B(C7H6O). B on oxidation gives an acid C(C7H6O2) which on treatment with a mixture of conc. HNO3 and H2SO4 gives a compound D(C7H5NO4). B on treatment with conc. NaOH gives a compound E(C7H8O) and C6H5COONa. Deduce the structures of A, B, C, D and E. (5)

30(a)State Henry Law and mention its two applications(5)

(b)Henry law constant for CO2 dissolving in water is 1.6 x 108 Pa at 298K. Calculate the quantity of CO2 in 1L of soda water when packed under 2.5atm. CO2

pressure at 298KOr

(a)The depression in freezing point of water observed for the same molar concentrations of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order as stated above. Explain.(b)Calculate the temperature at which a solution containing 54g glucose in 250g water will freeze.(Kf for water = 1.86 K kg mole-1)

SAMPLE PAPER –III (Prepared by Gr-Copper)

BLUE PRINTS.

No.Unit VSA

(1 Mark)

SA I(2

Marks)

SA II(3.Marks)

LA(5

Marks)

Total

01 Solid State 4(2) 4(2)02 Solution 2(1) 3(1) 5(2)03 Electrochemistry 2(1) 3(1) 5(2)04 Chemical Kinetics 5(1) 5(1)05 Surface Chemistry 1(1) 3(1) 4(2)06 General Principles &

Processes of isolation of elements

3(1) 3(1)

07 P block elements 3(1) 5(1) 8(2)08 d- & f- block elements 2(1) 3(1) 5(2)09 Coordination

Compounds1(1) 2(1) 3(2)

10 Haloalkanes & Haloarenes

4(2) 4(2)

11 Alcohols, phenol & ethers

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

12 Aldehyde, ketones & carboxylic acid

1(1) 5(1) 6(2)

13 Organic compounds containing Nitrogen.

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

14 Biomolecules 1(1) 3(1) 4(2)15 Polymers 1(1) 2(1) 3(2)16 Chemistry in everyday

life1(1) 2(1) 3(2)

Total 8(8) 20(10) 27(9) 15(3) 70(30)

Model Question Set for class XIITime allowed: 3 Hours Maximum Marks: 70All questions are compulsory. Question nos. 1 to 8 are very short answer questions and carry 1 mark each. Question nos 9 to 18 are short answer questions and carry 2 marks each. Question nos 19 to 27 are short answer questions and carry 3 marks each. Question nos28 to 30 are long answer questions and carry 5 marks each. Use log table if necessary .

1. Why are powdered substances more effective adsorbents than their crystalline forms? 1.

2. Write the formulas for the coordination compound Tetraamineaquachloridocobalt(III) chloride 1.

3. Give IUPAC name of the compound 1.

4. Write the structures of products of the following reaction. 1.

5. Arrange the following in decreasing order of their basic strength in gas phase:

1.

6. Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain. 1.

7. What is monomer of PTFE? 1.8. With reference to which classification has the statement, “ranitidine is an

antacid” been given? 1.9. X-ray diffraction studies show that copper crystallizes in an fcc unit cell with cell

edge of 3.608×10-8 cm. In a separate experiment, copper is determined to have a density of 8.92 g/cm3, calculate the atomic mass of copper. 2. OrSilver 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. 2.

10. In terms of band theory, what is the difference (i) between a conductor and an insulator (ii) between a conductor and a semiconductor? 1+1=2

11.State Henry’s law and mention some important applications? 1+1=212.Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

213.What is meant by ‘disproportionation’ of an oxidation state? Give an example.

214.Write the IUPAC names of the following coordination compounds:

(i) [Pt(NH3)2 Cl(NO2 )] (ii) K3 [Cr(C2 O4 )3 ] 2

15. Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer. (i)CH3CH2CH2CH2Br & CH3CH2CH(Br)CH3 2.

(ii) 16.Draw the structures of major monohalo products in each of the following

reactions: 1/2x4=2

17.Write the names of monomers of the following polymers: 1+1=2

18. Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor. Why ?

