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Rao IIT Academy/ XII HSC - Board Exam Chemistry Set [E] / Paper Solutions

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XII HSC - BOARD - FEBRUARY - 2018Date: 28.02.2018 CHEMISTRY (255) - SOLUTIONS

SECTION - IQ. 1 Select and write the most appropriate answer from the given alternatives for each sub-question :

(i) The process in which the value of 0U is __

(a) Adiabatic (b) Isothermal (c) Isobaric (d) IsochoricAns (b) Isothermal

ConceptualTopic: Thermodynamics; Sub-topic:Types of process_ L- Easy_XII-HSC Board Test__Chemistry

(ii) An ionic crystal lattice has rr

radius ratio of 0.320, its co-ordination number is __

(a) 3 (b) 4 (c) 6 (d) 8Ans (b) 4

ConceptualTopic: Solid state; Sub-topic:Radius ratio rule_ L-Easy_ XII-HSC Board Test__Chemistry

(iii) In hydrogen-oxygen fuel cell the carbon rods are immersed in hot aqueous solution of __

(a) KCl (b) KOH (c) 2 4H SO (d) 4NH Cl

Ans (b) KOHConceptual

Topic: Electrochemistry; Sub-topic:Fuel cell_ L-Easy_ XII-HSC Board Test__Chemistry

(iv) The chemical formula of willemite is __

(a) ZnS (b) 3ZnCO (c) ZnO (d) 2 4Zn SiO

Ans (d) 2 4Zn SiO

ConceptualTopic: General principle and process of isolation; Sub-topic:Ores and minerals_ L- Easy_XII-HSC Board

Test__Chemistry


2 2

(v) The oxidation state of nitrogen in dinitrogcn trioxide is(a) + 1 (b) + 2 (c) + 3 (d) + 4

Ans (c)

2 3N O Oxidation state for 2O

Oxidation state of N x

2 3( 2) 0x

2 6 0x

62 6 32

x x

Topic: p-block element; Sub-topic: group-15_ L-Easy_ XII-HSC Board Test__Chemistry

(vi) Which of the following 0.1 M aqueous solutions will exert highest osmotic pressure?

(a) 2 4 3( )Al SO (b) 2 4Na SO (c) 2MgCl (d) KCl

Ans (a) 2 4 3( )Al SO

3 22 4 42 3 (5particles)Al SO Al SO

22 4 42 (3particles)Na SO Na SO

22 2 (3particles)MgCl Mg Cl

(2 particles)KCl K Cl

Topic: Solutions and colligative properties; Sub-topic:Osmotic pressure_L-Easy_XII-HSC BoardTest__Chemistry

(vii) The half-life period of zero order reaction A product is given by

(a) 0[ ]Ak (b)

0.693k (c) 0[ ]

2Ak (d) 02[ ]A

k

Ans (c) 0[ ]2Ak

Topic: Chemical Kinetics; Sub-topic:Zero order_L-Easy_XII-HSC Board Test__Chemistry

Q.2 Answer any SIX of the following :(i) Derive the relation between elevation of boiling point and molar mass of solute.

Ans bT m

b bT K m

where bK is called boiling point elevation constant

Then the molality (m) of the solution is given by


3 3

2

1 2

1000wmw M

2

1 2

1000b b

wT Kw M

where,w1 = Mass of a solvent in gram; w2 = Mass of a solute in gram; M2 = Molecular mass of the solute

Topic: Solutions and colligative properties; Sub-topic:Elevation in boiling point_ L- Medium_ XII-HSC Board Test__Chemistry

(ii) State third law of thermodynamics. Give ‘two’ uses.Ans The entropy of perfectly ordered crystalline substance is zero at absolute zero of temperature.

Usefulness :(a) It helps in calculating thermodynamic properties.(b) It is helpful in measuring chemical affinity.(c) It is used to determine the absolute entropy of any substance either in solid, liquid or gaseous state at

any desired temperature.Topic:Thermodynamics; Sub-Topic:Entropy_L-1_XII-HSC Board Test__Chemistry

(iii) Draw a neat and labelled diagram of lead storage battery.

