Aieee 2008 Paper

8/2/2019 Aieee 2008 Paper

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MAHESH JANMANCHI AIEEE 2008

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Maximum Marks: 105

Question paper format and Marking scheme:

1. This question paper has 35 questions.

2. Candidates will be awarded three marks each for indicating correct response for each question.

3. 1 mark will be deducted for indicating incorrect response.

4. No deduction from the total score will be made if no response is indicated for an item in theanswer sheet.

36. The organic chloro compound, which shows complete stereochemical inversion during a SN2reaction, is

(1) ( )C H CHCl2 5 2 (2) ( )CH CCl3 3

(3) ( )CH CHCl3 2 (4) CH Cl3

Sol. (4)10 alkyl halides undergo SN2 reaction.

Among given, options 1 and 3 are 20, 2 is 30 and 4 is 10 alkyl halide.

In SN2 reaction, the C atom must be least hindered towards the attack of nucleophile .

Nucleophile attacks from backside resulting in the inversion of the molecule.

As we move from 10

to 30

alkyl halide, crowding increases and + I effect increases which makes C

bearing halogen less positively polarized and hence less readily attacked by the nucleophile


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37. Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. Theproduct so obtained is diazotised and then heated with cuprous bromide. The reaction mixture soformed contains(1) mixture of o and pbromotoluenes (2) mixture of o and pdibromobenzenes(3) mixture of o and pbromoanilines (4) mixture of o and mbromotoluenes

Sol. (1)


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38. The coordination number and the oxidation state of the element E in the complex[E(en)2(C2O4)]NO2 (where (en) is ethylene diamine) are, respectively,(1) 6 and 2 (2) 4 and 2(3) 4 and 3 (4) 6 and 3

Sol. (4)

In the complex given, (en) and C2O42-( oxalate) are bidentate ligandsTherefore, Coordination no is 6 (2x2 + 1x2)

Oxidation no of E is x + (2 x 0) +(1 x (-2)) + (1) = 0 x = 3 (en is a neutral ligand)

39. Identify the wrong statements in the following:(1) Chlorofluorocarbons are responsible for ozone layer depletion.(2) Greenhouse effect is responsible for global warming.(3) Ozone layer does not permit infrared radiation from the sun to reach the earth(4) Acid rains is mostly because of oxides of nitrogen and sulphur

Sol. (3)

Ozone layer does not permit ultraviolet radiation from sun to reach earth.

40. Phenol, when it first reacts with concentrated sulphuric acid and then with concentrated nitricacid, gives(1) 2,4,6-trinitrobenzene (2) o-nitrophenol(3) p-nitrophenol (4) nitrobenzene

Sol. (None of the above)


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41. In the following sequence of reactions, the alkene affords the compound B

O H O3 2CH CH=CHCH A B

3 3 Zn

The compound B is

(1) CH CH CHO3 2

(2) CH COCH3 3

(3) CH CH COCH3 2 3

(4) CH CHO3

Sol. (4)

42. Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids,the main reason being(1) 4f orbitals more diffused than the 5f orbitals(2) lesser energy difference between 5f and 6d than between 4f and 5d orbitals

(3) more energy difference between 5f and 6d than between 4f and 5d orbitals(4) more reactive nature of the actinoids than the lanthanoids

Sol. (2)Actinoids show different oxidation states like +2 , +3, +4, +5, +6 and +7., with +3 as the most

common oxidation state

The reason for this can be attributed to the fact that the 5f, 6d and 7s energy levels are of comparable

energies i.e., have less energy difference and hence all these subshells participate.

43. In which of the following octahedral complexes of Co (at. no. 27), will the magnitude of0 be the

highest?

(1) ( )3

6Co CN

(2) ( )3

2 4 3C o C O

(3) ( )3

2 6C o H O

+

(4) ( )3

3 6Co NH

+

Sol. (1)CN is stronger ligand among given. Hence 0 is highest..


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44. At 80oC, the vapour pressure of pure liquid A is 520 mm Hg and that of pure liquid B is 1000

mm Hg. If a mixture solution of A and B boils at 80oC and 1 atm pressure, the amount of A in

the mixture is (1 atm = 760 mm Hg)(1)50 mol percent (2) 34 mol percent(2)48 mol percent (4) 50 mol percent

Sol. (4)0 0

T A A B BP = P X + P X

( )0A B A760 = 520 X + P 1 - X

AX = 0.5

Thus, mole% of A = 50%

45. For a reaction 1 A 2B2

rate of disappearance of A is related to the rate of appearance of B by

the expression

(1)[ ] [ ]d A d B1

=dt 2 dt

(2)[ ] [ ]d A d B1

=dt 4 dt

(3)[ ] [ ]d A d B

=dt dt

(4)[ ] [ ]d A d B

= 4dt dt

Sol. (2)

