Aieee 2009 Paper

8/2/2019 Aieee 2009 Paper

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MAHESH JANMANCHI AIEEE 2009

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Maximum Marks: 144

Question paper format and Marking scheme:

1. This question paper has 30 questions.

2. Question No. 31 to 39 and 46 to 60 consist FOUR (4) marks each and Question No. 40 to 45

consist EIGHT (8) marks each for each correct response.

3. (one fourth) of total marks allotted to each question 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 the

answer sheet.


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31. Knowing that the Chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of

the following statements in incorrect ?

(1) Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly

ionic in character.

(2) The ionic sizes of Ln (III) decrease in general with increasing atomic number.

(3) Ln (III) compounds are generally colourless.

(4) Ln (III) hydroxides are mainly basic in character.

Sol. (3)

Ln+3

compounds are mostly coloured in solid state aswellas in aqueous solutions.Colour appears due topresence of unpaired f electrons which undergo f - f transitions.

32. A liquid was mixed with ethanol and a drop of concentrated H2SO4 was added. A compound

with a fruity smell was formed. The liquid was :

(1) CH3OH (2) HCHO

(3) CH3COCH3 (4) CH3COOH

Sol. (4)

As the compound formed has fruity smell, it must be an ester.

Esterification reaction :

( ) ( ) ( ) ( )2 4H SOCH COOH + C H OH CH COOC H + H O

3 2 5 3 2 25l l l l

33. Arrange the carbanions, ( ) ( )CH C , CCl , CH CH , C H CH3 3 3 6 5 23 2 ,in order of their decresing stability:

(1) ( ) ( )C H CH > CCl > CH C > CH CH6 5 2 3 3 33 2 (2) ( ) ( )CH CH > CCl > C H CH > CH C3 3 6 5 2 32 3

(3) ( ) ( )CCl > C H CH > CH CH CH C3 6 5 2 3 32 3> (4) ( ) ( )CH C CH CH C H CH CCl3 3 6 5 2 33 2> > >

Sol. (3)

Generally + I effecting groups decrease the stability of carbanion, while I effecting groupsincrease their stability.

Benzyl carbanion is stabilized due to resonance.

20

carbanion is more stable than 3o

and Cl is I effecting group.


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34. The alkene that exhibits geometrical isomerism is :

(1) propene (2) 2-methyl propene

(3) 2-butene (4) 2- methyl -2- butane

Sol. (3)

35. In which of the following arrangements, the sequence is not strictly according to the propertywritten against it?

(1)CO2 < SiO2 < SnO2 < PbO2 : increasing oxidizing power

(2)HF < HCl < HBr < HI : increasing acid strength

(3)NH3 < PH3 < AsH3 < SbH3 : increasing basic strength

(4)B < C < O < N : increasing first ionization enthalpy.

Sol. (3)Correct basic strength is NH3 > PH3 > AsH3 > SbH3. This order is due to the fact that, on going down

the group availability of lone pair of electrons at the central atom decreases as the size of the atom

increases.

36. The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide

is :

(1) benzoic acid (2) salicylaldehyde

(3) salicylic acid (4) phthalic acid

Sol. (3)

The KolbeSchmidt reaction/Kolbe process is a carboxylation chemical reaction that proceeds by

heating sodium phenolate (the sodium salt of phenol) with carbon dioxide under pressure, then treating

the product with sulfuric acid. The final product is an aromatic hydroxy acid,which is also known

as salicylic acid

OH ONa OH OH

COOH

NaOHCO

2

06atm,140 C

COONa

+

3H O

Salicyclic Acid


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37. Which of the following statements is incorrect regarding physissorptions?

(1) It occurs because of vanderwaals forces.

(2) More easily liquefiable gases are adsorbed readily.

(3) Under high pressure it results into multi molecular layer on adsorbent surface.

(4) Enthalpy of adsorption Hadsorption

is low and positive.

Sol. (4)

Enthalpy of adsorption regarding physisorption is negative since it is an exothermic reaction.

For adsorption H, S and G all are -ve

38. Which of the following on heating with aqueous KOH, produces acetaldehyde ?

(1) CH3COCl (2) CH3CH2Cl(3) CH2ClCH2Cl (4) CH3CHCl2

Sol. (4)

Ethylidine chloride, a geminal halide on heating with aq KOH gives acetaldehyde

aq.KOHCH CHCl CH CH

3 2 3

2H O

CH CHO3

OH

OH

39. In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity

with which the position of the electron can be located is (h = 6.6 1034 kg m2s1 , mass of electron,

-31e =9.110 kg)m

(1) 41.5210 m (2) 35.1010 m

(3) 31.9210 m (4) 33.8410 m

Sol. (3)

Heisenbergs uncertainity principle,

hx. mv = 4

hx =

4mv

0.005v = 600 0.03

100 =

34

31

3

6.625 10x =

4 3.14 9.1 10 0.03

1.92 10 m

=


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40. In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer.

