Class11 12 Nsec Incho

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PATHFINDER-2013

FIITJEE ALL INDIA SCREENING TEST

(NSEC + INCho)

Time: 3 Hours Maximum Marks: 230.

A. General: 1. This Question Paper contains 2 Parts. Part-A & Part-B.

2. No additional sheets will be provided for rough work.

3. Blank papers, clipboards, log tables, slide rules, calculators, cellular phones, pagers and electronic gadgets in any form are not allowed.

4. The answer sheet, a machine-gradable Objective Response Sheet (ORS), is provided separately for Part-A.

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B. Filling the top-half of the ORS: 7. Write your Name & Registration No. in the space provided in the bottom of this booklet.

8. Darken the appropriate bubbles below your Registration No. with HB Pencil.

C. Question paper format and Marking scheme: 9. The question paper consists of 2 Parts.

(i) Part-A (01 to 60) contains 60 multiple choice questions which have only one correct answer. Each question carries +3 marks for correct answer and 1 mark for wrong answer.

(ii) Part-B (1 to 3) consists of 3 Subjective Questions each question containing sub parts with marks indicated making up total of 50 Marks for 3 questions.

Registration Number :

Name of the Candidate : ___________________________________________________________

Test Centre : ___________________________________________________________


P-AIST-2013-NSEC+INCho-SP-2

PART A

1. When NH4Cl is dissolved in H2O, the temperature of H2O decreases it means: (where, Hs = enthalpy of solution, Hi = enthalpy of ionisation and Hh = enthalpy of hydration of

ions) (A) Hs = +ve, Hi < Hh (B) Hs = ve, Hi < Hh (C) Hs = ve, Hi > Hh (D) Hs = +ve, Hi > Hh

2. An electron in an atom jumps in such a way that its kinetic energy changes from x to x4

, the

change in potential energy will be:

(A) 3 x2

+ (B) 3 x8

(C) 3 x4

+ (D) 3 x4

3. The most reactive among the following for electrophilic substitution reaction is:

(A)

(B)

F

(C)

Cl

(D)

Br

4. The product which is not formed during Kolbes electrolysis of propanoic acid. (A) CH3CH2CH2CH2CH3 (B) CH3CH2COOCH2CH3 (C) CH2 = CH2 (D) None of the above 5. 2-halohexane is treated with CH3ONa and CH3OH. The highest ratio of amount of 1-hexene to 2-

hexene will be obtained, if halogen is (A) F (B) Cl (C) Br (D) I

Rough work



6. Match the following: Reaction Product

(a) Borax (i) Na[B(OH)4] + H3BO3 (b)

4Borax NH Cl+ (ii) NaBO2 + B2O3

(c) HClBorax (iii) BN (d)

2Borax H O+ (iv) H3BO3 a b c d (A) (ii) (iii) (i) (iv) (B) (iii) (ii) (iv) (i) (C) (ii) (iii) (iv) (i) (D) (ii) (i) (iv) (iii)

7. The correct set of all the four quantum numbers for the 21st electron in scandium (Sc = 21) is:

(A) n = 3, l = 1, m = - 2, S = 12

+ (B) n = 3, l = 2, m = - 2, S = 12

+

(C) n = 4, l = 0, m = - 1, S = 12

+ (D) n = 4, l = 0, m = - 2, S = 12

8. Identify the incorrect statement of the following: (A) Carbondioxide is isostructural with laughing gas. (B) Borazole is isostructural with benzene. (C) Azide ion is isostructural with polyiodide ion ( )3I (D) Trimethyl amine is isostructural with trisilylamine. 9. In which of the following reactions, hybridization of underlined atoms do not undergo any change?

(A) ( ) ( )3 32 6 6Fe H O 6CN Fe CN+ + (B) 3 4LiH AlH LiAlH+ (C) 2 5 2 3N O NO NO

+ + (D) ( )3 2 4 4 422NH H SO NH SO+ 10. The total volume of 0.1 M KMnO4 solution that are needed to oxidize 100 mg each of ferrous

oxalate and ferrous sulphate in a mixture in acidic medium is: (A) 4.16 ml (B) 1.32 ml (C) 5.48 ml (D) 2.64 ml

Rough work



11. Identify the correct statement among the following: (A) The number of isomers of compound with molecular formula C4H6 are 6 (B) Total number of aliphatic ethers corresponding to the molecular formula C5H12O are 6 (C) Total number of optically active isomers of HOOC-CH(OH) CH (OH) COOCH3; are 4 (D) All of the above 12. Which of the following carbananion is most stable? (A) CH2

(B) CH2

(C)

HC CH3

(D) HC CH3

NO2

13. The temperature at which the reaction, ( ) ( ) ( )2 21Ag O s 2Ag s O g2 + is at equilibrium is , given 1 1 1H 30.5 kJmol and S 0.066 kJK mol = =

(A) 462.12 K (B) 362.12 K (C) 262.12 K (D) 562.12 K 14. Which of the following exist as dimer? (A) CH3Li (B) Si(CH3)4 (C) Al(CH3)3 (D) Be(CH3)2 15. Solubility of calcium phosphate (molecualr mass M) in m g per 100 mL at 25oC, its solubility

product at 25oC will be approximately.

