ASSOCIATION OF CHEMISTRY TEACHERS

NATIONAL STANDARD EXAMINATION IN CHEMISTRY 2012-2013Date of Examination : 24th November 2012

Time 12.30 to 14.30 Hrs

Q. P. Code No.

Instruction to candidates

1. On the answer sheet, fill up all the entries carefully in the space provided, ONLY In BLOCKCAPITALS. Use only BLUE or BACK BALL PEN for making entries and marking answer.Incomplete / incorrect / carelessly filled information may disqualify your candidature.

2. Write the Q.P. Code No. mentioned above on YOUR answer sheet (in the space provided).Otherwise your answer sheet will NOT be examined.

3. The question paper contain 80 multiple-choice question. Each question has 4 options, out ofwhich only one is correct. Choose the correct answer and mark a cross in the correspondingbox on the answer sheet as shown below :

4. A correct answer carries 3 marks and 1 mark will be deducted for each wrong answer.

5. All rough work may be done on the blank sheet provided at the end of the question paper.

6. PLEASE DO NOT MAKE ANY MARK OTHER THAN (X) IN THE SPACE PROVIDED ONTHE ANSWER SHEET.Answer sheets are evaluated with the help of a machine. Due to this, CHANGE OF ENTRY ISNOT ALLOWED.

7. Scratching or overwriting may result in wrong score.DO NOT WRITE ANYTHING ON THE BACK OF ANSWER SHEET.

8. Use of a nonprogrammable calculator is allowed.

9. Periodic table and log table are provided at the end of this question paper.

10. The answers / solutions to this question paper will be available on our website - www.iapt.org.inby 30th November 2012.

Certificates & Awards

i) Certificates to top 10% students of each centre.ii) Merit certificates to statewise Top 1% students.iii) Merit certificate and a prize in term to Nationwise Top 1% students.

11. Result sheets and the �centre top 10%� certificates of NSEC are dispatched to the Professor in charge of

the centre. Thus you will get your marks from the Professor in charge of your centre by January 2013 end.

12. 300 (or so) students are called for the next examination - Indian National Chemistry Olympiads (INChO).Individual letters are sent to these students ONLY.

13. Gold medals may be awarded to TOP 35 students in this entire process.

14. No querries will be entertained in this regard.

RESONANCE PAGE - 2

NSEC-2012-13

ASSOCIATION OF CHEMISTRY TEACHERS

NATIONAL STANDARD EXAMINATION IN CHEMISTRY 2012-2013NSEC-2012-2013

1. An electron releasing group will not stabilize which of the following groups?(A) Carbocation (B) Carbanion (C) free radical (D) any of the above

Ans. (B)Sol. Carbocations and free radicals are electron deficient and stabilized by electron releasing groups.

2. The bond order for a species with the configuration1s2 *1s2 2s2 *2s2 p

X1 will be

(A) 1 (B) 21

(C) Zero (D) 3/2

Ans. (B)

Sol. BO = 21

(Nb � N

a) =

21

= (5 � 4) = 21

3. The widest range over which electronic excitations in organic compounds occur, is(A) 200 nm - 780 nm (B) 220 nm- 500nm (C) 250 nm- 700 nm (D) 290 nm -1000nm

Ans. (A)Sol. The organic functional groups undergoes electronic excitations in the range of 200 nm-780 nm range.

4. Which of the following compounds has the least tendency to form hydrogen bonds between molecules?(A) NH

3(B) H

2NOH (C) HF (D) CH

3F

Ans. (D)Sol. In CH

3F, H is not directly connected to F.

5. The species in which the central atom uses sp2 hybrid orbitals is(A) PH

3(B) NH

3(C) CH

3+ (D) SbH

3

Ans. (C)

Sol. has 3-bonds.

6. -D(+) glucose and -D(+) glucose are(A) Enantiomers (B) Geometrical isomers (C) Epimers (D) Anomers

Ans. (D)Sol. D (+) glucose and -D(+) glucose are anomers.

