JEE Main 2021 March 18 – Shift 1 Chemistry Question Paper with Solutions

SECTION-A

Question 1. The ionic radius of Na⁺ ion is 1.02 Å. The ionic radii (in Å) of Mg²⁺ and Al³⁺, respectively are:

a. 0.72 and 0.54
b. 0.68 and 0.72
c. 1.05 and 0.99
d. 0.85 and 0.99

Answer: (a)

For iso-electronic system,

Question 2. Match List-I with List-II:

Choose the most appropriate match:

a. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
b. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
c. (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
d. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)

Answer: (d)

Alcoholic KOH ⇒ Used for beta – elimination reaction

Pd/ BaSO₄ ⇒ Lindlar’s catalyst

BHC (Benzene hexachloride) ⇒ Addition product of benzene and

chlorine.

Polyacetylene ⇒ Used in electrodes in batteries

Question 3. The statements that are TRUE:

(A) methane leads to both global warming and photochemical smog

(B) methane is generated from paddy fields

(C) methane is a stronger global warming gas than CO₂

(D) methane is a part of reducing smog.

Choose the most appropriate answer from the option given below:

a. (B), (C), (D) only
b. (A), (B), (C) only
c. (A), (B), (D) only
d. (A) and (B) only

Answer: (b)

Contribution of global warming gas 

CO₂ > CH₄ > CFC > O₃ > N₂O > H₂O

But CH₄ is 40 times stronger greenhouse gases than CO2. It has more heating effects.

Question 4. Compound with molecular formula C₃H₆O can show:

a. Both positional isomerism and metamerism
b. Metamerism
c. Positional isomerism
d. Functional group isomerism

Answer: (4)

C₃H₆O DOU = 1

CH₃ – CH₂ – CH = O & CH₃ – CO – CH₃ are functional isomer. Therefore, functional group isomerism is the most appropriate.

Question 5. Match List-I with List-II:

Choose the most appropriate answer from the option given below:

a. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
b. (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
c. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
d. (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)

Answer: (a)

Ca(OCl)₂ → Bleaching powder

CaSO₄. 1/2H₂O → Plaster of Paris

CaO → Cement 

CaCO₃ → Antacid

Question 6. In a binary compound, atoms of element A form a hcp structure and those of element M occupy 2/3 of the tetrahedral voids of the hcp structure. The formula of the binary compound is:

a. M₂A₃
b. MA₃
c. M₄A
d. M₄A₃

Answer: (d)

A → hcp

M → 2/3^rd of tetrahedral

M_{2 × 12/3} A₆ = M₈A₆ = M₄A₃

Question 7. Match List-I with List-II:

Choose the most appropriate answer from the option given below:

a. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
b. (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
c. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
d. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

Answer: (b)

Antacid — Cimetidine

Artificial sweetener — Alitame

Antifertility — Novestrol

Tranquillizers — Valium

Question 8. Reagent, 1-naphthylamine and sulfanilic acid in acetic acid is used for the detection of:

a. NO
b. N₂O
c. NO^–₃
d. NO^–₂

Answer: (d)

For detection of NO₂^–, the following test is used

Question 9. The correct structures of trans-[NiBr₂(PPh₃)₂] and meridional-[Co(NH₃)₃(NO₂)₃] respectively are:

Answer: (b)

Question 10. Match List-I with List-II:

Choose the most appropriate answer from the option given below:

a. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
b. (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
c. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
d. (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)

Answer: (c)

⇒ Cis – Platin [Pt (NH₃)₂Cl₂] used in the treatment of cancer

⇒ Chlorophyll is complex of Mg

⇒ Vitamin B₁₂ is a complex of CO

⇒ Grubbs catalyst is a series of catalysts containing ruthenium

Question 11. The number of ionisable hydrogens present in the product obtained from a reaction of phosphorus trichloride and phosphonic acid is:

a. 3
b. 1
c. 0
d. 2

Answer: (4)

PCl₃ + H₃PO₃ +H₂O → H₄P₂O₅ + HCl

pyrophosphorous acid

The structure of pyrophosphorous acid shows that it has two acidic or ionisable hydrogen. 