(A) 5

7 m10M (B)

57 m10

M

(C) 5

5 m10M (D)

53 m10

M

Rough work



16. 1 M solution of CH3COOH should be diluted to ...................... times so that pH is doubled. (A) four times (B) 5.55 104 times (C) 5.55 106 times (D) 102 times 17. The dissociation constants of HA and HB are 6 104 and 2.0 104 respectively. The pH of a

solution which is mixture of 0.5 M of HA and 0.5 M of HB is (A) 2.00 (B) 1.70 (C) 2.30 (D) 1.90 18. pKb of NH3 is 4.74 and pKb of A, B and C are 4, 5 and 6 respectively. Aqueous solution of 0.01

M has pH in the incrasing order. (A) NH4A < NH4B < MH4C (B) NH4C < NH4B < NH4A (C) NH4C < NH4A < NH4B (D) All have equal pH, being salt of weak acid and weak base 19. Which of the following can be separated by H2S in dil HCl? (i) Bi3+Mn2+ (ii) Al3+Hg2+ (iii) Zn2+Cu2+ (iv) Ni2+Cu2+ (A) (i) and (ii) (B) (i) and (iii) (C) (i), (ii), (iii) (D) All 20. Kc for the reaction [Ag(CN)2]-1 ZZZXYZZZ Ag+ + 2CN-, the equilibrium constant at 25oC is 4 10-19,

then the silver ion concentration in the solution which was originally 0.1 M in KCN and 0.03 M in AgNO3 is

(A) 7.5 1018 (B) 7.5 10-18 (C) 7.5 1019 (D) 7.5 10-19 21. Important ore of lead is: (A) bauxite (B) galena (C) cinnabar (D) malachite 22. The ratio of the radius difference between 4th and 3rd orbit of H-atom and that of Li2+ ion is (A) 1 : 1 (B) 3 : 1 (C) 3 : 4 (D) 1 : 3

Rough work



23. For the reaction 3 22 4 2 4 2FeS KMnO H Fe H SO Mn H O

+ + ++ + + + + , if the molar mass of FeS2 is M, then equivalent mass of FeS2 would be equal to

(A) M (B) M/10 (C) M/11 (D) M/15 24. Borane have general formula (A) BnHn+4 (B) BnHn+6 (C) BnH2n+2 (D) Both (A) and (B) 25. NaOCl on heating with NH3 at high temperature gives: (A) H2N NH2(hydrazine) (B) Dinitrogen (N2) (C) H2(Dihydrogen) (D) Sodiumamide(NaNH2) 26. S2 in vapour phase is (A) paramagnetic (B) diamagnetic (C) ferromagnetic (D) doesnt exist 27. Which of the following statements is correct? (A) B2H6 reacts with excess of NH3 at high temperature to form an ionic substance (B) B2H6 reacts with excess of NH3 at low temperature to form a white slippery solid (C) Boron nitride has layer structure (D) orthoboric acid is monoprotic acid 28. CH2 = CHCH2Cl and CH3CH = CH2Cl can be distinguished by the reaction with (A) Br2 in CCl4 (B) K2Cr2O7/H2SO4 (C) AgNO3 (D) NaOI 29. A sample of peanut oil weighing 1.576 g is added to 25 ml of 0.421 M KOH. After saponification is

complete, 8.46 ml of 0.273 M H2SO4 is needed to neutralize excess of KOH. The saponification number of peanut oil is (saponification number is defined as the milligrams of KOH consumed by 1 g of oil)

(A) 209.6 (B) 108.9 (C) 98.9 (D) 218 .9 30. At which of the following four conditions, the density of nitrogen will be the largest? (A) STP (B) 273 K and 2 atm (C) 546 K and 1 atm (D) 546 K and 2 atm

Rough work



31. Cl2 gas has been introduced in an 8 litre cubical container at pressure 5 atm and 300 K temperature. The following equilibrium has been achieved.

Al(s) + 3/2Cl2(g) ZZZXYZZZ AlCl3(s) If the thickness of inner side of walls has been decreased by 0.04 mm at equilibrium, then find out

Kp for the reaction ( Al = 1.13 g/cc) (A) (2.7)+3/2 (B) (2.7)3/2 (C) (5.4)+3/2 (D) (5.4)3/2 32. A sample of hard water contains 0.005 mole of calcium chloride per litre. What is the minimum

concentration of sodium sulphate which must be added for removing the Ca2+ ions from this water sample? (Ksp(CaSO4) = 2.4105)

(A) 4.8102 (B) 4.8103 (C) 2.4102 (D) 2.4103 33. 2 2 3N 3H 2NH heat.+ +ZZZXYZZZ The activation energy for the forward as well as backward reaction is

decreased by 100 J, then the equilibrium amount of NH3 will (A) increase (B) decrease (C) remain constant (D) cannot be predicted 34. The coefficient of S in the balanced reaction is 32 5 2 2 7 3 4As S H Cr O H AsO S Cr

++ + + (A) 5 (B) 10 (C) 15 (D) 1 35. The pH of a saturated solution of M(OH)2 is 10. The Ksp value at this temperature is (A) 5 1013 (B) 5 1010 (C) 5 108 (D) None of the above 36. Which of the following statement is not correct? (A) Hybridization is the mixing of atomic orbitals prior to their combining into molecular orbitals. (B) sp2 hybrid orbitals are formed from two p-atomic orbitals and one s-atomic orbital. (C) dsp3 hybrid orbitals are all at 90o to one another. (D) d2sp3 hybrid orbitals are directed towards the corner of regular octahedral.

