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JEE Main 2020 Chemistry Paper With Solutions Jan 9 Shift 1

JEE Main 2020 solved shift 1 Chemistry paper (9th January) is given here. Students can easily download these solutions in PDF format for free. Learning these solutions help students to get a knowledge of the type of questions asked for the JEE Main exam. Solutions are given in a step by step manner so that students can easily understand the problems.
January 9 Shift 1 – Chemistry
Question 1. The de Broglie wavelength of an electron in the 4th Bohr orbit is:
a) 4πa0
b) 42πa0
c) 8πa0
d) 6πa0

n=4

Z = 1 , λ= ?

JEE Main 2020 Paper With Solutions Chemistry Shift 1 Jan 9

Circumference (2πr)= nλ

2πa0n2/z = n

On solving, we get

= 8πao

Answer:(c)

Question 2. If the magnetic moment of a dioxygen species is 1.73 B.M, it may be:
a) O2, O2, or O2+
b) O2, or O2+
c) O2 or O2,
d) O2, O2+

JEE Main 2020 Paper With Solutions Chemistry Shift 1 Jan 9

Answer: (b)

Question 3. If enthalpy of atomisation for Br2(l) is x kJ/mol and bond enthalpy for Br2 is y kJ/mol, the relation between them:
a) is x > y
b) is x < y
c) is x = y
d) does not exist

JEE Main 2020 Papers With Solutions Chemistry Shift 1 Jan 9

ΔHatomisation = ΔHvap + y

x − y = ΔHvap

Answer: (a)

Question 4. Which of the following oxides are acidic, basic and amphoteric, respectively?
a) MgO, Cl2O, Al2O3
b) N2O3, Li2O, Al2O3
c) SO3, Al2O3, Na2O
d) P4O10, Cl2O, CaO

Non-metallic oxides are acidic in nature, metallic oxides are basic in nature and Al2O3 is amphoteric in nature.

Answer: (b)

Question 5. Complex X of composition Cr(H2O)6Cln, has a spin only magnetic moment of 3.83 BM. It reacts with AgNO3 and shows geometrical isomerism. The IUPAC nomenclature of X is :
a) Hexaaqua chromium(III) chloride
b) Tetraaquadichlorido chromium(III) chloride dihydrate
c) Hexaaquachromium(IV) chloride
d) Tetraaquadichlorido chromium(IV) chloride dihydrate

Spin only magnetic moment = 3.8 B. M.

This implies, µ = √(n(n + 2)) B.M. (√16 = 4 implies that √15 should be less than four.)

This means, n = 3 as √15 = √(3(3 + 2))

Cr (24) = [Ar]4s1 3d5 (g.s)

For 3 unpaired electrons, the oxidation state of Cr should be +3

Cr3+ can be attained if the complex has a structure that looks like: [Cr(H2O)4Cl2]Cl. 2H2O [Cr(H2O)4Cl2]Cl. 2H2O has the IUPAC name : Tetraaquadichloridochromium(III) chloride dehydrate.

Answer: (b)

Question 6. The electronic configuration of bivalent europium and trivalent cerium, are: (Atomic Number : Xe = 54, Ce = 58, Eu = 63)
a) [Xe]4f 7, [Xe]4f 1
b) [Xe]4f 76s2 , [Xe]4f 26s2
c) [Xe]4f 2, [Xe]4f 7
d) [Xe]4f 4, [Xe]4f 9

Ce (58): [Xe] 6s24f 2 (g.s)

Ce3+: [Xe]4f1

Eu(63) ∶ [Xe]6s24f 7 (g.s)

Eu2+ ∶ [Xe]4f 7

Answer: (a)

Question 7. The Ksp for the following dissociation is = 1.6 × 10–5. PbCl2 (s) ⇌ Pb2 + (aq) + 2Cl(aq). Which of the following choices is correct for a mixture of 300 mL 0.134 M Pb(NO3)2 and 100mL
a) Q > Ksp
b) Q < Ksp
c) Q = Ksp
d) Not enough data provided

Pb(NO3)2: mmoles= 300 mL × 0.134 M = 40.2

NaCl: mmoles = 100 mL × 0.4 M = 40

This implies, [Pb]2+ = 40.2/400 ≈ 0.1M

Qsp = [Pb2+][2Cl]2 = 4 × 10−3 > Ksp

Answer: (a)

Question 8. The compound that cannot act both as oxidising and reducing agent is :
a) H2SO3
b) HNO2
c) H3PO4
d) H2O2

When the oxidation state is maximum it acts like a strong oxidising agent.

