Part I

(Answer all questions) (20 Marks)

1. (a) Fill in the blanks by choosing the appropriate word/words from those given in the

brackets: [5]

(iodo form, acetaldehyde, positive, greater, acidic, acetone, disaccharide,negative, increases, glucose, decreases, chloroform, polysaccharide, lactose,

lesser, basic, cationic hydrolysis, anionic, hydrolysis)

(i) Calcium acetate on heating gives ......... which gives ............ on heating with

iodine and sodium hydroxide solution.

(ii) On dilution of a solution, its specific conductance ........... while its

equivalent conductance ............

ISCSolved PAPER 2017

Fully Solved (Question-Answer)

Chemistry-XIITime : 3 hrs Max Marks : 70

General Instructions

1. Answer all questions in Part I and six questions from Part II, choosing two questions from Section A,

two from Section B and two form Section C.

2. Candidates are allowed additional 15 minutes for only reading the paper. They must not start

writing during this time.

3. All working including rough work, should be done on the same sheet as and adjacent to the rest of

the answer.

4. Balanced equations must be given whereever possible and diagrams where they are helpful.

When solving numerical problems, all essential working must be shown.

5. In working out problems, use the following data : Gas constant R = 1987. cal deg−1 mol−1

= 8 314. Jk−1 mol−1 = 00821. dm3 atm K−1 mol−1 11 atm = 1 3dm atm = 1013. J.

1 Faraday = 96500 Coulombs. Avogadro’s number = ×6023 1023.

6. The intended marks for questions or parts of question are given in brackets [ ].

(iii) Sucrose is a ........... and yields upon hydrolysis, a mixture of .............

and fructose.

(iv) More ........... the standard reduction potential of a substance, the

.............. is its ability to displace hydrogen from acids.

(v) An aqueous solution of CH3 COONa is ............ due to ...........

(b) Complete the following statements by selecting the correct alternative

from the choices given:

(i) In a face centered cubic lattice, atom (A) occupies the cornerpositions and at (B) occupies the face centre positions. If one atom of(B) is missing from om the face centered points, the formula of the

compound is:

(1) A B2 5 (2) A B2 3

(3) AB2 (4) A B2

(ii) The half life period of a first order reaction is 20 minutes. The timerequired the concentration of the reactant to change form 0.16 M to.0.02 M is :

(1) 80 minutes (2) 60 minutes

(3) 40 minutes (4) 20 minutes

(iii) For a spontaneous reaction ∆ G° and E° cell will be respectively:

(1) – ve and + ve (2) + ve and – ve

(3) + ve and + ve (4) – ve and – ve

(iv) The conjugate acid of HPO42− is:

(1) H PO3 3 (2) H PO3 4

(3) H PO42− (4) PO4

3−

(v) The polymer formed by the condensation of hexamethy lenediamine

and adipic is:

(1) Teflon (2) Bakelite

(3) Dacron (4) Nylon-66

(c) Answer the following questions:

(i) Why the freezing point depression ( )∆ T f of 0.4 M NaCI solution is

nearly twice than that of 0 4. M glucose solution?

(ii) Identify the order of reaction from each of the following units of rate

constant (k):

(a) mol L− −1 1sec (b) mol− −1 1sec

(iii) Specific conductivity of 0.20 M solution of KCI at 298 K is 0.025 S

cm−1. Calculate its molar conductivity.

(iv) Name the order of reaction which proceeds with a uniform rate

throughout.

(v) What are the products formed when phenol and nitrobenzene aretreated separately with a mixture of concentrated sulphuric acid and

concentrated nitric acid?

(d) Match the following: [5]

(i) Diazotisation (a) Bakelite

(ii) Argentite (b) Nernst equation

(iii) Thermosetting plastics (c) Aniline

(iv) Electrochemical cell (d) Ethylenediamine

(v) Bidentate ligand (d) Froth floatation process

PART II

Section A (50 Marks)

2. (a) (i) Determine the freezing point of a solution containing 0.625 g of

glucose ( )C H O6 12 6 dissolved in 102.8 g of water.

(Freezing point of water = 273 K, K f for water = 1 87. K kg mol−1, at.

wt. (C = 12, H = 1, O = 16)

(ii) A 0.15 M aqueous solution of KCl exerts an osmotic pressure of 6.8atm at 310 K. Calculate the degree of dissociation of KCl. (R = 0.0821

Lit. atm K mol− −1 1).