Rough work



37. Which of the following statement is/are not true for bond ? 1. It is formed by the overlapping of p p or p d orbitals. 2. It is stronger than sigma bond. 3. It is formed when bond already exists. 4. It is resulted from sideways overlapping of orbitals. (A) 1, 2, 3, 4 (B) 2, 3 and 4 (C) 2 and 4 (D) 1, 2 and 4 38. A sample of water from newly dug well read pH = 7 in the field study on the night of

31st Dec. 2005 in Shimla. The sample was (A) basic (B) acidic (C) neutral (D) cannot be predicted 39. The circumference of the 4th orbit of hydrogen is 5.32 nm. The wavelength of electron revolving in

the first Bohr orbit will be (A) 0.133 nm (B) 1.33 nm (C) 4.32 nm (D) None of the above 40. In which reaction, product formation occurs by Saytzeff rule?

(1) ONa

CH3

Br

CH3 , (2)

H2C

CH3CH3

OH

CH3

2 4conc. H SO

(3)

CH3

CH3Ph

Br

aq. KOH/ (4) ClF3C

Cl CH3

Cl / h2

Select the correct alternative from the codes given below: Codes: (A) 1, 2 and 3 (B) 2 and 3 (C) 1, 2 and 4 (D) 2, 3 and 4 41. What will be the molar solubility of molar solution of AgCl at 25oC if specific conductivity of

saturated AgCl solution is 2.3 106 S.cm1 at 25oC and molar conductivities of Ag+ and Cl at infinite dilution are 61.9 and 76.3 Scm2mol1 respectively?

(A) 2.382 103 M (B) 1.66 105 M (C) 5 108 M (D) 7.2 106 M

Rough work



42. Following reaction occurs at 25oC, 5 2 2 o22NO(g)(1 10 atm) Cl (g)(1 10 atm) 2NOCl(1 10 atm); the value of G is

+ ZZZXYZZZ (A) 45.65 kJ (B) 28.53 kJ (C) 22.82 kJ (D) 57.06 kJ 43. The approximate heat necessary to raise the temperature of 18.4 gm of ethyl alcohol from 45oC

to 55oC at 1 atm pressure (Cp = 26.8 cal/mole) is (A) 53.6 cal (B) 107.2 cal (C) 214.4 cal (D) 536 cal 44. Find H at 358 K for the reaction, 2 3 2 2Fe O (s) 3H (g) 2Fe(s) 3H O( )+ + A Given that, 298H 33.29 kJ/mole = and Cp for Fe2O3(s), Fe(s), 2 2H O( ) and H (g)A are 103.8, 25.1,

75.3 and 28.8 J/K mole. (A) 85.9 kJ/mole (B) 28.136 kJ/mole (C) +28.136 kJ/mole (D) 85.9 J/mole 45. Give the correct increasing order of acid strength.

OH OH

OMe

OH

NO2

OH

Cl(I) (II) (III) (IV)

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

Rough work



46. In the following reaction BH CH COOH3 33 2CH CH CH A B = A and B are,

(A) BCH3

CH3

CH3CH3

,

3

(B) ,CH3 CH2 CH2 B CH3O CH3

O3

(C) ,CH3 CH2 CH2 B CH3 CH33

(D) BCH3

CH3

CH3CH3

O

,

3 47. For the following compounds

N N

NH2

N

CH3

N

CN

(I) (II) (III) (IV) the correct order of basic strength (A) IV < III < II < I (B) IV < I < III < II (C) II < IV < I < III (D) I < II < III < IV 48. Borazene, B3N3H6, is isoelectronic and isostructural with benzene. Which of the following

statement(s) is/are true about borazene? (i) Borazene is aromatic. (ii) There are four isotopic disubstituted borazene molecules, B3N3H4X2. (iii) Borazene is more reactive towards addition reactions than benzene. (A) Only (i) (B) (i) and (ii) (C) (i) and (iii) (D) (i), (ii) and (iii)

Rough work



49. CH3

CH3

CH3

Cl

H5C2

HClR O2 2

A. A is

(A) CH3

CH3

CH3

Cl

Cl

C2H5

and formed through carbocation formation

(B) CH3

CH3

CH3

H

Cl

C2H5

Cl

and formed through carbocation formation

(C) CH3

CH3

CH3

Cl

Cl

C2H5

and formed through carbon free radical

(D) CH3

CH3

CH3

H

Cl

C2H5

Cl

and formed through carbon free radical

50. In Castners process some sodium is lost because (A) It reacts with dissolved O2 in molten NaOH (B) Due to its high reactivity it attacks other substance present in the system (C) At high temperature some of it vapourises (D) All of the above 51. One mole of a hydrocarbon (A), reacts with one mole of bromine in (CCl4), giving a dibromo

compound (C5H10Br2). Substance (A) on treatment with cold dil. Alkaline KMnO4 solution forms a compound (C5H12O2). On ozonlysis (A) gives equimolar quantities of propanone and ethanol. Hence compound (A) is

(A) 2-butene (B) 2-methyl but-2-ene (C) 1-methyl cyclobutene (D) 2-methyl but-1-ene 52. For the equilibrium at 298 K 2 32H O H O OH

