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Association of Chemistry TeachersNATIONAL STANDARD EXAMINATION IN CHEMISTRY 2011-2012
Date of Examination 27th November 2011
Time 12.30 to 14.30 Hrs
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. 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 :
3. A correct answer carries 3 marks and 1 mark will be deducted for each wrong answer.
4. All rough work may be done on the blank sheet provided at the end of the question paper.
5. 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.
6. Scratching or overwriting may result in wrong score.DO NOT WRITE ANYTHING ON THE BACK OF ANSWER SHEET.
7. Use of a nonprogrammable calculator is allowed.
8. Periodic table and log table are provided at the end of this question paper.
9. The answers / solutions to this question paper will be available on our website - www.iapt.org.inby 3rd December 2011.
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.
10. 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 2011 end.
11. TOP 300 (or so) students are called for the next examination - Indian National Chemistry Olympiads(INChO). Individual letters are sent to these students ONLY.
12. Gold medals may be awarded to TOP 35 students in this entire process.
13. No querries will be entertained in this regard.
RESONANCE PAGE - 2
NSEC-2011-12
INDIAN ASSOCIATION OF CHEMISTRY TEACHERS
NATIONAL STANDARD EXAMINATION IN CHEMISTRY 2011-2012NSEC-2011-2012
1. The number of water molecules present in 0.20 g sample of CuSO4 . 5H2O (Molar mass = 249.7) is
(A) 1.2 × 1021 (B) 2.14 × 10
21 (C) 2.14 × 1022 (D) 1.2 × 10
23
Ans. (B)
Sol. Number of waer molecules = 7.249
2.0N
A × 5 = 2.14 × 1021|
2. The group that has the species correctly listed in the order of decreasing radius is
(A) cu2+, Cu+, Cu (B) V, V2+, V3+ (C) F�, Br�, I (D) B, Be, Li
Ans. (B)
Sol. V > V+2 > V+3
Radius of same atom eargchtheofamount1
3. The number of isomers of dibromobiphenyl (Biphenyl - C6H5) is
(A) 8 (B) 10 (C) 12 (D) 4
Ans. (C)
Sol. Overall 12 isomers.
4 The enthalpies of decomposition of methane (CH4) (g) and ethane (C2H6)(g) are 400 and 670 kJ. mol�1,
respectively. The HC�C
in kJmol�1 is
(A) 270 (B) 70 (C) 200 (D) 240 mol
Ans. (B)
Sol. CH4(g) C(g) + 4H(g) H = 400
C2H
6(g) 2C(g) + 6H(g) H = 670
HC� H
= 100
HC�C
+ 6HC � H
= 670
HC�C
= 70
5. The correct formula for hexaaminecobalt (III) nitrate is
(A) [Co3(NH3)] (NO3)3 (B) [Co3(NH3)6](NO3)3 (C) [C0(NO3)3]. 6NH3 (D) [Co(NH3)6(NO3)3
Ans. (D)
Sol. Hexaamine cobalt () nitrate
[CO(NH3)6] (NO
3)3
it is fact
RESONANCE PAGE - 3
NSEC-2011-12
6. For the reaction PCl3(g) + Cl2(g) PCl5(g), kc is 26 at 250°. Kp at the same temperature is
(R = 8.314 JK�1mol�1)
(A) 4.6 × 103 (B) 5.7 × 10
3 (C) 6.0 × 10�3 (D) 8.3 × 10
�3
Ans. (C)
Sol. Kp = K
C (RT)ng
= 26 (8.314 × 523)�1
= 5.9 × 10�3
= 5.7 × 10�3
7. Curved arrows are used in Organic Chemistry to show the movements of electrons in the mechanism of
a reaction. The correct product of the following reaction is
(A) (B)
(C) (D)
Ans. (C)
Sol.
Given reaction is example of claisen rearrangement.
8. Denaturation of protein due to change in pH could be due to
(A) loss of van der Wall's interaction (B) hydrophobic interaction
(C) Change in ionic interaction (D) Breaking of covalent bonds
Ans. (C)
Sol. pH change brings change in ionic interactions, hence denaturation.