(iii) A solution containing 8.44 g of sucrose in 100 g of water has a vapourpressure 4.56 mm of Hg at 273K. If the vapour pressure of pure water

is 4.58 mm of Hg at the same temperature, calculate the molecularweight of sucrose.

(b) (i) When ammonium chloride and ammonium hydroxide are added to a

solution containing both Al3+ and Ca2+ ions, which ion is

precipitated first and why? [2]

(ii) A solution of potassium chloride has no effect on litmus whereas, a

solution of zinc chloride turns the blue litmus red. Give a reason.[2]

(c) How many sodium ions and chloride ions are present in a unit cell of

sodium chloride crystal? [1]

3. (a) (i) Lead sulphide has face centered cubic crystal structure. If the edgelength of the unit cell of lead sulphide is 495 pm, calculate the

density of the crystal. [1]

(at. wt. Pb = 207, S = 32)

(ii) For the reaction: 2 22H NO+ -2 2 2H O N+ the following rate data was

obtained:

S.No.

[ ]NO mol L−1 [ ]H mol L2−1 Rate : mol L− −1 1sec

1 0.40 0.40 4 6 10 3. × −

2 0.80 0.40 18 4 10 3. × −

3 0.40 0.80 92 10 3. × −

Calculate the following:

(1) The overall order of reaction.

(2) The rate law.

(3) The value of rate constant (k).

(b) (i) The following electrochemical cell is set up at 298 K:

Zn / Zn Cu Cu2+ 2+( )( ) / / ( )( ) /aq M aq M1 1

Given → ° +E Zn Zn2 /

= 0.761 V, E Cu Cu = + 0.339 V2+− ° /

(1) Write the cell reaction.

(2) Calculate the emf and free energy change at 298 K.

(ii) Answer the following:

(1) What is the effect of temperature on ionic product of water (Kw)?

(2) What happens to the ionic product of water (Kw) if some acid is

added to it?

(c) Frenkel defect does not change the density of the ionic crystal whereas,

Schottky defect lowers the density of ionic crystal. Give a reason. [2]

4. (a) (i) Name the law or principle to which the following observationsconform: [3]

(1) When water is added to a 1.0 M aqueous solution of acetic acid, the

number of hydrogen ion ( )H+ increases.

(2) When 9650 coulombs of electricity is passed through a solution ofcopper sulphate, 3 .175 g of copper is deposited on the cathode(at.wt. of Cu = 635. ).

(3) When ammonium chloride is added to a solution of ammonium

hydroxide, the concentration of hydroxyl ions decreases.

(ii) What is the difference between the order of a reaction and itsmolecularity? [2]

(b) (i) Explain why high pressure is ·required in the manufacture of

sulphur trioxide by contact process. State the law or principle used.[2]

(ii) Calculate the equilibrium constant (Kc) for the formation of NH3 in

the following reaction: [1]

N H NH2 2 33 2( ) ( ) ( )g g g+ −

At equilibrium, the concentration of NH H3 2, and N2 are

1 2 10 30 102 2. , .× ×− − and 1 5 10 2. × − M respectively.

(c) Explain the following: [2]

(i) Hydrolysis of ester (ethyl acetate) begins slowly but becomes fast

after sometime.

(ii) The pH value of acetic acid increases on addition of a few drops of

sodium acetate.

Section B

5. (a) Write the formula of the following compounds: [2]

(i) Potassium trioxalatoaluminate (III)

(ii) Hexaaquairon (II) sulphate.

(b) Name the types of isomerism shown by the following pairs of compounds:[1]

(i) Cu (NH )[PtCl ]43 4) and [ ][ ]Pt(NH ) CuCl3 4 4

(ii) [ ]Co(en) Cl2 2+ and [ ]Co(en) Cl2 2

+

(c) For the coordination complex ion [ ]Co(NH )3 63+

[2]

(i) Give the IUPAC name of the complex ion.

(ii) What is the oxidation number of cobalt in the complex ion?

(iii) State the type of hybridisation of the complex ion.

(iv) State the magnetic behaviour of the complex ion.

6. (a) Give balanced equations for the following reactions: [3]

(i) Potassium permanganate is heated with concentrated hydrochloric

acid.

(ii) Lead sulphide is heated with hydrogen peroxide.

(iii) Ozone is treated with potassium iodide solution.