Ans

Topic: Electrochemistry; Sub-topic:Galvanic cell_ L-Easy_XII-HSC Board Test__Chemistry

(iv) Ionic solids are hard and brittle. Explain.Ans Ionic solids are formed by the 3-D arrangement of cations and anions bounded by strong coulombic

(electrostatic) forces.Ionic solids are formed by the forces of attraction between ions of opposite changes and forces repulsionbetween ions of same changes making them hard and brittle.

Topic: Solid state; Sub-topic:Types of solids_ L-Medium_XII-HSC Board Test__Chemistry


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(v) A certain reaction occurs in the following steps -

(i) ( ) 3( ) ( ) 2( )g g g gCl O ClO O

(ii) ( ) ( ) ( ) 2( )g g g gClO O Cl O

(a) What is the molecularity of each of the elementary steps?(b) Identify the reaction intermediate and write the chemical equation for overall reaction.

Ans (a) (i) ( ) 3( ) ( ) 2( )g g g gCl O ClO O

Molacularity : Bimolecular

(ii) ( ) ( ) ( ) 2( )g g g gClO O Cl O

Molacularity : Bimolecular(b) Intermediate : ClO

( ) 3( ) ( ) 2( )g g g gCl O ClO O

( ) ( ) ( ) 2( )g g g gClO O Cl O

3( ) ( ) 2( )2g g gO O O Net reaction

Topic: Chemical kinetics; Sub-topic:Molecularity_L- Easy_ XII-HSC Board Test__Chemistry

(vi) Define : (a) Semipermeable membrane (b) Reference electrode

Ans (a) Semipermeable membrane: It is a membrane which allows the solvent molecules, but not the solutemolecules, to pass through it.

(b) Reference electrode : It is an electrode whose potential is arbitrarily taken as zero or is exactly known.Topic: Solution and colligative properties/Electrochemistry; Sub-topic:Osmotic pressure/Types of

electrode_ L- Easy_ XII-HSC Board Test__Chemistry

(vii) What is the action of chlorine on :

(a) 2CS

(b) Excess 3NH

Ans (a) 2 2 4 2 23CS Cl CCl S Cl

(b) 3 2 33 3NH Cl NCl HCl

Topic: p-block; Sub-topic:Reaction of chlorine_ L- Medium_ XII-HSC Board Test__Chemistry

(viii) Write the chemical equations involved in van Arkel method for refining zirconium metal.Ans This method is useful for removing impurities in the form of oxygen and nitrogen in metal like zirconium. The

crude metal is heated in an evacuated vessel with iodine. The metal iodine volatilizes.

2 4Impure Vapour

Zn I ZrI

The metal iodine is decomposed on tungsten filament, electrically heated to about 1800K. The pure metal isthus deposited on the filament.

4 2

Vapour2ZnI Zr I

Topic: General principle and process of isolation; Sub-topic:Vapour phase refining_ L- Medium_XII-HSC Board Test__Chemistry


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Q.3 Answer any THREE of the following :(i) Write balanced chemical equations for the following :

(a) Phosphorus reacts with magnesium.(b) Flowers of sulphur boiled with calcium hydroxide.(c) Action of ozone on hydrogen peroxide.

Ans Balanced equations:

(a) 4 2 26 2Mg P Mg P

(b) 2 5 2 3 23 ( ) 12 2 3Ca OH S CaS CaS O H O

(c) 2 2 3 2 22H O O H O O

Topic: p-block; Sub-topic:Chemical properties_L- Medium_XII-HSC Board Test__Chemistry

(ii) The density of iron crystal is 8.54 gram cm–3. If the edge length of unit cell is 2.8 A° and atomic mass is 56gram mol–1, find the number of atoms in the unit cell.

(Given : Avogadro's number = 23 86.022 10 ,1 1 10A cm )

Ans Given 82.8 2.8 10a Å cm Molar mass = 59 gmDensity = 8.54 g cm–3

Volume 3( )a

8 3(2.8 10 )

24 321.95 10 cm

Massof unit cellDensityof unit cellVolumeof unit cell

Mass of unit cell density volume

2429.98 10 8.54

24187.47 10

56 gm of Fe contains 236.022 10 atoms

23 2424 6.022 10 187.47 10187.47 10 contain

56gm Fe

120.15 10

2.01

2Number of atoms in unit cell is 2.