9. The initial activity of a radionuclide is 9750 counts per min and 975 counts after 5 min. the decay
constant of the radionuclide in min�1 is about
(A) 0.23 (B) 0.46 (C) 0.69 (D) 0.99
Ans. (B)
Sol. = t1n
NN0 =
51n 10 = 0.46
RESONANCE PAGE - 4
NSEC-2011-12
10. According to VSEPR theory the shape of IF5 molecule will be
(A) tetrahedral (B) trigonal bipyramid
(C) square pyramid (D) Trigonal planar
Ans. (C)
Sol. It is square pyramid for less p � bp & bp � bp repulsion.
11. The formal changes on the atoms underlined are
C6H5 � CN�O
(A) C = 0, N = �1, O =+ 1 (B) C = � 1, N = 1
(C) C = 0, N = 1 = �1 (D) C = +1, N = 0, O = �1
Ans. (C)
Sol. Formal charge on C = 28
4 = 0
Formal charge on N = 28
5 = + 1
Formal charge on O = 622
6 = � 1
12. The number of -particles emitted per second by a radioactive element reduces to 6.25% of the original
value of 48 days. Half- life period of the element in days is
(A) 3 (B) 8 (C) 12 (D) 6
Ans. (C)
Sol. 6.25 = n2
100
n = 4
2t1 =
448
= 12
13. The compound that does not have a bond is
(A) SO2 (B) SF6 (C) O2 (D) SO3
Ans. (B)
Sol. SF6
RESONANCE PAGE - 5
NSEC-2011-12
14. In mass spectrometry a compound is bombarded with high energy electrons to break it into smaller
fragments, which are recorded in the form of their masses (m/z). For example butane gives fragments
like m/s 58, 43, 29, 15, etc. The mass spectrum of an unknown compound is shown below.
The likely compound among the following is
(A) CH3COCl (B) CH
2=CH�CH
2CH
2OH (C) CH
3CH
2COOH (D) CH
3COCH
2CH
3
Ans. (D)
Sol. For butanone the radicals gives the peaks as in the graph (mentioned in question).
15. The solubility of calcium phosphate is S moldm�3. Hence, the solubility product is
(A) S5 (B) 27S3 (C) 54S4 (D) 108S5
Ans. (D)
Sol. Ca3 (PO
4)
2 3Ca2+ + 2PO
43�
3s 2s
(3s)3 (2s)2 = Ksp
Ksp
= 108 S5
16. The number of valence electrons in an atom with the configuration 1s2 2s2 2p6 3s2 3p2 is
(A) 6 (B) 5 (C) 4 (D) 2
Ans. (C)
Sol. Valence shell is n = 3
So valence e� is 4 (3s2 3p2)
17. The elevation in boling point of a solution containing 13.44 g of CuCl2 in 1 kg of water is
(Kb = 0.52 K kg mol�1)
(A) 0.05 (B) 0.10 (C) 0.16 (D) 0.21
Ans. (C)
Sol. Tb = iK
b m
= 3× 0.52 × 1
5.134/44.13 = 0.3× 0.52 = 0.16
18. The configurations of the carbon atoms C2 and C
3 in the following compound are respectively
(A) R, R (B) S, S (C) R, S (D) S, R
Ans. (A)
RESONANCE PAGE - 6
NSEC-2011-12
Sol.
Based on RS convention the given compound has RR configuration.
19. 0.1 dm3 of 0.1 M acetic acid is titrated against 0.1 M NaOH. When 50 cm3 of 0.1 M NaOH are added, the pH
of the solution will be (pKa = 4.74)
(A) 2.37 (B) 4.74 (C) 1.34 (D) 5.74
Ans. (B)
Sol. CH3COOH + NaOH CH
3Na + H
2O
10 milimol 5 mili mol
5 0 5 �
PH = 4.74 +log 55
= 4.74
20. The IUPAC name of complex [Cu(en)2 (H
2O)
2]+ is
(A) ethylene diamine Cu(II) dihydrate (B) diaquobis (ethylenediamine) Cu(II) ion
(C) diaquobisdiethylamine Cu(II) ion (D) diaquobis(ethylenediamine) cuprate(II)
Ans. (B)
Sol. There is a printing error in question copper (I) complex is given in formula and copper (II) complex is given
in options. If we change one of them, then only an answer is possible . it is (B)
21. Two protein molecules with the same average molecular mass (molecular weight) can absorb different amount
of ultraviolet radiation due to difference in the content of
(A) tyrosine (B) glutamic acid (C) lysine (D) methionine
Ans. (A)
Sol. In tyrosine lonepair of Oxygen is in conjugation with benzene ring therefore it has different properties in UV
light due to non bonding to * transition.