(b) Discuss the theory involved in the manufacture of sulphuric acid bycontact process. [2]

7. (a) (i) What are the types of hybridisation of iodine in interhalogen

compounds IF IF3 5, and IF7 , respectively?

(ii) Draw the structure of xenon hexafluoride ( )X Fe 6 molecule and state

the hybridisation of the central atom.

(b) Give the balanced equations for the conversion of argentite (Ag2 S) tometallic silver. [2]

Section C

8. (a) How can the following conversions be brought about:

(i) Acetaldehyde to propan-2-ol. [1]

(ii) Nitrobenzene to p-aminoazobenzene. [1]

(iii) Acetic acid to methylamine. [2]

(iv) Aniline to benzene. [1]

(b) (i) How will you distinguish between primary, secondary and tertiary

amines by Hinsberg‘s test? [1]

(ii) Why do alcohols possess higher boiling points as compared to those

of corresponding alkanes? [1]

(c) Identify the compounds A, Band C: [3]

(i) C H COOH A B C6 5

Pcl H Pd / BaSo

distil

KCNalc5 2 4

→ → →−

(ii) H C C H A BdilH SO H SO

H O

[Ni]

H

C2 4 g 4

2 2

− ≡ − → → →+ onc H SO

140 C

2 4

C°

9. (a) Give balanced equations for the following name reactions: [3]

(i) Friedel-Crafts reaction (alkylation)

(ii) Williamson‘s synthesis

(iii) Aldol condensation

(b) Give chemical test to distinguish: [3]

(i) Ethyl alcohol and see-propyl alcohol •

(ii) Acetaldehyde and acetic acid

(c) (i) Deficiency of which vitamin causes the following diseases: [4]

(1) Scurvy (2) Night blindness

(ii) Write two differences between globular and fibrous proteins.

10. (a) An aliphatic unsaturated hydrocarbon (A) when treated withH SO H SOg 4 2 4/ yields a compound (B) having molecular formula C H O63 .

(B) on oxidation with concentrated HNO3 gives two compounds (C) and

(D). Compound (C) when treated with PCl5 gives compound (E). (E) when

reacts with ethanol gives a sweet smelling liquid (F). Compound (F) is alsoformed when (C) reacts with ethanol in the presence of concentratedH SO2 4.

(i) Identify the compound A, B, C, D, E and F.

(ii) Give the chemical equation for the reaction of (C) with chlorine in

the presence of red phosphorous and name the reaction.

(b) Answer the following: [3]

(i) What is the common name. of the polymer obtained by the

polymerization of caprolactum? Is it addition polymer orcondensation polymer?

(ii) Name the two organic compounds which have the same molecularformula C H O2 6 . Will they react with PCI5? If they react, what are the

products formed?

(c) Give balanced equations for the following reactions: [3]

(i) Methyl magnesium bromide with ethyl alcohol.

(ii) Acetic anhydride with phosphorous pentachloride.

(iii) Acetaldehyde with hydroxylamine.

Solutions

1. (a) (i) Acetone, iodoform

(ii) Decreases, increases

(iii) Disaccharide, glucose

(iv) Negative, greater

(v) Basic, arionic hydrolysis

(b) (ii) (2) Kt

= = = −0693 0693

20003465

12

1. .. min

For first order reaction,

Rate constant ( ).

log[ ]

[ ]K

t

A

A= 2 303 0

[ ]A 0 = initial concentration = 016. M

[ ]A = concentration left after time ‘t’ = 002. M

∴ t = =2 303

003465

016

00260

.

.log

[ . ]

[ . ]minutes.

(i) (3) Number of atom (A) at corner = × =1

88 1

Number of atom (B) at Face centre = ×1

26 = 3

One atom of (B) is missing,

∴ Number of B atom = − =3 1 2

∴ Formula of compound = AB2

(iii) (2) E° cell is positive - cell reaction is spontaneous ∆G° is negative -call reaction is spontaneous

(iv) (3) The conjugate acid of HPO42− is H PO2 4

−

(v) (4) nH N (CH ) 6 NH2 2 2

+ HO C

O

(CH ) C

O

OH2 4n

Hexamethylenediamine Adipic acid

→

-n2H O2

N

H

CNheat Pressure,

2 6

N

H

C

O

(CH ) C

O

4 2 n

(c) (i) Depression is freezing point is a colligative property which dependson the number of particles. Glucose is a molecular compound, itremains as a single molecule is solution. NaCl ionises is solution togive two particles

NaCl - Na Cl+ −

Thus, depression in freezing point ( )∆TF of 0.4 M .4 M NaCl is greaterthen 0.4 M glucose solution.

