paper-1 (b.e./b.tech.) jee main 2019 · ch4=12+4=16 c=12 percentage c in ch4= 12 16 ×100 =75% 10....

PAPER-1 (B.E./B.TECH.)

JEE MAIN 2019 Computer Based Test

Solutions of Memory Based Questions

Date: 8th April 2019 (Shift-2)

Time: 02:30 P.M. to 05:30 P.M.

Durations: 3 Hours | Max. Marks: 360

Subject: Chemistry

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JEE MAIN 8 APRIL 2019 SHIFT-1 CHEMISTRY

1. Which of the following alkenes will give Anti-Markovnikov’s product as the major product

(A) Cl − CH = CH2

(B) CH3 − O − CH = CH2

(C) NH2 − CH = CH2

(D) CF3 − CH = CH2

Solution: (D)

Markovnikov’s product is formed when a stable carbocation is formed on the more substituted carbon.

In CF3 − CH = CH2, the carbocation formed will be destabilized by strong −I group of CF3. Hence, the Anti-

Markovnikov’s product will be formed.

2. Glucose and Fructose can be distinguished by:

(A) Barford’s test

(B) Fehling solution

(C) Benedict solution

(D) Seliwanoff’s test

Solution: (D)

Seliwanoff’s test is used to distinguish between glucose and fructose. Seliwanoff’s reagent (0.5% resorcinol in 3N

HCl) gives red solution with fructose but no change in colour with glucose.

3. Which of the following is incorrect about interstitial compounds?

(A) very reactive

(B) high metallic conductivity

(C) very hard

(D) high melting point

Solution: (A)

Interstitial compounds are almost inert by nature. Hence, ‘very reactive’ is the wrong description of interstitial

compounds.

4. Henry constant for the gases W, X, Y and Z are 0.5, 2, 35, 40 bar respectively, then select correct graph.


(A)

(B)

(C)

(D)

Solution: (D)

Pgas = KH. Xgas

Where, Pgas is the partial pressure of gas.

KH is Henry’s constant

Xgas is mole fraction of gas

Pgas = KH(1 − XH2O)

Pgas = KH − KHXH2O


5. product

is

(A)

(B)

(C)

(D)

Solution: (A)


lp of N − a, b, e, d are in involved in delocalization so they are not while in c lp is not involved in delocalization and

so it is more reactive towards nucleophile.

6. Which of the following correct about [I Cl4]− and I Cl5 compound

(A) Both are isostructural

(B) [I Cl4]− is square planar and I Cl5 is square pyramidal

(C) [ICl4]− is square pyramidal and I Cl5 is square planar

(D) I Cl4− is tetrahedral and I Cl5 is pentagonal bipyramidal

Solution:

[I Cl4]− = lp + bp = 2 + 4 = sp3d2

square planar

I Cl5 = 1 lp + 5 bp = 6 = sp3d2

square pyramidal

7. A →k1

B →k2

C; if all the reactions are first order and d[B]

dt= 0; Determine [B]

(A) (k1 + k2)[A]

(B) (k1 − k2)[A]

(C) (k1 × k2)[A]


(D)

k1

k2[A]

Solution: (D)

d[B]

dt=Rate of formation of B − Rate of dissociation of B

d[B]

dt= k1[A] − k2[B]

d[B]

dt= 0; k1[A] − k2[B] = 0

[B] =k1[A]

k2

8. More covalent compound

(A) BeX2

(B) MgX2

(C) SrX2

(D) CaX2

Solution: ()

Beryllium Halides are more covalent in nature, this is due to small size and high charge density of Be+2 ion. Hence it

has high polarization power and more covalent character.

9. What is the percentage of C in CH4

(A) 75%

(B) 25%

(C) 50%

(D) 65%

Solution: (A)

Mass by mass percentage of C in CH4 is

CH4 = 12 + 4 = 16

C = 12

Percentage C in CH4 =12

16× 100

= 75%

10. If wavelength of particle of momentum P is equal to λ; then what will be its wavelength for momentum 1.5 P

(A) 2

3λ


(B)

4

3λ

(C) 3

2λ

(D) λ

Solution: (A)

De-Broglie wavelength 𝜆 =h

P

(λ =h

P) given ……(i)

λ2 =h

1.5 P ……(ii)

(ii)

