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
Solution:
-
n=4
Z = 1 , λ= ?
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+
Solution:
-
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
Solution:
-
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
Solution:
-
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
Solution:
-
p>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
Solution:
-
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
Solution:
-
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
Solution:
-
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)
Solution:
-
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
Solution:
-
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 × neg)Δ0
= (−0.4 × 5 + 0.6 × 0)Δ0
= -2Δ0
Answer: (a)
Question 11. According to the following diagram, A reduces BO2 when the temperature is:
- a) > 14000C
- b) < 14000C
- c) > 12000C
- d) < 12000C
Solution:
-
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
It was found that the ?? 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
Solution:
-
K = Ae(-Ea/RT)
Kcatalyst = Kwithout catalyst
\(Ae^{(-\frac{(Ea)c}{RT_{500k}})} = Ae^{(-\frac{(Ea)}{RT_{700k}})}\)\(e^{(-\frac{(Ea)c}{RT_{500k}})} = e^{(-\frac{(Ea)}{RT_{700k}})}\)\({-\frac{(Ea)c}{RT_{500k}}} = {-\frac{(Ea)}{RT_{700k}})}\)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
Solution:
-
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:
Solution:
-
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?
Solution:
-
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 :
Solution:
-
Answer: (a)
Question 17. The correct order of heat of combustion for following alkadienes is:
- a) C > B > A
- b) B > A > C
- c) A > B > C
- d) C > A > B
Solution:
-
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.
Answer (c)
Question 18. The increasing order of basicity for the following intermediates is (from weak to strong)
- a) A > B > D > E > C
- b) B > A > D > C > E
- c) A > B > E > D > C
- d) C > E > D > B > A
Solution:
-
As we know weaker the conjugate base, stronger the acid.
The order of stability of conjugate base:
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
Solution:
-
It has a free amine group and hence reacts with ninhydrin to give a purple colour known as
Ruhemann's purple.
It has Sulphur, therefore, it will give a positive test with sodium nitroprusside.
It has chlorine and hence it forms a precipitate with AgNO3 in the Lassaigne’s extract of the sugar.
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:
Solution:
-
is a methyl ketone, which gives positive Iodoform test.
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).
Solution:
-
%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)
Solution:
-
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)
Solution:
-
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
Solution:
-
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:
Solution:
-
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%.
Answer (37.84)
Video Lessons - January 9 Shift 1 Maths
JEE Main 2020 Chemistry Paper January 9 Shift 1
