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2. Section-A 30ONLY ONE Q.1 to Q.10 Q.11 to Q.3
Section-B 10ONE or MORE than ONE
2 marks.
-C 20 Q.41 to Q.50 1 MarkQ.51 to Q.60 2 Marks
1 mark 1/3 marks2 marks 2/3 marks
contains Multiple Choice Questions (MCQ). Each question has 4 choices (a), (b), (c) and (d), for its answer, out of which is correct. From carries 1 Marks and 0 carries 2 Marks each.
3. contains Multiple Select Questions(MSQ). Each question has 4 choices (a), (b), (c) and (d) for its answer, out of which is/are correct. For each correct answer you will be awarded
4. Section contains Numerical Answer Type (NAT) questions. From carries each and carries each. For each NAT type question, the value of answer in between 0 to 9.
5. In all sections, questions not attempted will result in zero mark. In Section–A (MCQ), wrong answer will result in negative marks. For all questions, will be deducted for each wrong answer. For all questions, will be deducted for each wrong answer. In Section–B (MSQ),there is no negative and no partial marking provisions. There is no negative marking in Section –C (NAT) as well.
Regn. No.
E: [email protected], W : www . careerendeavour.inJIA SARAI
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SECTION-A : MULTIPLE CHOICE QUESTIONS (MCQ’s)
Q.1 to Q.10 : Carry 1 Mark each.
1. Order of reactivity of following compounds towards aromatic nucleophilic substitution.
(I)
Cl
NO2
(II)
Cl
NO2
(III)
Cl
NO2
NO2
(IV)
Cl
(a) II > III > IV > I (b) I > IV > II > III (c) III > II > I > IV (d) IV > I > II > IIISoln. If EWG are present at ortho and para position in halobenzene derivative than rate of reaction increases
Correct option is (c)
2.
O
Me
O
Me
O
The correct reagent for above reaction(a) m-CPBA (b) H2O2/AcOH (c) H2O2/NaOH (d) Bu-O–O–H, HCl
Soln. H O O H + Na OH H O O
O
Me
O
Me
OO O H
O H
O
Me
O
Peracid in the presence of base gives 1, 4-addition reaction.Correct option is (c)
3. The IUPAC name of the following compound is
COOH
H
(a) (R)-3- (prop-2-enyl) hex-5-ynoic acid (b) (S)-3- (prop-2-enyl) hex-5-ynoic acid(c) (R)-3- (prop-2-enyl) hex-5-enoic acid (d) (S)-3- (prop-2-ynyl) hex-5- enoic acid
Soln. IUPAC name of above compound can be written using following keynotes(1) carboxylic group has to be given highest priority(2) keeping the chain double bond is parent chain.
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COOH
H
32 1
(S)(S)-3-(prop-2 ynyl) hex-5-enoic acid
45
6.