19.Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2)at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g ofCHCl3 and 40 g of CH2Cl2 at 298 K and, (ii) mole fractions of each component in vapour phase. OrTwo elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol–1 Calculate atomic masses of A and B.

20. (i)Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.(ii) Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

21. (i)What is the difference between physisorption and chemisorption? (ii)Give two examples of heterogeneous catalysis. (iii)What is shape selective catalysis?

22. (i) What is the role of depressant in froth floatation process? 1(ii) Write down the reactions taking place in different zones in the blast furnace during the extraction of iron. 2

23.Write balanced equations for the following: 1x3= 3

(i) NaCl is heated with sulphuric acid in the presence of MnO2.(ii) Chlorine gas is passed into a solution of NaI in water.(iii) Ammonia reacts with a solution of Cu2+?

24.Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron (II) ions (ii) SO2? Write the ionic equations for the reactions. 1+2 = 3

25. (a) Write the equations involved in the following reactions: 2+1 =3(i) Reimer - Tiemann reaction (ii) Kolbe’s reaction

(b)Explain why propanol has higher boiling point than that of the hydrocarbon butane?

26.How will you convert

(i)Benzene into aniline (ii) Benzene into N, N dimethylaniline (iii) Cl–(CH2)4–Cl into hexan-1,6-diamine? 27. (i) What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?

(ii)When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?

(ii) What are monosaccharides?

28. (i) Determine the order of reaction represented by following time/ concentration graph. 1+2+2=5

Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t ) of the reaction.The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

Determine the rate law and the rate constant for the reaction.Or 2+2+1=5

(i) The decomposition of hydrocarbon follows the equation

Calculate E a.(ii) The following data were obtained during the first order thermal decomposition

of SO2Cl2 at a constant volume.

Calculate the rate constant of the reaction(iii) What is the order of I st order reaction29. Conc H2SO4 placed in the five test tubes as shown below

With the help of chemical equation explain in which test tube (i) The Black residue formed (ii) A pungent smelling gas produced which change the acidified potassium

dichromate solution green.(iii) Yellow powder disappears to produce pungent smelling gas which change

the acidified potassium dichromate solution green.(iv) Orange red gas produced.(v) Black powder disappears.

ORGive reason for the following :-(i) Bond angle in PH4+ is higher than that in PH3.(ii) PCl3 fume in moisture.(iii) O3 act as a powerful oxidising agent.(iv) In vapour state sulphur shows paramagnetic behavior.(v) Halogens have maximum negative electron gain enthalpy in the

respective periods of the periodic table. 30. (a) An organic compound with the molecular formula C9H10O forms 2,4-DNP

derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.(b) How will you bring about the following conversions in not more than two steps?(i) Propanone to Propene (ii) Benzoic acid to Benzaldehyde Or

(a) An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene.

Write equations for the reactions involved. (b) Arrange the following compounds in increasing order of their property as

indicated: (i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength)

ANSWER KEY

1. Due to large surface area of powdered solid. 1 mark

2. [Co(NH3)4 H2O Cl]Cl2 1 mark

3. 2,6-Dimethyl phenol. 1 mark

4.C6H5COC2H5 1 mark

5. Aniline < Ammonia < Ethanamine < Di ethyl amine. 1 mark

6. Sucrose or Glucose contain many polar O-H groups hence these are soluble in water. 1 mark

7. Tetra floro ethane 1 mark

8. On the basis of target molecule. 1 mark

9.Edge length =3.608×10-8 cm, density = 8.92 g/cm3, Z= 4 (FCC) Mol. Mass =? d = Z M/ a3NA 2 mark

M= d a3 NA/ Z = 8.92 g/cm3 x (3.608x10 -8 cm) 3 x 6.02x10 23 mol -1 4 = 62.9 gmol-1