Topic: Solid state; Sub-topic: Crystal packing_ L- Medium_ XII-HSC Board Test__Chemistry


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(iii) How many faradays of electricity are required to produce 13 gram of aluminium from aluminium chloridesolution? (Given : Molar mass of Al = 27.0 gram mol–1)

Ans 33 3AlCl Al Cl

3 3Al e Al

1 mole of Al requires passage of 3 mole of electrons. Charge on 3 mole of e is 3 Faraday

mass ofMoles of producedmolar mass of

AlAlAl

1327

0.48moles

As 3F of electricity produces 1 mole of Al

No. of faradays of electricity required to produce 0.48 mole of Al0.48 3

1.44 FaradayTopic: Electrochemistry; Sub-topic: Faraday’s law_L- Medium_ XII-HSC Board Test__Chemistry

(iv) Calculate the internal energy at 298K for the formation of one mole of ammonia, if the enthalpy change atconstant pressure is – 42.0 kJ mol–1.(Given: R = 8.314 J K–1 mol–1)

Ans Formation of 1 mole of ammonia

2( ) 2( ) 3( )1 32 2g g gN H NH

(no.of moles of gaseous product) (no.of moles of gaseous reactant)n

1 312 2

1

H U P V

P V nRT

( 1) 8.314 298

2477 J

2.477 KJ

42.0H KJ

42.0 2.477KJ U

42 2.477U

44.47 KJTopic: Thermodynamics; Sub-topic:First law of thermodynamics_ L- Medium_XII-HSC Board

Test__Chemistry


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Q.4(i) Define : (a) Enthalpy of atomization

(b) Enthalpy of vaporization

Ans (a) Enthalpy of atomization ( )atoH : The enthalpy change accompanying the dissociation of all the moleculesin one mole of a gas-phase substance into gaseous atoms is called enthalpy of atomization.

(b) Enthalpy of vaporization ( )vap H : The enthalpy change that accompanies the vaporization of one moleof liquid without changing its temperature at constant pressure is called enthalpy of vaporization.

Topic: Thermodynamics; Sub-topic:Enthalpy of physical changes_ L- Easy_XII-HSC BoardTest__Chemistry

(ii) Draw the structure of IF7. Write its geometry and the type of hybridization.

Ans The only known interhalogen compound of the 7 7XX ' is IF . It is formed by sp3d3 hybridization of thecentral I atom its third excited state. The molecule has a pentagonal bipyramical structure as shown in

(Ground state)

5s 5p 5d

(Second excited state)

(Hybridised state)

5s 5p 5d

5d

spd hybrisation3 3

Structure of XX7(IF)7

I

F

F

F

F

F

FF

7

4

3

6

2

1

5

Topic: p-block; Sub-topic:Interhalogen compounds_ L- Medium_ XII-HSC Board Test__Chemistry

(iii) (a) State Henry’s law.

Ans According to Henry, the solubility of a gas in a liquid at constant temperature is proportional to the pressureof that gas above the solution.

Consider S is the solubility of the gas in mol dm–3 then according to Henry’s LawS P i.e. S= KPWhere P is the pressure of the gas in atm and K is the constant of proportionality and has the unit of moldm–3 atm–1.