Correct answer is (d).4. The major product (P) is
SC
O
O
HNO3/AcOH (P)
(a) S
C
O
O
NO2
(b) S
C
O
O NO2
(c) S
C
O
OO2N(d)
SC
O
O
O2N
Soln.S
C
O
O
HNO3/AcOH
SC
O
O NO2
Furan gives electrophilic substitution reactions faster than thiophene.Correct option is (b)
5. The product (P) is
OEt
OEt
O
O
(i) NaOEt
(ii) Ph–CH2–I(iii) KOH(iv) PBr3/Br2(v)(vi) NH3
(P)
(a) Tyrosine (b) Valine (c) Leucine (d) Phenylalanine
Soln.OEt
OEt
O
O
NaOEtH
OEt
OEt
O
OPh CH2 I
OEt
OEt
O
O
H2CPh
KOH
OH
OH
O
O
H2CPh
PBr3
Br2OH
OH
O
O
H2CPh
Br
Ph CH2 CH COOH
Br
–CO2
NH3
Ph CH2 CH COOH
NH2
Correct option is (d)
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6. Which of the following statement are correct(a) All real gas are less compressible than ideal gas at high pressure(b) H2 and He are more compressible than ideal gas for all value of pressure(c) Except H2 and He, the value of z > 1 for all gases at low pressure(d) Compressibility failor of real gas is independent of temperature
Soln. All real gas are less compressible than ideal gas at high pressureCorrect option (a)
7. The oxides that has the inverse spinel structure is(a) FeCr2O4 (b) MnCr2O4 (c) CoAl2O4 (d) Fe2CoO4
Soln. Fe2CoO4 is an inverse spinel because in this case divalent cation (Co2+) occupy the octahedral void whereashalf of trivalent cation (Fe3+) occupy the tetrahedral void.Correct option is (d)
8. The Si–O–Si bond angle in Me3SiOSiMe3 is(a) ~ 120º (b) ~ 180º (c) ~ 90º (d) ~ 109º
Soln.O
Me3Si SiMe3
•• ••
109º
Correct option is (d)9. The ground state energy of hydrogen atom is –13.598 eV. The expectation value of potential energy in units
of eV is(a) –27.196 (b) 27.196 (c) –13.598 (d) 13.598
Soln. For hydrogen atom2 T V
2 13.590 V
V 27.196 Correct option is (a)
10. Which of the following is correct ?(a) Every odd order skew symmetric matrix has non zero determinant(b) Every odd order skew symmetric matrix has zero determinant(c) Eigen value of real orthogonal matrix is not of unit modulus(d) None of these
Soln. Property of skew symmetric matrix(1) odd order skew symmetric matrix has always zero determinant.Hence option (b) is correct and (a) is incorrectEigen value of real orthogonal matrix is always unit modulus.Correct option is (c)
Q.11 to Q.30 : Carry 2 Marks each.11. The product of the following reaction
OH
NH2NaNO2, HCl
77% yield(P)
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(a) CHO (b)
O
(c) O
(d) O
Soln.
OH
NH2
NaNH2.HCl(HNO2)
O
N2
HO
O
Correct option is (c)
12. The major product (P) is
Me3SiO
+
NO2(i)
(ii) H3O+
(iii) TiCl3, H2O
(P)
(a)
NO2
Me3SiO (b)
O
O (c)
O O(d)
Me3SiO NO2
Soln. Me3SiO
+NO2
(4+2) C.A.
NO2
Me3SiO
H3O+
NO2
OO
O
TiCl3/H2O
Correct option is (b)
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13. Consider the following entity
Br
Br Br
(A) (B) (C)
Rate of solvolysis in Ethanol (Aq) kA kB kC
If polarity of solvent is sufficiently increased, then what will happen?(a) Rate of solvolysis will increased at equal pace(b) Rate of solvolysis will be increased at unequal pace(c) Rate of solvolysis will not increased at all(d) Rate of increment will be in the order kC > kA > kB.
Soln. Rate of solvolysis in aqueous ethanol (weak base) will proceed preferantially by SN1 mechanism. Since rateof SN1 reaction stability of carbocation. Here, stability order of carbocation will be in the order of
(A) (B) (C)
• 3º carbocation• very stable
3º carbocation lesser chance of formation, because ofpresence of carbocation atbridgehead position
• 3º carbocataion• Least chance, becausehere lesser flexibility availableto form carbocation as comparedto B, because of its structure
Thus, rate of solvolysis will be in the order of kA > kB > kC and will not increased at equal pace.Correct option is (b)
14. Stabilized ylides react with aldehyde form(a) Kinetically controlled syn oxaphosphetane(b) Kinetically controlled anti-oxaphosphetane(c) Thermodynamically controlled anti-oxaphosphetane(d) Thermodynamically controlled syn oxaphosphetane
Soln. Ylides containing anion stabilising substituents adjacent to the negative charge are known as stabilised ylieds.