1A 2B

2

Rate of reaction w.r.t A and B can be given by,

[ ] [ ]d A d B1 1= +

1/ 2 dt 2 dt

[ ] [ ]d A d B1=

dt 4 dt

46. The equilibrium constants Kp1 and Kp2 for the reactions 2X Y and Z P Q+ , respectively

are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of totalpressure at these equilibria is

(1)1 : 36 (2) 1 : 1 (3) 1 : 3 (4) 1 : 9

Sol. (1)Consider, a general reversible reaction,

( ) ( ) ( ) ( )aA bB cC dD

g g g g+ +

( ).

.

nC Dn n P gc d TKP A B Nn n

a b

=

( )( ) ( )

( )( )

( )( ). . Pds. Reactantsn c d a b i e n n

g g g

= + +


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1 02X Y

(1-x) 2x

( )

( )1

2 1

12

1 1P

x PK

x x

=

+

( )

( )1

2 1

12

1 1P

x PK

x x

=

+

1 0 0

Z P Q+

(1-x) x x

( )2

12

2

1 1P

PxK

x x

=

+

1 1 1

2 2 2

4 1 1

9 36

Kp P P

Kp P P

= = =

47. Oxidising power of chlorine in aqueous solution can be determined by the parameters indicatedbelow:

( ) ( ) ( ) ( )

01 00 H H H1 eg hyddiss2Cl g Cl g Cl g Cl aq

22

The energy involved in the conversion of ( )1

Cl g22

to ( )Cl g (using data

e -1 e -1 e -1 H = 240 kJmol , H = - 349 kJmol , H = - 381 kJmol

diss Cl eg Cl hyd Cl2

Will be(1)+152 kJmol

-1(2) -610 kJmol

-1

(3) -850 kJmol-1

(4) +120 kJmol-1

Sol. (2)

For the process ( )212 aqCl g Cl

2

1

2diss eg hyd

H H of Cl Cl Cl

= + +

240349 381

2= +

1610kJ mol =


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48. Which of the following factors is of no significancefor roasting sulphide ores to the oxides and notsubjecting the sulphide ores to carbon reduction directly?(1) Metal sulphides are thermodynamically more stable than CS2(2) CO2 is thermodynamically more stable than CS2(3) Metal sulphides are less stable than the corresponding oxides(4) CO2 is more volatile than CS2

Sol. (4)

Oxidizing roasting is a common type of roasting in metallurgy and is carried out to remove S and

As in the form of their volatile oxides. CS2 is more volatile than CO2The reduction processis based on the thermodynamic stability of the products and not on their volatility.

49. Bakelite is obtained from phenol by reacting with

(1) ( )CH OH2 2 (2) CH CHO3 (3) CH COCH

3 3(4) HCHO

Sol. (4)

Bakelite is a thermosetting polymer which is made by reaction between phenol and formaldehyde

+HCHO

OH

CH OH2

CH OH2

Polymerize

CH2

CH2

O

n

OH

50. For the following three reactions a, b and c, equilibrium constants are given

(a) ( ) ( ) ( ) ( )2 2 2 1;CO g H O g CO g H g K + +

(b) ( ) ( ) ( ) ( )4 2 2 23 ;CH g H O g CO g H g K + +

(C) ( ) ( ) ( ) ( )4 2 2 2 32 4 ;CH g H O g CO g H g K + +

Which of the following relations is correct?

(1) 1 2 3K K K= (2) 2 3 3K K K=

(3) 3 1 2K K K= (4)3 2

3 2 1.K K K=

Sol. (3)Equation (c) = equation (a) + equation (b)

Thus K3 = K1.K2


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Note :

(a) If an equation is reversed;1

K =2

K1

(b)If an equation is multiplied by n; K =K2 1

n

(c) If an equation is divided by n; K =2 1

n K

(d)If an equation (c) is obtained by adding equations (a) and (b), then K =K .3 1 2

K

(e) If an equation (c) is obtained by subtracting equation (b) from (a) thenK

1K =3

2K

51. The absolute configuration of

is(1)S, S (2) R, R(3) R, S (4) S, R

Sol. (2)

Both C1 and C2 have R configuration. 3

Simply,

In the half where H is with dash, take the configuration as it is (from 1 to 3 , it is R)

In the half where H is with wedge. take the oppositeconfiguration as it is (from 1 to 3 , it is S buttake R)

52. The electrophile, E attacks the benzene ring to generate the intermediate -complex. Of the

following, which -complex is of lowest energy?