7. The chemical formula of 'laughing gas' is(A) NO (B) N

2O (C) N

2O

4(D) N

2O

5

Ans. (B)Sol. Nitrous oxide (N

2O) is a laughing gas.

8. The enzyme which hydrolyses triglycerides to fatty acids and glycerol is(A) lipase (B) maltase (C) pepsin (D) zymase

Ans. (A)Sol. Lipase is the enzyme which hydrolyses triglycerides to fatly acids and glycerol.

9. In which of the following ion/molecule , the 'S' atom does not assume sp3 hybridization?(A) SO

42� (B) SF

4(C) SF

2(D) S

8

Ans. (B)

RESONANCE PAGE - 3

NSEC-2012-13

Sol. SF4 steric no. of S is 5 i.e. sp3d hybridisation.

10. The most stable free radical which can be isolated is(A) Trityl radical (B) Diphenyl methyl radical(C) 2,4,6-Tri-ter-butylphenoxy radical (D) tert-butyl radical

Ans. (C)Sol. The most stable free radical which can be isolated is 2,4,6-Tri-ter-butyl phenony radical. It is unable to

dimerise due to steric reasons.

11. Phosphine is prepared by the reaction of(A) P and HNO

3(B) P and H

2SO

4(C) P and NaOH (D) P and H

2S

Ans. (C)

Sol.)White(

4P + 3NaOH + 3H2O PH

3 + 3NaH

2PO

2

12. Pheromones are chemical substances which are(A) formed by fermentation process of fungi (B) secreted by endocrine glands of man(C) secreted by insects (D) plant growth hormones.

Ans. (C)Sol. Pheromones are chemical substanes that are secreted by insects.

13. Which of the following does not reduce Benedict's solution?(A) Glucose (B) Fructose (C) Sucrose (D) Aldehyde

Ans. (C)Sol. Sucrose has not any aldehydic group hence does not reduce Benedict solution.

14. The inorganic precipitate which acts as a semipermeable membrane is(A) Calcium phosphate (B) Nickel phosphate (C) Plaster of paris (D) Copper ferrocyanide

Ans. (D)Sol. Copper ferrocyanide acts as a s.p.m.

15. The genetic material of a cell is made of(A) nucleic acids (B) proteins (C) carbohydrates (D) fats

Ans. (A)Sol. The genetic material of the cell is made up of nucleic acids (RNA, DNA).

16. Lanthanide contraction is caused due to(A) the appreciable shielding on outer electrons by 4f electrons from the nuclear charge(B) the appreciable shielding on outer electrons by 5d electrons from the nuclear charge(C) the same effective nuclear charge from Ce to Lu.(D) the imperfect shielding on outer electrons by 4f electrons from the nuclear charge

Ans. (D)Sol. 4f electrons does imperfect shielding on outer electrons there fore effective nuclear charge increase.

17. Which of the following contain maximum number of electrons in the antibonding molecular orbitals(A) O

22� (B) O

2(C) O

2�1 (D) O

2+

Ans. (A)Sol. O

22� has 4 antibonding electrons whereas O

2, O

2� and O

2+ have respectively 2, 3 and 1 antibonding electrons.

RESONANCE PAGE - 4

NSEC-2012-1318. Lattice energy for an ionic compound is calculated by using

(A) Kirchoff's equation (B) Markownikoff's rule (C) Born Haber cycle (D) Carnot cycleAns. (C)Sol. Lattice energy for an ionic compound is calculated by using Born Haber cycle.