C
O
R CH PPh3 C
O
R CH PPh3
Stabilised ylidesstabilised ylides form thermodynamically controlled anti-diastereomer of the oxaphosphetane. This step isreversible. In this diastereomer the bulky group are on opposite sides of the ring. Stereospecific eliminationof this anti-oxaphosphetane gives E alkene.
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C
O
R CH PPh3 + R' CHO
Ph3P O
R'CO
RMore stable
anti-oxaphosphetane
Fast formationof E alkene R
C
O
R'E-alkene
Hence, option (c) is correct
15. The increasing order of acid strength of the aqua complex ions is
(a) 2 2 3 2
2 2 2 26 6 6 6V H O Fe H O Mn H O Hg H O
(b) 2 2 2 3
2 2 2 26 6 6 6V H O Fe H O Hg H O Mn H O
(c) 2 2 2 3
2 2 2 26 6 6 6Hg H O V H O Fe H O Mn H O
(d) 2 2 2 3
2 2 2 26 6 6 6V H O Hg H O Fe H O Mn H O
Soln. More the positive charge on complex = more the acidity of complex.
In case of 2
2 6Hg H O
due to more polarization power of Hg2+ it has more acidity than 3
2 6Mn H O
Correct option is (a)
16. The half life of zero order reaction A P is given by (k = rate constant)
(a) 0
1/2 2A
tk
(b) 1/ 22.303t
k (c)
01/2
At
k (d) 1/2
0
1tk A
Soln. For a zero order reaction
A Pk
0d Ak A k
dt
On integration and putting proper finit we get
0A A kt ... (1)
012 and
2A
t t A
from equation (1) we get 0
1/22A
k t
01/2 2
At
k
Correct option is (a)
17. The angular momentum operator zL i
has eigen functions of the form exp [iB ]. The condition that a
full rotation leaves such an eigen function unchanged is satisfies for all the values of B(a) 0, 1/3, 2/3, 1, 4/3 (b) 0, 1, 2, 3 (c) 0, 1/2, 1, 3/2, . . . (d) 0,1/2, 3/2, 5/2, . . .
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Soln. In exp [iB ] B not a fractional (B = magnetic quantum number)Correct option is (b)
18. For a simple cubic crystal xray diffraction shows intense reflections for angles 1 and 2 . Which are assigned
to [101] and [111] planes respectively. The ratio 1
2
sinsin
is
(a) 1.5 (b) 1.22 (c) 0.82 (d) 0.67
Soln. 2 2 2nsin h k2a
1 2n nsin 1 0 1 sin 1 1 12a 2a
1
2
sin 2 1.414 0.816 0.82sin 1.7323
Correct option is (c)
19. The molecule with the smallest rotational constant (in the microwave) among the following is spectrum(a) N CH (b) HC CCl (c) CCl CF (d) B C Cl
Soln. 2
hB8 Ic
1BI
where I = moment of inertia
Since, I fore B would be smallest for ClC CFCorrect option is (c)
20. Using cuvettcs of 0.8 cm path length a 10–6 M solution of a chromophore shows 60% transmittance at certainwave length. The molar extinction coefficient of the chromophore at this wave length is(a) 4 1 127.75 10 m M (b) 4 1 120.75 10 m M
(c) 4 1 120.75 10 cm M (d) 4 1 127.75 10 cm M
Soln.1 1 100log ; log ; log ; log10 log6 ; 1 0.778 0.222
60/100 60 A A A A A
T6 4
6 4 1 10.222 10 222 10; 0.222 10 0.8 ; 27.75 10 cm M800 8
A c E E
Correct option is (d).