22. Each of the following options contains a structure and a description indicating the existence of given structure.
The correction option for methyl 3-hydroxypent-2-enoate is
(A) YES (B*) NO
(C) NO (D) YES
Ans. (A)
Sol. In our opinion given name has correct structure but it is unstable form but official answer given B.
RESONANCE PAGE - 7
NSEC-2011-12
23. Major product of mononitration of the following compound is
(A) (B)
(C) (D)
Ans. (D)
Sol. Nitration takes place at activated benzene ring.
24. If a b c and = = = 90º , the crystal system is
(A) monoclinic (B) triclinic (C) hexagonal (D) orthorhombic
Ans. (D)
Sol. Orthorhomonic
(by fact)
25. The electronic spectrum of [Ni(H2O)
6]++ shows a band at 8500 cm�1 due to d-d transition. [Ph
4As]
2[NiCl
4] will
have such a transition in cm�1 at
(A) 3778 (B) 8500 (C) 4250 (D) 850
Ans. (A)
Sol. because t = 94
0
26. In the conductometric titration of nitrate against KCI, the graph obtained is
(A)
Con
duct
ance
volume of KCl
(B)
Con
duct
ance
volume of KCl
(C)
Con
duct
ance
volume of KCl
(D)
Con
duct
ance
volume of KCl
Ans (B)
Sol. AgNO3 + KCl AgCl + KNO
3
initially Ag+ is replaced by K+
after eqvivalence point K+ and Cl� conc. will increase in solution.
RESONANCE PAGE - 8
NSEC-2011-12
27. The product obtained from the following sequence of reactions is
(A) propanal (B) 2-propanol (C) 1-propanol (D) propane
Ans. (B)
Sol. A = 33 CH�C�CH
||O
,B = 33 CH�CH�CH
|OH
Step 1 is Markownikov addition of water and Step 2 is reduction of ketone formed in step-I to 2-propanol.
28. The compound in which Mn has oxidation state of +3 is
(A) KMnO4
(B) K2 [Mn(CN)
6]
(C) MnSO4
(D) CsMn(SO4)
2.12H
2O
Ans. (D)
Sol. Cs[Mn(SO4)
2.12H
2O]
x � 4 = � 1
x = + 3
29. The SI units of viscocity is
(A) Nsm2 (B) Ns2m (C) Nsm-2 (D) Ns-2m
Ans. (C)
Sol. Nsm�2 viscocity = AreaForce
time
30. If titration of an amino acid present in the solution yielded pl (isoelectric point) value of 10.80, the amino
acid present in the solution may be
(A) glycine (B) arginine (C) histidine (D) proline
Ans. (B)
Sol. Arginine is basic amino acid due to presence of guanadine group and its pI is 10.80.
31. In the coordination compound, Na2[Pt(CN)
4] the Lewis acid is
(A) [Pt(CN)4]2- (B) Na+ (C) Pt2+ (D) CN-
Ans. (C)
Sol. Na2 [Pt (CN)
4]
Pt2+ is Lewis acid
* Because it is e� pair acceptor is above compound.
RESONANCE PAGE - 9
NSEC-2011-12
32. The product (P) of the following reaction is
CH3
+ lCl Lewis acid
P
(A)
Cl
(B)
l
H C3
(C)
Cl
H C3
(D)
CH3
Ans. (B)
Sol. It is simple example of electrophilic aromatic substitution.
33. The correct order of dipole moment for the following molecules is
(I)
OH
NO2
(II)
Cl
Cl
(III)
CH3
CH3
(A) = = (B) < < (C) > > (D) < <
Ans. (C)
Sol. I > II > III
High dipole moment of (I) is due to ionic structure of resonating form.
34. Lead dissolves most readily in dilute
(A) acetic acid (B) sulphuric acid
(C) phosphoric acid (D) sodium hydroxide
Ans. (A)
Sol. Lead dissolves both in NaOH and CH3COOH but more soluble in CH
3COOH.