Or 2mark Edge length = 4.07 × 10–8 cm density = 10.5 g cm–3 Z=4 Mol. Mass =? d = Z M/ a3NA M= d a3 NA/ Z =[10.5 g cm–3 x (4.07 × 10–8 cm)3 x 6.02x1023mol-1]/4 = 106.98 g mol-1

10. (i) In Conductor no energy gap present whereas in insulator a large energy gap present between valance band and conduction band 1mark (ii) In Conductor no energy gap present whereas in semiconductor a small energy

gap present between valance band and conduction band. 1 mark11. Henry’s law:- Solubility (mole fraction) of a gas in a liquid is directly

proportional to its (gas) partial pressure. 1 mark

P = KH χ Applications of Henry’s law:- (a) To determine solubility of a gas (b) In cold drinks

industry 1 mark12. rusting of iron :- 1 mark Fe - Fe 2+ + 2e- , (Anode reaction) 4H+ + O2 + 4e- 2 H2O (Cathode reaction) 1 mark Fe 2+ + H2O + O2 Fe2O3 . x H2O13. Disproportionation’ of an oxidation state:- If oxidation state of an element

in a compound disproportionate ( increase and decrease) In the reaction to form two different products then it is called Disproportionation’ of an oxidation state. 1 mark Cl2 + OH- (Dil & Cold) Cl- + ClO- + H2O 1 mark14. (i) [Pt(NH3)2 Cl(NO2 )] Diaminechloridonitro(o)platinum(II) 1 mark (ii) K3 [Cr(C2 O4 )3 ] Potassium trioxalatochromate(III) 1 mark15 i) Out of CH3CH2CH2CH2Br & CH3CH2CH(Br)CH3 , CH3CH2CH2CH2Br is more reactive

since it is primary halide. 1 mark

(ii) Out of Cyclohexile methyl chloride is more reactive since it is primary halide 1 mark

17 Each reaction ½ mark18. (I) Hexamethylene di amine and adipic acid (ii) Caprolactum. 1+1=2marks over dose of sleeping piles may harm the patient so it must be taken under proper guidance of a doctor. 1 mark19. P0 (CHCl3) = 200 mm Hg P0 (CH2Cl2) = 415 mm Hg 1 markMass of(CHCl3) = 25.5g Mass of(CH2Cl2) = 40 gn(CHCl3) = 25.5/119.5 = 0.213mole n(CH2Cl2) =40/85 =0.47mole X(CHCl3) = n(CHCl3) / n(CHCl3) + n(CH2Cl2) =0.313X(CH2Cl2) =0.687.P = P0 (CHCl3) X(CHCl3) + P0 (CH2Cl2) X(CH2Cl2) = 347.9 1 markp = 0.688 × 415 mm Hg = 285.5 mm Hg(CHCl3)p = 0.312 × 200 mm Hg = 62.4 mm Hg(CH2Cl2) y = 285.5 mm Hg/347.9 mm Hg = 0.82(CHCl3) y = 62.4 mm Hg/347.9 mm Hg = 0.18. 1 markOrAns – A =25.58u B= 42.64u 3 mark

20.(i) 1.5 mark

1.5 mark

(ii) 21. (i) Wrtite difference mark 1 (ii) Four equations mark 1 (iii) Write definition mark 1

22 (i) Role of depressant 1 mark(ii) Reactions in blast furnace 2 marks23. Each reactions 1 mark. 24. Preparation 1 mark, reactions 2 marks.25. Reaction 2marks , reasoning 1 mark26. Three conversions 1 mark each.27. Hydrolysis product 1 mark, each Definitions 1 mark28. Zero order reaction 1 mark, Solution of numerical 2 marks each.

or(i) Numerical – Formula writing ½ mark, comparison 1/2 mark, Calculation 1mark. (ii) Numerical – Formula writing ½ mark, comparison 1/2 mark, Calculation 1mark. (iii) Definition 1mark

29. Five reactions ofone mark each, or

Reasoning questions 1 mark each30. (a) Identification of compounds 1 mark , Reaction 2 marks

(b) Each conversion 1 Or

(a) Identification of compound 2 marks, Reactions involved 2 marks.

