final jee–main examination – april,2019

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Final JEE -Main Exam April,2019/08-04-2019/Evening Session

1. Calculate the standard cell potential in(V) of thecell in which following reaction takes place :Fe2+(aq) + Ag+(aq) ® Fe3+(aq) + Ag(s)Given that

oAg / Ag

E xV+ =

2oFe / Fe

E yV+ =

3oFe / Fe

E zV+ =

(1) x + 2y – 3z (2) x – z(3) x – y (4) x + y – zOfficial Ans. by NTA (1)

Sol. Fe+2(aq) + Ag+(aq) ® Fe+3(aq) + Ag(s)Cell reactionanode : Fe+2(aq) ® Fe+3(aq) + e1 ;

2 3oFe /Fe

E mV+ + =

cathode : Ag+ (aq) + e1 ® Ag(s) ;

oAg /Ag

E xV+ =

Þ cell standard potential = (m + x)V\ to find 'm';Fe+2 + 2e1® Fe ;

o o1 1E yV G – (2Fy)= Þ D =

Fe+3 + 3e1® Fe ;

o o2 2E zV G – (3Fz)= Þ D =

Fe+2(aq) ® Fe+3(aq) + e1;

o o3 3E mV G – (1Fm)= Þ D =

o o o3 1 2G G – GD = D D = (–2Fy + 3Fz) = –Fm

Þ m = (2y – 3z)

Þ ocellE = (x + 2y – 3z)V

FINAL JEE–MAIN EXAMINATION – APRIL,2019(Held On Monday 08th APRIL, 2019) TIME : 2 : 30 PM To 5 : 30 PM

CHEMISTRY TEST PAPER WITH ANSWER & SOLUTION

2. The major product in the following reaction is :

NH2

N

NNH

N + Base

3CH I ¾¾¾®

(1)

NH2

NCH3

NNH

N +

(2)

NH2

N

NNCH3

N

(3)

NH2

N

NNH

N

CH3

+(4)

NHCH3

N

NNH

N

Official Ans. by NTA (2)ALLEN Ans. (Bonus)

Sol. because one double bond is missing in all givenoption. So aromaticity is lost in both the ring.

3. For the following reactions, equilibriumconstants are given :S(s) + O2(g) � SO2(g); K1 = 1052

2S(s) + 3O2(g) � 2SO3(g); K2 = 10129

The equilibrium constant for the reaction,2SO2(g) + O2(g) � 2SO3(g) is :(1) 10181 (2) 10154 (3) 1025 (4) 1077

Official Ans. by NTA (3)Sol. S(s) + O2(g) � SO2(g) K1 = 1052 ...(1)

2S(s) + 3O2(g)�2SO3(g) K2 = 10129 ...(2)2SO2(g) + O2(g)�2SO3(g) K3 = xmultiplying equation (1) by 2;

2SO(s) + 2O2(g)�2SO2(g) 1041K 10¢ = ...(3)

Þ Substracting (3) from (2); we get

2 2 32SO (g) O (g) 2SO (g)+ �� ;

Keq = 10(129 – 104) = 1025

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4. The ion that has sp3d2 hybridization for thecentral atom, is :

(1) –2[ICI ] (2) –

6[IF ]

(3) –4[ICI ] (4) –

2[BrF ]Official Ans. by NTA (3)

Sol. Chemical species Hybridisation ofcentral atom

–2ICl sp3d

–6IF sp3d3

–4ICl sp3d2

–2BrF sp3d

5. The structure of Nylon-6 is :

(1) (CH ) – C – N2 6

O H

n

(2) (CH ) – C – N2 4

O H

n

(3) C – (CH ) – N2 5

O H

n

(4) C – (CH ) – N2 6

O H

nOfficial Ans. by NTA (3)

Sol. C – (CH ) – N2 5

O H

n

Nylon-6

6. The major product of the following reaction is:

O

Cl t

2 4

(1) BuOK(2) Conc. H SO / D¾¾¾¾¾¾¾®

(1)

O

(2) O

(3)

O

(4) O

Official Ans. by NTA (4)

Sol.