RESONANCE PAGE - 10
NSEC-2011-12
35. The degrees of freedom for the system CaCO3(s) CaO(s) + CO
2(g) are
(A) 1 (B) 2 (C) 3 (D) 4
Ans. (A)
Sol. In our opnion
For reactive system phase rule is stated as
F = C� � P + 2
Where the no. of components, C� = C � r � Z
C = no. of chemical conotituents
r = number of independent reactions
Z = independent restrictive conditions
Hence, for CaCO3 CaO + CO
2,
P = 3, C� = 3 � 1 = 2
F = 2 � 3 + 2 = 1
But official answer given A.
36. Semipermeable nature of the cell membrane can be attributed to the presence of
(A) protein and DNA (B) lipid and protein
(C) polysaccharide and lipid (D) DNA and lipid
Ans. (B)
Sol. It is fact.
37. Th emf of the cell (Zn|ZnSO4(0.1M) || CdSO
4 (0.01M) | Cd) is
(E0Zn
+2|Zn
= � 0.76 V, E0 Cd
2+|Cd
= 0.40 V at 298 K)
(A) +0.33 V (B) +0.36 V (C) +1.13 V (D) �0.36 V
Ans. (C)
Sol. In our opinion E = Eº � n59.0
logQ
= 1.16 � 259.0
log 01.
1.0
= 1.16 � 259.0
= 1.13 V
but official deleted
38. The nitrogen compound formed when Ca(CN)2 reacts with steam or hot water is
(A) N2O (B) NO (C) NO
2(D) NH
3
Ans. (D)
Sol. In our opinionCa(CN)2 is not likely to react significantly with hot water. The only possible reaction would be.
Ca(CN)2 + 2H�OH Ca(OH)
2 + 2HCN
HCN might have been hydrolysed to HCOOH + NH4+ in acidic medium. But the given reaction proceeds basic
medium, hence this is unlikely.
It seems that the examiner might have intended the hydrolysis of CaNCN (Calcium cyanamide), also written
as CaCN2. The bracket {Ca(CN)
2} was placed by mistake since oxides of nitrogen are simply impossible to
obtain by such a hydrolysis reaction.
formation of nitrogen oxides would involve oxidation of nitrogen (�3 in CN� to +1 in N2O) H
2O is no oxidant.
Hence D is only possible answer but official answer given A.
RESONANCE PAGE - 11
NSEC-2011-12
39. The order of acidities of the H-atoms underlined in the following compounds is in the order -
(I) (II) (III) (IV)
(A) IV>II>I>III (B) II>IV>III>I (C) III>IV>I>II (D) I>III>II>IV
Ans. (A)
Sol. IV > II > I > III
Structure IV is most acidic as the conjugate base is aromatic.
40. The half time for a second order reaction with equal concentrations of the reactants is 35 seconds. 99%
reaction will be completed in
(A) 69s (B) 138s (C) 1733s (D) 3465s
Ans. (D)
Sol.0t C
1C1 + kt t1/2 =
0kC1
11
= 3500
1100
1 t k =
0C351
t = .99 × 3500
= 3465 sec.
41. The 'd' orbitals will be split under square planar geometry into
(A) two levels (B) three levels (C) four levels (D) five levels
Ans. (C)
Sol. In square planer geometry �d� split in forus level
42. Rotational spectra of molecules are observed in
(A) UV region (B) Visible region (C) Near infrared region (D) Far infrared region
Ans. (D)
Sol. The energy gap between rotational energy levels is very-very small.
RESONANCE PAGE - 12
NSEC-2011-12
43. The pair of cations which cannot be separated by H2S in a 0.3N acid solution is -
(A) Al+++, Hg++ (B) Bi+++, Pb++ (C) Zn++, Cu++ (D) Ni++, Cd++
Ans. (B)
Sol. Both will be precipitated
44. Structural features of proteins secreted outside the cells may be stabilised by presence of -
(A) hydrogen bond (B) disulfide bond (C) hydrophobic force (D) phospho-diester bond
Ans. (B)
Sol. It is fact.
45. The C-O-C bond angle in dimethylether is
(A) 109°28 (B) 110° (C) 120° (D) 180°
Ans. (B)
Sol. It is experimental observation.