Topic: Solution and colligative properties; Sub-topic: Solubility of gases_ L- Medium_ XII-HSC BoardTest__Chemistry

(b) 22.22 gram of urea was dissolved in 300 grams of water. Calculate the number of moles of urea and molality of the urea solution. (Given : Molar mass of urea = 60 gram mol–1)

Ans22.22Molesof urea

60

0.370 moles

Moles of ureaMolality ( ) 1000Massof water

m


8 8

0.370 1000

300

11.23moles kg

Topic: Solution and colligative properties; Sub-topic: Concentration_ L-Easy_XII-HSC BoardTest__Chemistry

OR(i) What is the action of carbon on the following metal oxides :

(a) 2 3Fe O in blast furnace

(b) ZnO in vertical retort furnace

Ans (a) Action of carbon on 2 3Fe O

2 3 ( )3 2 3 gFe O C Fe CO

(b) Action of Carbon on ZnO

( )1673 gkZnO C Zn CO

Topic: General principles and process of isolation; Sub-topic:Reduction by carbon_ L- Medium_ XII-HSC Board Test__Chemistry

(ii) Write the molecular and structural formulae of :(a) Thiosulphuric acid(b) Dithionous acid

Ans (a) Thiosulphuric acid 2 2 3( )H S O ||

— —||

S

HO S OH

O

(b) Dithionous acid 2 2 4( )H S O || ||— — —

O O

HO S S OHTopic: p-block; Sub-topic:Oxoacids of sulphur_ L- Medium_ XII-HSC Board Test__Chemistry

(iii) The reaction A + B products is first order in each of the reactants.(a) How does the rate of reaction change if the concentration of A is increased by factor 3?

Ans Reaction, ProductA B

Rate = [ ] [ ]K A B

(a) If conc. of A is increased by factor 3 then rate also increases 3 times.Topic: Chemical kinetics; Sub-topic:First order reaction_ L- Medium_ XII-HSC Board Test__Chemistry

(b) What is the change in the rate of reaction if the concentration of A is halved and concentration of B isdoubled?

Ans If conc. of A is halved and concentration of B is doubled then rate remains unchanged.Topic: Chemical kinetics; Sub-topic:First order reaction_ L- Medium_ XII-HSC Board Test__Chemistry


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SECTION - IIQ. 5 Select and write the most appropriate answer from the given alternatives for each sub-question :

[7](i) A polymer used in paints is

(a) Nomex (b) Thiokol (c) Saran (d) GlyptalAns (d)

Conceptual.Topic: Polymers; Sub-topic:Uses of polymers_ L-Medium __ XII-HSC Board Test__Chemistry(ii) The number of primary and secondary hydroxyl groups in ribose are -

(a) 1, 3 (b) 2, 3 (c) 3, 1 (d) 3, 2Ans (a)

O

H H

OH OH

H H

OHHOH C2

5

4 1

23

Topic: Biomolecules; Sub-topic:Carbohydrate_ L-Easy_ XII-HSC Board Test__Chemistry(iii) The ligand diethylene triamine is -

(a) monodentate (b) bidentate (c) tridentate (d) tetradentateAns (c)

Topic: Coordination compound_; Sub-topic:Ligands_ L-Easy_XII-HSC Board Test__Chemistry(iv) Propene on oxidation with diborane in presence of alkaline hydrogen peroxide gives

(a) propan-1- ol (b) propan-2-ol (c) allyl alcohol (d) propan-1, 2-diolAns (a)

3 2 2 6 3 2 2 36 2

Propene Tripropylborane

CH CH CH B H CH CH CH B

3 2 2 2 2 3 2 2 333 3

Propane - 1 - ol

OHCH CH CH B H O CH CH CH OH B OH

Topic: Alcohols, phenols and Ether; Sub-topic:Preparation of alcohol_ L-Easy_XII-HSC BoardTest__Chemistry


1010

(v) Baeyer’s reagent is -(a) acidified potassium dichromate (b) alkaline potassium dichromate(c) alkaline potassium permanganate (d) acidified potassium permanganate

Ans (c)Cold and dilute alkaline KMnO4

Topic: d & f block; Sub-topic:KMNO4 _L-Easy __ XII-HSC Board Test__Chemistry(vi) Identify ‘A’ in the following reaction -

Dryether2 2,2,5,5 Tetramethylhexane + 2A Na NaBr

(a) 2- Bromo-2 methylbutane (b) 1 -Bromo-2,2-dimethylpropane(c) 1 - Bromo - 3 -methylbutane (d) 1 - Bromo- 2 -methylpropane

Ans (b)

H C – C – CH Br + 2Na + Br3 2

CH3

CH3

H C – C – CH2 3

CH3

CH3

dryether

H C – C – CH3 2

CH3

CH3

H C – C – CH2 3

CH3

CH3

+ 2NaBr

Topic: Halogen derivative of alkanes; Sub-topic:Wurtz reaction_ L-Medium__ XII-HSC BoardTest__Chemistry

(vii) An antifertility drug is -(a) Novestrol (b) Histamine (c) Veranal (d) Equanil

Ans (a)Conceptual.