19. If the radius of the first Bohr orbit is r, then the deBroglie wavelength in the third Bohr orbit is(A) 2r (B) 9r (C) r/3 (D) 6r

Ans. (D)

Sol.3

1

rr

= 3r

r = 2

2

3

)1(

r3 = 9r

now nr2

Or 2 × ×9r = 3Or = 6r

20. The IUPAC name of [Co(ONO)(NH3)

5Cl

2] is :

(A) pentamminenitrocobalt(ll)chloride (B) pentamminenitrosocobalt(IIl)chloride(C) pentamminenitritocobalt(lll)chloride (D) pentammineoxo-nitrocobalt(Ill)chloride

Ans. (Bonus)Sol. Ionic sphere is not given in the question.

21. In the Vander waal equation of state for a non ideal gas the term that accounts for intermolecular force is

(A) (V- b) (B) RT (C) )a

P(2

(D) 1/RT

Ans. (C)

Sol.

2V

aP terms accounts for inter molecular force.

22. The structure given below represents :

(A) Isoprene Rubber (B) Bakelite (C) PVC (D) Nylon 6,6Ans. (B)Sol. The given structure represent bakelite, which is made up of phenol with formaldehyde.

23. The maximum amount of CH3CI that can be prepared from 20g of CH

4 and 10g of Cl

2 by the following reaction,

is :CH

4 + Cl

2 CH

3CI + HCI, (presume that no other reaction is taking place)

(A) 3.625 mole (B) 0.141 mole (C) 1.41 mole (D) 0.365 moleAns. (B)Sol. CH

4+ Cl

2 CH

3CI + HCI

mol1620

mol7110

0 0

As Cl2 is Limiting reagent there fore

7110

mole i.e. 0.141 mol of CH3Cl will be formed.

RESONANCE PAGE - 5

NSEC-2012-1324. Which isomer of xylene can give three different monochloro derivatives?

(A) o-xylene (B) m-xylene(C) p-xylene (D) xylene cannot give a monochioro derivative

Ans. (B)Sol. m-xylene on monochlorination given three different products.

25. The most Carbocations, carbanions, free radicals and radical cation are reactive carbon intermediates. Theirhybrid orbitals respectively are(A) sp2, sp2, sp3, sp (B) sp2,sp2, sp, sp3 (C) sp2, sp3, sp2, sp (D) sp3, sp2, sp, sp2

Ans. (C)Sol. Carbocations, carbonions, free radicals and radical cations are sp2, sp3, sp2 and sp hybrid respectively.

26. effective electrolyte to cause the flocculation of a negatively charged arsenium sulphide colloid is:(A) NaCl (B) BaCl

2(C) K

3Fe(CN)

6(D) AICI

3

Ans. (D)Sol. Al+3 has maximum positive charge amongs these.

27. The electronegativities of acetylene, ethylene and ethane are in the order :(A) ethylene >.acetylene > ethane (B) acetylene > ethylene > ethane(C) ethane > acetylene > ethylene (C) acetylene > ethane > ethylene

Ans. (B)Sol. Electronagativity of sp > sp2 > sp3 carbon.

28. A catalyst accelerates a reaction primarily by stabilizing the(A) substrate (B) product (C) intermediate (D) transition state

Ans. (D)Sol. A Catalyst provide alternate path thereby decreasing the transition state and accelerates a reaction.

29. The number of transition states in a unimolecular nucleophilic substitution (SN1) reaction is

(A) 0 (B) 1 (C) 2 (D) 3Ans. (C)Sol. SN1 reaction is two step process with 1 carbocation intermediate and 2-transition state.

30. The dipole moments of halo compounds are in the orderA) CHCl

3> CCI

4> CH

2Cl

2 > cis-CHCI=CHCI (B) cis-CHCI=CHCI > CHCI

3> CH2Cl

2> CCI

4

(C) cis-CHCI=CHCI > CH2Cl

2> CHCI

3> CCI

4(D) CHCl

3> CH

2Cl

2> cis-CHCI=CHCI > CCl

4

Ans. (C)

Sol. In it dipole moments are additive CCl4 has zero dipole moment.