O

Clt–BuOK

E2

O

H / D+

O – H+

O

OH+

H

TautomerOH

7. The major product of the following reaction is:

CH3

Cl

2

2

(1) Cl / h(2) H O,

nD

¾¾¾¾®

(1)

CH2OH

Cl

(2)

CHCl2

Cl

(3)

CO2H

Cl

(4)

CHO

Cl


Sol.

CH3

Cl

CHCl2

Cl

CHO

Cl

Cl / 2 D H O2D

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8. The percentage composition of carbon by molein methane is :(1) 80% (2) 25% (3) 75% (4) 20%Official Ans. by NTA (4)

Sol. CH4

% by mole of carbon = 1molatom

1005molatom

´

= 20%9. The IUPAC symbol for the element with atomic

number 119 would be :(1) unh (2) uun (3) une (4) uueOfficial Ans. by NTA (4)

Sol. Symbol Atomic numberunh 106uun 110une 109uue 119

10. The compound that inhibits the growth oftumors is :(1) cis-[Pd(Cl)2(NH3)2](2) cis-[Pt(Cl)2(NH3)2](3) trans-[Pt(Cl)2(NH3)2](4) trans-[Pd(Cl)2(NH3)2]Official Ans. by NTA (2)

Sol. cis–[PtCl2(NH3)2] is used in chemotherapy toinhibits the growth of tumors.

11. The covalent alkaline earth metal halide(X = Cl, Br, I) is :(1) CaX2 (2) SrX2 (3) BeX2 (4) MgX2

Official Ans. by NTA (3)Sol. All halides of Be are predominantly covalent

in nature.12. The major product obtained in the following

reaction is :

CN O

NH2

3

2

(i)CHCl / KOH(ii)Pd / C/H

¾¾¾¾¾¾®

(1)

H N2 OH

NCH3

H

(2)

CN O

NCH3

H

(3)

CN OH

NCH3

H

(4)

CN OH

NCHCl2

H

Official Ans. by NTA (1)Sol.

NH2

NH2

OH

H

CN O

NC

OCN

CHCl / KOH3

Pd/C/H2

NCH3

Carbylaminereaction

13. The statement that is INCORRECT about theinterstitial compounds is :(1) They have high melting points(2) They are chemically reactive(3) They have metallic conductivity(4) They are very hardOfficial Ans. by NTA (2)

Sol. Generally interstitial compounds are chemicalyinert.

14. The maximum prescribed concentration ofcopper in drinking water is:(1) 5 ppm (2) 0.5 ppm(3) 0.05 ppm (4) 3 ppmOfficial Ans. by NTA (4)

Sol. The maximum prescribed concentration of Cuin drinking water is 3 ppm.

15. The calculated spin-only magnetic moments(BM) of the anionic and cationic species of[Fe(H2O)6]2 and [Fe(CN)6], respectively, are :(1) 4.9 and 0 (2) 2.84 and 5.92(3) 0 and 4.9 (4) 0 and 5.92Official Ans. by NTA (3)

Sol. Complex is [Fe (H2O)6]2 [Fe(CN)6]

Complex ion

Configuration No. of unpaired electrons

Magnetic moment

[Fe(H2O)6]2+ t2g

4 eg2 4 4.9 BM

[Fe(CN)6]4– t2g6eg

0 0 0

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16. 0.27 g of a long chain fatty acid was dissolved

in 100 cm3 of hexane. 10 mL of this solution

was added dropwise to the surface of water in

a round watch glass. Hexane evaporates and a

monolayer is formed. The distance from edge

to centre of the watch glass is 10 cm. What is

the height of the monolayer?