Topic:Biomolecules; Sub-topic:Drugs_L-Medium _XII-HSC Board Test__ChemistryQ.6 Answer any SIX of the following : [12](i) Write balanced chemical equations for the conversion of 2 2

4 2 7 to CrO Cr O in acidic medium and2 2

2 7 4 to Cr O CrO in basic medium.

Ans 2 2 7 2 4 22 2K Cr O KOH K CrO H O

2 4 2 2 7 22 2 2K CrO HCl K Cr O KCl H O

Topic: d & f block; Sub-topic:Properties of K2Cr2O7 _ L-Medium __ XII-HSC Board Test__Chemistry

(ii) Explain the geometry of 3

3 6Co NH

on the basis of hybridisation.

(Z of Co = 27)

Ans Octahedral complex, 3

3 6Co NH

, In this complex ion, oxidation state of cobalt is +3. It has electronic

configuration as 3d6. This complex involves the d2sp3 hybridisation.


1111

Orbitals of Co ion3+

Co undergoing d sp hybridisation3+ 2 3

Inner orbitalor low spin complex

33 6( )Co NH

The six pairs of electrons, one from each NH3 molecule, occupy the six hybrid orbitals. It proves thatcomplex has octahedral geometry. Absence of unpaired electron makes this complex diamagnetic in nature.

Topic: Coordination compound; Sub-topic:VBT_ L-Medium __ XII-HSC Board Test__Chemistry(iii) Why ethanol has higher boiling point than ethane ?Ans (a) Alcohol have higher boiling points than their corresponding alkanes due to the presence of intermolecular

hydrogen bonding which is absent in the alkanes.(b) Hydrogen bonding arises due to presence of electronegative atom oxygen in –OH group of alcohol.(c) Oxygen atom attracks electron density of O–H bond towards itself and hence it gets partial negative

charge, while H atom gets partial positive change.

R – O – H– +

(d) Hence in alcohol, R–OH molecules becomes polar.

CH3

CH2 OH

O CH2 CH3

H(e) There arises strong intermoleculer attraction between oxygen atom of one molecule of alcohol and ‘H’

atom of another alcohol molecule, giving rise to strong hydrogen bonding which is not present in alkanes.(f) Hence higher thermal energy is required to separate or evaporate alcohol molecule. Therefore alcohols

have higher boiling points than their corresponding alkanes.

Topic: Alcoholes, Phenols and Ethers; Sub-topic:Alcohol_ L-Medium _XII-HSC Board Test__Chemistry(iv) Write only reactions for the preparation of benzophenone from benzonitrile.Ans

Topic: Aldehyde, ketones and carboxylic acids; Sub-topic:Preparation_ L-Medium__ XII-HSC BoardTest__Chemistry


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(v) What is the action of p-toluenesulphonychloride on ethylamine and diethylamine ?Ans

H C3 S – Cl + H – N – C H2 5

O

O

H

S – N – C H2 5

O

O

H

CH3

N-ethyltoluenesulphonylamide

+ HCl

H C3 S – Cl + H – N – C H2 5

O

O

C H2 5

S – N – C H2 5

O

O

CH3 + HCl

C H2 5

N,N-diethyltoluenesulphonylamideIt help to distinguish between primary and secondary amines.

Topic: Compounds containing nitrogen; Sub-topic:AminesL-Medium __ XII-HSC BoardTest__Chemistry

(vi) What are amino acids ? Write the correct reaction for formation of peptide bond between amino acids.Ans Amino acids are bifunctional compounds containing a carboxylic acid group and an amino group on -

carbon.