21. If equal volume of BaCl2 and NaF solution are mixed, then which of the combination will gives a precipitate.
7sp 2K of BaF 1.7 10
(a) 2 2210 BaCl and 2 10 M NaF (b) 3 2
210 M BaCl and 1.5 10 M NaF
(c) 2 221.5 10 M BaCl and 10 M NaF (d) 2 2
22 10 M BaCl and 2 10 M NaF
Soln. 22BaF Ba 2F
22spK Ba F
(a)2 2
2 210 2 10Ba F 102 2
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I.P 22 2 60.5 10 10 0.5 10
spI.P K precipitation occur
(b) 22 2 6I.P 0.5 10 0.75 10 0.28 10
spI.P K precipitation occur
(c) 23 2 7I.P 0.75 10 0.5 10 0.187 10
spI.P K No precipitation
(d) 22 2 6I.P 10 10 10
spI.P K precipitation occurCorrect option is (c)
22. Which of the following conductometric tetration show plot as
Conductometric
Vol. of titration(a) Mixture of strong acid and strong acid with strong base(b) Mixture of weak acid and strong acid with strong base(c) Mixture of strong acid and strong acid with weak base(d) Mixture of strong acid and weak acid with weak base
Soln. Mixture of strong acid and weak acid with weak baseCorrect option is (d)
23. The complexation of Fe2+ with the chelating agent dipyridyl has been studied kinetically in both forwardand reverse directions.
223Fe 3dipy Fe dipy
Rate (forward) = (1.45×1013) [Fe2+] 3dipy
Rate (reverse) = (1.22×10–4) 23Fe dipy
The stability constant for the complex is
(a) 170.841 10 (b) 171.19 10 (c) 178.841 10 (d) 1711.9 10
Soln. 2R3 f
3RbFe 3dipy Fe dipy
Stability constant (Keq) = f
b
kk =
1317
41.45 10 1.19 101.22 10
Correct option is (b)
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24.0ºC P
+
MoCO OC
OC
OMe
The major product ‘P’ in the above reaction is:
(a) Mo
CO OMeOC
(b) MoCO
OC O
OMe (c) Mo
CO OCOC
MeO H
(d)
Mn(CO)3(OMe)
Soln.
+
MoCO
COC
OMe
O
MoCO
OC O
OMe
Correct option is (b)
25. Which of the following set(s) show(s) substaintial Jahn-Teller distortion
(I) 22 6Co H O
and 3
2 6Mn H O
(II) 2CrF and 22 6Cu H O
(III) 22 6Cr H O
and 2
2 6Cu H O
(IV) 46Co CN
and 3MnF
(a) Only I and II (b) Only I, II and III (c) II, III and IV (d) Only IIISoln. Species having electronically degenerate eg level undergo substantial Jahn-Teller distortion.
Correct option is (c)26. Which one of the following statements for Hb (Haemoglobine) is not TRUE?
(I) the binding with O2 is weaker in comparision with Myoglobin(II) Iron is five co-ordinated(III) Iron is coplaner with porphyrin ring in absence of oxygen(IV) the oxidation state of iron is +2(a) I, IV (b) I, II (c) I, III (d) III only
Soln. In deoxy form due to large size of Fe2+ iron ion lie above the plane of porphyrin ring.Correct option is (d)
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27. Number of BH
B and B
BB unit in B5H9 are respectively..
(a) 4, 2 (b) 4, 1 (c) 4, 0 (d) 3, 2Soln. STYX code for B5H9 is 4120
Hence, number of BH
B = 4 and number of BB
B= 1
Correct option is (b)
28. Antimony pentafluoride SbF5 reacts with XeF4 and XeF6 to form ionic compounds 3 6XeF SbF and 5 6XeF SbF
respectively. The geometries of the cations and anions in these compounds are respectively.(a) Trigonal planar, square pyramidal and octahedral(b) T-shape, trigonal bipyramidal and octahedral.(c) Trigonal planar, trigonal bipyramidal and octahedral.(d) T-shape, square pyramidal and octahedral.