The reaction is

( ) ( ) ( ) ( )3 2 2 23

CH OH + O g CO g +2H O2l l

At 298K standard Gibbs energies of formation for ( ) ( )3 2CH OH ,H Ol l and ( )2CO g are

-166.2, -237.2 and -394.4 kJ mol-1

respectively. If standard enthalpy of combustion of methanol

is -726kJ mol1

, efficiency of the fuel cell will be

(1)80% (2) 87%

(3) 90% (4) 97%

Sol. (4)

( ) ( ) ( ) ( ) -13 2 2 23

CH OH + O g CO g + 2 H O ;H 726kJ mol2

l l =

Also ( )0 -1f 3G CH OH 166.2kJ moll =

( )0 -1f 2G H O 237.2 kJ moll =

( )0 -1f 2G CO 394.4 kJ moll = 0

fG = GQ (products)0

f- G (reactants)

( )[ 394.4 2 237.2 ] ( 166.2)= + 1702.6kJ mol=

% efficiency of fuel cell =G

100H

= 702.6 100726

= 97%

41. Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1

mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to

this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg)

of X and Y in their pure states will be respectively :

(1) 200 and 300 (2) 300 and 400

(3) 400 and 600 (4) 500 and 600

Sol. (3)

0 0

T X X Y YP = P X + P X

XX = mol fraction of X

YX = mol fraction of Y

0 0

X Y

1 3550 = P + P

1+3 1+3

0 0

X YP 3P= +4 4


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( ) ( )0 0X Y550 4 = P + 3P ............. 1

Further 1 mol of Y is added and total pressure increases by 10 mm Hg.

0 0

X Y1 4560= P + P

1+4 1+4

( ) ( )0 0X Y560 5 = P + 4 P ............. 2

Solving (1) and (2)

We get, 0XP = 400mm Hg

0

YP = 600mmHg

42. The half life period of a first order chemical reaction is 6.93 minutes. The time required for the

completion of 99% of the chemical reaction will be (log 2=0.301) :

(1) 230.3 minutes (2) 23.03 minutes

(3) 46.06 minutes (4) 460.6 minutes

Sol. (3)

For a first order reaction,

0.693 0.693 1min6.93

1/ 2t

= =Q

Also[ ]

[ ]

02.303 logA

t

A

=

[ ]0A = initial concentration (amount)

[ ]A = final concentration (amount)

2.303 6.93 100t log

0.6932 1

=

= 46.06 minutes

(Or)

Using simplified formula, 1 1

2 2

log( / )log( / )

t a a xt a a x

=

50

99

log(100 /100 50)

log(100 /100 99)

t

t

=

99

6.93 log 2

log100t=

t99 = 46.06 minutes


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43. Given0 0E = - 0.036V, E = - 0.439V

3+ 2+Fe /Fe Fe /Fe. The value of standard electrode potential

for the change,( )

3 2+ e Fe( ) aq

Feaq

+ + will be:

(1) -0.072 V (2) 0.385 V

(3) 0.770 V (4) -0.270

Sol. (3)

3 03 ; 0.036Fe e Fe E V + + = Q

( )0 01 1 3 0.036G nFE F = =

=+0.108 F

Similarly,2 0

2 ; 0.439Fe e Fe E V +

+ = 0 0

2 2G nFE =

( )2 0.439F=

= +0.878 F0E for

( ) ( )3 2

aqFe e Fe aq+ ++ :

0 0G nFE = 0 0 0

1 2G G G= Q

0 0.108 0.878G F F= Q 0 0.108 0.878FE F F = + Q

0 0.878 0.108E = Q = 0.77 V

44. On the basis of the following thermochemical data :

( )( )0 0

aqG H+ =

( ) ( )2 ; 57.32H O l H OH aq H kJ+

+ =

( ) ( ) ( )2 2 21

; 286.22

H g O g H O l H kJ+ =

The value of enthalpy of formation of OH-ion at 25

0C is:

(1) -22.88 kJ (2) -228.88 kJ(3) +228.88 kJ (4) -343.52 kJ

Sol. (2)

Adding the two equations given,

( ) ( ) ( )2 2

1; 57.32 ( 286.2)

2g g aq

H O H OH H kJ+

+ + = + (Here 0fH of ( ) 0aqH+

= )

0H of OH-= - 228.88 kJ


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45. Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom?