22

H OH N CH C OH H N CH C OH

1R H 2R O

Amino acid Amino acid

2H N CH C NH CH COOH

1R2R

Peptide linkageProtein molecule (dipeptide)

O

O

Topic: Biomolecules; Sub-topic:ProteinsL- Easy_XII-HSC Board Test__Chemistry(vii) Define :

(1) Antiseptics(2) Antioxidants

Ans (1) Antiseptic : is a compound that kills bacteria or prvent the growth of micro-organism in living tissues.(2) Antioxidant: is a substance which when added to food, retards or prevents oxidative deterioration of

food. Fats and oils are oxidized easily, turn rancid and becomes unpalatable.Topic: Chemistry in everyday life; Sub-topic: Drugs_L-Easy_XII-HSC Board Test__Chemistry(viii) Explain only reaction mechanism for alkaline hydrolysis of tert-butylbromide.Ans — —

3 3 3 3(Nuclophile) (t-butylalcohol) (Bromide ion)(t-butyl bromide)(CH ) C Br : OH (CH ) C OH : Br

The rate of the above reaction depends upon the concentrations of the substrate, methyl bromide and thenucleophilic, OH ion.


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Mechanism of 1SN reaction: The alkaline hydrolysis of tertiary butyl bromide (a tertiary alkyl halide)proceeds through 1SN reaction mechanisum. This reaction is a two step process. The first step is a slow stepwhile the second one is a fast step. The hydrolysis reaction can be written as follows.Step-1 Formation of carbocation

H C – C – Br3

CH3

CH3

H C – C + Br3

CH3

CH3

+

Step-2 Attack of nucleophile on carbocation

CH3

CH3

+H C – C3OH H C – C – OH3

CH3

CH3

As carbocation has planar geometry therefore attack of nucleophile takes place from both the side to giveretention inversion in configuration therefore product formed is optically inactive racemic mixture.

Topic: Halogen derivative of alkanes; Sub-topic:Electrophilic substitution reactions_ L-Medium __ XII-HSC Board Test__Chemistry

Q.7 Answer any THREE of the following :(i) Complete and rewrite the balanced chemical equations -

(a) 473 ,PressureChlorobenzene ?NaCN CuCNK

(b) 50%KOHIsobutyraldehyde ?

(c) Butanone + 2, 4-dinitro-phenyl hydrazine ?H

Ans (a)

Cl

NaCN + CuCN473K, Pressure

CN

+ Cl—

Chlorobenzene Cyanobenzene

Topic: Halogen Derivative of alkanes; Sub-topic:Haloarenes_ L-Medium __ XII-HSC BoardTest__Chemistry(b)

CH – CH – CHO + CH – CH – CHO3 3

CH3 CH3

50% KOH CH – CH – CH – OH + CH – CH – COOK3 2 3

CH3 CH3

Isobutyraldehyde 2-methylpropanol

Topic:Aldehydes, Ketones and Carboxylic acids; Sub-topic: Cannizarro reaction_ L-Medium_ XII-HSCBoard Test__Chemistry


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(c) CH – CH – C – CH3 2 3

O

+ H N – NH2 NO2

NO2

H+

CH – CH – C = N – NH3 2 NO2+ H O2

NO2

Butanone 2,4-dinitrophenyl-hydrazine

2,4-dinitrophenyl-hydrazone

Topic:Aldehydes, Ketones and Carboxylic acids; Sub-topic:Chemical reaction_L-Medium_ XII-HSC BoardTest__Chemistry

(ii) Prepare carbolic acid from benzene sulphonic acid.Write a chemical equation for the action of neutral ferric chloride on phenol.

Ans (i) From Benzene Sulphonic Acid: Carbolic acid (phenol) is obtained from Benzene sulphonicacid in the following 3 steps.

Step- I : NeutralizationBenzene sulphonic acid is neutralized by the addition of dilute NaOH when sodium benzene sulphonate isobtained.

Benzene sulphonic acid

SO H3

+ NaOH

dil. Sodiumbenzene sulphonate

SO H3

+ H O2

Step- II: Fusion at 623 KDry sodium benzene sulphonate is fused with excess of solid sodium hydroxide at about 623 K when sodiumphenate (i.e. sodium phenoxide) is obtained.