[Density of fatty acid = 0.9 g cm–3, p = 3]

(1) 10–8 m (2) 10–6 m

(3) 10–4 m (4) 10–2 m

Official Ans. by NTA (2)Sol. Radius of watchglass= 10 cm

Þ surface area = pr2 = 3 × (10 cm)2

= 300 cm2

mass of fatty acid in 10 ml solution

=10 0.27

0.027gm100´

=

volume of fatty acid = 0.027g

0.9g / ml = 0.03 cm3

Þ Height = volume of fatty acid

surface area of watch glass

= 3

–62

0.03 cm0.0001 cm 10 m

300 cm= =

17. Among the following molecules / ions,

2– 2– 2–2 2 2 2C ,N ,O ,O

which one is diamagnetic and has the shortestbond length?

(1) 2–2C (2) 2–

2N (3) O2 (4) 2–2O


Sol.Chemical Species

Bond Order

Magnetic behaviour

C22–

3 diamagnetic N2

2– 2 paramagnetic O2 2 paramagnetic

O22– 1 diamagnetic

B.O. 1

bond lengthµ

18. 5 moles of an ideal gas at 100 K are allowed

to undergo reversible compression till its

temperature becomes 200 K.

If CV = 28 JK–1mol–1, calculate DU and DpV for

this process. (R = 8.0 JK–1 mol–1]

(1) DU = 14 kJ; D(pV) = 4 kJ

(2) DU = 14 kJ; D(pV) = 18 kJ

(3) DU = 2.8 kJ; D(pV) = 0.8 kJ

(4) DU = 14 kJ; D(pV) = 0.8 kJ


Sol. n = 5; Ti = 100 K; Tf = 200 K;

CV = 28 J/mol K; Ideal gas

DU = nCVDT

= 5 mol × 28 J/mol K × (200 – 100) K

= 14,000 J = 14 kJ

Þ Cp = Cv + R = (28 + 8) J/mol K

= 36 J/mol K

Þ DH = nCpDT = 5 mol × 36 J/mol K × 100 K

= 18000 J = 18 kJ

DH = DU + D(PV)

Þ D(PV) = DH – DU = (18 – 14) kJ = 4 kJ

19. Which one of the following alkenes when

treated with HCl yields majorly an anti

Markovnikov product?

(1) F3C – CH = CH2

(2) Cl – CH = CH2

(3) CH3O – CH = CH2

(4) H2N – CH = CH2


Sol. CF3–CH=CH2HCl¾¾¾® CF – CH – CH3 2

H

CF – CH – CH3 2

H Cl

Due to higher e– withdrawing nature of CF3

group.It follow anti markovnikoff product

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20. For a reaction scheme 1 2k kA B C¾¾® ¾¾® , if

the rate of formation of B is set to be zero thenthe concentration of B is given by :

(1)1

2

k[A]

k

æ öç ÷è ø

(2) (k1 + k2) [A]

(3) k1k2[A] (4) (k1 – k2) [A]Official Ans. by NTA (1)

Sol. 1 2K KA B C¾¾® ¾¾®

1 2d[B]

0 K [A] K [B]dt

= = -

Þ [B] = 1

2

K[A]

K

21. Which of the following compounds will showthe maximum enol content?(1) CH3COCH2COCH3

(2) CH3COCH3

(3) CH3COCH2CONH2

(4) CH3COCH2COOC2H5

Official Ans. by NTA (1)Sol. Solution

O

CH – C – CH3 2 – C – CH3

O

KetoCH – C3

OH

O

C – CH3CHEnol

Due to intramolecular H-bonding andresonance stabilisation enol content ismaximum

22. The correct statement about ICl5 and –4ICl is

(1) ICl5 is trigonal bipyramidal and –4ICl is

tetrahedral.

(2) ICl5 is square pyramidal and –4ICl is

tetrahedral.

(3) ICl5 is square pyramidal and –4ICl is square

planar.(4) Both are isostructural.Official Ans. by NTA (3)

Sol.