(ii) (a) mol L− − →1 1sec Zero order reaction

(b) mol L sec− − →1 1 Second order reaction.

(iii) Given,

Specific conductivity (K) = −0025 1. S cm

Concentration (C) =0.20 M

Now, Molar conductivity (∆m)1000

C× K

= ×1000

020025

..

= −125 2 1Scm mol

(iv) zero order reaction.

(v) When nitration of phenol is carried out with conc. nitric acid in thepresence of conc. sulphuric acid then picric acid (2, 4,6-trinitrophenol) is obtained.

(d) (i) Diazotisation—Aniline

(ii) Argentite—froth Floatation process

(iii) Thermosetting plastice—Bakelite

(iv) Electrochemical cell—Nernst equation

(v) Bidentate ligand—Ethylenediamine

Part-II

Section-A

2. (a) (i) Given, Freezing point of water = 273 K

Weight of solute (glucose) w = 0.625 g

Weight of solvent, W = 102.8 g

Molecular weight of solute, m = (C H O )6 12 6

= × + × + ×12 6 1 12 16 6

= −180 1g mol

K for water = −1.87 K kg mol 1

For glucose as solute, i = 1

∴ ∆TK w

m WF=

× ××

1000

= × ××

=1000

180

1.87 0.625

102.80.063 K

∴ Freezing point of solution = +T TF F∆(where, TF is freezing point of water = ° =0 273C K)

= + =273 0.063 273.063 K

Conc.H SO2 4

OH

+ 3HNO3

Picric acid(2, 4, 6-trinitrophenol)

OH

NO2O N2

+ 3H O2

NO2

→(conc.)

(ii) Osmotic pressure ( )π =CRT

Given, C R= = − −0.15 M, 0.0821 L atm K mol1 1

T = 310 K

∴ π = CRT

= × ×− −0.15 M 0.0821 L atm K mol K1 1 310

= 3.82 atm

Observed π = 6.8 atm

∴ Van’t Hoff Factor,

i = Observed magnitude of

Normal magnitude of

ππ

= ×6.8

3.821.78

If degree of dissociation α, then for KCl, n = 2as the number of soluteparticles in its aqueous solution is almost double the number of NaClmolecules.

∴ α = −−

= −−

=i

n

1

1

1

2 1

1.780.78

or α = 78%

(iii) Given, Vapour pressure of pure water

( )p° = 4.58mm of Hg

Vapour pressure of solution

( )p = 4.56mm of Hg

Mass of sucrose ( )w = 8.44 g

Mass of water ( )W = 100 g

Molecular mass of H O2 ( )M = 18 g

Molecular mass of sucrose, ( ) ?m =According to Raoult’s law,

p p

p

wM

W m

° −°

=⋅

4.58 4.56

4.58

8.44− = ××

18

100 m

∴ m = × ××

8.44 4.58

0.02

18

100

= −347.89 g mol 1

(b) (i) Al3+ is precipitated first notCa 2+ . This is because K sp for Al (OH)3 is

lesser than that of K sp for Ca (OH)2 and in presence of NH Cl4 ,dissociation of ammonium hydroxide is suppressed due to common

ion effect and the concentration of OH− ions is not enough to crossthe K sp for Ca(OH)2 .

(ii) Potassium chloride is a salt of strong acid, HCl and strong base, KOH.When dissolved in water it does not get hydrolysed and the solutionremains neutral.

Zinc chloride is a salt of strong acid, HCl and weak base, Zn(OH)2 .When it is dissolved in water, it undergoes hydrolysis and resultingsolution becomes acidic so, it turns blue litmus red.

ZnCl + H O Zn 2 OH 2HCl2 2 → + ++ −2

(c) 4Na + and 4Cl− are present in unit cell of sodium chloride crystal.

3. (a) (i) Density of a cubic crystal,

ρ = ×=

Z M

a N A3

Given, Z = 4 (for fcc crystal structure),

M = + =207 32 239, a = = × −495 495 10 10pm cm

N A = ×6.02 1023

∴ ρ = ×× × ×−

4 239

495 10 1010 3 23( ) 6.02

= −13.09 g cm 3

∴ Density of PbS is 13.09 g cm−3

(ii) Given, 2H + 2NO * * 2H O + N2 2 2

Let, the given reaction is of order q with respect to H2 and p withrespect to NO.

