2009 P1

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MAHESH JANMANCHI IIT JEE 2009 PAPER 1

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PAPER-1

Maximum Marks: 80

Question paper format and Marking scheme:

1. Section I contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its

answer, 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 thecorrect answer and zero mark if no bubble is darkened. In case of bubbling of incorrect answer, minus one

(-1) mark will be awarded.

2. Section II contains 4 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 to the

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 III contains 2 groups of questions. Each group has 3 questions based on a paragraph. Each questionhas 4 choices (A), (B), (C) and (D) for its answer, out of which only one is correct.

For each question in Section III, you will be awarded 4 marks if you darken the bubble(s) corresponding to

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

awarded

4. Section IV contains 2 questions. Each question has four statements (A, B, C and D) given in column I and

five statements (p, q, r, s and t) in Column II. Any given statement in column I can have correct matching with

one or more statements(s) given in column II

For each question in Section IV, 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 of 8marks. There is no negative marking for incorrect answer(s) for this section.

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SECTION-I

Single Correct Choice Type

This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) f or its

answer, out which ONLY ONE is correct.

1. The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 ×××× 105

atm. The mole

fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at

298 K and 5 atm pressure is

(A) 44.0 10−× (B) 54.0 10−

×

(C) 45.0 10−× (D) 64.0 10−

×

Sol. (A)

According to Henry’s law,

2Np = KH N

2

χ

(2N

p is partial pressure of N2 in gaseous phase,N

2

χ is the molefraction of N2 in solution phase)

Given, total pressure = 5 atmMole fraction of N2 = 0.8

Partial pressure of N2 = 0.8 x 5 = 4 atm

50.8 5 1 10

2 N

χ × = × ×

2

2

54 1010

N

N

n

n

−× =

+

210 N n + ≈ 10

2

44 10 N n−⇒ = ×

2. The correct acidity order of the following is

OH OH

Cl

COOH COOH

CH3

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

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

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Sol. (A)

Acids are stronger than phenols.i.e III and IV are stronger than I and II.Out of III and IV, III is stronger, due to +I and Hyperconjugation of – CH 3 group

Out of I and II , II is stronger, due to – I effect of - Cl group.

OH OH

Cl

COOH COOH

CH3

(I) (II) (III) (IV)pKa=9.98 pKa=9.38 pKa=4.17 pKa=4.37

Decreasing order of acidic strength : III > IV > II > I

3. The reaction of P4 with X leads selectively to P4O6. The X is

(A) Dry O2 (B) A mixture of O2 and N2 (C) Moist O2 (D) O2 in the presence of aqueous NaOH

Sol. (B)

2N

4 2 4 6P +3O P O → (exclusively)

N2 prevents further oxidation of P4O6 to P4O10.

4. Among cellulose, poly(vinyl chloride), nylon and natural rubber, the polymer in which the

intermolecular force of attraction is weakest is

(A) Nylon (B) Poly(vinyl chloride)

(C) Cellulose (D) Natural Rubber

Sol. (D)

Natural rubber has weak vander Waal’s forces , which are weakest forces of attraction.

Cellulose and nylon are fibres. So, their intermolecular forces are strongest.

Polyvinyl chloride is a thermoplastic polymer. So, intermolecular forces are intermediate to

elastomers and fibres

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5. Given that the abundances of isotopes54

Fe,56

Fe and57

Fe are 5%, 90% and 5% respectively, the

atomic mass of Fe is

(A) 55.85 (B) 55.95(C) 55.75 (D) 56.05

Sol. (B)

The average isotopic mass or atomic massi i

i

x=

x

m∑∑

(where mi is the mass of ith isotope and xi is the abundance of ith isotope)

54 5 56 90 57 5

100

× + × + ×=

= 55.95

6. The IUPAC name of the following compound is

CN

OH

Br

(A) 4-Bromo-3-cyanophenol (B) 2-Bromo-5-hydroxybenzonitrile

(C) 2-Cyano-4-hydroxybromobenzene (D) 6-Bromo-3-hdyroxybenzonitrile

Sol. (B)

-CN is given highest priority.

Least sum rule is to be followed. Do not go for –OH.

Therefore, IUPAC name of the compound is 2-Bromo-5-hydroxybenzonitrile

(should be 2-Bromo-5-hydroxybenzenecarbonitrile, as per latest IUPAC recommendation)

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7. Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective coagulating agent

for Sb2S3 sol is(A) Na2SO4 (B) CaCl2

(C) Al2(SO4)3 (D) NH4Cl

Sol. (C)

Sb2S3 is a negative sol. So, a positive ion can cause coagulation.