31. Which of the following information is not provided by a reaction mechanism?(A) Which bonds are formed and which bonds are broken(B) Which intermediates and transition states are formed(C) Energy content of the reacting species(D) Which is the slowest step

Ans. (C)Sol. Reaction mechanism does not provide the information regarding the energy content of the reacting species.

32. Tollen's reagent is(A) Cu

2O (B) [Cu(OH)

4]2� (C) Ag

2O (D) [Ag(NH

3)2]+

Ans. (D)Sol. Tollen�s reagent is ammonical silvernitrate, which has the species [Ag(NH

3)

2]+.

RESONANCE PAGE - 6

NSEC-2012-1333. The R/S designation for the following stereoisomer of 1,3-dibromo-2-methylbutane is :

(A) 2R, 3R (B) 2R, 3S (C) 2S, 3R (D) 2S, 3SAns. (A)Sol. The given structure has 2R, 3R designation.

34. The bond energy of B-F bond in BF3 is 646 kJ mol-1, while that of N-F bond in NF

3 is 280 kJ mol�1 this is

because :(A) N is more electronegative than B(B) The atomic mass of N is higher than that of B(C) The B-F bond gets a partial double bond character due to p-p overlap(D) N has a lone pair of electrons while B does not nave

Ans. (C)Sol. In BF

3 p�p

back bonding occurs which increases bond energy.

35. The amino acid that cannot be obtained by hydrolysis of proteins is

(A)

�32 COO)NH(CHHOOCCH (B)

(C) (D)

Ans. (B)Sol. Hydrolysis of protein (natural molecules), yields -aminoacids, but the option (B) is -aminoacid.

36. When equal volumes of the following solutions are mixed precipitation of AgCl (Ksp

= 1.8 × 10�10) will occuronly with :(A) 10�4 M Ag+ and 10�4 M Cl� (B) 10�5 M Ag+ and 10�5 M Cl�

(C) 10�6 M Ag+ and 10�6 M Cl� (D) 10�10 M Ag+ and 10�10 M Cl�

Ans. (A)Sol. On mixing equal volume new concentration becomes half of the previous.

In (A) IP = sp

44

K2

102

10

or 2.5 × 10�9

RESONANCE PAGE - 7

NSEC-2012-13

37. The quantum numbers for the 19th electron of Cr (Z = 24) are :(A) n = 3, l = 0, m = 0, s = + ½ (B) n = 4, l = 0, m = 0, s = + ½(C) n = 3, l = 2, m = 2, s = + ½ (D) n = 4, l = 2, m = 2, s = + ½

Ans. (C)Sol. 19th e� in Cr enters in 4s subshell.

38. The oxidation of SO2 by O

2 is an exothermic reaction. The yield of SO

3 can be maximized if :

(A) temperature is increased and pressure is kept constant(B) temperature is decreased and pressure is increased(C) both temperature and pressure are increased(D) both temperature and pressure are decreased

Ans. (B)Sol. 2SO

2(g) + O

2(g) 2SO

3(g) + heat exothermic therfore low temperature and as molecules are decreasing

from L R therefore high pressure are suitable conditions.

39. Which of the following ion is colourless :(A) Mn2+ (B) Cu+ (C) Cr3 (D) Fe2+

Ans. (B)Sol. Cu+ = [Ar]3d10

All electrons are paired therefore diamagnetic and colourless.

40. Which of the following has a positive entropy change?(A) H

2O

(g) H

2O

(l)(B) BF

3(g) + NH

3(g) F

3B.NH

3(s)

(C) 2SO2(g)

+ O2(g)

2SO3(g)

(D) 2NH4NO

3(g) 2N

2(g) + 4H

2(O)

(l) + O

2(g)

Ans. (D)Sol. In it gaseous substance are increasing in reaction.

