JEE Main 2020 Paper with Solutions Chemistry - Shift 2 is a perfect resource for the candidates who are appearing for JEE Main to have an idea of the difficulty level of the exam. The answer keys are prepared in such a way that the candidates can have a crystal clear understanding of the JEE Main question paper.
The solutions uploaded at BYJU’S will enable the students to obtain a fair idea about the distribution of marks with respect to each topic. Practising these questions will help them understand the amount of preparation required to face the examination.
September 4 Shift 2 - Chemistry
Question 1: The reaction in which the hybridisation of the underlined atom is affected is:
Solution:
Question 2: The process that is NOT endothermic in nature is :
- 1) H(g) + e– → H(g)-
- 2) Na(g) → Na(g)- → e–
- 3) Ar(g) + e– → Ar(g)-
- 4) O-(g) + e– → O(g)2-
Solution:
-
Answer: (1)
H(g) + e– → H(g)- is an exothermic reaction.
Question 3: 
Solution:
Question 4: A sample of red ink (a colloidal suspension) is prepared by mixing eosin dye, egg white, HCHO and water. The component which ensures the stability of the ink sample is :
- 1) HCHO
- 2) Water
- 3) Eosin dye
- 4) Egg white
Solution:
-
Answer: (4)
Surface theoretical egg white
Question 5: The one that can exhibit the highest paramagnetic behaviour among the following is: gly = glycinato; bpy = 2, 2’-bipyridine
- 1) [Ti (NH3)6]3+
- 2) [Co (OX)2 (OH)2]-
- 3) [Pd (gly)2]
- 4) [Fe (en) (bpy) (NH3)2]2+
Solution:
-
Answer: (2)
[1] [Ti (NH3)6]3+ ⇒ Ti3+ (3d1) ⇒ ? = √3
[2] [Co (OX)2 (OH)2]- (Δ0 > P) ⇒ Co+5 (3d4) ⇒ t2g4 eg0
[3] [Pd (gly)2] ⇒ pd2+ (4d8) → Square planar
n = 0, ? = 0, diamagnetic
[4] [Fe (en) (bpy) (NH3)2]2+
Fe2+ ⇒ 3d6 (t2g6 eg0) ⇒ n = 0, ? = 0
Question 6: Which of the following compounds will form the precipitate with aq. AgNO3 solution most readily?
Solution:
-
Answer: (2)
Rate of reaction ∝ stability of carbocation.
Question 7: Five moles of an ideal gas at 1 bar and 298 K is expanded into the vacuum to double the volume. The work done is:
- 1) zero
- 2) Cv [T2 - T1]
- 3) – RT (V2 – V1)
- 4) – RT ln (V2/V1)
Solution:
-
Answer: (1)
As it is free expansion against zero external pressure, work done = zero.
Question 8: 250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO3 and 0.1 M AuCl. The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electrodeposited will be:
- 1) Silver and gold in proportion to their atomic weights
- 2) Silver and gold in equal mass proportion
- 3) only silver
- 4) only gold
Solution:
-
Answer: (1)
Here the current is the same. Both metals are univalent and of the same concentrations. So, both will be deposited in proportions of their equivalent weight or atomic weight.
Question 9: The mechanism of action of “Terfenadine” (Seldane) is:
- 1) Helps in the secretion of histamine
- 2) Activates the histamine receptor
- 3) Inhibits the secretion of histamine
- 4) Inhibits the action of histamine receptor
Solution:
-
Answer: (4)
The mechanism of action of “Terfenadine” (Seldane) is to inhibit the action of histamine receptors.
Question 10: The shortest wavelength of the H atom in the Lyman series is λ1. The longest wavelength in the Balmer series of He+ is:
- 1) 9λ1/5
- 2) 27λ1/5
- 3) 36λ1/5
- 4) 5λ1/9
Solution:
-
Answer: (1)
(1/λ1) = R4 × (1)2 × {1 - [1/∞2]} = RH
(1/λ2) = R4 × (2)2 × {[1/4] - [1/9]} = RH {5/9}
(λ2/λ1) = 9/5
λ2 = (9/5) λ1
Question 11: The major product [B] in the following reactions is:
Solution:
Question 12: The major product [C] of the following reaction sequence will be:
Solution:
Question 13: The Crystal Field Stabilization Energy (CFSE) of [CoF3(H2O)3] (Δ0 < P) is:
- 1) – 0.8Δ0
- 2) – 0.8Δ0 + 2P
- 3) – 0.4Δ0 + P
- 4) – 0.4Δ0
Solution:
-
Answer: (4)
[CoF3(H2O)3] (Δ0 < P)
CO3 + (3d6) = t2 g4 eg2
CFSE = [(-2/5) * 4 + (3/5) * 2] Δ0
= – 0.4 Δ0
Question 14: Among the following compounds, which one has the shortest C – Cl bond?
