JEE Main 2021 Question Paper Chemistry Feb 24 Shift-1

Question 1. What is the reason for the formation of a meta product in the following reaction?

a. Aniline is ortho/para directing
b. Aniline is meta directing
c. In acidic medium, aniline is converted into anilinium ion, which is ortho/para directing
d. In acidic medium, aniline is converted into anilinium ion which is meta directing

Answer: (d)

In an acidic medium, aniline is converted into anilinium ion, which is meta directing.

Question 2. The missing reagent P is:

Answer: (a)

(Pink Colour)

Question 3. Which force is responsible for the stacking of the a-helix structure of protein? Hydrogen (¹H, ²H, ³H) is ________

a. H-bond
b. Ionic bond
c. Covalent bond
d. Van der Waals forces

Answer: (a)

Hydrogen bond is responsible for the stacking of a-helix structure of protein. a-helix is helical in shape because of hydrogen bonding between >C=O and -NH group.

Question 4. The gas evolved due to anaerobic degradation of vegetation causes:

a. Global warming and cancer
b. Acid rain
c. Ozone hole
d. Metal corrosion

Answer: (a)

Anaerobic degradation means the breakdown in the absence of oxygen. During anaerobic degradation of vegetation, CH₄ is produced, which is responsible for global warming as it absorbs sun’s heat. It is also known to cause cancer.

Question 5. Match the following:

(i) Caprolactam

(ii) Acrylonitrile

(iii) 2-chlorobuta-1,3-diene

(iv) 2-Methylbuta-1,3-diene

p>(a) Neoprene

(b) Buna N

(c) Nylon – 6

(d) Natural rubber

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

Answer: (c)

(i) Caprolactum is a monomer of nylon-6.

(ii) Acrylonitrile is one of the monomers of BUNA-N. BUNA-N is a polymer of Buta-1,3-diene and acrylonitrile.

(iii) 2-Chlorobuta-1,3-diene also known as chloroprene, is a monomer of neoprene.

(iv) 2-Methyl-buta-1,3-diene upon polymerisation gives natural rubber.

Hence, the correct matches are (i) ? (c), (ii) ? (b), (iii) ? (a), (iv) ? (d).

Question 6. What is the major product of the following reaction

Answer: (a)

1. The lone pair of oxygen attacks H⁺ from HCl

2. Removal of H₂O takes place, forming a 2^o carbocation intermediate.

3. 1,2 Hydride shift takes place, giving a more stable 3^o carbocation.

4. Cl^– from HCl acts as a nucleophile and attacks the 3^o carbocation to form the product.

Question 7. What is the major product of the following reaction?

Answer: (c)

Since HI is an acid, the pi bond attacks H⁺ and a carbocation intermediate is formed. Since this is a secondary carbocation, 1,2 methyl shift takes place to give a more stable tertiary carbocation. Since I^– is a weak base, it will act as a nucleophile and forms the product (c).

Question 8. Identify the major product:

Answer: (b)

Question 10. Which reagent (A) is used for the following given conversion?

a. Cu / ? / high pressure
b. Molybdenum Oxide
c. Manganese Acetate
d. Potassium Permanganate

Answer: (b)

Molybdenum Oxide acts as a catalyst and catalyzes the conversion of propane to propanal.

This is an oxidative dehydrogenation.

Question 11. Find A and B.

Answer: (c)

Question 12. Which of the following pairs are isostructural

a) TiCl₄, SiCl₄

b) SO^2-₃, CrO^2-₃

c) NH₃, NO^–₃

d) ClF₃, BCl₃

a. A, B
b. A, C
c. B, C
d. A, D

Answer: (a)

Isostructural compounds are those that have the same structure as well as same hybridisation.

In case (B), the structure of SO^2-₄ is tetrahedral as it has 4s bonds and zero lone pairs on the central atom. Similarly, CrO^2-₄ also has 4s bonds and zero lone pairs on the central atom. According to VSEPR, the structures of SO^2-₄ and CrO^2-₄ are:

Hence, we can say, SO^2-₄ and CrO^2-₄ are isostructural.

In case (A), TiCl₄ has 4s bonds and zero lone pairs on the central atom. Hence, it is tetrahedral. Similarly, SiCl₄ is also tetrahedral.

Hence, TiCl₄ and SiCl₄ are also isostructural.

In case (C), NH₃ has 3s bonds and 1 lone pair of electrons around nitrogen. Thus, according to VSEPR, NH₃ has a pyramidal structure. In NO^–₃, the nitrogen atom has 3 bonds and no lone pair of electrons. Thus, it is trigonal planar. According to VSEPR, the structures of NH₃ and NO^–₃ are:

Hence, NH₃ and NO^–₃ are not isostructural.

In case (D), ClF₃ has 3 s and 2 lone pairs on the central atom. Its structure is distorted T shaped (but geometry is trigonal bipyramidal). BCl₃ has 3 s bonds and no lone pair on the B atom. Thus, it is trigonal planar.

Hence, ClF₃ and BCl₃ are not isostructural.