41. Equal volumes of two solutions of pH = 2 and pH = 4 are mixed together the pH of the resulting solution willbe :(A) 2.0 (B) 3.1 (C) 4.2 (D) 2.3

Ans. (D)Sol. pH = 2 or [H+] = 10�2 M

pH = 4 or [H+] = 10�4 MN

1V

1 + N

2V

2 = N

fV

f

10�2 × V + 10�4 × V = Nf × 2V

or Nf =

21010 42

~ 210 2

[H+] = CN

× x = N

f × x = 1

210 2

pH = 2.3

42. A first order reaction is 20% complete in 600 s. The time required to complete 75% of the same reaction willbe :(A) 3120 s (B) 3720 s (C) 4320 s (D) 4920 s

Ans. (B)

Sol.xa

alog

t303.2

K

25100

logt

303.280

100log

600303.2

K

or t = 3720 sec.

43. The vapour density of gas A is four times that of B. If the molecular mass of B is M then molecular mass ofA is :(A) M (B) 4M (C) M/4 (D) 2M

Ans. (B)

Sol. VDA = 4VD

Bor

2

M4

2

M BA , or MA = 4M

B

RESONANCE PAGE - 8

NSEC-2012-1344. Among the isomers of dimethylcyclohexanes, the chiral ones are

(A) 1, 2-trans and1, 3-cis (B) 1, 2-cis and 1,3-trans(C) 1, 3-trans and 1, 4-trans (D) 1, 2-trans and 1,3-trans

Ans. (D)Sol. trans-1,2-dimethylcyclohexanes and trans-1,3-dimethylcyclohexanes is chiral.

45. The relative basics strengths of NH3, CH

3NH

2 and NF

3 are in the order :

(A) CH3NH

2 > NH

3 > NF

3(B) NH

3 > CH

3NH

2 > NF

3

(C)NF3 > CH

3NH

2 > NH

3(D) CH

3NH

2 > NF

3 > NH

3

Ans. (A)Sol. CH

3 � show + I effect which increases base strength. In NF

3 F shows � I effect which reduces base strength.

46. The outer most electronic configuration of the most electronegative element is :(A) ns2, np3 B) ns2,np6 (n� 1) d2 (C) ns2, np5 (D) ns2,np6

Ans. (C)Sol. F has ns2p5 outer most electric configuration.

47. The conductivity of a metal decreases with increase in temperature because :(A) the kinetic energy of the electrons increases (B) the movement of electrons becomes haphazard(C) the ions start vibrating (D) the metal becomes hot and starts emiting radiation

Ans. (B)Sol. In metals by increases of temperature conductivity decreases and resistance increases as movement of

electrons becomes haphazard.

48. The lanthanide compound which is used as a most powerful liquid laser after dissolving in selenium oxychlo-ride is :(A) Cerium oxide (B) Neodynium oxide(C) Promethium sulphate (D) Cerium sulphate

Ans. (B)Sol. It is neodynium oxide.

49. The solubility of SrF2 in water at 303 K is 9.55 × 10�5 mol.dm�3. The solubility product of the salt is :

(A) 8.7 × 10�17 (B) 9.1 × 10�11 (C) 9.55 × 10�5 (D) 3.48 × 10�12

Ans. (D)

Sol.

Ksp

= 4 s3

= 4 × (9.55 × 10�5)3

= 3483.9 × 10�15

= 3.484 × 10�12

50. The amount of electricity required to deposit 1.0 mole of aluminium from a solution of AlCl3 will be :

(A) 1 faraday (B) 3 faradays (C) 0.33 faraday (D) 1.33 faradayAns. (B)Sol. AlCl

3 Al3+ + 3Cl�

Al3+ + 3e� Al Amount of electricity required for 1 mole of aluminium = 3 Faraday.