Solution:
Question 15: The major product [R] in the following sequence of reactions is:
Solution:
-
Answer: (2)
Question 16: The molecule in which hybrid AOs involve only one d-orbital of the central atom is:
- 1) [CrF6]3-
- 2) XeF4
- 3) BrF5
- 4) [Ni (CN)4]2-
Solution:
-
Answer: (4)
(1) (CrF6)3– – d2Sp3
(2) XeF4 – Sp3d2
(3) BrF5 – Sp3d2
(4) [Ni(CN)4]2– → dsp2
Question 17: In the following reaction sequence, [C] is:
Solution:
Question 18: The processes of calcination and roasting in metallurgical industries, respectively, can lead to:
- 1) Photochemical smog and ozone layer depletion
- 2) Photochemical smog and global warming
- 3) Global warming and photochemical smog
- 4) Global warming and acid rain
Solution:
-
Answer: (4)
Environmental
Calcination Releases → CO2 → Global warming
Roasting Releases → SO2 → Acid Rain
Question 19: The incorrect statement(s) among (a) - (c) is (are):
(a) W(VI) is more stable than Cr(VI).
(b) in the presence of HCl, permanganate titrations provide satisfactory results.
(c) some lanthanoid oxides can be used as phosphors.
- 1) (a) only
- 2) (b) and (c) only
- 3) (a) and (b) only
- 4) (b) only
Solution:
-
Answer: (4)
Fact
Question 20: An alkaline earth metal ‘M’ readily forms water-soluble sulphate and water-insoluble hydroxide. Its oxide MO is very stable to heat and does not have a rock-salt structure. M is :
- 1) Ca
- 2) Be
- 3) Mg
- 4) Sr
Solution:
-
Answer: (2)
Fact
Question 21: The osmotic pressure of a solution of NaCl is 0.10 atm and that of a glucose solution is 0.20 atm. The osmotic pressure of a solution formed by mixing 1 L of the sodium chloride solution with 2 L of the glucose solution is x × 10-3 atm. x is _______. (nearest integer)
Solution:
-
Answer: 167
[0.1 × 1 + 0.2 × 2]/3 = 0.5/3 = (500/3) × 10-3 = 167
Question 22: The number of molecules with energy greater than the threshold energy for a reaction increases fivefold by a rise of temperature from 27°C to 42 °C. Its energy of activation in J/mol is ________. (Take ln 5 = 1.6094 ; R = 8.314 J mol-1K-1)
Solution:
-
Answer: 84297.47
(1/5) = e-Ea/300R/e-Ea/315R
(5) = e[Ea/R][(1/300) - (1/315)]
(Ea/R) [(15/(300 × 315)] = ln 5
Ea = 1.6094 × 315 × 20 × 8.314
Ea = 84297.47 J/mol
Question 23: A 100 mL solution was made by adding 1.43 g of Na2CO3. xH2O. The normality of the solution is 0.1 N. The value of x is ________. (The atomic mass of Na is 23 g/mol).
Solution:
-
Answer: 10
(0.1/2) × (100/1000) = (1.43/[160 + 18x])
106 + 18x = 286
18x = 180
x = 10
Question 24: Consider the following equations:
2 Fe2+ + H2O2 → x A + y B (in basic medium)
2 MnO4- + 6 H+ + 5 H2O2 → x 'C + y 'D + z'E (in acidic medium).
The sum of the stoichiometric coefficients x, y, x’,y’ and z’ for products A, B, C, D and E, respectively, is _________.
Solution:
-
Answer: 19
2Fe2+ + H2O2 → xA + yB → 2Fe3+ + 2OH–
2MnO4 – + 6H+ + 5H2O2 → x`C + y`D + Z`E → 2Mn+2 + 5O2 + 8H2O
x = 2 ; y = 2; x’ = 2, y’ = 5, z’ = 8
2 + 2 + 2 + 5 + 8 = 19
Question 25: The number of chiral centres present in threonine is ________.
Solution:
Answer: 2
Video Lessons - September 4 Shift 2 Chemistry
JEE Main 2020 Chemistry Paper With Solutions Shift 2 September 4
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