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MAHESH JANMANCHI IIT JEE 2011 PAPER 1

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PAPER-1

Maximum Marks: 80

Question paper format and Marking scheme:

1. In Section I ( Total Marks: 21), for each question you will be awarded 3 marks if you darken ONLY the

bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases,

minus one (1) mark will be awarded.

2. In Section II (Total Marks: 16), for each question you will be awarded 4 marks if you darken ALL the

bubble(s) corresponding to the correct answer(s) ONLY and zero marks other wise. There are no negativemarks in this section.

3. In Section III (Total Marks: 15), for each question you will be awarded 3 marks if you darken ONLY the

bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases,

minus one (1) mark will be awarded.

4. In Section IV (Total Marks: 28), for each question you will be awarded 4 marks if you darken ONLY the

bubble corresponding to the correct answer and zero marks otherwise. There are no negative marks in this

section.

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SECTION I (Total Marks : 21)

(Single Correct Answer Type)

This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct.

1. Extra pure N2 can be obtained by heating

(A)3NH with CuO (B) 4 3NH NO (C) ( )4 2 72NH Cr O (D) ( )3 2Ba N

Sol. (D)

Very pure nitrogen can be obtained by the thermal decomposition of sodium or barium azide.

( )3 22 3Ba N Ba N +

2. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. Themolarity of the solution is(A) 1.78 M (B) 2.00 M (C)2.05 M (D) 2.22 M

Sol. (C)

Total mass of solution = 1000 + 120 = 1120 g

Total volume of solution in (L) = 31120

101.15

( )3

1

120 1.15 1060 1120

2.05

wMolarity

M V in L

M

=

=

=

3. Bombardment of aluminium by -particle leads to its artificial disintegration in two ways, (i) and(ii) as shown. Products X, Y and Z respectively are,

(A) proton, neutron, positron (B) neutron, positron, proton

(C) proton, positron, neutron (D) positron, proton, neutron

Sol. (A)

( )27 4 30 113 2 14 1

Al He Si P X+ +

( )27 4 30 113 2 15 0

Al He P n Y+ +

( )30 30 015 14 1

P Si e Z + +

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4. Geometrical shapes of the complexes formed by the reaction of Ni2+ with Cl , CN and H2O,

respectively, are(A) octahedral, tetrahedral and square planar (B) tetrahedral, square planar and octahedral

(C) square planar, tetrahedral and octahedral (D) octahedral, square planar and octahedral

Sol. (B)

2

4NiCl

. :

2+2Ni +4Cl NiCl4

Nickel in + 2 Oxidation state has 3d8

configuration

Cl-being weak field ligand does not forcibly pair up the electrons.

So, this complex has tetrahedral geometry with sp3

hybridisation

( )4

2Ni CN

:

( )2+2Ni +4CN Ni CN

4


configuration

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.

CN-. being strong field ligand forcibly pairs up the electrons.

So, the complex has square planar geometry with dsp2hybridisation

( )2

2 6Ni H O

+

:

( )+2

+2Ni +6H O NI H O2 2 6


configuration

With 3d8 configuration, two d-orbitals are unavailable for d2sp3 hybridisation.

So, hybridisation of Ni (II) is sp3d2 and Ni (II) with six co-ordination has octahedral geometry.

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5. The major product of the following reaction is

(A) (B)

(C) (D)

Sol. (A)

6. Among the following compounds, the most acidic is(A) p-nitrophenol (B) p-hydroxybenzoic acid

(C) o-hydroxybenzoic acid (D) p-toluic acid

Sol. (C)

Due to ortho effect, o-Hydroxy benzoic acid( Salicylic acid) is strongest acidCorrect decreasing order of Ka is

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7. AgNO3(aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution

was measured. The plot of conductance ()()()() versus the volume of AgNO3

is

(A) (P) (B) (Q) (C) (R) (D) (S)

Sol. (D)

Precipitation titration may be carried out by conductometric methods.