(1) 108 pm (2) 127 pm

(3) 157 pm (4) 181 pm

Sol. (2)

For FCC lattice,

2 4a r= (Particles touch each other along the face- diagonal)

2 2 361

4 4

ar

= =

= 127 pm

46. Which of the following has an optical isomer?

(1) ( )+

3 3Co NH Cl (2) ( ) ( )

2+

3 2Co en NH

(3) ( ) ( )3+

2 4Co enH O (4) ( ) ( )

3+

32 2Co en NH

Sol. (4)

It is an octahedral complex of the type ( ) 22AAM X

Where AA is bidentate ligand.

47. Solid Ba (NO3)2 is gradually dissolved in a 1.0 104

M Na2CO3 solution. At what concentration of

Ba2+

will a precipitate begin to form ?(Ksp for BaCO3 = 5.1 10-9

).

(1) 54.1 10 M (2) 55.1 10 M

(3) 88.1 10 M (4) 78.1 10 M

Sol. (2)


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( )3 3 3 32 2Ba NO CaCO BaCO NaNO+ +

[ ]2 43 2 3 10CO Na CO M = =

( )

2 23

9 2 45.1 10 10

2 55.1 10

K Ba COsp

Ba

Ba

+ =

+ =

+ =

At this concentration , precipitation just starts.

48. Which one of the following reactions of Xenon compounds is not feasible?

(1) XeO +6 HF XeF +2 H O

3 6 2

(2)2 3 2

3 XeF +6 H 2Xe+XeO 12 1.54

O HF O + +

(3) 2 22 XeF + 2 H 2Xe + 4HF + O2O

(4) ( )7XeF + RbF Rb XeF6

Sol. (1)

This reaction is not feasible, as XeF6 formed will further produce XeO3 by hydrolysis.

6 2 3XeF + 3 H XeO + 6HFO

49. Using MO theory predict which of the following species has the shortest bond length?

(1) 22

O+ (2)

2O

+

(3) 2O

(4) 22O

Sol. (1)

Bond length1

bond order

Bond order =no.of bonding e-no.of antibonding e

2

E.C of O2 molecule is

( ) 2 2 2 2 2 1 121

2 2 21 2 2 2 2 2 216

z x y x ysps s s p p p p

O e

= = =

( ) 2 2 2 2 2 0 021

2

2 2 21 2 2 2 2 2 214


O e +

= = =

Bond order =10 4

32

= .


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( ) 2 2 2 2 2 1 021

2 2 21 2 2 2 2 2 215


O e +

= = =

Bond order = 10 5 2.52 = .

( ) 2 2 2 2 2 2 121

2 2 21 2 2 2 2 2 217


O e

= = =

Bond order =10 7

1.52

= .

( ) 2 2 2 2 2 2 221

2

2 2 21 2 2 2 2 2 218


O e

= = =

Bond order =10 8

12

= .

So, O22+ has shortest bondlength.

50. In context with the transition elements, which of the following statements is incorrect?

(1) In addition to the normal oxidation states, the zero oxidation state is also shown by these

elements in complexes.

(2) In the highest oxidation states, the transition metal show basic character and form cationic

complexes.

(3) In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d

electrons are used for bonding.

(4) Once the d5

configuration is exceeded, the tendency to involve all the 3d electrons in bonding

decreases.

Sol. (2)

In higher oxidation states transition elements, they no longer have tendency to give away electrons.So,

they show acidic nature and form anionic complexes.

51. Calculate the wavelength (in nanometer) associated with a proton moving at 3 11.0 10 ms

(Mass of proton = 271.67 10 kg and h = 346.63 10 Js

):

(1) 0.032 nm (2) 0.40 nm

(3) 2.5 nm (4) 14.0 nm

Sol. (2)

De Broglies equation,34

27 3

6.63 10

1.67 10 10

0.40 nm

h

mv

= =

=


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52. A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following

statements is correct regarding the behaviour of the solution ?

(1) The solution formed is an ideal solution

(2) The solution is non-ideal, showing +ve deviation from Raoults law.

(3) The solution is non-ideal, showing ve deviation from Raoults law.

(4) n-heptane shows +ve deviation while ethanol shows ve deviation from Raoults law.

Sol. (2)

The solution is non-ideal, showing +ve deviation from Raoults law.The interactions between n

heptane and ethanol are weaker than that in pure components. Originally, ethanol molecules are held

together by strong H bonds. These forces become weaker on mixing n-Heptane.