When the oxidation state is minimum it acts like a strong reducing agent.

When the oxidation state is between its maximum and minimum, it acts like both an oxidizing and as a reducing agent.

In H3PO4, P has a +5 oxidation state and hence can act like a strong oxidising agent. In the rest, the oxidation state is between their maximum and minimum.

Answer: (c)

Question 9. B has a smaller first ionization enthalpy than Be. Consider the following statements:

(i) It is easier to remove 2p electron than 2s electron

(ii) 2p electron of B is more shielded from the nucleus by the inner core of electrons than the 2s electron of Be

(iii) 2s electron has more penetration power than 2p electron

Atomic radius of B is more than Be (Atomic number B=5, Be=4)

The correct statements are:


a) (i), (ii), and (iii)
b) (i), (iii), and (iv)
c) (ii), (ii), and (iii)
d) (i), (ii), and (iv)

Be (4): 1s22s2

B (5): 1s22s22p1

The electron in 2p1 can easily be extracted.

The penetrating power is of the order: s > p > d > f The shielding power order: s > p > d > f

As we move along the period, the size decreases, as Zeff increases. Hence the radius of B is smaller than the radius of Be.

Answer: (a)

Question 10. [Pd(F)(Cl)(Br)(I)]2−,has n number of geometrical Isomers. Then, the spin-only magnetic moment and crystal field stabilisation energy [CFSE] of [Fe(CN)6]n−6 , respectively, [Note: Ignore pairing energy].
a) 1.73 BM and – 2Δ0
b) 2.84 BM and – 1.6Δ0
c) 0 BM and – 2.4Δ0
d) 5.92 BM and 0

JEE Main 2020 Paper With Solution Chemistry Shift 1 Jan 9

Number of geometrical isomers (n) = 3

[Fe(CN)6]n−6 = [Fe(CN)6]3−6 = [Fe(CN)6]−3

This implies, that Iron is in its +3 oxidation state.

Fe3+(26): [Ar]3d5

CN is a strong ligand in [Fe(CN)6]−3 and causes pairing. Hence, according to CFT, the configuration will be t2g5 e0g.

Hence, there is only 1 unpaired electron, i.e, n=1 in √n(n + 2) = √3 =

1.73 B.M

CFSE = (−0.4 × nt2g + 0.6 × neg0

= (−0.4 × 5 + 0.6 × 0)Δ0

= -2Δ0

Answer: (a)

Question 11. According to the following diagram, A reduces BO2 when the temperature is:

JEE Main Chemistry 2020 Solved Paper For Shift 1 Jan 9


a) > 14000C
b) < 14000C
c) > 12000C
d) < 12000C

Solution: In Ellingham’s diagram, the line of the element that lies below can reduce the oxide of the element which lies above it. Therefore, for A to reduce BO2 , the temperature when the line for element A is below that of BO2, according to the graph when T > 1400 ℃.