In reactions of the type,

AgNO3+ KCl (aq) AgCl (s) + KNO3 (aq)Where a precipitate is formed, one salt is replaced by an equivalent amount of another, e.g.

potassium chloride by KNO3and here the conductance remains almost constant before the end point.

After the equivalence point is passed, however, the excess KNO3 causes a sharp rise in the conductance

.

SECTION II (Total Marks : 16)

(Multiple Correct Answers Type)

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

which ONE OR MORE may be correct.

8. Amongst the given options, the compound(s) in which all the atoms are in one plane in all the

possible conformations (if any), is (are)

(A) (B)

(C) (D)

Sol. (B, C)

In (B) and (C) all atoms lie in same plane

Along C-C single bond conformations are possible in A in which all the atoms may not lie in sameplane.

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9. Extraction of metal from the ore cassiterite involves(A) carbon reduction of an oxide ore (B) self-reduction of a sulphide ore

(C) removal of copper impurity (D) removal of iron impurity

Sol. (A, D)Cassiterite or Tinstone (SnO2) is the ore of Tin.

SnO2 is reduced to metal using carbon at 1200 to 1300C in an electric furnace.

SnO2 + 2C Sn + 2CO

The product generally contains traces of Fe impurities , which are removed by blowing air through the

molten mixture to oxidise FeO.

Fe + O2 FeO

10. According to kinetic theory of gases(A) collisions are always elastic

(B) heavier molecules transfer more momentum to the wall of the container

(C) only a small number of molecules have very high velocity

(D) between collisions, the molecules move in straight lines with constant velocities

Sol. (A, C, D)

11. The correct statement(s) pertaining to the adsorption of a gas on a solid surface is (are)(A) Adsorption is always exothermic(B) Physisorption may transform into chemisorption at high temperature

(C) Physiosorption increases with increasing temperature but chemisorption decreases with increasing

temperature

(D) Chemisorption is more exothermic than physisorption, however it is very slow due to higher energy

of activation

Sol. (A, B, D)

(A) In case ofadsorption H is negative, S is negative and G is also negative

(B) A physical adsorption at low temperature may transform into chemisorption as the temperature is

increased.

For example, dihydrogen is first adsorbed on nickel by van der Waals forces. Molecules of hydrogen

then dissociate to form hydrogen atoms which are held on the surface by chemisorption..(D) Chemical bonds are stronger than physical or vander waals forces. So chemical adsorption or

activated adsorption is more exothermic.

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SECTION-III (Total Marls : 15)(Paragraph Type)

This section contains 2 paragraphs. Based upon one of paragraphs 2 multiple choice questions andbased on the other paragraph 3 multiple choice questions have to be answered. Each of these questions

has four choices (A), (B), (C) and (D) out ofwhich ONLY ONE is correct.

Paragraph for Question Nos. 12 and 13

An acyclic hydrocarbon P, having molecular formula C6H10, gave acetone as the only organic

product through the following sequence of reactions, in which Q is an intermediate organic

compound.

12. The structure of compound P is

(A)3 2 2 2

CH CH CH CH C C H (B)3 2 2 3

H CH C C C CH CH

(C)

H-C-C C-CH3

H C3

H C3 (D)

Sol. (D)

13. The structure of the compound Q is

(A)

H-C-C-CH CH2 3

H C3

H C3

OH

H

(B)

(C) (D)

Sol. (B)

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Paragraph for Question Nos. 14 to 16

When a metal rod M is dipped into an aqueous colourless concentrated solution of compound N,

the solution turns light blue. Addition of aqueous NaCl to the blue solution gives a white

precipitate O. Addition of aqueous NH3 dissolves O and gives an intense blue solution.

14. The metal rod M is(A) Fe (B) Cu

(C) Ni (D) Co

Sol. (B)

( )2 23 3 2Cu AgNO Cu NO AgMN Blue

+ +

Cu partially oxidizes to Cu(NO3)2 and remaining AgNO3 reacts with NaCl.