51. In the reaction, 2KClO3 2KCl + 3O

2 when 36.75 g of KClO

3 is heated, the volume of oxygen evolved at

N.T.P. will be :(A) 9.74 dm3 (B) 8.92 dm3 (C) 10.08 dm3 (D) 22.4 dm3

Ans. (C)Sol. 2KClO

3 2KCl + 3O

2

nKClO3

= 5.122

75.36 = 0.3

By mole-mole analysis 3

n

2

n23 OKClO

or nO2

= 3.023 = 0.45

RESONANCE PAGE - 9

NSEC-2012-13or volume of O

2= 0.45 × 22.4

= 10.08 lit.= 10.08 dm3

52. The pKa volume of H2O of picric acid, acetic acid and phenol are in the order :

(A) Picric acid 0.4, acetric acid 4.75, phenol 10.0(B) acetic acid 0.4, picric acid 4.75, phenol 10.0(C) picric acid 0.4, phenol 4.75, acetic acid 10.0(D) phenol 0.4, acetic acid 4.75, picric acid 10.0

Ans. (A)Sol. Presence of 3-nitro group on phenol makes picric acid a stronger acid, whereas acetic acid is moderately

acidic due to resonance stabilization on carboxylate ion. Phenol is least acidic.

53. The correct IUPAC name of the following compound is :

(A) 2-Bromo-5-methylbicyclo[5:4:0]heptanes (B) 3-Bromo-7-methylbicyclo[3.2.0]heptanes(C) 3-Bromo-6-methylbicyclo[3.2.0]heptanes (D) 2-Methyl-6-bromobicyclo[2.3.0]heptane

Ans. (C)

Sol.

IUPAC name : 3-Bromo-6-methylbicyclo[3.2.0]heptane.

54. The first ionisation potential of Na, Mg, Al and Si are in the order :(A) Na < Mg > Al < Si (B) Na > Mg > Al > Si (C) Na < Mg < Al > Si (D) Na > Mg > Al < Si

Ans. (A)Sol. Mg = 1s2, 2s2p6, 3s2. In it last e� has to be removed from 3s orbital which has more penetration power than

last electron of Al which is in 3p orbital.IE

mg > IE

Al

55. The first four ionization energy values of a metal are 191,587,872 and 5962 kcal/mol respectively. The numberof valence electrons in the element is :(A) 1 (B) 2 (C) 3 (D) 5

Ans. (C)Sol. E

4 is abnormally higher than E

3 which indicates that it has become very diffcult to remove 4th electron (due

to attainment of inert gas configuration). Hence number of valence electrons must be 3.

56. An aqueous of a salt �X� gives white precipitate with dilute H2SO

4. The same solution with a few drops of aq.

KI gives golden yellow precipitate which dissolves on heating. The salt �X� is :

(A) Ba(NO3)2

(B) Sr(NO3)2

(C) Pb(NO3)2

(D) Zn(NO3)2

Ans. (C)Sol. PbSO

4 is white

Pb2 is yellow.

57. The rate of the reaction MnO4�

(aq.) + 8H+

(aq.) + 5Fe2+

(aq.) Mn2+

(aq.) + 5Fe3+

(aq.) + 4H

2O can be best measured

by monitoring colorimetrically the concentration of :(A) MnO�

4(aq.)(B) Mn2+

(aq.)(C) Fe2+

(aq.)(D) Fe3+

(aq.)

Ans. (A)Sol. MnO

4� is used as standard solution in titrations and has intense colour.

58. Which of the following observation indicates colligative properties?I. A 0.5 M NaBr solution has a higher vapour pressure than 0.5 M BaCl

2.

II. A 0.5 M NaOH solution freezes at a lower temperature than pure water.III. Pure water freezes at a higher temperature than pure ethanol.(A) Only I (B) Only II (C) Only III (D) I and II

Ans. (D)Sol. Relative lowering of vapour pressure

RESONANCE PAGE - 10

NSEC-2012-13 Depression in freezing point. Comparision of melting point of two different compounds.& are indicating colligative properties.