(b) Arrangement of compound 1 mark

SAMPLE PAPER – IV (Prepared By Group-Silver)

BLUE PRINTCLASS XII SUBJECT: CHEMISTRY MAX. MARKS : 70

TIME : 3 HOURS

UNIT NO.

NAME OF THE UNIT

KNOWLEDGE UNDERSTANDING APPLICATION AND SKIL TOTAL

VSA1

SA 2

SA 3

LA 5

VSA1

SA 2

SA 3

LA 5

VSA1

SA 2

SA 3

LA

5

MARKS

NO. OF

QUESTIONS

1 Solid State 2(1) 2(1) 4 22 Solution 2(1) 3(1) 5 2

3Electrochemistry 2(1) 3(1) 5 2

4Chemical Kinetics 5(1) 5 1

5Surface Chemistry 1(1) 3(1) 4 2

6

General Principles & Processes ofIsolation of Elements

3(1) 3 1

7The p-Block Elements 3(1) 5(1) 8 2

8

The d- and f-Block Elements

2(1) 3(1) 5 2

9Coordination Compounds 1(1) 2(1) 3 2

10

Haloalkanes and Haloarenes

2(1) 2(1) 4 2

11

Alcohols, Phenols and Ethers

1(1) 3(1) 4 2

12

Aldehydes, Ketones and Carboxylic Acids

1(1) 5(1) 6 2

13

Organic Compounds Containing Nitrogen

3(1) 1(1) 4 2

14 Biomolecules 1(1) 3(1) 4 215 Polymer 1(1) 2(1) 3 2

16 Chemistry in Everyday Life

2(1) 1(1) 3 2

SUB TOTAL6(6) 6(3) 9(3) 2(2

) 6(3) 12(4) 15(3) 8(4) 6(2)

70 30GRAND TOTAL 21(12) 35(12) 14(6)

NOTE : - 1. Number of questions are written within the bracket and their marks out side the bracket. 2. Internal choice maybe given for all LA (5 marks) , one SA (3 marks) and one SA (2 marks). 3. Numercial problems may be given for 8 to 10 marks.

WEIGHTAGE TO TYPE OF QUESTIONS

VSA TYPE QUESTIONS OF IMARK :08 SA TYPE QUESTIONS OF 2 MARKS : 10 SA TYPE QUESTIONS OF 3 MARKS : 09 LA TYPE QUESTIONS OF 5 MARKS : 03

WEIGHTAGE TO LEARNING OUTCOMES

KNOWLEDGE : 21 MARKS (30 %) UNDERSTANDING : 35 MARKS (50 %) APPLICATION : 10 MARKS (14 %) SKILL : 04 MARKS (06 %)

WEIGHTAGE TO

DIFFICULTY LEVEL

EASY : 15 %

AVERAGE : 70 %

DIFFICULT : 15 %

SAMPLE PAPER-IVCLASS : XII SUB: CHEMISTRY

TIME: 3 HOURS MAX. MARKS: 70General Instructions:1. All questions are compulsory.2. Question numbers 1 to 8 are very short answer questions of one mark each. Answer these in one word or about one sentence each.3. Question numbers 9 to 18 are short answer question of two marks each. Answer these in about 30 words each.4. Question numbers 19 to 27 are short answer question of three marks each. Answer these in about 40 words each.5. Question numbers 28 to 30 are long answer question of five marks each. Answer these in about 70 words each.6. Use log tables, if necessary.

1. Name the catalyst and the promoter used in Haber’s process for manufacture of ammonia. 1

2. Give the IUPAC name of : [Co(NH3)Cl(en)2] Cl2 1

3. Write the structure of : 1

4. Give the IUPAC name of the organic compound 1 (CH3)2 C═ CH C CH3

│ OH

5. Name the type of bonding which stabilizes -helix structure in proteins. 1

6. What do you mean by homopolymers? 1

7. How will you distinguish between primary aliphatic amine and secondary aliphatic amine? 1

8. Give one example of an artificial sweetener used by the diabetic patients. 1

9. (a)Name the type of point defect that occurs in a crystal of zinc sulphide. 1x2=2 (b)How many octahedral voids are there in l mole of a compound having cubic close packed structure?