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


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Sol. (2)It is of low energy due to resonance stabilization of +ve charge

In other structures, - NO2 is electron withdrawing which will destabilize - complex and hence they

will have greater energy

53. -D-(+)-glucose and -D-(+)-glucose are(1) conformers (2) epimers(3) anomers (4) enantiomers

Sol. (3)A pair of stereoisomers which differ in configuration at C -1 are called anomers and the carbon is called

anomeric carbon

- D (+) glucose and - D (+) glucose are anomers.

54. Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK1

mol1

, respectively. For the reaction,

1 3X + Y XY

2 2 32 2 , 30kJH = , to be at equilibrium, the temperature will be

(1) 1250 K (2) 500 K(3) 750 K (4) 1000 K

Sol. (3)

1 3X + Y XY

2 2 32 2

0 0 0

reaction products reactants

0 1

reaction

S S S

3 1S 50 40 60 40Jmol

2 2

=

= + =

G = H TS

At equilibrium G = 0

H = TS 330 10 T - 40 =

750KT =


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55. Four species are listed below

(i) 3HCO (ii) 3H O

+

(iii) 4HSO

(iv) 3HSO F Which one of the following is the correct sequence of their acid strength?(1) iv < ii < iii < I (2) ii < iii < i < iv(3) i < iii < ii < iv (4) iii < i < iv < ii

Sol. (3)HSO3F is the super acid. Its acidic strength is greater than any given species .Pka values of other species are

3HCO = 10.25 3H O

+ = - 1.744

HSO = 1.92

Lesser the pka , greater is the acidic strength

Therefore, order of acidic strength is 3HSO F > 3H O

+

> 4HSO

> 3HCO

56. Which one of the following constitutes a group of the isoelectronic species?

(1) 2C ,O ,CO, NO2 2

(2) 2N , C ,CN and N2 2

+

(3)2 2CN , N , O ,C

2 2 2 (4) N , O , NO , CO

2 2 +

Sol. (2)No of e- in the species given

N2 = 14 , CN-= 14, O2

-= 17, 2C

2

= 14, NO+ =14,2O

2

=18, CO = 14, NO = 15

2N , C ,CN and N2 2

+ have fourteen electrons.

57. Which one of the following pairs of species have the same bond order?

(1) +CN and NO (2) +CN and CN

(3) O and CN2 (4) NO and CN+ +

Sol. (1)Both are isoelectronic ( have 14 e- each )and have same bond order ( 3)

Bond order = 10 4 32 = .

58. The ionization enthalpy of hydrogen atom is 1.312 106 Jmol1. The energy required to excite theelectron in the atom from n = 1 to n = 2 is

(1) 5 18.51 10 Jmol (2) 5 16.56 10 Jmol

(3) 5 17.56 10 Jmol (4) 519.84 10 Jmol

Sol. (4)


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IE is the energy required to remove an e from ground state of an atom

TE = - IE = - 1.312 106

Energy required to excite the electron from n = 1 to n = 2 is6 6

2 1 2

1.31210 1.31210E = E E

2 1

=

5 19.84 10 Jmol=

59. Which one of the following is the correct statement?(1) Boric acid is a protonic acid(2) Beryllium exhibits coordination number of six(3) Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase(4) B2H6.2NH3 is known as inorganic benzene

Sol. (3)BeCl2 and AlCl3 have bridged polymeric structure in solid phase

60. Given 0 0.72 V3 /

ECr Cr

= +

, 0 0.42 V2 /

EFe Fe

= +

.

The potential for the cell( ) ( )3+ 2+Cr | Cr 0.1M ||Fe 0.01M | Fe is

(1) 0.26 V (2) 0.399 V(3) -0.339 V (4) -0.26 V

Sol. (1)

0E =-0.72 V3+Cr/Cr

and2

0E =-0.42 VFe /Fe+

Overall cell reaction can be written as,

3+

2+

2+ 3+

(Cr Cr 3 )2

(Fe 2 Fe)3

2Cr + 3Fe 3Fe + 2Cr

e

e

+

+

Acc to Nernst eqn,

3 20

2 3

0.0591 [ ]log

6 [ ]cell cell

CrE E

Fe

+

+=


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( )( )

( )

( )

( )

2

3

2

3

0.10.05910.42 0.72 log

6 0.01

0.10.05910.30 log

6 0.01

= +

=

2

6

4

0.0591 100.30 log

6 10

0.05910.30 log10

6

=

=

0.2606 VEcell

=

61. Amount of oxalic acid present in a solution can be determined by its titration with KMnO4solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out inthe presence of HCl, because HCl(1) gets oxidised by oxalic acid to chlorine(2) furnishes H

+ions in addition to those from oxalic acid

(3) reduces permanganate to Mn2+

(4) oxidises oxalic acid to carbon dioxide and water

Sol. (3)

HCl being stronger reducing agent reduces 4MnO to Mn

2+and itself gets oxidized to Cl2.