Soln. 4 6 3 6XeF SbF XeF SbF
6 6 5 6XeF SbF XeF SbF
33
8 3 1XeF 5 sp d2
Xe
F
F
F••
••
+
T-shaped
35
8 5 1XeF 6 sp d2
F
F F
F
Xe
F
••
+
Square pyramidal
3 26
5 6 1SbF 6 sp d2
F
F F
F
Sb
F –
OctahedralF
Correct option is (d)
29. + Br2 Product
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The product will be
(a)
Br
Br
(b)
Br
Br
(c)
Br
Br
(d)
Br
Soln. Phanathracene electrophilic addition at, 9th and 10th position.Correct option is (a)
30. The adsorption of a gas is described by langmuir isotherm with the equilibrium constant 10.5 kPak at25 ºC. The pressure (in kPa) at which the fractional surface coverage is 0.25 is(a) 0.66 (b) 0.55 (c) 0.33 (d) 0.45
Soln. 1 0.251 1 0.25
kP PkP k
1 0.250.5 0.75
kP kP P
1 1 1 10 0.661 0.5 3 1.5 15
kP
Correct answer is (a).
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SECTION-B : MULTIPLE SELECT QUESTIONS (MSQ’s)
Q.31 to Q.40 : Carry 2 Marks each.
31. In which of the following elimination reaction(s) Hoffmann product will be formed as major product
(a) N(CH3)3OH
(b) O
O
Pyrolysis
(c) Br
NaOEt/EtOH (d) Br
O K /t-BuOH
Soln. Example A is case of Hoffmann elimination example B is case of E1 and pyrolysis of esters in case ofexample D, base is bulky. So, less strically hindered proton will eliminate.Correct options are (a), (b) and (d).
32. Which of the following product will not formed in following reaction
OH
(i) PCC(ii) CH3Li
(iii) H+ (P)
(a) O
(b) H
O
(c) OH
(d)
OH
Soln.OH PCC H
O
+ CH3Li OHH+
OH
Correct options ar (a), (b) and (d)
33. Consider the following reaction,
Ag4 4 220ºC rtS N A B
and select the correct statement regarding above
(a) A is isoelectronic with 24S and A is S2N2.(b) B–O in A is 1.5 and B is (SN)X(c) B has metallic properties and is one dimensional psuedometal(d) B is first discovered non-metal superconductor (below 0.26K)
Soln.
Ag4 4 2 2 x220ºC rt
A BS N S N SN
S
N S
N
(S2N2) hence, B.O. is 1.5
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B has metallic property due to presence of electron in * . Hence, behave as metallic conductor. * behaveas conduction bond.
(SN)x at low temperature behave as non-metal super conductor.Hence, all the option is correct i.e. (a, b, c, d)
34. Consider the following reaction sequence
(i) O3
(ii) Zn/H2O(iii) Na2CO3, (iv) LiAlH4, (v) Pd-C,
(A)
Mark the correct statement about the resulting product A(a) A is a polar compound having considerable dipole moment(b) A only gives electrophilic aromatic subtitution(c) A is an aromatic compound having diamagnetic ring current a characteristics feature(d) A gives electrophilic as well as nucleophilic aromatic substitution reaction.
Soln. (i) O3
(ii) Zn/H2O
O
O
Na2CO3
Intramolecularaldol reaction
OH
O O
LiAlH4 HO
Pd-C
Azulene
Azulene is a polar compound which exist in dipolar form. It gives electrophilic as well as nucleophilicaromatic substitution reaction. Since it i an aromatic compound, it gives diamagnetic ring current.Hence, (a), (c) and (d) are correct options.
35. Compound A and B exhibit two singlets each in their 1H NMR spectra. The correct statement aboutexpected chemical shift is/are
(A)
O Me
OOMe
O(B)
OMe
OMe
O
O
(a) 7.7 ppm for compound AA (b) 3.9 ppm for compound AA
(c) 6.9 ppm for compound B (d) 2.1ppm for compound B
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Soln.O
O
CMe
O
C
O
Me
O
O
CMe
O
C
O
Me
A
ring H’s are shielded due to resonance.
Ring becomes electron rich , proton get shielded; appears at 6.9 ppm.