10. An element X with an atomic mass of 60g/mol has density of 6.23g cm -3. If the edge length of its cubic unit cell is 400 pm, identify the type of cubic unit cell. 2

11. Give reasons for the following:(i)At higher altitudes, people suffer from a disease called anoxia. In this disease, they become weak and cannot think clearly.(ii) Write the unit of molal elevation constant 2x1=2

12. Write the chemical equations for all the steps involved in the electrochemical theory of rusting of iron. Give any one method to prevent rusting. 2

13. Explain the following facts(a) Chromium group elements have the highest melting points in their respective series.(b) Transition metals form coloured complexes. 2x1=2

14. (a)Give the electronic configuration of the d- orbitals of Ti in [Ti (H2O)6] 3+ ion in an octahedral crystal field. (b)Why is this complex coloured? Explain on the basis of distribution of electrons in the d- orbitals.

2x1=2OR

(b) Name the isomerism exhibited by the following pair of coordination compounds:[Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br

(c) Give one chemical test to distinguish between these two compounds. 2x1=2

15. (a)Vinyl chloride is less reactive than ethyl chloride towards nucleophilic substitution reaction. Explain why?(b)Haloalkanes react with KCN to give alkyl cyanide as main product while with AgCN they form isocyanides as main product. Give reason. 2x1=2

16. Write the formula of main product formed in the following chemical reactions. 4x ½=2 Na

(i) (CH3)2 CH-C1 ──────→ Dry ether ∆

(ii) CH3Br + AgF──────→

Dry acetone(iii) CH3CH2Br + Nal ───────→

Cu /HCl(iv) C6H5N2Cl ───────→

17. Write name of monomer of the following polymers and classify them as addition or condensation polymers.

(a) Teflon (b) Natural Rubber 2x1=2

18. Define the following :- (a) Antiseptic (b) Disinfectant with one example of each. 2x1=2

19. Ethylene glycol (molar mass = 62 g mol¯ 1) is a common automobile antifreeze. Calculate the freezing point of a solution containing 12.4g of this substance in 100 g of water. Given : Kf for water = 1.86K kg/mol. 3

20. Calculate the equilibrium constant for the reaction 3

ORCalculate the standard free energy change for the following reaction at 250C

21. (a) Adsorption of a gas on the surface of solid is generally accompanied by a decrease in entropy. Still it is a spontaneous process. Explain.(b) How does an increase in temperature affect both physical as well as chemical adsorptions? 1+2=3

22. (a) Name the method used for refining of (i) Nickel (ii) Zirconium. 2 x ½ +2=3(b)Write the chemical equation of the reactions involve in the extraction of Au by leaching with NaCN.

23. Account for the following: 3x1=3(a) N2 is inert at room temperature. (b) catenation tendency of nitrogen lesser than that of phosphorus(c) PCl3 fume in atmosphere

24. Give reasons for the following: 3x1=3(a) Among the lanthanoids, Ce(III) is easily oxidised to Ce(IV). (b) Fe3+ / Fe2+ redox couple has less positive electrode potential than Mn3+ / Mn2+couple. (c) The second and third transition series elements have almost similar atomic radii.

