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MAHESH JANMANCHI IIT JEE 2009 PAPER 2

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PAPER-2

Maximum Marks: 80

Question paper format and Marking scheme:

1.Section I contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for itsanswer, out of which only one is correct.For each question in Section I you will be awarded 3 marks if you darken the bubble corresponding to the

correct answer and zero mark if no bubbles is darkened. In case of bubbling of incorrect answer, minus one(-1) mark will be awarded.

2. Section II contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its

answer, out of which one or more is/are correct.

For each question in Section II, you will be awarded 4 marks if you darken the bubble(s) corresponding tothe correct choice(s) for the answer, and zero mark if no bubble is darkened. In all other cases, Minus one (-1)

mark will be awarded.

3. Section II contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its

answer, out of which one or more is/are correct.

For each question in Section III, you will be awarded 2 marks for each row in which you have darkened thebubble(s) corresponding to the correct answer. Thus, each question in this section carries a maximum of8

marks. There is no negative marking for incorrect answer(s) for this section.

4. Section IV contains 8 questions. The answer to each of the questions is a single digit integer, ranging from

0 to 9.

For each question in Section IV, you will be awarded 4 marks if you darken the bubble corresponding to the

correct answer and zero mark if no bubble is darkened. In all other cases, minus one (-1) mark will be

awarded.

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SECTION-I

Single Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for itsanswer, out which ONLY ONE is correct.

1. In the following carbocation, H/CH3 that is most likely to migrate to the positively charged carbon

is

(A)CH3 at C - 4 (B) H at C - 4

(B) CH3 at C - 2 (D) H at C - 2

Sol. (D)

In the given carbocation , H-shift takes place from C 2 to C 3 because of two reasons.

1) The positive charge on C 2 will be in conjugation with OH group.

2) + I effect of CH3 group at C 2 stabilises positive charge that is formed.

H C C C C CH3 3

H H

HOH CH3

12

3

4 5H shift from C toC

2 3

2 2H C C CH CH CH3

OH CH3

1

2

43

2. The correct stability order of the following resonance structures is

( )

+H C= N= N

2

I

( )

+H C N = N

2

II

( )

H C N = N2

IV

+

(A) (I) > (II) > (IV) > (III) (B) (I) > (III) > (II) > (IV)

(C) (II) > (I) > (III) > (IV) (D) (III) > (I) > (IV) > (II)

Sol. (B)

Greater the no of covalent bonds , greater is the stability. So I and III are more stable than II and IV.

Out of I and III, II is less stable as less electronegative C atom has ve charge and more electronegativeN atom has + ve charge

Out of II and IV, IV is less stable as less stable as less electronegative C atom has ve charge and more

electronegative N atom has + ve charge

Therefore the overall stability order is (I) > (III) > (II) > (IV)

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3. The spin only magnetic moment value (in Bohr magneton units ) of[Cr(CO)6] is

(A) 0 (B) 2.84(C) 4.90 (D) 5.92

Sol. (A)

[Cr(CO)6]

Cr(24) = [Ar] 3d5

4s1

CO is strong field ligand. So, forcible pairing occurs.

In Cr(CO)6 , there are no unpaired electrons. So spin only magnetic moment is zero.

4. For a first order reaction A P, the temperature (T) dependent rate constant (k) was found to

follow the equation log k = ( )1

2000 6.0T

+ . The pre-exponential factor A and the activation

energy Ea, respectively, are

(A) 6 1 11.0 10 9.2S and kJ mol (B) 1 16.0 16.6S and kJ mol

(B) 6 1 11.0 10 16.6S and kJ mol (D) 6 1 11.0 10 38.3S and kJ mol

Sol. (D)

According to Arrhenius equation,

log K = log AEa

2.303RT

Given, log K =2000

6T

Comparing the above two equations,

log A = 6. So , A = 106

sec-1

Ea

2.303RT =

2000

T

So, Ea = 2.303 x 8.314 x 2000 = 38294 J / mole = 38.3 kJ / mole

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SECTION-II

Multiple Correct Choice Type

This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for itsanswer, out which ONE OR MORE is/are correct.

5. The nitrogen oxide(s) that contain(s) NN bond(s) is(are)

(A) N2O (B) N2O3

(C) N2O4 (D) N2O5

Sol. (A,B,C)

N O2

: ( Nitrous oxide Laughing gas)

N O2 3

: (Dinitrogen trioxide)

N2O4 : (Dinitrogen tetroxide)

N2O5 : (Dinitrogen pentoxide)

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6. In the reaction ,

( ) [ ]2 6 2 422X B H BH X BH+

+

The amine(s) X is (are)

(A)NH3 (B) CH3NH2

(C) (CH3)2NH (D) (CH3)3N

Sol. (A,B,C)Trimethylamine does not react due to its bulkiness.

7. The correct statement(s) about the following sugars X and Y is(are)

(A) X is a reducing sugar and Y is a non-reducing sugar

(B) X is a non-reducing sugar and Y is a reducing sugar(C) The glycosidic linkages in X and Y are and , respectively

(D) The glycosidic linkages in X and Y are and , respectively

Sol. (B, C)

In X , reducing ends of both the moieties are involved in bonding i.e they are in acetal form.

So, it is non reducing.

The glycosidic linkage in X is .In Y, reducing end at C 1 is free( hemiacetal form). So it is reducing.

The glycosidic linkage in Y is .

8. Among the following, the state function(s) is(are)(A) Internal energy (B) Irreversible expansion work

(C) Reversible expansion work (D) Molar enthalpy

Sol. (A, D)

Internal energy and Molar enthalpy depend on the initial and final states of the system but not on the

path adopted to bring about the change. So, they are state functions.