C
O OMe
O
OMe
B
C
O OMe
O
OMe
ring H’s are deshielded
Ring becomes electron defficient, proton get deshielded, appears at 7.7 ppm.C
O
Me appears at 2.1 ppm, C
O
OMe appears at 3.9 ppm.
Correct options are (a),(b),(c),(d)
36. The CORRECT set(s) regarding bond order for 2 2 2 2N , O , N , O is/are
(a) 2 2N N (b) 2 2N O (c) 2 2O N (d) 2 2N O
Soln. When we draw the MOT of N2, N2–, O2, O2
–, the bond order comes out to be 3, 2.5, 2, 1.5.Correct options are (a) , (b), (d)
37. For the first order consecutive reactive P Q R , under steady state approximately to [Q]. Thevariation of [P], [Q] and [R] with time are given below the incorrect representation is/are
(a) conc
entra
tion [Q]
[R]
Time
[P]
(b) conc
entra
tion [Q]
[R]
Time
[P]
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(c) conc
entra
tion
[Q]
[R]
Time
[P]
(d) conc
entra
tion
[Q]
[R]
Time
[P]
Soln. The correct representation is conc
entra
tion
[Q]
[R]
Time
[P]
Correct options are (a), (b) and (d)
38. Which one of the following plots not represents an acceptable ware function
(a)
x
(b)
x
(c)
x
(d)
x
Soln. The acceptable wave function is
x
Correct options are (a), (b) and (c)
39. Activation energy graph are represented regarding transeffect, select the correct set of graph corresponding tonature of ligands.
E
(I)
E
(II)
( )Ea
( )Ea
E
(III)
( )Ea
(a) I and II corresponds to weak -donor and stronger -donor respectively..(b) II and III corresponds to weak donor and strong -acceptor respectively(c) I and II corresponds to strong and weak weak donor respectively(d) I and III corresponds to weak donor and strong -acceptor
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Soln. Stronger -donor leads to increase in energy of ground state and strong -acceptor decrease the valueof activation energy along with increase in energy of ground state.Correct options are (a) and (d)
40. A given mass of gas expand from state A to B by three paths 1, 2 and 3 as shown in the figure
A
B
3
21
P
VWhich of the following relation does not hold for 1 2w , w and 3w
(a) 1 2 3w w w (b) 1 2 3w w w (c) 1 2 3w w w (d) 1 2 3w w w Soln. Work done is equal to area under P-V graph As area under 3 is maximum and area under 1 is minimum so
relation between 1 2w , w and 3w is
1 2 3w w w So relation which does not hold areCorrect option are (a, c, d)
SECTION-C : NUMERICAL ANSWER TYPE (NAT’s)
Q.41 to Q.50 : Carry 1 Mark each.
41. The number of D-glucose-Subunit/(s) present in lactose is/are _______________
Soln.
OO
O
OHOHOH
HOOH
HOOH OHD-galactose D-glucose
Structure of lactose. Hence, only one D-glucose-Subunit/(s) present in Lactose.Correct answer is (1)
42. How many of the following will give diamagnetic ring current, when subjected to external magnetic field
(A) (B) (C)
(D) B
H
(E)
NC CN
(F)
N
N
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Soln. (A) Anti-aroamtic
(B)
Aromatic
(C)
Non-aromatic
(D)
B
H
Aromatic
(E)
NC CN
Aromatic
(F)
N
N
Aromatic
Hence, 4 compound among given are aromatic and thus they will give diamagnetic ring current.Hence, correct answer is (4).