53. The number of stereoisomers possible for a compound of the molecular formula

CH3 CH = CHCH OH Me is :

(1) 3 (2) 2

(3) 4 (4) 6

Sol. (3)

Given compound has a C=C bond and one chiral carbonAbout the double bond, two geometrical isomers are possible and the compound is having one chiral

carbon. So each geometrical isomer will have d and l isomers

Therefore total stereoisomers = 4.

54. The IUPAC name of neopentane is

(1) 2-methylbutane (2) 2, 2-dimethylpropane

(3) 2-methylpropane (4) 2,2-dimethylbutane

Sol. (2)

H C C CH3 3

CH3

CH3

Neopentane is

IUPAC name is 2, 2-Dimethylpropane

55. The set representing the correct order of ionic radius is(1) 2 2Li Be Na Mg

+ + + +> > > (2)

2 2Na Li Mg Be+ + + +> > >

(3) 2 2Li Na Mg Be+ + + +

> > > (4)2 2Mg Be Li Na+ + + +> >

Sol. (2)

In a period, from left to right, ionic radii decrease due to increase in effective nuclear charge as the

additional electrons are added to the same principal energy level.

In a group, from top to bottom, ionic radii increase because the additional electrons are added to samenew principal energy level.

Li+

and Mg2+

are diagonally related.


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56. The two functional groups present in a typical carbohydrate are:

(1) OH and COOH (2) CHO and COOH

(3) > C = O and OH (4) OH and CHO

Sol. (3)

Carbohydrates are polyhydroxy carbonyl compounds.So, the two functional groups present in

carbohydrates are aldehyde and ketone

57. The bond dissociation energy of B F in BF3 is 646 kJ mol1

whereas that of C-F in CF4 is 515 kJ

mol1

. The correct reason for higher B-F bond dissociation energy as compared to that of C- F is:

(1) smaller size of B-atom as compared to that of C- atom

(2) stronger bond between B and F in BF3 as compared to that between C and F in CF4

(3) significant p - p interaction between B and F in BF3 whereas there is no possibility of such

interaction between C and F in CF4 .

(4) lower degree of p - p interaction between B and F in BF3 than that between C and F in CF4.

Sol. (3)

In BF3, B is sp2 hybridised and has vacant 2p orbital which overlaps laterally with a filled 2p

orbital of F forming strong p p bond

In CF4, C doesnot have vacant p orbitals to undergo p p bonding.So, bond energy of B F bond is greater than that of C - F bond.

58. In Cannizzaro reaction given below

( )( )..

22 2

OHPh CHO Ph CH OH Ph C O

+

the slowest step is:

(1) the attack of( )OH

at the carboxyl group

(2) the transfer of hydride to the carbonyl group

(3) the abstraction of proton from the carboxylic group

(4) the deprotonation of Ph CH2OH

Sol. (2)

Hydride transfer is the slowest step.


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Mechenism of Cannizaro reaction :

Step 1: Addition of hydroxide ion to carbonyl group (Fast)

OH C6H5 C

H

O C6H5 C

H

O

OH

Hydroxy alkoxide ion

Step 2: Transfer of hydride ion to another aldehyde molecule. This is RATE DETERMINING STEP

Step 3: Transfer of proton from acid to phenoxide (Rearrangement)

59. Which of the following pairs represents linkage isomers?

(1) ( ) [ ]3 44Cu NH PtCl and ( ) [ ]3 44NH CuClPt

(2) ( ) ( )3 2 2P Ph NCSPd and ( ) ( )3 2 2P Ph SCNPd

(3) ( )3 3 45CO NH NO SO and ( )3 4 35CO NH SO NO

(4) ( )2 3 24PtCl NH Br and ( )2 3 24PtBr NH Cl

Sol. (2)

Linkage isomerism is exhibited by complexes containing ambidentate ligand

NCS- is ambidentate ligand and it can be linked through N (or) S

( ) ( )3 2 2P Ph NCSPd (linked through N) and

( ) ( )3 2 2P Ph SCNPd (linked through S)


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60. Buna-N synthetic rubber is a copolymer of:

(1)

Cl|

H C=CH C CH2 2

and H C=CH CH-CH2 2

(2) H C=CH CH CH2 2

and H C -CH=CH5 6 2

(3) H C=CH-CN2

and H C=CH CH=CH2 2

(4) H C=CH CN2

and H C=CH C =CH2 2|

CH3

Sol. (3)

Buna-N in a copolymer of

Buta- 1,3-diene and acrylonitrile

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