25.(a) Write the formation of diethyl ether from ethanol in the presence of concentrated sulphuric acid. (b) How do you account for the miscibility of ethoxyethane with water? 2+1=3

26. (i)Complete and name the following reactions:

(ii) How will you distinguish between: C6H5NH2 and CH3NH2 ? 2+1=3

27. (a)Draw the Haworth structure of -D-(+)-Glucopyranose (b) Name the two components of starch. How do they differ from each other structurally? 1+2=3

28. (i)

Answer the following questions on the basis of the above curve for a first order reaction A → P:- (a) What is the relation between slope of this line and rate constant? (1) (b) Calculate the rate constant of the above reaction if the slope is 2 x 10-4 s-1

(ii)The decomposition reaction of ammonia gas on platinum surface has a rate constant k = 2.5 x 10–4

mol L–1 s–1.What is the order of the reaction? (iii) For an elementary reaction 2A + B → 3C, the rate of appearance of C at time‘t’ is 1.3 x 10-4 mol L-1 s-1.Calculate at this time

(a) Rate of the reaction.(b) Rate of disappearance of A. 2x1+1+2x1=5

OR

For a certain chemical reaction A + 2B → 2C + D,the experimentally obtained information is tabulated below.

For this reaction(i) Derive the order of reaction w.r.t. both the reactants A and B.(ii) Write the rate law.(iii) Calculate the value of rate constant k(iv) Write the expression for the rate of reaction in terms of A . 2+1+1+1=5

29. (i) Give reasons for the following(a) Fluorine exhibits only – 1 oxidation state whereas other halogens exhibit higher positive

oxidation states also. (b) CN¯ ion is known but CP¯ ion is not known.(c) NO2 dimerises to form N2O4

(d) ICl is more reactive than I2

(ii) Predict the shape of CIF3 on the basis of VSEPR theory. 4x1+1=5

OR(a)What is the covalence of nitrogen in N2O5?(b) Explain why both N and Bi do not form pentahalides while phosphorus does.(c)Why does chlorine water lose its yellow colour on standing?(d) What happens when Cl2 reacts with cold dilute solution of sodium hydroxide? Write equation only. (e) Write down the equations for hydrolysis of XeF4. 5x1=5

Experiment [A]o in mol/L [B]o in mol/L Initial rate of the reactionin moll-1s-1

1234

0.300.600.300.60

0.300.300.600.60

0.0960.3840.1920.768

30. (a) An organic compound ‘A’ with molecular formula C5H8O2 is reduced to n-pentane on treatment

with Zn-Hg/HCl. ‘A’ forms a dioxime with hydroxylamine and gives a positive iodoform test and Tollen’s test. Identify the compound ‘A’ and deduce its structure.

(b) Write the chemical equations for the following conversion: (i) Ethyl benzene to benzoic acid (ii) Acetaldehyde to 3-hydroxy butanal

(iii) Acetone to propan-2-ol 2+3=5

OR

(a)An organic compound (A) having molecular formula C8H8O gives positive DNP and iodoform test. It does not reduce Tollen’s or Fehling’s reagent and does not decolorize bromine water also. On oxidation with chromic acid, it gives a carboxylic acid (B) with molecular formula C7H6O2. Deduce the structures of ‘A’ and ‘B’.

(b) Complete the following reactions by identifying A, B and C. Pd/BaSO4

(ii) A + H2(g) ────────→ (CH3)2CH-CHO

(i) 2HCHO + conc. NaOH ────────→ B + C 3+2=5************************************************************************************

Marking Scheme1. Promotor-Mo/Al2O3, Catalyst-Fe ½+ ½ = 1

2. amminechloridobis-(ethane-1,2-diamine)cobalt(III) chloride 1

3. 1

4. 4-methylpent-3-en-2-ol 15. intra molecular hydrogen bonding 16. polymers made up of only one kind of monomers 17. isocyanide test (carbyl amine reaction) or Hinsberg’s test 18. Sucrolose 19. (a) Frenkel defect………….1

(b) 1 mole voids …………...1

10. Given ; d= 6.23g cm-3

M=60g/mol a = 400 pm = 4 x 10-8 cm z = ?.................................................................................1/2

we know that d= Mz/a3 NA

z= d. a3. NA/M…………………………………1/2 = 6.23 x 64 x 10-24 x 6.023 x 1023/ 60

= 4.0018…………………………………….1/2 Z = 4 ( fcc structure)…………………….1/2

11. (a) decrease in partial pressure of dioxygen………………………………………………..1 (b) K Kg mol-1 1