43. Consider the following well known named reaction. How many of those can be used for the preparationof C–C bond in organic synthesis(A) Reformatsky reaction (B) Claisen ester condensation(C) Wittig reaction (D) Knoevenagel reaction
Soln. All of the above mentioned named reaction will lead to formation of C–C bond in organic synthesis.Hence, correct answer is (4)
44. Borazine (B3N3H6) is an inorganic benzene, the number of complete delocalising electrons are ___________Soln. In B3N3H6, there is partial delocalisation not complete delocalisation. So, number of complete delocalising
electron are zero.Correct answer is (0)
45. Consider a two state system at thermal equilibrium with equal degeneracy where the excited state is higherin energy than the ground state by 0.1 eV. The ratio of the population of the excited state to that of the
ground state at a temperature for which Bk T 0.05 eV is ________________
Soln.0 0
B
Ek TJ JN g e
N g
/
0
E k TJ BN eN
0.10J 0.05
0
Ne
N
2J
0
N eN
J
0
Nn 2N
J
0
N 0.13 – 0.14N
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46. An engine has an efficiency of 25% takes in 200 J of heat during each cycle. The amount of work performby engine is _______ J.
Soln. Hq 200J
H200w 25% of q 50 J4
Correct answer is (49 to 51)
47. In acidic medium,2
4 2MnO 8H 5e Mn 4H O
If H ion concentration is doubled, electrode potential of the half cell 24MnO , Mn /Pt will increased by
______________ mV.
Soln.
20
1 8
4
Mn0.0591E E log5 MnO H
... (1)
20
2 8
4
Mn0.0591E E log5 MnO 2H
20
2 8 8
4
Mn0.0591 0.0591 1E E log log5 5 2MnO H
or 2 1 80.0591 1E E log
5 2 {using equation (1)}
2 10.0591 8E E log 2
5
E 0.0284 V
E 28.4 mV
Correct answer is (28 to 29)
48. The number of unpaired electron in deoxyhemerythin is/are _________________Soln. Deoxyhemerythrin is diamagnetic due to antiferromagnetic coupling. Hence, number of unpaired electron is
zero.Correct answer is (0)
49. The number of microstates in B2 molecule is __________________
Soln.2 g u g u u uB
Number of microstate = n! 4! 4 3 2! 12 6
r! n r ! 2! 4 2 ! 2! 2! 2
Correct answer is (6)
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50. The distribution coefficient of an organic compound A for benzene and water is 10. The amount of A extractedif 1.0 gm of it dissolved in 100 mL of water is equilibrated in a separatory funnel with 100 mL of benzeneis_________ gm.
Soln. Suppose amount of A extracted with 100 mL of benzene = xg. The amount of A left in 100 mL water = 1–xg
Now, benzene
water
C 10 (given)C
or,
/100 10 or 11 101 /100
x xx
or, 10 0.909gm11
x
The amount of A extracted with 100 mL of benzene is 0.909 gm.Correct answer is (0.85 to 0.95)
Q.51 to Q.60 : Carry 2 Marks each.51. How many T.S. for the following conversion energy profile diagram?
OHH+/H2SO4
Soln. OH H+
(1)
1, 2-hydrideshift
H
H
–H+
(2)
(3)
Number of T.S. = number of intermediate + 13 + 1 = 4
Correct answer is (4)
52. Among the following compounds which does not under-go [3, 3]-sigmatropic rearrangement
(1) (2)
O
(3)
O
(4) (5) (6)
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Soln. For 3, 3-sigmatropic rearrangement a compound should be in 1, 5-hexadiene form.Two unsaturated centres are trans to each other, they will not undergo 3, 3-sigmatropic rearrangement.Correct answer is (3)
53. In the given compound ,O
OThe total number of 1H NMR signal is _________________
Soln. H3CO
O
CH2
CH3
H
H
H He
ca
bd
f
Correct answer is (6)
54. In structure of Al2(CH3)6, the number two center two electron bond are ________________Soln. In struxture of Al2(CH3)6, two centre two electron bonds are 22.