The rate of formation H O2 and N2 can also be written as,

(a) 4.6 × =−10 3 K A Bp q[ ] [ ] = K p q[ ] [ ]0.4 0.4 ...(i)

(b) 18.4 0.8 0.4× =−10 3 K p q( ) ( ) ...(ii)

(c) 9.2 0.4 0.8× =−10 3 K p q( ) ( ) ...(iii)

From eqn. (i) and (ii), we get

18.4

4.6

0.8 0.4

0.4 0.4

××

=−

−10

10

3

3

K

K

p q

p q

( ) ( )

( ) ( )

4 2= ( ) p ⇒ ( ) ( )2 22 = p

Hence, p = 2

From Eqs. (i) and (iii), we get

9.2

4.6

0.4 0.8

0.4 0.4

××

=−

−10

10

3

3

K

K

p q

p q

( ) ( )

( ) ( )

2 2= ( )q

∴ q = 1

(1) Overall order of reaction is ( )2 1 3+ =(2) Rate law equation, r K= [ ] [ ]NO H2

2

(3) r K= [ ]H [NO]22

Rate constant,

Kr= = × −

[H ] [NO]

4.6

0.4 0.422

10 3

2[ ] [ ]

= × − − −7.18 mol L s10 2 2 2 1

(b) (i) Electrical energy or the maximum work is the measure of EMF of

cell. W nFEmax cell= °

According to thermodynamics, free energy change ( )∆G° is equal tomaximum work.

Hence, − = ° = − °W G nFEmax cell∆

(i ) Cell reaction

Zn Cu Cu + Zn( ) ( ) ( ) ( )s aq s aq+ →+ +2 2

i.e. at cathode

Cu + 2 Cu2+ e− → ; E = 0.339 V

At anode

Zn Zn 2→ ++ −2 e E = 0.761V

(ii ) E E Ecell cathode anode° = ° − °

Given, EZn, Zn2+ 0.761° = + V

ECu ,Cu2+ 0.339 V= +

or EZn , Zn2+ 0.761° = −

Hence, E E Ecell Cu ,Cu Zn ,Zn2+ 2+° = ° − °

= − −0.339 0.761)(

= 1.1V

Putting the value of Ecell° in ∆G nFE° = − °

cell

where, n = 2, F = 96500 C

∆G° = − × ×2 96500 1.1 = − 212.3 kJ

(ii) (1) With increase in temperature, the dissociation of ater increases,

increasing [H ]+ and [ ]OH− and consequently, K w increases.

(2) If some acid or base is added to water then [ ] [ ]H OH+ −≠ however, K w

remains constant at the constant temperature.

(c) Frenkel defect is due to shifting of an ion from its normal positioin to aninterstitial site in the lattice. Due to this, density remains same.

Schottky defect arises due to missing of equal number of cations andanions from their position leaving some of its lattice points unoccupied.Hence, its density decreases.

4. (a) (i) (1) Recall the Ostwald’s dilution law.

According to Ostwald’s dilution law, the extent of ionisation of a weakelectrolyte is directly proportional to the square root of the volume ordilution. So, when 1 L of water, is added to 1 M acetic acid which is aweak electrolyte, the ionisation of acetic acid increases which causesincrease in the number of hydronium ions.

(2) Faraday’s first law of electrolysis confirmed from the given statementwhich states that ‘the amount of substance deposited at a particularelectrode as a result of electrolysis is directly proportional to thequantity of electricity passed’.

(3) Common on effect

when NH Cl4 (strong electrolyte) is added to a solution of NH OH4

(weak electrolyte), concentration of NH4+ ions increases.

Due to common ion effect, dissociation of NH OH4 is suppressed and

the concentration of OH− ions is decreased.

NH OH s NH OH4Weak electrolyte

4+ −+ (feebly ionised)

NH Cl s NH Cl4Strong electrolyte

4+ −+ (highly ionised)

(ii) Differences between order and molecularity of a reaction

Order of a reaction Molecularity of a reaction

It is the sum of the powers of concentration terms on whichthe rate of reaction actually depends or it is the sum of theexponents of the concentrations in the rate law equation.

It is the number of atoms, ions ormolecules that must collide with oneanother simultaneously so as to resultinto a chemical reaction.