+ +ZZZXYZZZ G is approximately: (A) 100 KJ (B) 80 KJ (C) 80 KJ (D) 100 KJ

Rough work



53. NCl3 on hydrolysis will give NH3 and HOCl but PCl3 on hydrolysis gives H3PO3 & HCl because (A) N is more electronegative than phosphorus (B) size of P is larger than nitrogen (C) P is having vacant d-orbital (D) A & C are correct 54. Calculate resonance energy of N2O from the following data: ( )0 1f 2H N O 82KJ mol = BE of N N = 946 KJ mol1 BE of N = N = 418 KJ mol1 BE of O = O = 498 KJ mol1 BE of N = O = 607 KJ mol1 (A) 68 KJ mol1 (B) 42 KJ mol1 (C) 88 KJ mol1 (D) 56 KJ mol1 55. Consider the following single step conversion:

CH3

OH

( )Reagent x COOH

OH The reagent (X) in the above reaction is (A) MnO2 (B) PbO2 (C) KMnO4/OH (D) K2Cr2O7/H+ 56. The standard reduction potential values of three metallic cations x, y, z are 0.52, 3.03 and 1.18

V respectively. The order of reducing power of the corresponding metal is (A) y > z > x (B) x > y > z (C) z > y > x (D) z > x > y

Rough work



57.

3

2 2

BH .THFH O / OH

A, A is

O

C C H

(A)

O

O

H

(B)

O

CH3

O

(C)

O

OH

O

(D)

O

H

O

58. In the following there are three carbon-oxygen bonds denoted by x, y and z:

CH3 C

O

O CH3x y z

Their lengths are in order (A) x = y = z (B) x < y < z (C) x < y = z (D) y < x < z

Rough work



59.

HO CH3

O

H

The correct IUPAC name of the above compound: (A) 4-(3-methyl-5-hydroxylphenyl)-3-methylpent-2-enal (B) 4-(3-hydroxy-5-methylphenyl)-3-methylpent-2-enal (C) 4-(3-methylphenol)-pent-2-enal (D) 4-(3-hydroxy-5-methylphenyl)-3,4-dimethylbut-2-enal 60. When acetylene is passed through conc. solution of Cu2Cl2 and ammonium chloride containing

some dil HCl and the substance formed reacts with dil HCl at 160C, if forms another substance which then polymerizes. The polymer formed is

(A) Benzene (B) Polyisoprene (C) Neoprene (D) PMMA

Rough work



PART B

Problem 1: [16 marks] Amount of 50 g of a 4 % sodium hydroxide solution and 50 g of a 1.825 % solution of hydrochloric acid were mixed in a heat insulated vessel at a temperature of 20 C. The temperature of the solution obtained in this way increased to 23.4 C. Then 70 g of a 3.5 % solution of sulphuric acid at a temperature of 20 C were added to the above solution.

1.1 Calculate the final temperature of the resulting solution.

[1/2+1/2 +1/2+1 marks]

1.2 Determine the amount of a dry residue that remains after evaporation of the solution. In calculating the first problem use the heat capacity value c = 4.19 J g1 K1. Relative atomic masses: Ar(H) = 1; Ar(O) = 16; Ar(Na)= 23; Ar(S) = 32; Ar(Cl) = 35.5.

[2 marks]

1.3 The gas escaping from a blast furnace has the following composition: 12.0 volume % of CO2 28.0 volume % of CO 3.0 volume % of H2 0.6 volume % of CH4 0.2 volume % of C2H4 56.2 volume % of N2

(i) Calculate the theoretical consumption of air (in m3) which is necessary for a total combustion of 200 m3 of the above gas if both the gas and air are measured at the same temperature. (Oxygen content in the air is about 20 % by volume).

[2 marks]



(ii) Determine the composition of combustion products if the gas is burned in a 20 % excess of air.

[3 marks]

1.4

A mixture of two solid elements with a mass of 1.52 g was treated with an excess of hydrochloric acid. A volume of 0.896 dm3 of a gas was liberated in this process and 0.56 g of a residue remained which was undissolved in the excess of the acid.

In another experiment, 1.52 g of the same mixture were allowed to react with an excess of a 10 % sodium hydroxide solution. In this case 0.896 dm3 of a gas were also evolved but 0.96 g of an undissolved residue remained.

In the third experiment, 1.52 g of the initial mixture were heated to a high temperature without access of the air. In this way a compound was formed which was totally soluble in hydrochloric acid and 0.448 dm3 of an unknown gas were released. All the gas obtained was introduced into a one litre closed vessel filled with oxygen. After the reaction of the unknown gas with oxygen the pressure in the vessel decreased by approximately ten times (T = const).

Write chemical equations for the above reactions and prove their correctness by calculations. In solving the problem consider that the volumes of gases were measured at STP and round up the relative atomic masses to whole numbers.

[1+1+1+1/2+1/2 marks]

1.5

(i) A mixture of metallic iron with freshly prepared iron (II) and iron (III) oxides was heated in a closed vessel in the atmosphere of hydrogen. An amount of 4.72 g of the mixture when reacted, yields 3.92 g of iron and 0.90 g of water.

When the same amount of the mixture was allowed to react with an excess of a copper(II) sulphate solution, 4.96 g of a solid mixture were obtained.

Calculate the amount of 7.3 % hydrochloric acid ( = 1.03 g cm-3) which is needed for a total dissolution of 4.72 g of the starting mixture.