12. Anode : 2Fe→ 2Fe2+ + 4 e- 1/2

Cathode O2 + 4 H+ + 4e- → 2H2O 1/2 The overall reaction is 2Fe + O2 + 4H+ → 2Fe2+ + 2H2O 2Fe2+(aq.) + H2O (l) + ½ O2 → Fe2O3 + 4H+ 1/2

Fe2O3 then combine with requisite amount of water to form Fe2O3.xH2O that is rust. Method of electroplating may be used to prevent rusting. 1/213. (a) presence of half filled (n-1)d subshell 1 (b) d-d transition of (n-1)d electrons 1

14. (a)[Ar]183d1 1

(b)due to d-d (draw splitting of d orbitals) 1

OR

(a) ionisatios isomerism 1 (b) BaCl2 test and AgNO3 test 1

15. (a) resonance stabilization of vinyl chloride makes it less reactive than ethyl chloride towards nucleophilic substitution reaction. 1

(b)because ionic bond is present in KCN whereas the bonding in AgCN is covalent. 1

16. Write the formula of main product formed in the following chemical reactions. Na

(a) (CH3)2 CH-C1 ──────→ (CH3)2 CH-CH(CH3)2 1/2

Dry ether ∆

(b) CH3Br + AgF──────→CH3F 1/2

Dry acetone(c) CH3CH2Br + Nal ───────→ CH3CH2I 1/2

Cu /HCl(d) C6H5N2Cl ───────→C6H5Cl ½

17. (a) Teflon- tetra fluoro ethane (addition polymer) ½ + ½ =1 (b) Natural Rubber – isoprene (add. Polymer) ½ + ½ =1

18. (a) Antiseptic – chemical compound which are used to prevent or kill the harmful micro organisms and are safe to apply on living tissue example dettol ½ + ½ =1 (b) Disinfectant- chemical compound which are used to prevent or kill the harmful micro organisms but are not safe to apply on living tissue. They are used in drains, toilets, wash basins etc. example-phenol 2 % solution ½ + ½ =119. Given;

M2= 62 g mol-1 W2=12.4g W1= 100 g Kf for water = 1.86K Kg/mol. 1We know that ∆Tf= Kf.W2.1000/M2.W1 1 = 1.86K Kg/mol x 12.4g x 1000/62 g mol-1 x 100g 1/2 = 3.72 K 1/2

20. equilibrium constant for the reaction is E0 cell = E0 r – E0 l 1/2

= -0.403-(-0.763) V = 0.36 V 1

E0 cell = (0.0591/n) log Kc 1/2

log Kc = 0.36 V x 2 / 0.0591 1/2=12.18 Kc = antilog of 12.18 1/2 OR

Calculate the standard free energy change for the following reaction at 250CE0 cell = E0 r – E0 l 1/2 = - 2.87 – 1.5 = - 4.37 V 1/2

We know that∆G0 = - nF E0 cell 1

= -2 x 96500 C x (-4.37V) 1/2

=8.43 x 105 J 1/2

21. (a) Adsorption of a gas on the surface of solid is a exothermic process 1

(b) Physical adsorption decreases with rise in temperature whereas chemisorption initially increases and then decreases with rise in temperature 1 + 1

OR

(a) (i) Nickel- Mond’s process (ii) Zirconium- Van – Arkel method ½ + ½ =1

(b) (i) 4Au(s) + 8CN–(aq)+ 2H2O(aq) + O2(g) →4[Au(CN)2]– (aq) +4OH–(aq) 1

(ii) 2[Au(CN)2]– (aq) + Zn → [Zn(CN)4]2- + 2Au 1

23. (a) because of high bond enthalpy due tothe presence of triple bond in N2. 1 (b) because of weaker N-N bonds whereas P-P bonds are much stronger with higher bond energy. 1(c) PCl5 PCl3 + Cl2 1