It can be fractional as well as zero. It is always a whole number.

It can be determined experimentally only and cannot becalculated theoretically.

It can be calculated by adding themolecules of the slowest step.Therefore, it is a theoretical concept.

(b) (i) In the manufacture of SO2 by contact process, the following chemicalreaction occurs.

2SO O r 2SO2 2 3( ) ( ) ( )g g g+2 moles 1 mole 1 moles

3 moles

According to Le-Chaterlier’s principle, the equilibrium must shift inthe forward direction to oppose the increase in number of moles perunit volume. Therefore, more SO2 and O2 combine to form more SO3 .Hence, high pressure is used in the manufacture of SO3 by contactsprocess.

Le-Chatelier’s principle is used for the formation of SO3 . Accordingto Le-Chatelier’s principle, if a system in equilibrium is subjected tochange in temperature, pressure or concentration of any componentin the system, equilibrium automatically shifts in such a direction ofthe reaction so as to reduce the effect caused by that change.

(ii) Equilibrium constant [ ]Kc =[NH ]

[N ] [H ]

32

2 23

Given, [ ]NH 1.2 M3210= × −

[ ]N 1.5 M2210= × −

[ ]H M223 10= × −

∴ Kc = ×× ×

−

− −[ ]

[ ] [ ]

1.2

1.5

10

10 3 10

2 2

2 2 3

= −355.6 M 2

(c) (ii) The suppression of the degree of ionisation of a weak electrolyte bythe addition of a strong electrolyte which has an ion common with theweak electrolyte is known as common ion effect.

e.g. acetic acid (weak acid) ionises to a small extent as follows :

CH COOH s CH COO H3Weak electrolyte

3+− +

If the salt of this weak acid with a strong base, i.e. CH COONa3 (strongelectrolyte) is added to this solution. This salt ionises completely.

CH COONa s CH COO Na3Strong electrolyte

3+− +

As a result, concentration of CH COO3− ions increases and that of H+

ions decreases as CH COO3− ions combine with H+ ions to form

undissociated CH COOH3 molecules.

Due to decreases of H+ ion concentration, pH value increases.

Ester undergo slow hydrolysis with water. However, in presence ofacids or alkalies, hydrolysis takes place more rapidly.

Section-B

5. (a) (i) K [Al(OX) ] or K [Al(C O ) ]3 3 3 2 4 3

(ii) [Fe(H O) ] SO2 6 4

(b) (i) Coordination isomerism

(ii) Geometrical or cis-trans isomerism

(c) (i) Hexaammine cobalt (III) ion

(ii) Oxidation number of Co in the complex [Co(NH ) ]3 63+ is

x × × = +6 0 3( )

x = +3

(iii) Electronic configuration of Co is [Ar]3 7 2d s4 .

Co atom

Co3+ ion

NH3 is a strong field ligand, it causes pairing of3 6d electrons.

Thus,

(i i i ) Coordination number of the complex is 6. Hence, hybridisation is

d sp2 3 .

(iv ) All electrons are paired. So, it is diamagnetic.

6. (a) The balanced chemical equations are

(i) + 8H O + 5Cl2 2

(ii) PbS + 4H O PbSO + 4H OLead sulphide

2 2 4Lead sulphate

2→∆

(iii) 2KI H O OPotassium iodide Ozone

( ) ( ) ( )aq l g+ + →2 3 2KOH IIodine

( ) ( ) ( )aq s O g+ +2 2

(b) H SO2 4 is commercially prepared by contact process.

2SO + O 2SO ( );2 2

- K

atm V O

3

2 5

( ) ( )

,

,

g g g

670 720

2

(∆H = 196.6 kJ)−The reaction is reversible, exothermic and involves a decrease in thenumber of moles. Therefore, decrease in temperature, increase in pressure

3d7

4s2 4p

0

4s0 4p

03d

6

3d 4s 4p

Six empty orbitals for NH

hybridisation

32 3d sp

and increase in the concentration of reactants favours the forwardreaction.

SO3 obtained is dissolved in concentrated H SO2 4 to get oleum (H S O )2 2 7 .H SO2 4 of desired concentration can be obtained by diluting oleum withwater.