Question 13. Which of the following ores are concentrated by cyanide of group 1^st element?

a. Sphalerite
b. Malachite
c. Calamine
d. Siderite

Answer: (a)

Cyanide acts as a depressant and is therefore used to concentrate the galena-sphalerite ore through the froth floatation process. Hence, option (a) is the correct answer.

Question 14. S-1: Colourless cupric metaborate is converted into cuprous metaborate in a luminous flame.

a. S-1 is true and S-2 is false
b. S-1 is false and S-2 is true
c. Both are false
d. Both are true

Answer: (c)

Cupric metaborate (Cu(BO₂)₂ is blue-green in color and is not colorless, but it does convert into cuprous metaborate in a luminous flame(reducing flame)

2Cu(BO2)₂ + C ? 2CuBO₂ + B₂O₃ + CO?

Thus, S-1 is incorrect.

Cuprous metaborate is formed by reacting copper sulphate with boric anhydride in a luminous flame.

CuSO₄ + B₂O₃ ? Cu(BO₂)₂ + SO₃?

Hence, S-2 is also incorrect.

Question 15. In the given reactions,

1) I₂ + H₂O₂ + 2OH^– ? 2I^– + 2H₂O + O₂

2) H₂O₂ + HOCl ? Cl^– + H₃O⁺ + O₂

a. H₂O₂ acts as an oxidising agent in both the reactions
b. H₂O₂ acts as a reducing agent in both the reactions
c. H₂O₂ acts as an oxidising agent in reaction (1) and as a reducing agent in reaction (2)
d. H₂O₂ acts as a reducing agent in reaction (1) and as an oxidizing agent in reaction (2)

Answer: (b)

In the first reaction, I is reduced from 0 to -1.

In the second reaction, Cl is reduced from +1 to -1

Hence, in both the reactions, H₂O₂ acts as a reducing agent.

Question 16. E^o_M²⁺_{/ M} has a positive value for which of the following elements of 3d transition series?

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

Answer: (a)

According to the electrochemical series, elements present above H have positive E^o value. For M²⁺_(aq)/ M_(s), out of the given options, Cu²⁺_(aq)/Cu_(s) will be the answer.

Cu²⁺_(aq) + 2e^– ? Cu_(s) ; E^o = +0.34 V

Zn²⁺_(aq) + 2e^– ? Zn_(s) ; E^o = -0.763 V

Co²⁺_(aq) + 2e^– ? Co_(s) ; E^o = -0.28 V

Cr²⁺_(aq) + 2e^– ? Cr_(s) ; E^o = -0.9 V

Question 17. Identify X, Y, Z in the given reaction sequence.

a. X = Na[Al(OH)₄] ; Y = CO₂ ; Z = Al₂O₃.xH₂O
b. X = Na[Al(OH)₄] ; Y = SO₂ ; Z = Al₂O₃.xH₂O
c. X = Al(OH)₃ ; Y = CO₂ ; Z = Al₂O₃
d. X = Al(OH)₃ ; Y = SO₂ ; Z = Al₂O₃

Answer: (a)

Thus, X = Na[Al(OH)₄] ; Y = CO₂ ; Z = Al₂O₃.xH₂O

Question 18. The slope of the straight line given in the following diagram for adsorption is:

a. 1/n (0 to 1)
b. 1/n (0.1 to 0.5)
c. log n
d. log (1/n)

Answer: (a)

According to Freundlich adsorption isotherm, log (x/m) = log K + 1/n log P.

So, the slope of the graph of log (x/m) vs log P is 1/n. Where, 1/n is from 0 to 1.

Question 20. Arrange Mg, Al, Si, P and S in the correct order of their ionisation potentials.

Answer: P > S > Si > Mg > Al

In the third period of the periodic table,

Ideally, from left to right, ionization enthalpy increases. But we see exceptions in (Mg, Al) & (P, S) where the element to the left, Mg and P, have higher ionization enthalpies compared to their respective immediate neighbors due to stable electronic configuration.

So, the order will be: P > S > Si > Mg > Al

Section B

Question 1. Cl₂(g) ? 2Cl(g)

For the given reaction at equilibrium, moles of Cl₂(g) is equal to the moles of Cl(g) and the equilibrium pressure is 1atm. If K_p of this reaction is x ×10^–1, find x.

Answer: 5

According to the question: Cl₂ ? 2Cl [P_total = 1 atm].

Moles of Cl₂ = Moles of Cl (at equilibrium).

Given: K_p = x ×10^-1

n_Cl2 = n_Cl.

Therefore, P_Cl2 = P_Cl

Hence, P_Cl2 = P_Cl = 0.5 atm

k_p= (PCl)2/ PCl2 = 0.5 = 5 x 10-1

So, x = 5

Question 2. S₈ + bOH^– ? cS^2- + sS₂O^2-₃ + H₂O. Find the value of c.

Answer: 4

Let us look at the half reactions i.e. oxidation and reduction separately.