[2 marks]



(ii) What volume of a gas at STP is released? Relative atomic masses:

Ar(O) = 16; Ar(S) = 32; Ar(Cl) = 35.5; Ar(Fe) = 56; Ar(Cu) = 64

[1/2 mark]



Problem 2: [16 marks]

The sample A participates in the transformations in scheme 1. Only the products containing A are shown in the scheme 1. Scheme 1:

( ) ( ) ( )2H O KOH electrolysis3 4 5C D E F

2O catalyst+( )2( ) ( ) ( ) ( )2O KOH A D1 6 7 8A B G H A B E + +

( )9 2H with heating( ) ( ) ( )KOH A D10 11 12I J K A I E + +

(a) Substance A is a solid and is insoluble in water. (b) Substances B and I are gases soluble in water. (c) Substances E, F, J and K are solid and soluble in water. (d) Aqueous solutions of B, G, H, I, J and K react with F, the products in all cases being E and D. (e) The following transformations occur during the interaction with an aqueous solution of iodine:

( )19B D ; 20G E ; ( )21H L ; ( )22I A ; ( )23J A ; ( )24K A Write the chemical equations for the following interactions and balance them.

2.1

[1 mark] 2.2

[1 mark) 2.3

[1 mark] 2.4

[1 mark]

A B

B C

C D

D E



2.5

[2 marks]

2.6

[1 mark] 2.7

[1 mark]

2.8

[2 marks]

2.9

[1 mark)

2.10

[1 mark] 2.11

[2 marks]

2.12

[2 marks]

E F

B G

G H

H A + B + E

A I

I J

J K

K A + I + E



Problem 3: [18 Marks]

A. Structure Elucidation of Organic Compound

3.1 Compound A (C9H10O) shows a strong IR absorption at 1690 cm1). It shows the following signals in its 1H NMR: 1.25 (3H, triplet), 3.0 (2H, quartet), 7.5 (3H, multiplet) and 8.0 (2H, multiplet). A on oxidation with acidic potassium permanganate gives benzoic acid.

[Refer to the Table 1 and Table 2 give at the end of the problem] (i) From the IR spectrum the compound A is expected to have functional group/s. [1 mark] (ii) The structure of A. [3 marks) ( ) ( )

2 4

9 119 9

Br /H NaBH

C H BrOC H BrOA C D

+

3.2 (i) Draw the structures of C and D. [2 marks] (ii) Draw the Fischer projection formulae of the isomers of D.

[2 marks] 3.3 Compound B, isomeric with A shows a strong IR absorption at 1730 cm1 and a medium peak at

2700 cm1. It shows the following signals in its 1H NMR spectrum: 2.7 (2H, triplet), 2.95 (2H, triplet), 7.3 (5H, multiplet) and 9.7 (1H, singlet).

[Refer to the Table 1 and Table 2 give at the end of the problem] (i) From the IR spectrum, the compound B is expected to have functional group/s. [1 mark]



(ii) The structure of B. [3 marks]

B. Organic Reactions, Mechanisms and Stereochemistry 3.4 The reaction of cyclopentanol with sodium metal gives product A. A reacts with bromoethane to

give B. (i) Identify A and B.

[2 marks]

(ii) Bromocyclopentane, on heating with sodium ethoxide, gives C. Identify C.

[1 mark]

(iii) In the above reaction (ii), ethoxide functions as: (a) Electrophile (b) Nucleophile

(c) Base

(d) Lewis acid [1 mark] When C is reacted with DBr (in dark), a mixture of products is obtained. (iv) Draw the structures of the products. [2 marks]

A B

C



Table 1: Some Characteristic Proton Chemical Shifts

Types of Proton Chemical shift, (ppm) p,s,t-C-H 0.9-1.5 C = C-H 4.6-5.9 C = C-H 2.0-2.3 Ar H 6.0-8.5

Ar C H 2.2-3.0 O C H 3.3-4.0

O = C C H 2.0-2.7 ROOC C H 2.0-2.6

O = C H 9.0-10.0 R COOH 610-12

Table 1: Some Characteristic Infrared Absorption Frequencies

Compound Type Frequency Range (cm1) Alkanes C H 2850 - 2960 Alkenes C H 1350 1470

3020 3080 Aromatic C H 675 1000

3000 3100 Olefin C = C 675 870

1640 1680 Alkyne C C

C O C = O O H C N

2100 2260 1080 1300 1650 1800 3200 3600 2210 - 2260



PATHFINDER-2013

FIITJEE ALL INDIA SCREENING TEST

(NSEC + INCho)

ANSWERS, HINTS & SOLUTIONS

PART- A

1. D 2. A 3. B 4. D 5. A 6. C 7. B 8. D 9. D 10. C 11. D 12. D 13. A 14. C 15. B 16. B 17. B 18. B 19. D 20. B 21. B 22. B 23. D 24. D 25. B 26. A 27. C 28. C 29. A 30. B 31. B 32. B 33. C 34. C 35. A 36. C 37. C 38. B 39. D 40. B 41. B 42. A 43. B 44. B 45. A 46. C 47. B 48. D 49. A 50. A 51. B 52. C 53. D 54. C 55. B 56. A 57. A 58. B 59. B 60. C



HINTS & SOLUTIONS PART- A

2. Change in P.E. = ( )2x 2x 8x 32x x4 4 2

+= = = 3. F has dominating +M effect. 5. F being the worst leaving group among halogens forms Hoffmann product. 7. 21st electron in Sc is present in 3dxy orbital. Hence, (B) is correct answer. 8. (A) O C O & N N O= = , both are linear. (b)

N

B

N

B B

N

H

H

HH

H

H

Borazole

Benzene

Hence, both are isostructural. (C) Azide ion ( )3N is linear (sp-hybridisation) while ( )3I ion is also linear (AX2E3) geometry.