H SO + SO H S O2 4 3 2 2 7Oleum

→ ; H S O + H O 2H SO2 2 7 2 2 4→

7. (a) (i)

(ii) XeF6 has sp d3 3 hybridisation and distorted octahedral geometry due

to the presence of one lone pair of electrons

(b) Ag S 4NaCN s 2Na[ (CN) ]2 2Sodium dicyano argentate (I

+ Ag)

2Na S+

The solution containing Na[Ag(CN) ]2 is treated with Zn-scrap so that Aggets precipitated.

2Na[Ag(CN) ] Zn 2Ag

ppt

Na2

[Zn (CN)4

]

Sodium tetrac

2 + → ↓ +yanozincate (II)

(b) (i) Hinsberg’s Test

(i ) Primary amine when reacts with benzenesulphonyl chloride(Hinsberg’s reagent), it yields N-ethylbenzene sulphonamide.Hydrogen attached to nitrogen is strongly acidic due to the presenceof strong electron withdrawing sulphonyl group, hence it is solublein alkali.

(ii ) Secondary amine when reacts with benzenesulphonyl chloride, ityields N,N-diethylbenzene sulphonamide. It does not have any

I F

F

F

F

F F

F F

F

F

F F

F F

F

F

F

IF –sp d33 IF –sp d3

3 2 IF –sp d73 3

XeF

F

F F

F

F

N-ethylbenzenesulphonamide

(soluble in alkali)

1° amine

S Cl+ H

O

O H

N C H2 5 S

O

O H

N C H + HCl2 5→–HCl

H-atom attached to N-atom. Thus, it is not acidic and henceinsoluble in alkali.

(iii ) Tertiary amines do not react with benzene sulphonyl chloride.

(ii) The boiling points of alcohols are considerably higher than those ofcorresponding hydrocarbons, haloalkanes, ethers, etc. Due to stronghydrogen bonding between their molecules, the boiling points of isomericalcohols follow the order 1 2 3°> °> °.

(c) (i)

(ii)

(c) (i) (1) Scurvy—Vitamin C (Ascorbic acid)

(2) Night blindness—Vitamin A (Retinol)

(ii) Table type

Fibrous protein Globular protein

(a) Polypeptide chains consist ofthread like structures which tendto lie side by side to form fibres.

The chain of polypeptide coilaround to give a spherical shape.

(b) The molecules are held togetherby H-bonds in some cases.

The interaction present in theseare H-bonds, ionic or salt bridges.

C H COOH6 5

Benzoic

acid

PCl5

–POCl

–HCl3

C H COCl6 5

Benzoyl

chloride

(A)

H/Pd,BaSO4C H CHO6 5

Benzaldehyde

(B)

KCN

(alc.) distil

C H —CH—C—C H6 5 6 5

OH O

Benzoyl

(C)

(ii) HC

Benzoic

acid

H O2

Dil.H SO +HgSO2 4 4

CH CHO3

Ethanal

(A)

H2CH CH OH3 2

Ethanol

(B)

140ºCcone.

H SO2 4

CH CH —O—CH CH3 2 2 3

Diethylether

+

H O2

CHNi

N,N-diethylbenzene sulphonamide(insoluble in alkali)

S Cl +H

O

O C H2 5

N C H2 5 S

O

O C H2 5

N C H + HCl2 5→

Section-C

8. (a)

(i) CH CHO

Acetaldehyde

CH MgBrCH CH

OMgBr

CH

Ad3 3 3

3 →

dition Product

CH CH

OH

CH

Propan Ol

H / H O

Mg OH Br3 3

20

2

− −

+←

− ( )

(ii)

(iii) CH COOH

Acetic acid

NHCH COOHNH

Ammonium H3 3 4

3

2

→ → −

∆O

CH CONH

Acetamide3 2

CH NH

Methanamine

Br KOH3 2

2 / →

9. (a) (i)

NH2

NaNO+Di.HCl

273-278K

N NCl≡ –

H PO /Cu+3 2

Benzenediazonium

chloride

BenzeneAnilineDiazotisation

NH2

CH Cl/anyhd.AlCl3 3

Anisole

OCH3

P-methylanisole

(major)

CH3

+

OCH3

O-methylanisole

(minor)

CH3

CH3

NO2

Sn+HCl

Reduction

NH2

NaNO , Dil.HCl

273-278k2

Diazotisation

Nitrobenzene Aniline

NH2

Benzenediazonium

Coupling with

aniline pH,4–5—N==N—NH—

Diazominobenese

—N==N—

P-Aminoazobenzene

H ,+ ∆Rearrangement

—NH2

(ii) By Williamson’s synthesis This is one of the best method for thepreparation of ethers.