Oxidation:

S₈ ? S₂O^2-₃

S₈ + 24OH^– ? 4S₂O^2-₃ + 12H₂O +16e^– ….(1)

Reduction:

S₈ ? S^2-

S₈ + 16e^– ? 8S^2- ….(2)

Adding both the reactions (1) and (2),

2S₈ + 24OH^– ? 4S₂O^2-₃ + 8S^2- + 12H₂O

Dividing the whole equation by 2,

S₈ + 12OH^– ? 2S₂O^2-₃ + 4S^2- + 6H₂O

So, c = 4

Question 3. Calculate the time taken in seconds for 40% completion of a first order reaction, if its rate constant is 3.3× 10^-4 sec^-1.

Answer: 1518

= 30 x 10³ x 0.506 = 1518 sec

Question 4. For a chemical reaction, K_eq is 100 at 300K, the value of ?G^o is –xR Joule at 1 atm pressure. Find the value of x. (Use ln 10 = 2.3)

Answer: 1382

Given: K_eq= 100 at 300K and ?G^o = –xR Joule 

?G^o = -RTln(K_eq) = -2.303RTlog(K_eq) = -1381.8R

Therefore, x = 1381.8 or 1382

Question 5. Cu²⁺ + NH₃ ? [Cu(NH₃)]²⁺ K₁ = 10⁴

[Cu(NH₃)]²⁺ + NH₃ ? [Cu(NH₃)]²⁺ K₂ = 1.58 × 10³

[Cu(NH₃)₂]²⁺ + NH₃ ? [Cu(NH₃)₃]²⁺ K₃ = 5 × 10²

[Cu(NH₃)₃]²⁺ + NH₃ ? [Cu(NH₃)₄]²⁺ K₄ = 10²

If the dissociation constant of [Cu(NH₃)₄]²⁺ is X × 10¹². Determine X.

Answer: 1.26

Overall reaction constant (ß):

ß = K₁ × K₂ × K₃ × K₄

= 10⁴ (1.58 ×10³) × 5 × 10² × 10² = 7.9 × 10¹¹

Cu²⁺ + 4NH₃ ? [Cu(NH₃)₄]²⁺ ; ß = 7.9 × 10¹¹

So, the dissociation constant (K_Disso.) will be:

K_Disso = 1 / ß = 1.26 × 10^-12

Hence, the value of X = 1.26

Question 6. 9.45g of CH₂ClCOOH is dissolved in 500 mL of H₂O solution and the depression in freezing point of the solution is 0.5°C. Find the percentage dissociation.

(K_f)_H2O = 1.86K kg mole^-1.

Answer: 34.4%

CH₂Cl COOH CH₂ClCOO^– + H⁺

Initial 1 0 0

Dissociated a a a

Left (1-a) a a

i = final moles / initial moles = 1 – a + a + a / 1 = 1 + a

?T_f = i x K_f × m

0.5 = (1 + a) × 1.86 × m

Molality (m) = n_B / W_A (kg) = W_BMB x WA (kg) = 9.45 × 1000 / 94.5 × 500 = 0.2

Here, A is for solute and B is for solvent

W_A=W_H2O = 500g (Density of H₂O = 1g/mL)

0.5 = (1+ a) ×1.86 × 0.2

a = 0.344

Percentage of dissociation = 34.4%

Question 7. What is the coordination number in Body Centered Cubic (BCC) arrangement of identical particles?

Answer: 8

The easiest way is to look at the atom at the body center. It lies at the center of the body diagonal and touches the eight corner atoms. So, coordination number = 8.

Question 8. Among the following compounds, how many are amphoteric in nature?

Be(OH)₂, BeO, Ba(OH)₂, Sr(OH)₂

Answer: 2

The oxide and hydroxide of Be are amphoteric. Hence BeO and Be(OH)₂ is amphoteric. Ba(OH)₂ and Sr(OH)₂ are basic.

Question 9. 4.5 g of a solute having molar mass of 90 g/mol is dissolved in water to make a 250 mL solution. Calculate the molarity of the solution.

Answer: 0.2

W_B (given weight of solute) = 4.5 g

M_B (Molar mass of solute) = 90 g/mol

V_S (Volume of solution) = 250 mL

= 250 / 1000 L = 1/4 L

Molarity (M) = nB / V_S(L) = W_B / M_B.V_S(L = 4.5 / (90 ×1/4) = 0.2 molar

Question 10. The mass of Li³⁺ is 8.33 times the mass of a proton. If Li³⁺ and proton are accelerated through the same potential difference, then the ratio of de Broglie’s wavelength of Li³⁺ to proton is x ×10^–1. Find x

m (_Li³⁺) = 8.33 × m _p+ (given)

Debroglie’s wavelength (?) = h / p

KE = ½ mv²

Multiplying by m on both sides

m. KE = ½ mv²

2m. KE = (mv)² = p²

P = v2m. KE

Also KE= q × V

So, P = 2mq.V

Comparing with x × 10 ^-1 = 2 × 10 ^-1

x = 2