46. Dimethyl glyoxime forms a square planar complex with Ni2+. This complex should be
(A) diamagnetic (B) paramagnetic having 1 unpaired electron
(C) paramagnetic having 2 unpaired electrons (D) ferromagnetic
Ans. (A)
Sol. [Ni(DMG)2]
DMG form strong field ligand complex
DMG is bidentate ligand
It is diamagnetic
47. A 0.056 M Solution of benzoic acid, C6H
5COOH, is titrated with a strong base. [H+] of the solution when half
of the solution is titrated before the equivalence point is (Ka of benzoic acid = 6.3×10�5)
(A) 6.3×10�5 M (B) 1.8×10�3 M (C) 7.9×10�3 M (D) 2.6×10�2 M
Ans. (A)
Sol. ka =
]COOHHC[
]H[]COOHC[
56
56
OHCOONaHC
NaOHCOOHHC
256
56
6.3 × 10�5 = C
]H[C
[H+] = 6.3 × 10�5 Ans. A
48. The formula of the isothiocyanate ion is
(A) OCN� (B) SCN� (C) ONC� (D) CN�
Ans. (B)
Sol. SCN is called isothiocyanate ion
RESONANCE PAGE - 13
NSEC-2011-12
49. The compound that is chiral is
(A) 3-methyl-3-hexene (B) 4-chloro-1-methylcyclohexane
(C) 2-phenylpentane (D) 1,3-diisopropylbenzene
Ans. (C)
Sol. Only (C) has asymmetric carbon atom.
322*
3 CH�CH�CH�HC�CH|Ph
50. The monomer/s of the following polymer is/are
(-CH2-CH(CH
3)-CH
2-CH(CH
3)-CH
2-CH(CH
3)-)n
(A) ethylene (B) propylene (C) 2-butene (D) ethylene + propylene
Ans. (B)
Sol.CHCH|CH
2
3
51. Of the interhalogen compounds, CIF3 is more reactive than BrF
3 has higher conductance in the liquid state.
The reason is that
(A) BrF3has higher molecular weight (B) CIF
3 is volatile
(C) BrF3 dissociates into BrF2� more easily (D) CIF
3 is most reactive
Ans. (C)
Sol. But error in dissociation of BrF3. It dissociates into BrF
2+ and BrF
4�
52. An element X is found to combine with oxygen to form X4O
6. If 8.40 g of this element combine with 6.50 g of
oxygen, the atomic weight of the element in grams is
(A) 24.0 (B) 31.0 (C) 50.4 (D) 118.7
Ans. (B)
Sol.616
M45.64.8
M = 45.6
4.8616
M = 31
53. synthesis of RNA in a cells would be affected adversely due to shortage of
(A) sulfate (B) acetate (C) oxalate (D) phosphate
Ans. (D)
Sol. It is fact.
54. The most abundant element in the earth's crust is :
(A) aluminium (B) oxygen (C) silicon (D) iron
Ans. (B)
Sol. Most abundant element is earth must is oxygen
RESONANCE PAGE - 14
NSEC-2011-12
55. A beaker is heated from 27°C to 127°C, the percentage of air originally present in beaker that is expelled is:
(A) 50% (B) 25% (C) 33% (D) 40%
Ans. (B)
Sol. n1T
1 = n
2T
2
n × 300 = n2 × 400
n2 =
43
n
expelled air = n
n41
× 100 = 25%
56. The product (C) of the following sequence of reactions is :
(A) (B) (C) (D)
Ans. (D)
Sol. In our opinion there are no methyl group in above reaction
A = B = C =
but official answer given A.
57. The strongest, but the most reactive bond among the following is :
(A) C = N (B) C = C (C) C C (D) C = O
Ans. (C)
Sol. The reactivity of naturally depend on nature of reactant e.g.
(i) �CC� or respond to addition of Br2/ H
2
while will not former is more reactive
(ii) will react with grinardes reagent and given nucleophilic addtion reaction while C = C and CC wil not.
(iii) is most readily protonated while other there are not.
Hence, the question sounds more like " I have through of any one interion in my mind. guess it & I will mark
only that answer correct. " This does not sound like scince.
However on basis of bond strenght �CC is the strogest.a
RESONANCE PAGE - 15
NSEC-2011-12
58. Radioactive inert gas is :
(A) technetium (B) radon (C) xenon (D) curium
Ans. (B)
59. The IUPAC name of the following compound is :
(A) 3-methoxy ethylpropanoate (B) ethyl 4-methoxybutanoate
(C) 1,4-diethoxybutane (D) ethoxy 3-methoxybutyrate
Ans. (B)
Sol.