59. A 500 g toothpaste sample has 0.4 g fluoride concentration. The fluoride concentration in terms of ppm willbe:(A) 200 (B) 400 (C) 500 (D) 800

Ans. (D)Sol. 500 gm toothpaste contains 0.4 g fluoride

106 g toothpaste will contain 500

4.0× 106

= 800 g fluorideppm of fluoride = 800

60. Among the following carbon centered reactive intermediates, the carbon that has octet of electrons is :(A) Carbocation (B) Carbanion (C) Carbine (D) Radical

Ans. (B)Sol. Carbanion has complete octet.

61. The molecule that has maximum covalent character :(A) NaH (B) Na

2S (C) CaCl

2(D) SnCl

4

Ans. (D)Sol. According to fajan�s rule.

Charge Polarising power of cation Covalent character.

SnCl4 (Sn4+) has maximum covalent character.

62. The mode of expression in which the concentration remains independent of temperature is :(A) Molarity (B) Normality (C) Formality (D) Molality

Ans. (D)

Sol. Molality = )Kgin(SolventofMassSoluteofmolerof.No

Since no. of moles and mass both are independent of temperature, hence molality does not depend ontemperature.

63. The enthalpy changes for the following reactions are :C

diamond + O

2(g) CO

2(g)H = � 395.3 kJ mol�1

Cgraphite

+ O2(g)

CO2(g)

H = � 393.4 kJ mol�1

The enthalpy change for the transitionC

diamond C

graphite will be :

(A) � 3.8 kJ mol�1 (B) + 3.8 kJ mol�1 (C) � 1.9 kJ mol�1 (D) + 1.9 kJ mol�1

Ans. (C)Sol. C

diamond + O

2(g) CO

2(g) H = � 395.3 KJ/mol.

Cgraphite

+ O2(g) CO

2(g) H = � 393.4 KJ/mol.

Substracting eq. (2) from eq. (1)C

diamond C

graphite H = ?

H = � 395.3 � (�393.4)

= � 1.9 KJ/mole

64. The sequence of steps involved in aromatic nucleophilic substitution involving a benzyne intermediate is :(A) Addition-elimination (B) Elimination-addition(C) Addition-rearrangement (D) Elimination-rearrangement

Ans. (B)Sol. Aromatic nuclephilie substition is a two step process.

Step-1 : Elimination of HX and formation of benzyne.Step-2 : Addition of NH

3 on benzyne.

RESONANCE PAGE - 11

NSEC-2012-13

65.

The �product� in the above reaction is :

(A) (B)

(C) (D) This reaction cannot take place

Ans. (B)

Sol.

66. The commercial of calcium hydride is(A) Lime (B) Hydrolyth (C) Slaked lime (D) Calgon

Ans. (B)Sol. Calcium hydride is called hydrolyth.

67. The number of moles of KMnO4 that will be needed to react completely with one mole of ferrous oxalate

[Fe(C2O

4)] in acidic solution is

(A) 1 (B) 2/5 (C) 3/5 (D) 4/5Ans. (C)

Sol.

3KMnO4 + 5FeC

2O

4 Products

by mole-mole analysis.

5

OFeC

3

KMnO 42n

4n

nKMnO4 = 1

53 moles.

68. Protein and DNA being charged molecule, can be separated by(A) Electrophoresis (B) Centrifugation (C) Filtration (D) Spectrophotometry

Ans. (A)Sol. The charged particles/molecules can be seperated by electrophoresis.

69. The biomolecule which does not have a secondary structure is(A) Protein (B) Lipid (C) DNA (D) RNA

Ans. (B)Sol. Lipids does not have a secondary structures.

RESONANCE PAGE - 12

NSEC-2012-13

70.

The rate of o-nitration of the above compounds, (I) toluene, (II) 2-D-toluene and (III) 2, 6-D2-tolueneare is in

the following order(A) I > II > III (B) II > I > III(C) III > I > II (D) The rate of the same for all the three compounds

Ans. (D)Sol. Nitration reaction is independent of leaving group ability hence all the three have nearly equal rate.