15. The compound N is(A) AgNO3 (B) Zn(NO3)2(C) Al(NO3)3 (D) Pb(NO3)2

Sol. (A)

16. The final solution contains

(A) ( )2

3 4Pb NH

+

and [ ]2

4CoCl

(B) ( )3

3 4Al NH

+

and ( )2

3 4Cu NH

+

(C) ( )3 2Ag NH+

and( )

2

3 4Cu NH

+

(D) ( )3 2Ag NH+

and ( )2

3 6Ni NH

+

Sol. (C)

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( )3 3

AgNO NaCl AgCl NaNO

ON

+ +

[ ]3 3 22 ( )AgCl NH Ag NH Cl+

+

( ) ( )2

3 4 32 44Cu NO NH OH Cu NH

+

+ ( Deep blue coloured)

SECTION-IV (Total Marks : 28)(Integer Answer Type)

This section contains 7 questions. The answer to each of the questions is a single digit integer, ranging

from 0 to 9. The bubble corresponding to the correct is to be darkened in the ORS.

17. The difference in the oxidation numbers of the two types of sulphur atoms in Na2S4O6 is

Sol. (5)

+5 ( For two S atoms bonded to Oxygen atoms) &

0 ( For two S atoms bonded to S atoms)Difference in oxidation numbers = 5

18. A decapeptide (Mol. Wt. 796) on complete hydrolysis gives glycine (Mol. Wt. 75), alanine and

phenylalanine. Glycine contributes 47.0% to the total weight of the hydrolysed products. The

number of glycine units present in the decapeptide is

Sol. (6)

Molecular weight of decapeptide = 796 g/mol

No of peptide bonds to be hydrolysed = (10 1) = 9 per molecule

( A decapeptide contains 9 peptide bonds)Total weight of H2O added = 9 18 = 162 g/mol

Total weight of hydrolysis product = 796 + 162 = 958 g

Total weight % of glycine = 47% of the hydrolysed product

Total weight of glycine in the hydrolysed product = (958 x 47) / 100

= 450.26 gm/mol

Molecular weight of glycine = 75 g/mol

Number of glycine molecules = 450/75 = 6.

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19. The work function ()()()() of some metals is listed below. The number of metals which will show

photoelectric effect when light of 300 nm wavelength falls on the metal is

Sol. (4)

The energy associated with incident photon =hc

34 8

9

6.625 10 3 10

300 10E J

=

E in eV =

34 8

9 19

6.625 10 3 10

4.14300 10 1.602 10 eV

= =

Or

Using the simplified formula0

12375( )

( )E ev

A=

12375( )

3000E ev = = 4.13

Photoelectric effect can take place only if Ephoton So, number of metals showing photo-electric effects will be (4), i.e., Li, Na, K, Mg

20. The maximum number of electrons that can have principal quantum number, n = 3, and spin

quantum number, ms =1

2 , is

Sol. (9)

For principal quantum number (n = 3)

Number of orbitals possible = n2

= 9

Number of electrons possible = 18

So, number of electrons with1

2sm = will be 9

(Half of the total will have +1/2 and remaining half will have -1/2 spin)

21. Reaction of Br2 with Na2CO3 in aqueous solution gives sodium bromide and sodium bromate with

evolution of CO2 gas. The number of sodium bromide molecules involved in the balanced chemical

equation is

Sol. (5)

2 2 3 3 23 3 5 3Br Na CO NaBr NaBrO CO+ + +

So, number of NaBr molecules = 5

Metal Li Na K Mg Cu Ag Fe Pt W

( )eV 2.4 2.3 2.2 3.7 4.8 4.3 4.7 6.3 4.75

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22. To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mol of He and

1.0 mol of an unknown compound (vapour pressure 0.68 atm. at 0oC) are introduced. Considering

the ideal gas behaviour, the total volume (in litre) of the gases at 0oC is close to

Sol. (7)

For He, n = 0.1, P = 1- 0.68 = 0.32 atm., V = ?, T = 273For an ideal gas, PV = nRT

0.32 V = 0.1 0.0821 273

V = 7 litre (unknown compound X does not follow ideal gas equation)

23. The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane

using alcoholic KOH is

Sol. (5)

Total no. of alkenes possible by dehydrobromination is 5

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2011 P1