JEE Main 2021 Chemistry Paper With Solutions Feb 24 Shift 2

SECTION A

Question 1. The correct order of the following compounds showing increasing tendency towards nucleophilic substitution reaction is:

a. (iv) < (i) < (iii) < (ii)
b. (iv) < (i) < (ii) < (iii)
c. (i) < (ii) < (iii) < (iv)
d. (iv) < (iii) < (ii) < (i)

Answer: (c)

Reactivity ∝ – M group present at o/p position.

Question 2. Match List-I with List-II

List- I List-II

(Metal) (Ores)

(a) Aluminum (i) Siderite

(b) Iron (ii) Calamine

(c) Copper (iii) Kaolinite

(d) Zinc (iv) Malachite

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

Answer: (c)

Siderite FeCO₃

Calamine ZnCO₃

Kaolinite Si₂Al₂O₅(OH)₄ or Al₂O₃.2SiO₂.2H₂O

Malachite CuCO₃.Cu(OH)₂

Question 3. Match List-I with List-II

List- I List-II

(Salt) (Flame colour wavelength)

(a) LiCl (i) 455.5 nm

(b) NaCl (ii) 970.8 nm

(c) RbCl (iii) 780.0 nm

(d) CsCl (iv) 589.2 nm

Choose the correct Answer from the options given below:

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

Answer: (b)

Range of visible region: 390 nm – 760 nm

VIBGYOR

Violet – Red

LiCl Crimson Red

NaCl Golden yellow

RbCl Violet

CsCl Blue

So, LiCl which is crimson have wavelengths close to red in the spectrum of visible region which is as per given data.

Question 4. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Hydrogen is the most abundant element in the Universe, but it is not the most abundant gas in the troposphere.

Reason R: Hydrogen is the lightest element.

In the light of the above statements, choose the correct Answer from the given below

(1) A is false but R is true

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(2) Both A and R are true and R is the correct explanation of A

(3) A is true but R is false

(4) Both A and R are true but R is NOT the correct explanation of A

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

Answer: (b)

Hydrogen is the most abundant element in the universe because all luminous bodies of the universe i.e. stars and nebulae are made up of hydrogen which acts as nuclear fuel and fusion reaction is responsible for their light.

Question 5. Given below are two statements:

Statement I: The value of the parameter ”Biochemical Oxygen Demand (BOD)” is important for survival of aquatic life.

Statement II: The optimum value of BOD is 6.5 ppm.

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

a. Both Statement I and Statement II are false
b. Statement I is false but Statement II is true
c. Statement I is true but Statement II is false
d. Both Statement I and Statement II are true

Answer: (c)

For survival of aquatic life dissolved oxygen is responsible for its optimum limit 6.5 ppm and optimum limit of BOD ranges from 10-20 ppm & BOD stands for biochemical oxygen demand.

Question 6. Which one of the following carbonyl compounds cannot be prepared by addition of water on an alkyne in the presence of HgSO₄ and H₂SO₄?

Answer: (a)

Reaction of Alkyne with HgSO₄ & H₂SO₄ follow as

Hence, by this process preparation of CH₃CH₂CHO can’t be possible.

Question 7. Which one of the following compounds is non-aromatic?

Answer: (b)

(Not planar)

Hence, it is non-aromatic.

Question 8. The incorrect statement among the following is:

a. VOSO₄ is a reducing agent
b. Red color of ruby is due to the presence of CO³⁺
c. Cr₂O₃ is an amphoteric oxide
d. RuO₄ is an oxidizing agent

Answer: (b)

Red color of ruby is due to presence of CrO₃ or Cr⁺⁶ not CO³⁺

Question 9. According to Bohr’s atomic theory:

(a) Kinetic energy of electron is ∝ Z² / n²

(b) The product of velocity (v) of electron and principal quantum number (n). ‘v_n’ ∝ z²

(c) Frequency of revolution of electron in an orbit is ∝ Z³ / n³

(d) Coulombic force of attraction on the electron is ∝ Z³ / n⁴

Choose the most appropriate Answer from the options given below

a. (c) only
b. (a) and (d) only
c. (a) only
d. (a), (c) and (d) only

Answer: (b)

(a) KE = –TE = 13.6 × Z² / n² eV

KE α Z² / n²

(b) V = 2.188 × 10⁶ × z / n m/s

So, V_n ∝ Z

Frequency = V / 2πr

F α Z² / n² [∴ r α Z² / n² and v α Z / n ]

(d) Force ∝ Z² / r²

So, F ∝ Z³ / n⁴

So, only statement (A) is correct.

