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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.



3

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)



5

(a) CHO (b)

O

(c) O

(d) O

Soln.

OH

NH2

NaNH2.HCl(HNO2)

O

N2

HO

O


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)



6

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.



7

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, . . .



8

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


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



9

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




10

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


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)



11

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


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



12

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).



13

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



14

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



15

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]



16

(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



17

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



18

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



19

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


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)



20

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


52. Among the following compounds which does not under-go [3, 3]-sigmatropic rearrangement

(1) (2)

O

(3)

O

(4) (5) (6)



21

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


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


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



22

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



23

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



24

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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