71. In which of the following reaction is Kp > K

c

(A) H2 + I

2 2 HI (B) N

2 + 3H

2 2 NH

3

(C) 2SO3 2 SO

2 + O

2(D) PCl

3 + Cl

2 PCl

5

Ans. (C)Sol. K

p = K

c(RT)n

for reaction ,n is + ve. Hence K

p > K

c.

72. The preferred sites of protonation in the following compounds are

(A) 1 and 3 (B) 2 and 4 (C) 1 and 4 (D) 2 and 3Ans. (A)Sol. Protonation at site 1 and 3 is supported by resonance stabilization.

73. Which of the following vibrational modes show no IR absorption bands ?(A) Symmetric CO

2 stretch (B) Antisymmetric CO

2 stretch

(C) Symmetric S = C= O stretch (D) Antisymmetric S=C=O stretchAns. (A)Sol. Symmetrical CO

2 stretching does not change the molecular dipole, therefore shows no IR absorption bands.

74. The crimson colour imparted to flame is due to a salt of(A) Barium (B) Copper (C) Calcium (D) Strontium

Ans. (D)Sol. Sr2+ gives crimson colour to flame.

75. Which of the following weighs less when weighed in magnetic field ?(A) ScCl

3(B) FeCl

3(C) TiCl

3(D) VCl

3

Ans. (A)Sol. It is for diamagnetic substance (Sc+3). It has all e� s paired.

76. Essential vitamin required for the production of RBCs is(A) Folic acid (B) Nicotinic acid (C) Pantothenic acid (D) None of the above

Ans. (A)Sol. Folic acid is the basic requirement for the production of RBCs and deficiency of which cases anaemia.

RESONANCE PAGE - 13

NSEC-2012-1377. For the reaction NH

4+ + NO

2� N

2 + 2H

2O the following data was recorded

Set NH4+ / M NO2

� Rate / MS�1

1 0.010 0.020 0.020 2 0.015 0.020 0.030 3 0.010 0.010 0.005

(A) rate = K [NH4

+] [NO2�] (B) rate = K [NH

4+]2 [NO

2�]

(C) rate = K[NH4

+] [NO2�]2 (D) rate = K[NH

4+]2 [NO

2]2

Ans. (C)Sol. When [NH

4+] is increased by 1.5 times, keeping [NO

2�] constant, rate increases by 1.5 times.

r [NH4

+]when [NO

2�] is doubled keeping [NH

4+] same, the rate becomes 4 times.

r [NO2�]2

r [NH4

+] [NO2�]2

or r = K [NH4

+] [NO2�]2

78. In a nitration experiment, 10.0g of benzene gas 13.2 g of nitrobenzene. The percentage yield is :(A) 83.5% (B) 62.7% (C) 88.9% (D) 26.7%

Ans. (A)

Sol.

78 gm benzene gives 123 gm nitrobenzene

10 gm benzene = be) should(gm77.151078

123

% age yield = 100 yieldlTheoretica

yieldActual

= 10077.152.13 _~ 83.5%

79. The rate constant of a reaction increases by 5% when the temperature is increased from 27°C to 28°C.

Therefore, the energy of activation of the reaction is(A) 36.6 kJ mol�1 (B) 46.6 kJ mol�1 (C) 16.6 kJ mol�1 (D) 26.6 kJ mol�1

Ans. (A)

Sol.

211

2

T1

T1

R303.2Ea

K

Klog

or

301

1

300

1

314.8303.2

E

100

105log a

or301300

1314.8303.2

E021.0 a

or Ea = 36636 J mole�1

= 36.636 kJ mole�1

RESONANCE PAGE - 14

NSEC-2012-1380. Which one of the following compounds has R configuration ?

I. II. III. IV.

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

Ans. (D)

Sol.

RESONANCE PAGE - 15

NSEC-2012-13