60. Excess of silver nitrate is added to a water sample to determine the amount of chloride ion present in the
sample. 1.4 g of silver chloride is precipitated. The mass of chloride ion present in the sample is :
Molar masses (g-mol�1) : AgNO3
169.91, AgCl 143.25
(A) 0.25 g (B) 0.35 g (C) 0.50 g (D) 0.75 g
Ans. (B)
Sol. AgNO3 + Cl� AgCl + NO
3�
1.4 g
= 5.143
4.1mol
= 25.143
4.1× 35.5 gm Cl = 0.35 gm
61. The best nucleophile among the following is :
(A) H2O (B) CH
3SH (C) Cl� (D) NH
3
Ans. (B)
Sol. S atom has high polarizibility.
62. The wavelength of a moving body of mass 0.1 mg is 3.31 × 10�29 m. The kinetic energy of the body in J would
be :
(A) 2.0 × 10�6 (B) 1.0 × 10�3 (C) 4.0 × 10�3 (D) 2.0 × 10�3
Ans. (A)
Sol. In our opinion = mvh
mv =
h
m2v2 = 2
2
m
h
21
mv2 = 2
2
m2
h
RESONANCE PAGE - 16
NSEC-2011-12
KE = 2
2
m2
h
= 2297
234
)1031.3(102
)106.6(
= 2 × 10�3.
but official answer given is correct that is D.
63. Secondary structures could be formed in nucleic acid similar to protein due to formation of :
(A) covalent bond (B) ionic bond (C) co-ordinate bond (D) hydrogen bond
Ans. (D)
Sol. It is fact
64. The following titration curve represents the titration of a _______ acid with a _______ base.
(A) strong, strong (B) weak, strong (C) strong, weak(D) weak, weak
Ans. (A)
Sol. Initially pH is very low so strong acid finally pH = 9 to 10
so base is weak.
65. The element with the lowest electronegativity is :
(A) S (B) (C) Ba (D) Al
Ans. (C)
Sol. The lowest electronegative atom is because it is move electropositive metal.
66. Oxalic acid, H2C
2O
4, reacts with paramagnet ion according to the balanced equation 5H
2C
2O
4 (aq) + 2MnO
4�
(aq) 2 Mn2+ (aq) + 10 CO2 (g) + 8 H
2O (l). The volume in mL of 0.0162 M KMnO
4 solution required to
react with 25.0 mL of 0.022 M H2C
2O
4 solution is :
(A) 13.6 (B) 18.5 (C) 33.8 (D) 84.4
Ans. (A)
Sol. Equivalent of KMnO4 = Eq. of H
2C
2OH
0.0162 × V × 5 = .022 × 2 × 25
V = 0162.
252022.0
= 13.6 ml
67. The element that has the highest tendency to catenate is :
(A) silicon (B) germanium (C) sulphur (D) boron
Ans. (C)
Sol. However B and Si also exist as covalent network solid which demonstrates their excellent tedency to cat-
enate (compare B12
with S8; or diamond -like Si)
RESONANCE PAGE - 17
NSEC-2011-12
68. The isotope of carbon which is used in carbon dating (a method to estimate the ageof an ancient sample
containing carbon) is :
(A) carbon-12 (B) carbon-13 (C) carbon-14 (D) carbon-15
Ans. (C)
69. Electronic configuration for the atoms of four elements are given below. The configuration that indicates
colourless aqueous solution is :
(A) 2,8,14,2 (B) 2,8,16,2 (C) 2,8,18,2 (D) 2,8,13,1
Ans. (C)
Sol. 2,8,18,2
it is Zn atom which is diamagnetic & colourless is atom & ion form
(Zn) (Zn+2)
70. The number of stereoisomers of compound CH3�CH=CH�CH(Br)CH
3 is :
(A) 2 (B) 3 (C) 4 (D) 6
Ans. (C)
Sol. CH3�CH=CH�CH(Br)CH
3 has 4 stereo isomers, with (Z, R), (Z, S), (E, R) and (E, S) configuration.
71. At 445º C, Kc for the following reaction is 0.020.
2HI(g) H2(g) + I
2(g)
A mixture of H2, I
2 and HI in a vessel at 445º C has the following concentration :
[HI] = 2.0M, [H2] = 0.50M and [I
2] = 0.10M. The statement that is true concerning the reaction quotient, Q
c is:
(A) Qc K
c ; the system is at equilibrium
(B) Qc less than K
c ; more H
2 and I
2 will be produced
(C) Qc less than K
c ; more HI will be produced
(D) Qc is greater than K
c; more H
2 and I
2 will be produced
Ans. (B)
Sol. Qc = 22
1.05.0
= 405.