Dry sodiumbenzene sulphonate

SONa Na – OH3

+ NaO H

excesssolid

Sodiumphenoxide or

sodium phenate

ONa

+ NaSO+ HO2 3 2

Fusion(623K)

Sodium sulphite

Step III : Decomposition by dil. H2SO4:Sodium phenate (Sodium phenoxide) on boiling with dil. H2SO4 is decomposed to form carbolic acid (orphenol)

SodiumPhenate

ONa

+ H SO2 4

dil. Carbolic acidOR

Phenol

+ Na SO2 42

OH

2

Topic:Alcohols, Phenols and Ethers; Sub-topic:Preparation of phenol L-Easy_XII-HSC BoardTest__Chemistry


1515

Chemical equation for the action of neutral ferric chloride on phenol :

3C H OH + FeCl6 5 3

Phenol

(C H O) Fe + 3HCl6 5 3

Ferric phenoxide (Violet)

Topic:Alcohols, Phenols and Ethers; Sub-topic:Distinguishing reaction between alcohol and phenol_L-_Medium_ XII-HSC Board Test__Chemistry

(iii) Explain the preparation and uses of nylon-2-nylon-6.Ans Nylon - 2 - nylon - 6

It is a copolymer and contains polyamide linkages. It is obtained by condensation polymerisation of themonomers, glycine and amino caproic acid.

22 2 2 2 5 2 2 5

H OnH N CH COOH NH CH COOH C CH NH C CH NH

O Oglycine -aminocaproicacidNylon 2nylon 6

Biodegradable polymers are used as orthopaedic devices, implants, sutures and rug release matrices.They are degraded by bacteria in the environment.

Topic: Polymers; Sub-topic:Prepartion of polymers_ L-Medium_ XII-HSC Board Test__Chemistry(iv) How glucose is prepared from cane sugar ?

Write the formula of the complex copper (II) hexacyanoferrate (II).Ans (i) Glucose is prepared in the laboratory by hydrolysis of sucrose with alcoholic conc. HCl solution.

Sucrose is heated at about 323K or water bath for two hours, with alcoholic conc. HCl solution, afterhydrolysis it gives glucose and fructose.

Alcoholic conc. HCl12 22 11 2 6 12 6 6 12 6323 KC H O H O C H O C H O

Sucrose Glucose Fructose

(ii) On cooling the reaction mixture, glucose separate out because it is less soluble in alcohol and fructoseremains in the solution, Crystals of glucose are then filtered and purified by recrystallization

Formula for complex copper (II) hexacyanoferrate (II) is 2 6Cu Fe CN .

Topic: Biomolecules; Sub-topic:Preparation of glucose_ L-Medium __ XII-HSC Board Test__Chemistry

Q.8 [7]What is lanthanide contraction ?

Ans. (i) Definition of Lanthanoid Contraction: The Lanthanoid contraction may be defined as gradual(or small) decrease in atomic and ionic radii of lanthanoids with the increase in atomic number.

(ii) In Lanthanides, after Lanthanum 57 La , the electrons are added to anti-penultimate shell i.e. 4f orbital.

(iii) There are 14 Lanthanoides from 58Ce to 57 .Lu(iv) For each electron, one proton is also added to the nucleus of the atom of the element. Hence from

58Ce to 71 Lu as atomic number increases, nuclear charge increases, therefore nuclear attraction increases.(v) Due to this as atomic number increases, atomic volume or radius decreases as observed with all the

elements along the period.(vi) But in case of Lanthanoids this decreases in atomic volume or radius is very slow or comparatively

small. This is explained in terms of Lanthanoid contraction.Topic: d & f block; Sub-topic:Lanthanoids_ L-Medium _XII-HSC Board Test__Chemistry


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Explain the cause of lanthanide contraction.Ans. (i) In the lanthanide series with increasing atomic number, the atomic and ionic radii decreases from one

element to another but decrease is very small.(ii) Increase in nuclear charge in a period increases force of attraction, towards outer electrons. Hence

atomic size or radius decreases.(iii) Filling of the electrons in the inner shells increase atomic size as inner electrons repel outer electrons.