(i)=

λ2

λ=

P

1.5 P

λ2 =10

15λ =

2

3λ

11. Fe+2 + Ag+ → Fe+3 + Ag

E0 Ag+

Ag= xv E0 Fe+3

Fe= zv E0 Fe+2

Fe= yv

(A) x + 2y − 3z

(B) x − y

(C) x − z

(D) 2x + y − 3z

Solution: (A)

Fe+2A

+ Ag+C

→ Fe3+ + Ag

Ecell0 = SRP of cathode − SRP of anode

= E0Ag+

Ag− E°

Fe+3

Fe+2

= x − E0Fe3+

Fe+2

Since we cannot add E°cell,

(i) Fe+3 → Fe → Z = E10; ΔG1 = −3 × F × Z

(ii) Fe+2 → Fe y = E20; ΔG2 = −2 × F × Y

Fe+3 → Fe+2 (i-ii) ΔG3 = ΔG1 − ΔG2

(E0 = −2y + 3z)

Ecell0 = x − (−2y + 3z)

x + 2y − 3z


12. product is

(A)

(B)

(C)

(D) All of these

Solution: (A)


H2| Pd − C reduced the −C ≡ N into −CH2 − NH2 and into 2𝑜 Alcohol.

13. Polysubstitution is the drawback of which reaction.

(A) Friedel-Craft’s acylation

(B) Friedel-Craft’s alkylation

(C) Reimer Tiemann Reaction

(D) Wurtz Reaction

Solution: ()

Alkylation is followed by polyalkylation as alkyl group is highly activating

While other are not followed by polyalkylation.

14.

(A)


(B)

(C)

(D)

Solution: (A)

Intramolecular aldol

15. Hybridisation of I Cl4− is

(A) sp3

(B) sp3d

(C) sp3d2

(D) sp3d3


Solution: (C)

Steric number = σ + tp

= 4 + 2 = 6

sp3d2 hybridised.

16. Mond's Process purification is used for

(A) Mn

(B) Ni

(C) Fe

(D) Al

Solution: ()

Mond's Process

Ni(s) + 4Co(g) ⇌ Ni(Co)g

In the Mond Process for the purification of Nickel carbon mono oxide is reacted with nickel to produce Ni(Co)4;

which is a gas and can therefore be separated from solid Impurities.

17. Which species has highest bond length and diamagnetic

(A) C2−

(B) N22−

(C) O22−

(D) O2

Solution: (C)

O2(16) = σ1s2σ∗1s2 σ 2s2 σ∗2s2 σ 2pz2 π 2px

2 = π2py2

π∗ 2px1 = π2py

1

Paramagnetic; B.O = 2

O22−

(18)= σ1s2σ∗1s2 σ 2s2 σ∗2s2 σ 2pz

2 π 2px2 = π2py

2

π∗ 2px2 = π∗2py

2


Diamagnetic; B.O= 1 ;

N22−

(16)= σ1s2σ∗1s2 σ 2s2 σ∗2s2 π2px

2 π2py2 σ2pz

2π∗2px1 = π∗2py

1

Paramagnetic; B.O = 2

C2−

(13)= σ1s2σ∗1s2 σ 2s2 σ∗2s2 π2px

2 = π2py2 σ2pz

1

Paramagnetic; 1

2(9 − 4) = 2.5

16. Good reducing nature of H3PO2 is due to the presence of

(A) One P − OH bond

(B) One P − H bond

(C) Two P − OH bond

(D) Two P − H bond

Solution: (D)

The reducing nature of H3PO2 is due to 2 P − H band.

17. Which of the following compound is used for treatment of cancer.

(A) Cis [Pt(NH3)2Cl2]

(B) trans (Pt(NH3)2Cl2)

(C) k[Pt Cl3(η2 − C2H4)]

(D) Fe(n5 − (C5H5)2)

Solution: (B)

Cis chloroplatin = Cis [Pt (NH3)2Cl2]

is used for cancer treatment

18. The spin-magnetic moment of anionic and cationic part of complex [Fe(H2O)6]2[Fe(CN)6] is

(A) 2.9 and 4.9 BM

(B) Zero and 4.9 BM

(C) 4.9 and Zero BM


(D) Zero and 2.9 BM

Solution: ()

In both cases iron exhibits +2 oxidation state

Electronic Configuration = [Ar]3d6

In [Fe(H2O)6]2+, H2O is a weak field ligand and hence no pairing up.