24. (a) vacant (n-2) f subshell in Ce(IV). 1 (b) extra stability of Fe3+ than Mn3+ ion 1 (c) Due to lanthanoid contraction 1

25. (a) C2H5OH -------------→ C2H5OC2H5 2 (b) intermolecular hydrogen bonding makes ethoxyethane miscible with water 1

26. (i) Carbyl ammine reaction RNH2 + CHCl3 + 3 KOH → RNC + 3KCl + 3H2O ½ + ½ =1(ii) Hoffman bromamide reaction

RCONH2 + Br2 + 4NaOH → RNH2 + 2NaBr + Na2CO3 + 2H2O ½ + ½ =1 (iii) Azodye test 1

27. (a) Haworth structure of -D-(+)-Glucopyranose 1

(b) amylase and amylopectine.(c) Amylase is a linear polymer whereas amylopectine is a branched polymer 1+1

28. (i) (a) slope = k/2.303 1 (b) k = slope x 2.303 = 2 x 10-4 s-1 x 2.303

= 4.6 x 10-4 s-1 1

(ii) zero order reaction 1

413 K/ conc. H2SO4

(iii) (a) rate of reaction = 3 x is 1.3 x 10-4 mol L-1 s-1

= 3.9 x 10-4 mol L-1 s-1 1

(b) Rate of disappearance of A = 3/2 x 1.3 x 10-4 mol L-1 s-1

= 1.95 x 10-4 mol L-1 s-1 1

(i) order of reaction w.r.t. A=2 and B = 1 1+1(ii) rate = k [A]2[B] 1.(iii) rate constant k = rate/[A]2[B] 1

(iv) rate of disappearance of A = -∆[A]/ ∆t 1

29. (i) a. Fluorine is the most electronegative element 1b. Due to the large size of P atom it can not form multiple bond with carbon. 1c. NO2 contains odd number of valence electrons. It behaves as a typical odd molecule. On dimerisation, it is converted to stable N2O4 molecule 1d. I-Cl bond is weaker than I-I bond 1

(ii) bent ‘T’ shape 1

OR(a) covalence is 5 1(b) N can not expand its covalence to five due to absence of d-subshell whereas Bi do not form pentahalides due to inert pair effect. 1(c) Chlorine water on standing loses its yellow colour due to the formation of HCl and HOCl. 1(d) 2NaOH + Cl2 NaCl + NaOCl + H2O 1(e) 6XeF4 + 12 H2O 4Xe + 2Xe03 + 24 HF + 3 O2 1

30. (a) A is a straight chain organic compound with carbonyl functional group. Formation of dioxime suggests the presence of two carbonyl groups. ½ It is positive towards iodoform test which indicates presence of CH3CO- group 1/2. Positive tollens reagent test indicates the presence of –CHO group. 1/2 Therefore the organic compound is 4-oxo pentanal 1/2

(b) (i) n-propyl benzene to benzoic acid 1

(ii) Acetaldehyde to 3-hydroxy butanal 1

(ii) Acetone to propan-2-ol 1

CH3COCH3 + H2 ────────→ CH3CH(OH)CH3

OR

(a)Compound C8H8O gives positive DNP and iodoform test indicates the presence of CH3CO- group. 1 It does not reduce Tollen’s or Fehling’s reagent which indicates the presence of ketonic functional group. 1 On oxidation with chromic acid, it gives a carboxylic acid (B) with molecular formula C7H6O2. Therefore compound A is C6H5COCH3 and compound B is C6H5 COOH. 1

(c) Complete the following reactions by identifying A, B and C. Pd/BaSO4

(iii) (CH3)2CH-COCl + H2(g) ────────→ (CH3)2CH-CHO 1

(iv) 2HCHO + conc. NaOH ────────→ HCOONa + CH3OH ½ + ½

************************************************************************************

Ni/Pt

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