Chemical species

Hybridisation Shape

ICl5 sp3d2 Square pyramidal

ICl4– sp3d2 Square planar

I

ClCl

Cl Cl

Cl

I

ClCl

Cl Cl

–ICl4

–ICl 5

23. The major product obtained in the followingreaction is

CH3

OHC

O NaOH

D¾¾¾®

(1)

O

CH3

(2) O CH3

HH3C

(3) O CH2

HH3C

(4)

O

CH3

Official Ans. by NTA (4)Sol.

CH3

OHC

O

O

CH3

NaOHD

Intramolecularaldol condensation

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24. Fructose and glucose can be distinguished by :

(1) Fehling's test

(2) Barfoed's test

(3) Benedict's test

(4) Seliwanoff's test


Sol. Seliwanoff's test is used to distinguished aldoseand ketose group.

25. If p is the momentum of the fastest electronejected from a metal surface after the irradiationof light having wavelength l, then for 1.5 pmomentum of the photoelectron, thewavelength of the light should be:

(Assume kinetic energy of ejectedphotoelectron to be very high in comparisonto work function)

(1)12

l (2) 34

l

(3) 23

l (4) 49

l


Sol. h KEn -f=

incident

hcKEæ öÞ = +fç ÷lè ø

incident

hcKEæ ö

ç ÷lè ø;

KE = 2

incident

p hc hc2m

= =l l ...(1)

Þ2 2p (1.5) hc

2m´

=¢l

...(2)

divide (1) and (2)

(1.5)2 = l¢l

49l¢Þl =

26. Consider the bcc unit cells of the solids 1 and2 with the position of atoms as shown below.The radius of atom B is twice that of atom A.The unit cell edge length is 50% more in solid2 than in 1. What is the approximate packingefficiency in solid 2?

A

A

A A

A

A

A

AA

Solid 1

A

A

A A

A

A

A

A

B

Solid 2

(1) 45% (2) 65% (3) 90% (4) 75%Official Ans. by NTA (3)

Sol. p.f. =

3 3eff A eff B

A B3

4 4z r z r

3 3

a

æ ö æ ö´ p + ´ pç ÷ ç ÷è ø è ø

2(rA + rB) = 3a

Þ 2(rA + 2rA) = 3a

ÞA2 3 r a=

Þ p.f.=( )3 3

A A

3A

4 4 41 r 8r 9

3 3 38 3 3 r 8 3 3 2 3

´ p + p ´ p p= =

´ ´

p. efficiency = 100 90%2 3

p´ ;

27. Polysubstitution is a major drawback in:(1) Reimer Tiemann reaction(2) Friedel Craft's acylation(3) Friedel Craft's alkylation(4) Acetylation of anilineOfficial Ans. by NTA (3)

Sol. In Friedal crafts alkylation product obtainedis more activated and hence polysubtitutionwill take place.

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28. The Mond process is used for the

(1) extraction of Mo

(2) Purification of Ni

(3) Purification of Zr and Ti

(4) Extraction of Zn


Sol. Mond's process is used for the purification ofNickel.

29. The strength of 11.2 volume solution of H2O2

is : [Given that molar mass of H = 1 g mol–1

and O = 16 g mol–1]

(1) 13.6% (2) 3.4%

(3) 34% (4) 1.7%


Sol. Volume strength = 11.2 × molarity = 11.2

Þ molarity = 1 M

Þ strength = 34 g/L

Þ % w/w = 34

100 3.4%1000

´ =

30. For the solution of the gases w, x, y and z inwater at 298K, the Henrys law constants (KH)are 0.5, 2, 35 and 40 kbar, respectively. Thecorrect plot for the given data is :-

(1)

(0,0) mole fractionof water

x w

yz

partialpressure

(2)


z

partialpressure

x

w

y

(3)


z

partialpressure

x w

y

(4)


partialpressure

z

y

x

w


Sol.2

gasH

H O gas

np k

n n

æ ö= ´ç ÷ç ÷+è ø

2

2

H OH

H O gas

nk 1 –

n n

æ ö= ç ÷ç ÷+è ø

Þ p = kH – kH × 2H Oc

p = (–kH) × 2H Oc + kH

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final jee–main examination – april,2019

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