Al
C
C
H
H
H
HH
C
Al
H HC
H
H H
C
H HH
C
H
H
HH 17 16
18
19
20 2122
1514
1312
11
H
910 8 7
65
4
3 2
1
55. The arrhenius parameter for the thermal decomposition of 2NOCl, 2NOCl 2NO(g) Cl (g) are7 1 1A 10 M s , aE 250 kJ/mol and RT 4.5 kJ/mol . The enthalpy (in kJ/mol) of the activated com-
plex is _________
Soln. #a gH E n 1 RT
#gn 1 Bimolecular reactiv
H 250 kJ/mol 1 1 4.5 kJ/mol
H 250 9 kJ/mol 241 kJ/mol
Correct answer is (240 to 242)
56. The Gº for the reaction is ________ kJ
2 21CO g O CO g Hº 282.84 kJ2
Given : 2
0 1 1COS 213.8 JK mol
2
0 1 1OS 205 JK mol
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0 1 1COS 197.9 JK mol
Soln.1Sº 213.8 197.9 2052
1 3 1Sº 86.6 Jk 86.6 10 kJ mol
3Gº Hº T Sº 282.84 298 86.6 10
282.84 25.87 Gº 257 kJ
Correct answer is (–256 to –258)
57. 3 02MnO Mn E 0.95V
3 2 0Mn Mn E 1.51V
The standard potential of 22MnO Mn is _______ V
Soln. 2 0 0 02MnO 2e Mn E ?? G 2FE
3 0 02 1 1MnO e Mn G FE 0.95F ... (1)
3 2 0 02 2Mn e Mn G FE 1.51F ... (2)
On adding (1) and (2) we obtained desired equation
02FE 0.95 1.51 F
0E 1.23VCorrect answer is (1 to 1.5).
58. The number of bridging carbonyl groups present in 3 12Fe CO is ____________
Soln.Fe Fe
OC
OC
OC
CO
CO
CO
FeOC CO
COOC
OO
Correct answer is (2)59. Half life of 24Na radionuclide is 15h the time in which the activity of a sample of 24Na radionuclide will
decrease by 87.5% is ___________________h.
Soln. ½ 15 hourst ; 0.69315
02.303 logt
NtN
2.303 10015log0.693 100 87.5
t
2.303 15 100log0.693 12.5
t
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t = 45 hours.Correct answer is (45)
60. The fundamental mode of HCl occurs at a 3886 cm–1, predict the frequency of the corresponding mode of DClis __________ cm–1.
Soln. The difference in frequency arises due to difference in mass of the isotopic molecules, we have
osc.osc.
osc.
1 k ; (say) ;2 i
(1 35)(1 35)HCl(2 35)
DCl (2 35)
35 37 37 0.71680.36 70 72
10.7168 3886 2785.48 cm . Correct answer is (2780 – 2790).
END OF QUESTION PAPER
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IIT-JAM CHEMISTRY(CY) 22-01-2017
TEST SERIES - 4FULL LENGTH TEST - (I)
ANSWER KEY
SECTION-A
1. (c) 2. (c) 3. (d) 4. (b) 5. (d)6. (a) 7. (d) 8. (d) 9. (a) 10. (c)11. (c) 12. (b) 13. (b) 14. (c) 15. (a)16. (a) 17. (b) 18. (c) 19. (c) 20. (d)21. (c) 22. (d) 23. (b) 24. (b) 25. (c)26. (d) 27. (b) 28. (d) 29. (b) 30. (a)
SECTION-B
31. (a),(b),(d) 32. (a),(b), (d) 33. (a),(b),(c), (d) 34. (a),(c),(d)35. (a),(b),(c),(d) 36. (a),(b), (d) 37. (a),(b),(d) 38. (a),(b),(c)39. (a),(d) 40. (a),(c),(d)
SECTION-C
41. (1) 42. (4 to 4) 43. (4 to 4) 44. (0) 45. (0.13 to 0.14)46. (49 to 51) 47. (28 to 29) 48. (0) 49. (6) 50.(0.85 to 0.95)51. (4) 52. (3) 53. (6 to 6) 54. (22 to 22) 55.(240 to 242)56. (–256 to –258) 57. (1 to 1.5) 58. (2) 59. (45 to 45)60. (2780 to 2790)
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