Work is a path function, depends on path adopted.

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9. For the reduction of 3NO ion in an aqueous solution, E0 is +0.96 V. Values of E0 for some metal

ions are given below

( )2+ 0V aq + 2 e V ; E = 1.19V

( )3+ 0Fe aq + 3 e Fe ; E = 0.04V

( )3+ 0Au aq + 3 e Au ; E = + 1.40 V

( )2+ 0Hg aq + 2 e Hg ; E = + 0.86V

The pair(s) of metals that is (are) oxidized by 3NO in aqueous solution is(are)

(A)V and Hg (B) Hg and Fe

(C) Fe and Au (D) Fe and V

Sol. (A,B,D)

Given are std reduction potentials (SRP)

Greater the SRP, greater is the tendency to undergo reduction and greater is the tendency to act as

oxidant and vice versa

( )3

0

NO aq(SRP)

E =0.96V

V, Fe, Hg have less SRP compared to3NO .

So, V, Fe, Hg can be oxidized by3

NO in aqueous solution.

SECTION III

Matrix Match Type

This section contains 2 questions. Each question contains statements given in two columns, which have to be

matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are

labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE

statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be

darkened as illustrated in the following example:

If the correct matches are A p, s and t; B q and r; C p and q; and D s and t; then the correct darkening of

Bubbles.

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10. Match each of the compounds given in Column I with the reaction(s), that they can undergo given

in Column II.

Column I Column II

(A) (p) Nucleophilic substitution

(B) (q) Elimination

(C) (r) Nucleophilic addition

(D) (s) Esterification with acetic anhydride

(t) Dehydrogenation

Sol. (A p, q, t) (B p, s ) (C r, s) (D p))

A p, q, t

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B p, s

C r, s

D p

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11. Match each of the reactions given in Column I with the corresponding product(s) given in

Column II.

Column I Column-II

(A) 3Cu + dil HNO (p) NO

(B)3

Cu + conc HNO (q) NO2

(C)3

Zn + dil HNO (r) N2O

(D) 3Zn + conc HNO (s) Cu(NO3)2

(t) Zn(NO3)2

Sol. (A p, s) (B q, s) (C r, t) (D q, t))

( )3 3 223 Cu + 8 HNO (dil) 3 Cu NO + 4 H O + 2 NO ( )3 3 2 22Cu + 4HNO (conc) Cu NO + 2 H O + 2 NO

( )3 3 2 224 Zn + 10HNO (dil) 4Zn NO + 5 H O + N O

( )3 3 2 22Zn + 4HNO (conc) Zn NO + 2 H O + 2 NO

SECTION IV

Integer Answer Type

This section contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 0 to

9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For

example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then thecorrect darkening of bubbles will look like the following:

12. The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is

Sol. (6)

Alkaline oxidative fusion of MnO2 forms 2 4K MnO

( )2 2 2 4 22 MnO + 4KOH + O 2 K MnO + 2 H O

Potassium mangnate

Oxidation state of Mn in 2 4K MnO is +6.

13. The number of water molecule(s) directly bonded to the metal centre in CuSO4.5H2O is

Sol. (4)Hydrated Copper sulphate is blue in colour, called Blue vitriol.

This blue colour is due to the presence of ( )2

2 4Cu H O

+

( )4 2 2 4 24CuSO .5H O Cu H O SO .2H O

So, water molecules directly attached to Cu are 4.

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14. The coordination number of Al in the crystalline state of AlCl3 is

Sol. (6)Coordination number of Al in AlCl3 is 6.

AlCl3 exists in cubic close packed lattice with 6 coordinate layer lattice

15. In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess

oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to

298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ

K1, the numerical value for the enthalpy of combustion of the gas in kJmol

1is

Sol. (9)

Energy released at constant volume due to combustion of 3.5 gm of a gas = 2.5 0.45

So, energy released due to the combustion of 28 gm (i.e., 1 mole) of a gas

1

0.45= 28 2.5

3.5

= 9 JmolK

16. The dissociation constant of a substituted benzoic acid at 25oC is 1.0 10

4. The pH of a 0.01 M

solution of its sodium salt is

Sol. (8)- 4

aK = 1 10

pka = (- log ka ) = 4

C = 0.01 M

pH

of a salt formed from a salt of weak acid and strong base, substituted benzoic acid

1 17 log

2 2akpH p C= + +

1 17 (4) log(0.01)

2 2pH = + +

17 2 ( 2)

2pH = + +

pH = 8

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17. The total number of cyclic structural as well as stereo isomers possible for a compound with the

molecular formula C5H10 is

Sol. (7)

CH3H3C CH3H3C and its mirror image

Cis(meso) Trans

So, Total 7 isomers are possible.

18. The total number of and particles emitted in the nuclear reaction 238 21492 82U Pb is

Sol. (8)

No of particles =4

A( A = Mass no )

238 214

4

= 6=

No of particles = (2 - Z) ( Z Atomic no )

2(6) (92 82)= 2=

( ( )6 ,2 ,i.e., a total of 8 particles are emitted in the nuclear reaction 238 21492 82U Pb

19. At 400 K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most

probable speed of gas Y at 60 K. The molecular weight of the gas Y is

Sol. (4)

( )( ) ( )( )rms gas 400K mp gas 60KV =V

X Y

1 23 2

X Y

RT RT

M M=

3 400 2 60

40y

Rx Rx

M=

12030

YM=

4YM =

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2009 P2