Acc to Hardy – Schulze rule, the greater the valence of the flocculating ion added, the greater is its

power to cause precipitation.

Therefore, Al2(SO4)3 is the most effective coagulating agent for Sb2S3 sol.

Order of effectiveness of cations given is : 3+ ++ + +

4

Al >Ca >Na >NH

8. The term that corrects for the attractive forces present in a real gas in the van der Waals equation

is

(A) nb (B)2

2

an

V

(C)2

2

an

V− (D) – nb

Sol. (B)

In Vanderwaal’s equation, ( )

2

2anP+ V-nb =nRTV

, the pressure correction term

2

2a nV

is a measure

of force of attraction among the molecules.

SECTION-II

Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its

answer, out which ONE OR MORE is/are correct.

9. The compound(s) formed upon combustion of sodium metal in excess air is(are)

(A) Na2O2 (B) Na2O (C) NaO2 (D) NaOH

Sol. (A) or (A , B) ( By IITs)

On combustion in excess of air, sodium forms monoxide Na2O and peroxide Na2O2

4 Na +O2 → 2Na2 O (oxide)

2Na +O2 →Na2O2 (peroxide)

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10. The correct statement(s) about the compound H3C(HO)HC −−−− CH = CH – CH (OH) CH3 (X)

is(are)

(A) The total number of stereoisomers possible for X is 6(B) The total number of diastereomers possible for X is 3(C) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for

X is 4

(D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for

X is 2

Sol. (A, D)

Given molecule has 2 asymmetric centres and a double bond .

So, the molecule will have 6 stereo isomers.R cis R , S cis S, R cis S and R trans R, S trans S, R trans S.

If the stereochemistry about the double bond in X is cis or trans , it gives a pair of enantiomers.

11. The compound(s) that exhibit(s) geometrical isomerism is(are)

(A) ( ) 2Pt en Cl (B) ( ) 22Pt en Cl

(C) ( ) 2 22Pt en Cl Cl (D) ( )3 22

Pt Cl NH

Sol. (C, D)

Octahedral complexes of the type M(AA)2X2 , where AA is a symmetric bidentate ligand exhibit

geometrical isomerism.

Square planar complexes of the type MA2X2 exhibit geometrical isomerism.

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12. The correct statement(s) regarding defects in solids is(are) (A) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion

(B) Frenkel defect is a dislocation defect

(C) Trapping of an electron in the lattice leads to the formation of F-center(D) Schottky defects have no effect on the physical properties of solids

Sol. (B, C) or (C) (By IITs)Frenkel defect is shown by ionic substance in which there is a large difference in the size of ions.

Example, ZnS, AgCl, AgBr and AgI due to small size of Zn2+ and Ag+ ions.

Frenkel defect is also called dislocation defect. Because in this defect, the smaller ion (usually cation) is

dislocated from its normal site to an interstitial site It creates a vacancy defect at its original site and aninterstitial defect at its new location, a combination of vacancy and interstitial defects

In Metal excess defect due to anionic vacancies, anionic sites are occupied by unpaired electrons.

These sites are called F-centres (German word Farbenzenter for colour centre). They impart

colour to the crystals.

Schottky defect decreases the density of the substance due to paired cation and anion vacancies

(Schottky pair)

SECTION-III

Comprehension Type

This section contains 2 groups of questions. Each group has 3 multiple choice question based on a paragraph.

Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.

Paragraph for Question Nos. 13 to 15

A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed bydehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes

intramolecular aldol reaction to give predominantly S.

13. The structure of the carbonyl compound P is

(A) (B)

(C) (D)

Sol. (B)

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14. The structure of the product Q and R, respectively are

(A)

(B)

(C)

(D)

Sol. (A)

15. The structure of the product S is

(A) (B)

(C) (D)

Sol. (B)

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Solutions for question no’s. 13 to 15

Paragraph for Question Nos. 16 to 18

p-Amino- N , N -dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with

a few drops of aqueous solution of Y to yield blue coloration due to the formation of methylene blue. Treatment

of the aqueous solution of Y with the reagent potassium hexacyanoferrate (II) leads to the formation of an

intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, treatment of the

solution of Y with the solution of potassium hexacyanoferrate (III) leads to a brown colouration due to the

formation of Z.