In this method, alkyl halide is treated with a suitable sodium (orpotassium) alkoxide.

R O Na XR+

Sodium alkoxide Alkylhalide

+ ′ →− R O R NaXEther

′ +

Note This method is not applicable when the alkyl halide used in this synthesisis tertiary.

(iii) Aldehydes and ketones containing α H-atoms undergo aldolcondensation in presence of dilute alkali as catalyst, e.g.

2CH CHO3Ethanal

Dil. NaOH

CH CH CH CHO3

3-hydroxybutanal(aldol)

→

OH

H O2

2–

∆

CH CH3But -2-enal

(aldol condensation produc

== CH CHO

t)

(b) (i) Ethanol is primary alcohol while 2-propanol is secondary alcohol.So, the two can be distinguished by Victor Meyer’s test.

In Victor Meyer’s test, the given alcohol is treated with redphosphorus and iodine to convert it into the corresponding alkyliodide. This compound is then treated with silver nitrite ( )AgNO2 toobtain the corresponding nitroalkane which is further treated withnitrous acid ( )NaNO dil. H2 2+ SO 4 and then the resulting solution ismade alkaline. If blood red colour appears, then alcohol is primary, ifblue colour appears, the given alcohol is secondary and if thesolution remains colourless, the given alcohol is tertiary.

The reactions involved for the two alcohols are as follows :

CH CH OH3 2

4 2

Ethanol(1 )

P I

°

+→ CH CH I CH CH NO3 2

23 2 2

Ethyl iodide

AgNO

Nitroethane

→HNO2→

CH C

N OH

NO red colour3 2

Nitrolic acid

NaOH

→ Blood

CH C H

OH

C H (CH )3

4 2

Propan-2-ol

3

(2 )

P I

3 2

→

°

+

CHI (CH ) CHNO (H C) C

NO

NOAgNO

2

HNO

3 2 2

(Ps

2

3 2

2→ →

eudo nitrol)

NaOHBlue col→ our

(b) Tollen’s test—Acetaldehyde reduces Tollen’s reagent to give shining silvermirror but acetic acid does not.

CH CHO +2[Ag(NH ) ]3Acetaldehyde

3 2+

Tollen's reagent

+ →−3OH

CH COO 2Ag 2H O 4NH3Acetate ion Silver mirror

2− + ↓ + + 3 CH COOH No reaction3

Tollen's reagent

→

Acetic acid

10. (a) The reaction takes place as follows :

(b) (i) Nylon-6 or Perlon L, (Structure) is obtained by polymerisation ofcaprolactam. It is a condensation polymer.

(ii) Ethanol (CH CH OH)3 2 and dimethyl ether (CH — O — CH )3 3 have thesame molecular formula, C H O2 6 .

Yes, they but react with PCl5 .

CH CH OH + PCl CH CH Cl + POCl + HCl3 2 5 3 2 3→

N

H

OPolymerisation

Caprolactum

C NH (CH )2 5

O

Nylon-6 n

→

CH3

Propyne

HgSO4

H SO2 4

CH —C—CH3

Acetone

(B)

3 [O]

C H OH2 5

cone.

H SO2 4

CH —O—OC H3 2 5

(F)

CHConc.HNO3

—C

O

CH3

Ethyl acetate

(F)

C H OH2 5

–HClCH —C—CH3

Acid chloride

(E)

PCl5

–POCl ,

–HCl3

CHCOOC H2 5

O

CH COOH+HCOOH3

Ethanoic

acid

(C)

(D)

O

Ethanol Ethyl chloride Phosphorus oxychloride

CH — O — CH + PCl 2CH Cl3 3Demethyl ether

5 3Methyl chloride

→∆

+ POCl3

(c)(i) CH CH OH

Ethyl alcohol

CH MgBr CH

Methane

MgBr

OC H

3 2 42 5

3+ → +Ethyoxy Magnesium

bromide

(ii) CH C

O

O C

O

CH

Acetic anhydride

PCl

Phosphorus

pe

3 3 5

+

ntachloride

2CH

Acetyl

chloride

C

O

Cl+POCL→

3 3

(iii) CH CHO + NH OH CH — CH NOH +3Acetaldehyde

2Hydroxyl amine

3→ == H O2