Question 12. A certain orbital has no angular nodes and two radial nodes. The orbital is:

a. 2p
b. 3p
c. 2s
d. 3s

Answer: (d)

No angular nodes ⇒ ℓ = 0

Radial nodes = n – ℓ –1 = n – 0 – 1 = 2

n = 3

Answer: 3s

Question 13. Consider the above chemical reaction and identity product “A”:

Answer: (c)

Question 14. Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:

Assertion A: During the boiling of water having a temporary hardness, Mg(HCO₃)₂ is converted to MgCO₃.

Reason R: The solubility product of Mg(OH)₂ is greater than that of MgCO₃.

In the light of the above statements, choose the most appropriate answer from the options given below.

a. A is false but R is true
b. Both A and R are true and R is the correct explanation of A
c. Both A and R are true but R is NOT the correct explanation of A
d. A is true but R is false

Answer: (a)

Mg(HCO₃)₂ → Mg(OH)₂ ↓ + CO₂

Temporary Hardness

Ca(HCO₃)₂ → CaCO₃ ↓ + CO₂ + H₂O

K_sp Mg(OH)₂ > K_sp MgCO₃

And hence, Mg(OH)₂ precipitation first.

Question 15. The chemical is added to reduce the melting point of the reaction mixture during the extraction of aluminium is:

a. Cryolite
b. Calamine
c. Kaolite
d. Bauxite

Answer: (a)

For reducing the melting point of Alumina, Cryolite i.e. Na₃AlF₆ is added.

Question 16. Considering the below chemical reaction, identity the product “X”:

Answer: (c)

Question 17. Considering the below reaction, X and Y respectively are:

Answer: (d)

Question 18. Reaction of Grignard reagent, C₂H₅MgBr with C₈H₈O followed by hydrolysis gives compound “A” which reacts instantly with Lucas reagent to give compound B, C₁₀H₁₃Cl. The Compound B is:

Answer: (d)

Question 19. A non-reducing sugar ”A” hydrolyses to give two reducing monosaccharides. Sugar A is:

a. Glucose
b. Fructose
c. Sucrose
d. Galactose

Answer: (c)

C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆

Sucrose D-glucose D-fructose

(Non-reducing sugar) (reducing sugar)

A non-reducing sugar is a carbohydrate that is not oxidized by a weak oxidizing agent (an oxidizing agent that oxidises aldehydes but not alcohols, such as the Tollen’s reagent) in a basic aqueous solution.

Question 20. Match List-I with List-II:

Choose the most appropriate answer from the option given below:

a. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
b. (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
c. (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
d. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

Answer: (c)

Deacon’s process is used for industrial preparation of Chlorine gas

Contact process is used for industrial preparation of sulphuric acid & V₂O₅ in catalyst involved in the process.

SECTION – B

Question 21. 2 molal solution of a weak acid HA has a freezing point of 3.885 ^oC. The degree of dissociation of this acid is _________ × 10^–3. (Round off to the Nearest Integer). [Given: Molal depression constant of water = 1.85 K kg mol^–1Freezing point of pure water = 0⁰C]

Answer: 50

T_{f sol.} = – 3.885°C

ΔT_f = +3.885 = i × k_f × m

3.885 = i × 1.85 × 2

Answer: 50

Question 22. The total number of unpaired electrons present in the complex K₃[Cr(oxalate)₃] is _______.

Answer: (3)

K₃[Cr(ox)₃]

Chromium & in + 3 oxidation state

Cr → 3d⁵ 4s¹

Cr³⁺ → 3d³ 3 unpaired electron the hybridization of chromium in the complex is d²sp³ 

Question 23. AX is a covalent diatomic molecule where A and X are second-row elements of the periodic table. Based on Molecular orbital theory, the bond order of AX is 2.5. The total number of electrons in AX is __________. (Round off to the Nearest Integer). 