Question 10. Match List-I with List-II

List- I List-II

(a) Valium (iv) Tranquilizer

(b) Morphine (iii) Analgesic

(c) Norethindrone (i) Antifertility drug

(d) Vitamin B12 (ii) Pernicious anemia

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

Answer: (d)

(a) Valium (iv) Tranquilizer

(b) Morphine (iii) Analgesic

(c) Norethindrone (i) Antifertility drug

(d) Vitamin B12 (ii) Pernicious anemia

Question 11. The Correct set from the following in which both pairs are in correct order of melting point is

a. LiF > LiCl ; NaCl > MgO
b. LiF > LiCl ; MgO > NaCl
c. LiCl > LiF ; NaCl > MgO
d. LiCl > LiF ; MgO > NaCl

Answer: (b)

Generally,

M.P. ∝ Lattice energy =

∝ (packing efficiency)

Question 12. The calculated magnetic moments (spin only value) for species

a. 5.92, 4.90 and 0 BM

b. 5.82, 0 and 0 BM

c. 4.90, 0 and 1.73 BM

d. 4.90, 0 and 2.83 BM

Answer: (c)

Fe²⁺ 3d⁶ → 4 unpaired electrons. as Cl^– in a weak field liquid.

Μ_spin = √24 BM

= 4.9 BM

Question 13. Which of the following reagent is suitable for the preparation of the product in the following reaction?

a. Red P + Cl₂
b. NH₂-NH₂/ C₂H₅O^–Na⁺
c. Ni/H₂
d. NaBH₄

Answer: (b)

It is a wolf-kishner reduction of carbonyl compounds.

Question 14. The diazonium salt of which of the following compounds will form a coloured dye on reaction with β-Naphthol in NaOH?

Answer: (c)

Orange bright dye.

Question 15. What is the correct sequence of reagents used for converting nitrobenzene into m- dibromobenzene?

Answer: (d)

Question 16. The correct shape and I-I-I bond angles respectively in I^–₃ ion are:

a. Trigonal planar; 120^o
b. Distorted trigonal planar; 135^o and 90^o
c. Linear; 180º
d. T-shaped; 180º and 90º

Answer: (c)

I^–₃ has sp³d hybridization (2 BP + 3 LP) and linear geometry.

Question 17. What is the correct order of the following elements with respect to their density?

a. Cr < Fe < Co < Cu < Zn
b. Cr < Zn < Co < Cu < Fe
c. Zn < Cu < Co < Fe < Cr
d. Zn < Cr < Fe < Co < Cu

Answer: (d)

Fact Based

Density depends on many factors like atomic mass. atomic radius and packing efficiency.

Question 18. Match List-I and List-II.

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

Answer: (d)

Question 19. In polymer Buna-S: ‘S’ stands for

a. Styrene
b. Sulphur
c. Strength
d. Sulphonation

Answer: (a)

Buna-S is the co-polymer of buta-1,3-diene & styrene

Question 20. Most suitable salt which can be used for efficient clotting of blood will be:

a. Mg(HCO₃)₂
b. FeSO₄
c. NaHCO₃
d. FeCl₃

Answer: (d)

Blood is a negative sol, according to Hardy-Schulz’s rule, the cation with high charge has high coagulation power. Hence, FeCl₃ can be used for clotting blood.

Section B

Question 1. The magnitude of the change in oxidising power of the MnO4-/ Mn2+ couple is x × 10–4 V, if the H+ concentration is decreased from 1M to 10–4 M at 25°C. (Assume concentration of MnO4- and Mn2+ to be same on change in H+ concentration). The value of x is _____. (Rounded off to the nearest integer).

[Given: 230RT / F = 0.059]

Answer: 3776

5e^– + MnO^–₄ + 8H → Mn⁺² + 4H₂O

= 3776 × 10^–4

So, x = 3776

Question 2. Among the following allotropic forms of sulphur, the number of allotropic forms, which will show paramagnetism is ______.