= 1.25 × 10�2
Kc = 2 × 10�2
Kc > Q
cforward shiftment.
72. The order of decreasing stability is :
(A) IV > I > II > III (B) I > IV > III > II (C) I > II > IV > III (D) IV > II > I > III
Ans. (A)
Sol. Benzene is most stable as it is aromatic, I and II are more stable than III because of conjugation.
RESONANCE PAGE - 18
NSEC-2011-12
73. The number of amino acid residues found in a protein that is synthesized from a RNA molecule with 120
nucleotides is :
(A) 120 (B) 80 (C) 40 (D) 60
Ans. (C)
Sol. An amino acid is synthesised by a codon made up of 3-neculeotides,
120 nucleoides = 3
120 amino acid.
74. Hypochlorous acid ionizes as :
HOCl(aq) H+(aq) + OCI�(aq)
OCl�(aq) + H2O(l) HOCl(aq) + OH+(aq)
Ka for this reaction at 25ºC is 3.0 × 10�8 (Kw = 1.0 × 10�14 at 25º C)
Hence, Kh for HOCl is :
(A) 3.3 × 10�7 (B) 3.0 × 10�8 (C) 3.0 × 106 (D) 3.3 × 10�7
Ans. (B)
Sol. In our opinion Ka =
]HOCl[]OCl[]H[
Ans.
Kh = K
a K
h of HOCl will be, that case where HOCl reacts with H
2O
example HOCl + H2O H
3O+ + OCl�.
but official answer given A.
75. Einsteinium has 11 electrons in the 4f subshell. The number of unpaired electrons in the subshell is :
(A) 3 (B) 4 (C) 7 (D) 11
Ans. (A)
76. The order of reactivity of ammonia with the following compound is :
(I) CH2=CHBr (II) CH
3�CH
2�COCl (III) CH
3�CH
2�CH
2�Cl (IV) C(CH
3)3C-Br
(A) IV > II > I > III (B) II > IV > III > I (C) III > IV > II > I (D) I > IV > II > III
Ans. (B)
Sol. Acid chloride is more electrophilic than alkyl halide, hence more reactive.
77. The freezing point of a solution containing 8.1 g of HBr in 100 g of water, assuming the acid to be 90% ionized
is [H = 1, Br = 80, Kf for water = 1.86 K kg mol�1]
(A) 0.85ºC (B) �3.53ºC (C) 0ºC (D) �0.35ºC
Ans. (B)
Sol. Tf = (1 � + 2) 1.86 ×
1.081
1.8
= 1.9 × 1.86 = 3.53
Tf of water = � 3.53ºC
RESONANCE PAGE - 19
NSEC-2011-12
78. The reaction of 50% aq KOH on an equimolar mixture of 4-methylbenzaldehyde and formaldehyde followed
by acidification gives -
(A) (B)
(C) (D)
Ans. (B)
Sol. It is example of Canizzaro reaction.
79. Iodide ion is oxidized by acidified dichromate ions as shown in this equation.
Cr2O
72� (aq) + 9� (aq) + 14H+ (aq) 2Cr3+ (aq) + 3
3� (aq) + 7H2O )( . These data were obtianed
when the reaction was studied at a constant pH. The order of the reaction with respect to Cr2O
72� (aq) are :
Exeriment [Cr2O
72�], M [�], M Rate, M.s�1
1 0.0050 0.0125 0.00050
2 0.010 0.0125 0.0010
3 0.0150 0.0250 0.00660
(A) first order with respect to both Cr2O
72� and �
(B) second order with respect to both Cr2O
72� and �
(C) second order with respect to Cr2O
72� and first order with respect to �
(D) first order with respect to Cr2O
72� and second order with respect to �
Ans. (D)
Sol. R = K [Cr2O
72�] [�]2
80. The number of atoms per unit cell and number of the nearest neighbrous in a body centred cubic structure
are:
(A) 4, 12 (B) 2, 6 (C) 9, 6 (D) 2, 8
Ans. (D)
Sol. fact.
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NSEC-2011-12
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