Hence, isostructural. (D) Trimethylamine is pyramidal trisilyl while amine is triangular planar due to p d bonding

latter.

N

CH3CH3

CH3

sp3

Pyramidal

N

H3Si SiH3

SiH3

sp2

Triangularplanar 9. (A) ( ) ( )3 2 2 33 3

sp d d sp

2 6 6Fe H O 6CN Fe CN

+ +

(B) 2 3sp sp

3 4LiH AlH LiAlH+

(C) 2 2sp sp sp

2 5 2 3N O NO NO+ +

(D) 3 3sp sp

3 2 4 4 4

2

2NH H SO NH SO +

Hence, (D) is the correct answer. 10. Total meq. of KMnO4 = Total meq. of FeC2O4 + Total meq. of FeSO4

(0.1 5) V 100 100144 152

3 1

= +

V = 5.48 ml. Hence, (C) is the correct answer.



11. (A) C C C; C C C C

(B) | |C C

C O C C C C; C CO C C C; C O C C C; C O C C C

| |

|C

C CC O C C; C C O C C

(C) COOH

COOCH3

H OH

OH H

COOH

COOCH3

OH H

H OH

COOH

COOCH3

H OH

H OH

COOH

COOCH3

OH H

OH H

Hence, (D) is the correct answer. 12. Increasing order of stability is (C)



pKa = logC logKa = logC C = Ka = 1.8 105 M Dilution 41 5.55 10

C= = times

17. ( )

1 2

4 2a 1 a 2H K C K C 6 0.5 2 0.5 10 2 10 M

+ = + = + = pH = log(2 102) = log2 log102 = 0.3010 + 2.00 = 1.70 18. Salts (NH4A, NH4B, NH4C) are of weak acid and weak base.

a bpK pK

pH 72 2

= + Thus, greater the value of pKa of acid, greater the pH. pKb (A, B, C): A < B < C 4 5 6 pKa (HA, HB, HC): HC < HB < HA 8 9 10 Thus pH (NH4C< NH4B < NH4A) Hence (B) is the correct option. 19. Solution Bi3+ will form black precipitate where as Mn2+ does not in acidic medium due to higher

Ksp value. Hg2+ gets precipitate as HgS but Al3+ does not. Cu2+ gets precipitated as CuS(black) whereas Zn2+ gets precipitated but Ni2+ does not. 20. Kc = 4 10-19

( ) 12Ag 2CN Ag CN0.03 0.1 00.03

+ + ZZZXYZZZ

( ) 12Ag CN Ag 2CNt 0 0.3 0 0.04

0.03 x x 0.04 2x

+ + =

+

ZZZXYZZZ

( )2 19

c

x 0.04 2xK 4 10

0.03 x+= =

0.04 2x 0.04+ 0.03 x 0.03

( )2 19c

x. 0.04K 4 10

0.03= =

x = 7.5 10-18 21. Important ore of lead is galena (PbS).

22. 2

nnrZ

for H, 2 2

4 34 3r , r 16 9 71 1

= =

r4 r3 for Li 16 9 73 3 3 =



So, ratio 7 37 / 3

= Ions = 3 : 1 23. 3 22 4 2 4 2FeS KMnO H Fe H SO Mn H O

+ + ++ + + + + 32FeS Fe

+ Change of oxidation state of Fe is +1

22 2 4FeS H SO

1 2 6 22 12

=

Change is 14 Total change 14 + 1 = 15, n-factor = 15 Eq. wt. of FeS2 is M/15 25. 3 2 22NaOCl 2NH N 3H O NaCl+ + + 28. CH2 = CHCH2Cl reacts with AgNO3 form stable allylic carbocation and ppt. of AgCl but CH3CH =

CH2Cl form unstable vinyl carbocation so, reaction cannt occur in 2nd case. 29. Equivalents of KOH used by oil = [25 0.421 8.46 0.273 2 ]103 Moles of KOH used = 5.90 103 Mass of KOH used in milligrams = 5.90 103 56 1000 = 330.40 Saponification number = 330.40

1.576 = 209.64

30. Density of a gas is given = PMRT

. Obviously, the choice that has greater PT

would have greater

density. (B)

31. 2Cl

5 8n0.082 300

= = 1.63 Mass of Al(s) = 0.05101(20206)1.13 = 13.56 gm So, mass Al(s) dissociated = 0.5 moles So, equilibrium established Al(s) + 3/2 Cl2(g) ZZZXYZZZ AlCl3(s)

31.63 0.52

0.5

So, Kp = ( )2

3 / 23 / 2

Cl0.88 0.082 300p

8

= = (2.7)3/2

32. [Ca2+] = [CaCl2] = 0.005 (N) ( )

4

2 2sp 4CaSO

K Ca SO+ = 5 242.40 10 0.005 SO

=

( )524 32.4 10SO M5 10

= = 4.8103



35. 22M(OH) M 2OH+ +

pH = 14 pOH 10 = 14 pOH pOH = 14 10 = 4 log OH 4; = [OH] = 104 M 2spK M OH=

4

4 210 [10 ]2

=

= 5 1013 38. The value of K decreases with decreases in temperature. 14wK 10

< (at temperature > 25oC) H+ < 107 For neutral water, pH = log H+ pH > 7 So, pH 7 indicated must be acidic. 39. 2 r n = Circumference 42 r 5.32 4= = = 4

5.32 1.334

= = Now, 21 1 4 12 r , also r r 4 = = Hence, (D). 40. (1) proceeds through E2 path while (2) and (3) proceeds through E1 path and (4) proceeds

through E1CB path. Hence, only 2 and 3 takes place through Saytzeff rule.