(a) α-sulphur (b) β-sulphur (c) S₂-form

Answer: (a)

S₂ is like O₂ i.e. paramagnetic as per molecular orbital theory.

Question 3. C₆H₆ freezes at 5.5ºC. The temperature at which a solution of 10 g of C₄H₁₀ in 200 g of C₆H₆ freeze is ________ C. (The molal freezing point depression constant of C₆H₆ is 5.12C/m).

Answer: 1

ΔT_f = i × K_f × m

= 1 × 5.12 × 10 / 58200 × 1000

∆T_f = 5.12 × 50 / 58 = 4.414

T_f(soution) = T_k(solvent) – ΔT = 5.5 – 4.414 = 1.086^oC

≈ 1.09°C = 1 (nearest integer)

Question 4. The volume occupied by 4.75 g of acetylene gas at 50°C and 740 mmHg pressure is _______L. (Rounded off to the nearest integer)

(Given R = 0.0826 L atm K^–1 mol^–1)

Answer: 5

T = 50C = 323.15 K

P = 740 mm of Hg = 740 / 760 atm

V = ?

moles (n) = 4.75 / 26 atm

V = 4.75 / 26 × 0.0821 × 323.15 / 740 × 760

V = 4.97 5 Lit

Question 5. The solubility product of PbI₂ is 8.0 × 10^–9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10^–6 mol/L. The value of x is ________ (Rounded off to the nearest integer)

Given: √2 = 1.41

Answer: 141

PbI₂(s) ⇌ Pb²⁺ (aq) + 2I^– (aq)

S+0.1 2s

K_SP (PbI₂) = 8 × 10⁹

K_SP = [Pb⁺²][I^–]²

8 × 10^–9 = (S + 0.1) (2S)² ⇒ (8 × 10^–9 + 0.1) × 4S²

⇒ S² = 2 × 10^–8

S = 1.414 × 10^–4 mol/Lit

= x × 10^–6 mol/Lit

∴ x = 141.4 141

Question 6. The total number of amines among the following which can be synthesized by Gabriel synthesis is _______

Answer: 3

Only 1^o amines can be prepared by Gabriel synthesis.

Question 7. 1.86 g of aniline completely reacts to form acetanilide. 10% of the product is lost during purification. Amount of acetanilide obtained after purification (in g) is ____× 10–2.

Answer: 243

93 g Aniline produce 135 g acetanilide

1.86 g produce 135 × 1.86 / 93 = 2.70 g

At 10% loss, 90% product will be formed after purification.

∴ Amount of product obtained = 2.70 × 90 / 100 = 2.43 g = 243 × 10^–2 g

Question 8. The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O₂ for complete oxidation and produces 4 times its own volume of CO₂ is C_xH_y. The value of y is

Answer: 8

C_xH_y + 6O₂ 4CO₂ + y/2 H₂O

Applying POAC on ‘O’ atoms

6 × 2 = 4 × 2 + y/2 × 1

y/2 = 4 ⇒ y = 8

Question 9. Sucrose hydrolysis in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25ºC. After 9h, the fraction of sucrose remaining is f. The value of is _________× 10^–2 (Rounded off to the nearest integer)

[Assume: ln10 = 2.303, ln2 = 0.693]

Answer: 81

t_1/2 = 3.33h = 10 / 3h

Fraction of sucrose remaining = f =

Log (1/f) =

= x × 10^–2

x = 81

Question 10. Assuming ideal behavior, the magnitude of log K for the following reaction at 25ºC is x × 10^–1. The value of x is __________. (Integer Answer)

3HC ≡ CH(g) ⇌ C₆H₆ (l)

[Given: Δ_fG° (HC = CH) = – 2.04 × 10⁵] mol^–1; Δ_fG°(C₆H₆) = – 1.24 × 10⁵ J mol^–1;

R = 8.314 J K^–1 mol^–1] 

Answer: 855

ΔG°_r = ΔG°_f [C₆H₆ (l)] – 3 × ΔG°_f [HC = CH]

= [– 1.24 × 10⁵ – 3x (– 2.04 × 10⁵)]

= 4.88 × 10⁵ J/mol

ΔG°_r= – RT ln (K_eq)

log (K_eq) = ΔG° / 2.303RT