∴ n = 4 ⇒ Spin – magnetic moment = √4(4 + 2) = √24 = 4.9 BM

In [Fe(CN)6]4−; CN is a strong field ligand and hence, paring up happens ∴ n = 0 ⇒ spin

19. Name of the element having atomic number 119 is

(A) Uu

(B) Uun

(C) Une

(D) Uue

Solution: (D)

The element with atomic number 119 is called ununennium (Uue)

20. The structure of Nylon 6 is

(A)

(B)

(C)

(D)


Solution: (B)

Nylon 6 is obtained by heating caprolactam with water at high temperature

21. 28. product is

(A)

(B)


(C)

(D)

Solution: (D)

22. 0.27 g of fatty acid is dissolved in 100ml of solvent. 10 ml of such solution is taken and placed over round plate.

Distance from centre to edge is 10 cm. Now solvent is evaporated and only fatty acid is remaining. Density of fatty

acid = 0.9 g/cc determine height of fatty acid layer.


(A) 10−4 cm

(B) 10−6 cm

(C) 10−8 cm

(D) 10−2 cm

Solution: (A)

Mass of fatty acid = 0.027 g in 10 ml solution

Density of fatty acid 0.9 g/cc

Volume of fatty acid =0.027

0.9= 0.03 cc

Area of plate = πr2 = 3 × 102 = 300 cm2

Height of fatty acid layer =volume

area=

0.03

300= 10−4 cm

23. A crystallizer in a BCC lattice. If the atom at the body centre is replaced with atom B having twice the radius,

what will be the packing efficiency

(A) 68%

(B) 74%

(C) 82%

(D) 90.5%

Solution: (D)

• A

○ B

If radius of A is r radius of B will be 2r. Atoms in the body diagonal touch.

r + 2r + 2r + r = √3 a

6 r = √3 a or a =6r

√3

Packing Efficiency

=1 ×

43 πr3 + 1 ×

43 π(2r)3

a3=

43 πr3 × 9

(6r/√3)3


=4 × 9 × π × r3 × 3√3

3 × 216 × r3= 90.5 %

24. Enol content is maximum is

(A)

(B)

(C)

(D)

Solution: (C)

Enol contained in mono keto group is less then keto so D is least in (A), (B), (C) diketo group are present and active

methyl group so

enol % ∝ acidic character of ∝ active methylene group ∝ −m effect

−M effect of is highest so 𝐴 is answer.

25. 5 mole of ideal gas at 100K (CVm = 28 j/mole/K) It is heated upto 200K. 200K Calculate ΔU and Δ(PV) for

the process (R = 8 J/mole − k)

(A) ΔU = 28 kJ ; Δ(PV) = 8 kJ


(B) ΔU = 14 kJ ; ΔPV = 4 kJ

(C) ΔU = 14 kJ ; Δ(PV) = 8 kJ

(D) ΔU = 28 kJ ; Δ(PV) = 4 kJ

Solution: (B)

ΔU = nCVΔT

= 5 × 28 × 100 = 14000 J

= 14 KJ

Δ(PV) = nRΔT = 5 × 8 × 100 = 4000 J = 4 kJ

26. Find out percentage strength of 11.2 V H2O2

(A) 34%

(B) 3.4%

(C) 1.7 %

(D) 13.8%

Solution: (B)

Molarity of H2O =Volume strength

11.2

M =11.2

11.2= 1

∴ 1 molar means 1000 ml and solution contains 1 mole H2O2 = 1 × 34 gm

∴ 1000 ml solution contains =34

1000× 10 = 3.4%

27. S(s) + O2(g) ⇌ SO2(g); k1 = 1052

2 S(s) + 3O2(g) ⇌ 2SO3(g); k2 = 10129

Calculate Keq for; 2 SO2(g) + O2(g) ⇌ 2 SO3(g)

(A) 1025

(B) 1077

(C) 1070

(D) 1040

Solution: (A)

Keq =(SO3)2

(SO2)2(O2)

K1 =(SO2)

(O2)= 1052


K2 =(SO3)2

(O2)3= 10129

Keq =1012 g

(1052)2= 1025

paper-1 (b.e./b.tech.) jee main 2019 · ch4=12+4=16 c=12 percentage c in ch4= 12 16 ×100 =75% 10....

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