62. The vapour pressure of water at 20oC is 17.5 mm Hg. If 18 g of glucose (C6H12O6) is added to 178.2

g of water at 20oC, the vapour pressure of the resulting solution will be

(1) 17.675 mm Hg (2) 15.750 mm Hg(3) 16.500 mm Hg (4) 17.325 mm Hg

Sol. (4)

0P Ps X

solutePs

=

17.5 0.110

PsP

s

=

17.50.01

Ps

Ps

=

17.325Ps

= mm Hg


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63. Among the following substituted silanes the one which will give rise to cross linked siliconepolymer on hydrolysis is(1) R4Si (2) RSiCl3(3) R2SiCl2 (4) R3SiCl

Sol. (2)

64. In context with the industrial preparation of hydrogen from water gas (CO + H2), which of thefollowing is the correct statement?(1) CO and H2 are fractionally separated using differences in their densities(2) CO is removed by absorption in aqueous Cu2Cl2 solution(3) H2 is removed through occlusion with Pd(4) CO is oxidised to CO2 with steam in the presence of a catalyst followed by absorption of CO2in Alkali

Sol. (4)CO is oxidized to CO2 by passing the gases and steam over iron oxide or cobalt oxide or chromium

oxide catalyst at 673 k resulting in the production of H22 3

2 2 2673

2 2 3 22 2

Fe O

kCO H O CO H

CO NaOH Na CO H O

+ +

+ +

65. In a compound atoms of element Y from ccp lattice and those of element X occupy 2/3rd oftetrahedral voids. The formula of the compound will be(1) X4Y3 (2) X2Y3(3) X2Y (4) X3Y4

Sol. (1)No. of atoms of Y = 4 (CCP or FCC)

No of tetrahedral voids = 8

No of tetrahedral voids occupied by2 16

83 3

X = =

Therefore formula of the compound = X Y

16/3 4

So, it is X16 Y12 = X4Y3


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66. Gold numbers of protective colloids A, B, C and D are 0.50, 0.01, 0.10 and 0.005, respectively. Thecorrect order of their protective powers is(1) D < A < C < B (2) C < B < D < A(3) A < C < B < D (4) B < D < A < C

Sol. (3)

Lyophobic sols are less stable than lyophilic sols. However, their stability may be increased on addition

of lyophilic sols. This phenomenon of stabilizing lyophobic sols by the addition of lyophilic colloids is

known as protection. The lyophilic colloids used for this purpose are known as protective colloids.

The protective character of various lyophilic substances is expressed in terms of gold number.

Gold number of a lyophilic sol can be defined as the minimum amount of lyophilic colloid in

milligrams which prevents the coagulation of 10 ml gold sol on addition of 1 ml of 10% NaCl solution.

Higher the gold number lesser will be the protective power of colloid.

67. The hydrocarbon which can react with sodium in liquid ammonia is(1) CH3CH2CH2C CCH2CH2CH3 (2) CH3CH2C CH(3) CH3CH = CHCH3 (4) CH3CH2C CCH2CH3

Sol. (2)

Na/Liq.NH3CH CH C CH CH CH C CNa

3 2 3 2

It is a terminal alkyne, having acidic hydrogen.Terminal alkynes react with sodium in liquid ammonia to give sodium alkylides

68. The treatment of CH3MgX with CH3C CH produces

(1) CH CH CH=CH3 2

(2) CH C CH3 3

C

(3) C CCH

3CH

3

HH

(4) CH4

Sol. (4)

Grignard reagents react with compounds having acidic or active hydrogens to form alkane

CH MgX CH C C H CH3 3 4

+


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69. The correct decreasing order of priority for the functional groups of organic compounds in the `IUPAC system of nomenclature is(1) COOH, SO3H, CONH2, CHO(2) SO3H, COOH, CONH2, CHO(3) CHO, COOH, SO3H, CONH2(4) CONH2, CHO, SO3H, COOH

Sol. (1)COOH, SO3H, CONH2, CHO

70. The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueoussolution of the corresponding salt, BA, will be(1) 9.58 (2) 4.79(3) 7.01 (4) 9.22

Sol. (3)

It is a salt of weak acid and weak base

H

H

H

1 1p 7

2 2

1 1p 7 (4.8) (4.78)

2 2

p 7 2.4 2.39 7.01

a bK Kp p= +

= +

= + =

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