For T > 1400 ℃ , ΔGr < 0 for A + BO2-> B + AO2

Answer: (a)

Question 12. For following reactions

Shift 1 Jan 9 JEE Main 2020 Paper With Solutions Chemistry

It was found that the 𝐸𝑎 is decreased by 30 kJ/mol in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponenetial factor is same)


a) 75 kJ/mol
b) 135 kJ/mol
c) 105 kJ/mol
d) 198 kJ/mol

K = Ae(-Ea/RT)

Kcatalyst = Kwithout catalyst

Shift 1 Jan 9 Chemistry JEE Main 2020 Paper With Solutions

\(\begin{array}{l}Ae^{(-\frac{(Ea)c}{RT_{500k}})} = Ae^{(-\frac{(Ea)}{RT_{700k}})}\end{array} \)
\(\begin{array}{l}e^{(-\frac{(Ea)c}{RT_{500k}})} = e^{(-\frac{(Ea)}{RT_{700k}})}\end{array} \)
\(\begin{array}{l}{-\frac{(Ea)c}{RT_{500k}}} = {-\frac{(Ea)}{RT_{700k}})}\end{array} \)

Eac = Ea – 30

-(Ea-30)/T500k = -Ea/T700k

On solving Ea = 105 kJ mol-1

Answer: (c)

Question 13. ‘X’ melts at low temperature and is a bad conductor of electricity in both liquid and solid state. X is:
a) Mercury
b) Silicon Carbide
c) Zinc Sulphide
d) Carbon Tetrachloride

CCl4 is non polar and does not conduct in either solid or liquid state.

Answer: (d)

Question 14. The major product Z obtained in the following reaction scheme is:

Shift 1 Jan 9 JEE Main 2020 Chemistry Paper With Solutions

Shift 1 Jan 9 JEE Main 2020 Chemistry Paper With Solution


Shift 1 Jan 9 JEE Main 2020 Chemistry Papers With Solutions

Hence, major product formed is that of option c.

Answer: (c)

Question 15. Which of these will produce the highest yield in Friedel-Craft’s reaction?

Shift 1 Jan 9 JEE Main 2020 Solved Paper For Chemistry


Out of the four options given, only aniline and phenol show strong +R effects, but as we know, aniline is a Lewis base and can react with a Lewis acid that is added during the reaction. Hence, Phenol gives the highest yield in Friedel-Craft’s reaction.

Answer: (b)

Question 16. The major product (Y) in the following reactions is :

Shift 1 Jan 9 JEE Main 2020 Solved Paper Chemistry

Chemistry JEE Main 2020 Paper With Solutions For Shift 1 Jan 9


Chemistry JEE Main 2020 Paper Solutions For Shift 1 Jan 9

Answer: (a)

Question 17. The correct order of heat of combustion for following alkadienes is:

Chemistry Solved Paper JEE Main 2020 For Shift 1 Jan 9


a) C > B > A
b) B > A > C
c) A > B > C
d) C > A > B

Heat of combustion ∝ 1/ stability

The trans-isomer is more stable than the cis-isomer. More the number of trans forms in a structure, higher the stability.

Chemistry Solved Paper Shift 1 JEE Main 2020 Jan 9

Answer (c)

Question 18. The increasing order of basicity for the following intermediates is (from weak to strong)

Chemistry Shift 1 JEE Main 2020 Solved Paper Jan 9


a) A > B > D > E > C
b) B > A > D > C > E
c) A > B > E > D > C
d) C > E > D > B > A

As we know weaker the conjugate base, stronger the acid.

The order of stability of conjugate base:

Chemistry JEE Main 2020 Solved Paper Shift 1 Jan 9

Hence, the order of basicity or acidic strength is:

A > B > D> E> C

Answer (a)

Question 19. A chemist has 4 samples of artificial sweetener A, B, C and D. To identify these samples, he performed certain experiments and noted the following observations:

(i) A and D both form blue-violet colour with ninhydrin.

(ii) Lassaigne extract of C gives positive AgNO3 test and negative Fe4[Fe(CN)6]3 test.

(iii) Lassaigne extract of B and D gives positive sodium nitroprusside test.

Based on these observations which option is correct?


a) A – Alitame, B – Saccharin, C – Aspartame, D – Sucralose
b) A –Saccharin, B – Alimate, C – Sucralose, D – Aspartame
c) A – Aspartame, B – Alitame, C – Saccharin , D – Sucralose
d) A – Aspartame, B – Saccharin, C – Sucralose, D – Alitame

Chemistry Solved Paper Shift 1 JEE Main 2020 Jan 9

It has a free amine group and hence reacts with ninhydrin to give a purple colour known as

Ruhemann’s purple.