41. oAgCl

K 1000S =

62.3 10 1000

(61.9 76.3)

= + 51.66 10 M=

42. 2

2NOCl

eq 2NO Cl

PK

P P= = 10

8

eqG RT n K 2.303 8.314 298 8 45.65 kJ = = = A

43. 18.4n 0.446

= = pQ nC T 0.4 26.8 10 107.2 cal /mole= = = 44. pC 2 25.1 3 75.3 (103.8 3 28.8) 85.9 J /mole = + + = 358 298 p

2 1

H HC

T T =

358H 28.136 kJ/mole =



45. The decreasing order of electron withdrawing group is NO2 > Cl > OMe So have NO2 is most strong electron withdrawing group. 46. ( )3 3BH CH COOH3 2 3 2 2 3 2 33CH CH CH CH CH CH B CH CH CH = 51.

H3C CH C

CH3

CH3

( )2 methyl but 2 ene

A

3 2O /H O / ZnOzonolysis CH3 CHO C O

CH3

CH3Ethanol3-propanone

H3C C C

CH3

CH3

Br Br

HH3C C C

CH3

CH3

OH OH

H

2 4Br / CCl

( ) 41% cold alk. KMnOhydroxylation

52. G = 2.303 RT log Keq = 2.303 (8.314) 298 log 1014 = 79881.87 J = 79.9 KJ 80 KJ 54. From the given BE values

( ) ( ) ( ) ( )0f 2 N N O O N N N O1H N O Be BE BE BE2 = = = = + + = 498946 418 607 1195 1025

2 + + =

= 170 KJ Resonance energy = observed Hf calculated Hf = 82 170 = 88 KJ mol1

55. PbO2 selectively oxides alkyl benzene to benzoic acid even in the presence of an electron releasing group. If the same conversion is carried out by using KMnO4/OH, the benzene ring will undergo cleavage.

56. Lower the reduction potential, stronger is the reducing agent.

57. O

C C H

3BH , THF

O

C C B

H

3BH , THF

O

C C OH

H

O

H

O



60.

Polymerisation

2 24

Cu Cl

NH ClCHHC CHH2C C

CH

HCl

160 CHH2C C

CH2

Cl

CHH2C C

CH2

Cl nNeoprene

PART B

Problem 1: 1.1 (a) NaOH + HCl NaCl + H2O ( ) ( ) ( )( ) 1

m solution NaOH w NaOH 50 g 0.04N NaOH 0.05 molM NaOH 40 g mol

= = =

( ) 150g 0.01825n HCl 0.025 mol36.5g mol= =

unreacted: n(NaOH) = 0.025 mol (b) When 1 mol of water is formed, neutralization heat is:

( )

1 11

neutr2

mc t 10g 4.19Jg K 3.4KH 57000 J moln H O 0.025 mol

= = =

(c) NaOH + H2SO4 NaHSO4 + H2O The temperature of the resulting solution is calculated according to the equation: m1c1t1 + m2c2t2 = mct c1 = c2 = c m1t1 + m2t2 = mt

( ) ( ) 01 1 2 2 100 23.4 70 20.0m t m tt 22 C

m 170 + +

= = =

(d) The temperature increase due to the reaction of NaOH with H2SO4 is as follows:

( ) 12 neutr

1 1

n H O H 0.025 mol 57000 jmolt 2Kmc 170 g 4.19Jg K

= = =

The final temperature of the solution: t = 22 + 2 = 24 C 1.2 When the solution has evaporated the following reaction is assumed to take place: NaCl + NaHSO4 Na2SO4 + HCl Na2SO4 is the dry residue. m(Na2SO4) = n M = 0.025 mol 142 g mol-1 = 3.55 g 1.3 O2 _______ (i) 2CO + O2 2CO2 14 2H2 + O2 2H2O 1.5 CH4 + 2O2 CO2 + 2H2O 1.2 C2H4 + 3O2 2CO2 + 2H2O 0.6 _________ 17.3 parts 5 = 86.5 parts of the air



200 m3 of the gas ........2 86.5 = 173.0 m3 of the air + 20% 34.6 m3 ________________ 207.6 m3 of the air

(ii) 207.6 : 5 = 41.52 parts of O2 : 2 = 20.76 parts of O2 for 100 m3 of the gas 20.76 4 = 83.04 parts of N2 for 100 m3 of the gas Balance: CO2 H2O N2 O2 (Volume parts) 12.00 3.00 56.20 20.76 28.00 1.20 83.04 17.30 0.60 0.40 0.40 ___________________________________________ 41.00 4.60 139.24 3.46 Total: 41.00 + 4.60 + 139.24 + 3.46 = 188.30 of volume parts of the gaseous components.

24.60% H O 100 2.44

188.30= =

2139.24% N 100 73.95188.30

= =

23.46%O 100 1.84

188.30= =

1.4 (a) Reaction with hydrochloric acid: 1.52 g 0.56 g = 0.96 g of a metal reacted and 0.896 dm3 of hydrogen (0.04 mol) were formed.