The shielding effect of electrons decreases in the order s > p > d > f(iv) The new electrons are added to the same inner 4f - orbital. Hence lanthanide contraction is due to

(a) poor shielding effect due to 4f electrons.(b) Increase in nuclear charge.(c) Non-introduction of new outer shells.

Topic: d & f block; Sub-topic:Lanthanoids_ L-Medium_ XII-HSC Board Test__ChemistryDraw the structures of chloroxylenol and adenine.

Ans.

OH

CH3H C3

Cl

C

N

C

C CH

NH2

CN

HC

N

Structure of chloroxylenol Structure of adenine

Topic: Chemistry in everyday life / Biomolecules; Sub-topic:Antiseptic / Nucleic acid_ L-Easy __ XII-HSC Board Test__ChemistryHow are ethylamine and ethyl methyl amine distinguished by using nitrous acid ?

Ans (a) Action of nitrous acid on ethylamine

C H – NH + OH – N = O2 5 2

NaNO +HCl2 C H – OH + N + H O2 5 2 2

ethylamine nitrous acid ethanol(b) Action of nitrous acid on ethylmethylamine

C H – NH – CH + OH – N = O2 5 3

ethylmethylamine nitrous acid

NaNO +HCl2 C H – N – N = O + H O2 5 2

CH3

N-nitrosoethyl-methylamine

When nitrous acid reacts with primary amine like 2 5 2C H NH it gives ethanol while when nitrous acid reactionswith secondary amine like ethylmethylamine it gives yellow oily N-nitrosoamine.

Topic: Compounds containing nitrogen; Sub-topic:Chemical reaction L-Medium__ XII-HSC BoardTest__Chemistry

OR


1717

What is the action of the following reagents on ethanoic acid ? [7]

(a) 4 3/LiAlH H O

(b) 3, heatPCl

(c) 2 5 , heatP OIdentify ‘A’ and ‘B’ in the following reaction and rewrite the complete reaction :

2 53 2

NaC H OHCH CH Br AgCN A B

Explain Hoffmann bromamide degradation reaction.

Ans (a) 4 3/3 3 2

LiAlH H OCH COOH CH CH OH

ethanoic acid ethanol

Ethanoic acid when treated with reducing agent like 4 ,LiAlH it gives ethanol.

(b)3 3 3 3 33 3

ethanoic acid acetylcholorideCH COOH PCl CH COCl H PO

Ethanoic acid when treated with 3PCl gives acetylchloride.

(c) 2 5

3 3 3 22

two molecules of acetic anhydrideethanoic acid

P OCH COOH CH COOH CH CO H O

When two molecules of ethanoic acid heated with strong dehydrating agent P2O5, it removes watermolecules to for acetic anhydride.

Topic:Aldehyde, Ketone, Carboxylic acid; Sub-topic:Chemical reactions_L-Medium_ XII-HSC BoardTest__Chemistry

2 53 2

NaC H OHCH CH Br AgCN A B

2 5./3 2 3 2 3 2 3

ethylmethylamine

Na C H OHCH CH Br Ag CN CH CH NC CH CH NH CH

Hoffmann Bromamide degradation reacation : - When amide is treated with bromine and aqueous or alkoholicNaOH , it gives primary amine

2 2 2 2 3 24 2 2R CO NH Br NaOH R NH Na CO NaBr H O .

This reaction is useful for decreasing the length of carbon chain by one carbon atom. It is an example ofmolecular rearrangement and involves the migration of an alkyl group from the carbonyl carbon to theadjacent nitrogen atom. e.g. when propanamide is treated with bromine and aqueous or alcoholic NaOH, itgives ethanamine.

3 2 2 2 3 2 2 2 3 24 2 2Propanamide ethanamineCH CH CO NH Br NaOH CH CH NH Na CO NaBr H O

Topic: Compounds containing nitrogen; Sub-topic:Preparation of amines_ L-Easy__ XII-HSC BoardTest__Chemistry

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