Chemistry JEE Main 2020 Paper Solution For Shift 1 Jan 9

It has Sulphur, therefore, it will give a positive test with sodium nitroprusside.

Chemistry JEE Main 2020 Paper With Solution For Shift 1 Jan 9

It has chlorine and hence it forms a precipitate with AgNO3 in the Lassaigne’s extract of the sugar.

Solved Paper JEE Main 2020 Chemistry Shift 1 Jan 9

It has a free amine group and hence reacts with ninhydrin to give purple colour known as Ruhemann’s purple. Also, it has Sulphur, therefore, it will give positive test with sodium nitroprusside.

Answer: (d)

Question 20. Identify (A) in the following reaction sequence:

Solved Paper Chemisrty Shift 1 JEE Main 2020 Jan 9

Solved Paper 2020 JEE Main Chemistry Shift 1 Jan 9


Solved Paper 2020 JEE Main Shift 1 Chemistry Jan 9

is a methyl ketone, which gives positive Iodoform test.

Solved Paper 2020 Jan 9 JEE Main Shift 1 Chemistry

Answer: (b)

Question 21. The molarity of HNO3 in a sample which has density 1.4 g/mL and mass percentage of 63% is :(Molecular weight of HNO3= 63).

%w/w = 63%

ρ = 1.41g/mL

M = ((%w/w)× ρ× 10)/MM

= (63×1.4×10)/63

= 14 mol/L

Answer: (14)

Question 22.The hardness of a water sample containing 10-3 M MgSO4 expressed as CaCO3 equivalents (in ppm)is (molar mass of MgSO4 is 120.37 g/mol)

Hardness of water is measured in ppm in terms CaCO3.

nCaCO3 = nMgSO4

ppm is the parts (in grams) present per million i.e, 106

1000 mL has 10−3 moles of MgSO4.

Grams of CaCO3 in 1000 mL = 10−3 ×100 grams

Grams of CaCO3 in 1 mL = 10−3×100/1000 grams

Hardness = (10−3×100/1000)×106

= 100

Answer: (100.00)

Question 23. How much amount of NaCl should be added to 600 g of water (ρ = 1.00 g/mL) to decrease the freezing point of water to -0.2°C? (The freezing point depression constant for water = 2 K Kg mol-1)

NaCl is strong electrolyte and gives 2 ions in the solution. This implies, i = 2.

Molarity = (w×1000)/58.5×600

ΔTf = 0.20C

ΔTf = i ×kf ×m

On solving we get

w = 1.76 grams

Answer: (1.76)

Question 24. 108 g silver (molar mass 108 g mol-1) is deposited at cathode from AgNO3(aq) solution by a certain quantity of electricity. The volume (in L) of oxygen gas produced at 273K and 1 bar pressure from water by the same quantity of electricity is

On applying Faraday’s 1st law,

Moles of Ag deposited= 108/108= 1 mol.

Ag+ + e -> Ag

1Faraday is required to deposit 1 mole of Ag.

H2O -> 2H+ + ½ O2 + 2e

½ moles of O2 are deposited by 2F of charge.

This implies, 1F will deposit ¼ moles of O2.

Using PV = nRT

P= 1 bar

T= 273 K

R= 0.0823 Lbar mol−1 K−1

On solving we get,

V = 5.68 L

Answer: (5.8)

Question 25. The mass percentage of nitrogen in histamine is:

Molecular mass of Histamine= 111

In Histamine, 3 nitrogen atoms are present (42g)

The percentage of nitrogen by mass in Histamine = (42/111)×100 = 37.84%.

Solved Paper 2020 Jan 9 JEE Main Shift 1 Chemistry Jan 9

Answer (37.84)

Video Lessons – January 9 Shift 1 Chemistry

JEE Main 2020 Chemistry Paper January 9 Shift 1

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