Combining mass of the metal: 0.9611.2 12g0.896

= Possible solutions

Relative atomic mass of the metal

Oxidation number Element Satisfying

12 I C No 24 II Mg Yes 36 III Cl No

Reaction: 2 2Mg 2HCl MgCl H+ + (b) Reaction with sodium hydroxide:

1.52 g 0.96 g = 0.56 g of an element reacted, 0.896 dm3 (0.04 mol) of hydrogen were formed.

Combining mass of the metal: 0.5611.2 7g0.896

= Possible solutions

Relative atomic mass of the element

Oxidation number Element Satisfying

7 I Li No 14 II N No 21 III Ne No 28 IV Si Yess

Reaction: Si + 2 NaOH + H2O Na2SiO3 + 2 H2 (c) Combining of both elements:

0.96 g Mg + 0.56 g Si = 1.52 g of silicide MgxSiy

( ) 0.96gw Mg 0.631.52g

= = ( ) 0.56gw Si 0.371.52g

= =

0.63 0.37x : y : 2 :124 28

= =



(d) Reaction of the silicide with acid: 2 2 4Mg Si 4Cl 2MgCl SiH+ + ( )2 11.52gn Mg Si 0.02 mol76gmol= = ( ) 34 3 10.448dmn SiH 0.02mol22.4 dm mol= = (e) Reaction of silane with oxygen 4 2 2 2SiH 2O SiO 2H O+ + 3V 1dm= On the assumption that T = const: 22 1

1

np pn

=

( ) 31 2 3 11dmn O 0.0446mol22.4 dm mol= = Consumption of oxygen in the reaction: ( )2n O 0.04 mol= The remainder of oxygen in the closed vessel: ( )2 2n O 0.0446 mol 0.04 mol 0.0046 mol= = 2 1 1

0.0046 molp p 0.1p0.0446 mol

= = 1.5 (i) (a) Reduction by hydrogen:

FeO + H2 Fe + H2O n(Fe) = n(FeO); n(H2O) = n(FeO) Fe2O3 + 3H2 2Fe + 3H2O n(Fe) = 2n(Fe2O3); n(H2O) = 3n(Fe2O3) The mass of iron after reduction: 3.92 g The total amount of substance of iron after reduction:

( ) ( ) ( )2 3 13.92gn Fe n FeO 2n Fe O 0.07 mol56 gmol+ + = = (b) Reaction with copper(II) sulphate: 4 4Fe CuSO Cu FeSO+ + Increase of the mass: 4.96 g 4.72 g = 0.24 g After reaction of 1 mol Fe, an increase of the molar mass would be ( ) ( ) 1 1 1M Cu M Fe 64 g mol 56g mol 8g mol = = Amount of substance of iron in the mixture:

( ) 10.24gn Fe 0.03 mol8g mol= = (c) Formation of water after reduction:

0.90 g H2O, i.e. 0.05 mol 0.05 mol = n(Fe) + 3 n(Fe2O3) By solving equations (1), (2), and (3): n(FeO) = 0.02 mol n(Fe2O3) = 0.01 mol



(d) Consumption of acid: Fe + 2HCl FeCl2 + H2 FeO + 2HCl FeCl2 + H2O Fe2O3 + 6HCl 2FeCl2 + 3H2O n(HCl) = 2n(Fe) + 2n(FeO) + 6n(Fe2O3)

= 0.06 mol + 0.04 mol + 0.06 mol = 0.16 mol A part of iron reacts according to the equation: Fe + 2FeCl3 3FeCl2 n(Fe) = 0.5 n(FeCl3) = n(Fe2O3) n(Fe) = 0.01 mol It means that the consumption of acid decreases by 0.02 mol. The total consumption of acid: n(HCl) = 0.14 mol

( ) 1 33n M 0.14 mol 36.5 g molV 7.3%HCl 68cmw 0.073 1.03 g cm

= = =

(ii) Volume of hydrogen:

Fe + 2HCl FeCl2 + H2 Iron in the mixture: 0.03 mol Iron reacted with FeCl3: 0.01 mol Iron reacted with acid: 0.02 mol

Hence, 0.02 mol of hydrogen, i.e. 0.448 dm3 of hydrogen are formed.

Problem 2: 2.1 S + O2 SO2 2.2 2 SO2 + O2 2SO3 2.3 SO3 + H2O H2SO4 2.4 2 KOH + H2SO4 K2SO4 + 2 H2O 2.5 2 24 2 82SO 2e S O

2.6 SO2 + 2 KOH K2SO3 + H2O 2.7 K2SO3 + S K2S2O3 2.8 K2S2O3 + H2SO4 K2SO4 + S + SO2 + H2O 2.9 H2 + S H2S 2.10 H2S + 2 KOH K2S + 2 H2O 2.11 K2S + x S K2S(x+1) 2.12 K2S(x+1) + H2SO4 K2SO4 + xS + H2S A: S B: SO2 C: SO3 D: H2SO4 E: K2SO4 F: K2S2O8 G: K2SO3 H: K2S2O3 I: H2S J: K2S K: K2Sx L: K2S4O6 Problem 3: 3.1 (i) CO

(unsaturated ketone)



(ii) CO CH2 CH3

3.2 (i) CO CH

Br

CH3

C

C CH

Br

CH3

OH

D (ii) Ph

CH3

H OH

H Br

3.3 (i)

C

O

H group (ii)

H2C CH2 C

O

H

3.4 (i) ONa O CH2 CH3

A B(Willianson's synthesis)

(ii)

C (iii) Base right (iv) D

Br

D

Br

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