Answer: (15)

The compound AX is NO; its bond order is 2.5 & it has a total of 15 electrons.

Question 24. ___________ grams of 3-Hydroxy propanal (MW = 74) must be dehydrated to produce 7.8 g of acrolein (MW = 56) (C₃H₄O) if the percentage yield is 64. (Round off to the Nearest Integer).

[Given: Atomic masses: C : 12.0 u, H : 1.0 u, O : 16.0 u]

Answer: 16

7.8/ 56 = 0.14 mol

% yield =

Question 25. A reaction of 0.1 mole of Benzyl amine with bromomethane gave 23 g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are n × 10^–1, when n = ___________ . (Round off to the Nearest Integer). [Given: Atomic masses: C: 12.0 u, H : 1.0 u, N : 14.0 u, Br : 80.0 u]

Answer: (3)

Ph – CH₂NH₂ + 3CH₃Br → PhCH₂N⁺(Me)₃Br^–

0.1 mol 23 / 230 = 0.1 mol

∴ moles of CH₃Br = 0.3 = 3 × 10^–1 mol

Question 26. 2NO (g) + Cl₂ (g) → 2 NOCl (s): This reaction was studied at – 10⁰C and the following data was obtained.

[NO]₀ and [Cl₂]₀ are the initial concentrations and r₀ is the initial reaction rate. The overall order of the reaction is _________. (Round off to the Nearest Integer).

Answer: (3)

Exp. (I) 0.18 = K (0.1) ^x (0.1) ^+y …. (1)

Exp. (II) 0.35 = K (0.1) ^x (0.2) ^y …. (2)

Exp. (III) 1.40 = K (0.2) ^x(0.2)^y …. (3)

(2) ÷ (3)

(1) ÷ (2)

Question 27. For the reaction: 2Fe³⁺ (aq) + 2I^– (aq) → 2Fe²⁺ (aq) + I₂(s)

The magnitude of the standard molar free energy change, ΔrG^o_m = – _______ kJ (Round off the Nearest Integer).

Answer: 45 kJ

2Fe³⁺(aq) + 2I^–(aq) → 2Fe⁺²(aq) + I₂(s)

= – 0.108 + 0.

= 0.772 V

= 0.772 – 0.539 = 0.233 V

ΔG° nF E^o_cell

= +2 × 96500 × 0.233

= 44969 J = 44.9 KJ ≃ 45 KJ

Question 28. For the reaction: C₂H₆→ C₂H₄ + H₂ : The reaction enthalpy Δ_rH = __________ kJ mol^–1. (Round off to the Nearest Integer).

[Given: Bond enthalpies in kJ mol^–1; C – C: 347, C = C : 611; C – H : 414; H – H ; 436]

Answer: 131 kJ/mol

C₂H₆ → C₂H₄ + H₂

ΔH = ??

ΔH = [ (E c-c + 6 E C-H) – (E C=C + 4E C-H + E H-H)

ΔH= [ 347 + 6*414] –(611+ 4*414 + 433] = 2831 – 2700 

= 131 kJ/mol

Question 29. In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is 1.0 M, the concentration of sodium acetate in the buffer is _________ M. (Round off to the Nearest Integer).

[Given: pK_a (acetic acid) = 4.74]

Answer: 10

Buffer p^H = 5.74 = P(K_a) (acetic acid) + log sodium acetate / acetic acid

So, on solving, sodium acetate / acetic acid = 10

Sodium acetate = 10 M

Question 30. Complete combustion of 3 g of ethane gives x × 10²² molecules of water. The value of x is ________. [Round off to the Nearest Integer].

[Use: N_A = 6.023 × 10²³; Atomic masses in u : C : 12.0; O : 16.0 : H : 1.0 ]

Given: 18

Answer: 18

C₂H₆ + O₂ → 2CO₂ + 3H₂O

3gm 0.3 mol

0.1 mol 0.3 N_A

= 0.3 × 6.023 × 10²³ molecules of H₂O

= 1.8069 × 10²³

= 18.069 × 10²²