16. The compound X is(A) NaNO3 (B) NaCl

(C) Na2SO4 (D) Na2S

Sol. (D)

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17. The compound Y is

(A) MgCl2 (B) FeCl2

(C) FeCl3 (D) ZnCl2

Sol. (C)

18. The compound Z is

(A) ( )2 6Mg Fe CN (B) ( )

6Fe Fe CN

(C) ( )4 6 3Fe Fe CN (D) ( )2 3 6 2

K Zn Fe CN

Sol. (B)

Solutions for the question nos. 16 to 18

+ +

2 2Na S + 2 H H S + 2 Na

( ) X

→

Methylene blue

( ) ( )3 4 46 6 34 FeCl + 3 K Fe CN Fe Fe CN + 2 KCl

→

Blue ppt

( ) ( )3 3 6 6FeCl + K Fe CN Fe Fe CN + 3KCl

→

Brown coloration (Z)

( ) 2X - Na S

( ) 3Y - FeCl

( ) ( )6

Z - Fe Fe CN

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SECTION-IV

Matrix – Match Type

This section contains 2 questions. Each question contains statements given in two columns, which have to bematched. 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 will look like the following:

19. Match each of the compounds in Column I with its characteristic reaction(s) in Column II.

Column – I Column-II

(A) 3 2 2CH CH CH CN (p) Reduction with Pd−C/H2

(B) 3 2 3CH CH OCOCH (q) Reduction with SnCl2 /HCl(C)

3 2CH -CH=CH-CH OH (r) Development of foul smell on treatment with

Chloroform and alcoholic KOH

(D)3 2 2 2 2

CH CH CH CH NH (s) Reduction with diisobutylaluminium hydride

(DIBAL-H)

(t) Alkaline hydrolysis

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

A – p, q, s, t :

pd C −

2 H 3 2 2CH CH CH CN 3 2 2 2 2CH CH CH CH NH

3 2 2CH CH CH CN2SnCl

HCl3 2 2 4

CH CH CH CHO + NH Cl

3 2 2CH CH CH CN 3 2 2 2 2CH CH CH CH NHDIBAL-H

3 2 2CH CH CH CN2 / OH H O

−

3 2 2 3CH CH CH COOH + NH

B – s, t :

3 2 3CH CH OCOCH 3 22 CH CH OHDIBAL-H

2 / OH H O−

3 2 3CH CH OCOCH 3 3 2CH COOH + CH CH OH

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C – p :

pd C −

2 H 3 2

CH -CH=CH-CH OH3 2 2 2

CH CH CH CH OH

D – r : ( Carbyl amine reaction)

3 2 2 2 2CH CH CH CH NH3CHCl

alc KOH

3 2 2 2CH CH CH CH NC

20. Match each of the diatomic molecules in Column I with its property/properties in Column II.Column-I Column-II

(A) B2 (p) Paramagnetic

(B) N2 (q) Undergoes oxidation

(C)2

O− (r) Undergoes reduction

(D) O2 (s) Bond order 2≥

(t) Mixing of ‘s’ and ‘p’ orbitals

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

Note :

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

2

In case of species of elements upto N2, difference in energy between 2s and 2p is small. So they

intermix. Whereas in case of species of elements after N2, difference in energy between 2s and 2p islarge. So they cannot intermix.

A species undergoes oxidation,if removal of electron leads to stability. Generally removal of electron

from non bonding molecular orbitals leads to stability, as bond order increases.

A species undergoes reduction,if addition of electron leads to stability. Generally addition of electron tobonding molecular orbitals leads to stability,as bond order increases.

A – p, r, t :

( ) 2 2 2 1 12

12 1 2 2 2 2

10 x ys

s s s p p B e σ σ σ σ π π

− ∗ ∗= =

Bond order =6 4

12

−= .

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It has unpaired electrons. So, paramagnetic.

Undergoes mixing of s and p orbitals.

Undergoes reduction.

B – s, t :

( ) 2 2 2 2 22

12 2 21 2 2 2 2

14 z x ys

ps s s p p N e σ σ σ σ π π σ

− ∗ ∗= =

Bond order =10 4

32

−= .

Diamagnetic since it has paired electrons.

Undergoes mixing of s and p orbitals.Doesnot undergo oxidation or reduction since it is stable as such.

C – p, q :

( ) 2 2 2 2 2 2 121

2 2 21 2 2 2 2 2 217

z x y x ys ps s s p p p p

O e σ σ σ σ σ π π π π − − ∗ ∗

= = ∗ = ∗

Bond order =10 7

1.52

−= .

It has unpaired electrons. So, paramagnetic.Undergoes oxidation.

D – p, q, s :

( ) 2 2 2 2 2 1 121

2 2 21 2 2 2 2 2 216

z x y x ys ps s s p p p p

O e σ σ σ σ σ π π π π − ∗ ∗

= = ∗ = ∗

Bond order =10 6

22

−= .

It has unpaired electrons. So, paramagnetic

Undergoes oxidation.

.

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2009 P1