JEE Main 2021 Chemistry Paper With Solutions Feb 25 Shift 1

SECTION A

Question 1. Ellingham diagram is a graphical representation of:

a. ΔG vs T
b. (ΔG – TΔS) vs T
c. ΔH vs T
d. ΔG vs P

Answer: (a)

Ellingham diagram tells us about the spontaneity of a reaction with temperature.

Question 2. Which of the following equations depicts the oxidizing nature of H₂O₂?

a. Cl₂ + H₂O₂ → 2HCl + O₂
b. KlO₄ + H₂O₂ → KlO₃ + H₂O + O₂
c. 2l^– + H₂O₂ + 2H⁺ → I₂ + 2H₂O
d. I₂ + H₂O₂ + 2OH^– → 2I^– + 2H₂O + O₂

Answer: (c)

2l^– + H₂O₂ + 2H⁺ → I₂ + 2H₂O

Oxygen reduces from –1 to –2

So, its reduction will take place. Hence it will behave as oxidising agent or it shows

oxidising nature.

While in other options it changes from (–1) to 0.

Question 3. In Freundlich adsorption isotherm at moderate pressure, the extent of adsorption (x/m) is directly proportional to P^x. The value of x is:

a. ∞
b. 1
c. zero
d. 1/n

Answer: (d)

x / m = p^x

The formula is x / m = p^x

Hence, x= 1 / n. The value of ‘n’ is any natural number.

Question 4. According to molecular orbital theory, the species among the following that does not exist is:

a. He₂^–
b. He₂⁺
c. O₂^2-
d. Be₂

Answer: (d)

B.O. of Be₂ is zero, so it does not exist.

Question 5. Identify A in the given chemical reaction.

Answer: (d)

Aromatization reaction or hydroforming reaction.

Question 6. Given below are two statements:

Statement-I: CeO₂ can be used for oxidation of aldehydes and ketones.

Statement-II: Aqueous solution of EuSO₄ is a strong reducing agent.

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

Answer: (d)

CeO₂ can be used as an oxidizing agent like seO₂. Similarly, EuSO4 is used as a reducing agent.

Question 7. The major product of the following chemical reaction is:

a. (CH₃CH₂CO)₂O
b. CH₃CH₂CHO
c. CH₃CH₂CH₃
d. CH₃CH₂CH₂OH

Answer: (b)

Question 8. Complete combustion of 1.80 g of an oxygen-containing compound (C_xH_yO_z) gave 2.64 g of CO₂ and 1.08 g of H₂O. The percentage of oxygen in the organic compound is:

a. 63.53
b. 51.63
c. 53.33
d. 50.33

Answer: (c)

n_co2 = 2.64 / 44 = 0.06

n_c = 0.06

weight of carbon = 0.06 × 12 = 0.72 gm

n(H₂O) = 1.08 / 1.8 = 0.06

n_H = 0.06 × 2 = 0.12

weight of H = 0.12 gm

∴ Weight of oxygen in C_xH_yO_z

= 1.8 – (0.72 + 0.12)

= 0.96 gram

% weight of oxygen = 0.96/18 × 100 = 53.3%

Question 9.The correct statement about B₂H₆ is:

a. All B–H–B angles are of 120°.
b. Its fragment, BH₃, behaves as a Lewis base.
c. Terminal B–H bonds have less p-character when compared to bridging bonds.
d. The two B–H–B bonds are not of the same length.

Answer: (c)

The terminal bond angle is greater than that of bridge bond angle

Bond angle ∝ S-character

Question 10. In which of the following pairs, the outermost electronic configuration will be the same?

a. Fe²⁺ and Co⁺
b. Cr⁺ and Mn²⁺
c. Ni²⁺ and Cu⁺
d. V²⁺ and Cr⁺

Answer: (b)

Cr⁺ → Ar]3d⁵

Mn²⁺ → [Ar]3d⁵

Question 11. Which statement is correct?

a. Buna-S is a synthetic and linear thermosetting polymer
b. Neoprene is an addition copolymer used in plastic bucket manufacturing
c. Synthesis of Buna-S needs nascent oxygen
d. Buna-N is a natural polymer

Answer: (c)

Synthesis of Buna-S needs nascent oxygen.

Question 12. Given below are two statements:

Statement-I: An allotrope of oxygen is an important intermediate in the formation of reducing smog.

Statement-II: Gases such as oxides of nitrogen and Sulphur present in the troposphere contribute to the formation of photochemical smog.

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

Answer: (c)

Reducing smog acts as a reducing agent and the reducing character is due to presence of sulphur dioxide and carbon particles.

Question 13. The plots of radial distribution functions for various orbitals of hydrogen atom against ‘r’ are given below:

The correct plot for 3s orbital is

a. (4)
b. (2)
c. (1)
d. (3)

Answer: (a)

3s orbital

Number of radial nodes = n – λ – 1

For 3s orbital n = 3, λ = 0

Number of radial nodes = 3 – 0 – 1 = 2

It is correctly represented in graph of option 4

Question 14. Which of the glycosidic linkage galactose and glucose is present in lactose?

a. C-1 of glucose and C-6 of galactose
b. C-1 of galactose and C-4 of glucose
c. C-1 of glucose and C-4 of galactose
d. C-1 of galactose and C-6 of glucose

Answer: (b)

Question 15. Which one of the following reactions will not form acetaldehyde?

Answer: (a)

Question 16. Which of the following reaction/s will not give p-amino azobenzene?

a. 2 only
b. 1 and 2
c. 3 only
d. 1 only

Answer: (a)

Question 17. The hybridization and magnetic nature of [Mn(CN)₆]^4– and [Fe(CN)₆]^3–, respectively are:

a. d²sp³ and paramagnetic
b. sp³d² and paramagnetic
c. d²sp³ and diamagnetic
d. sp³d² and diamagnetic

Answer: (a)

Question 18. Identify A and B in the chemical reaction.

Answer: (d)

Question 19. Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are:

a. 2 and 3 only
b. 1 only
c. 2 only
d. 3 only

Answer: (a)

Compounds that are more acidic than H₂CO₃, gives CO₂ gas in reaction with NaHCO₃. Compound B i.e. Benzoic acid and compound C i.e. picric acid both are more acidic than H₂CO₃.

Question 20. The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is: [Assume: No cyano complex is formed; K_sp(AgCN) = 2.2 × 10^–16 and K_a (HCN) = 6.2 × 10^–10]

a. 0.625 × 10^-6
b. 1.6 × 10^-6
c. 2.2 × 10^-16
d. 1.9 × 10^-5

Answer: (d)

Let solubility is x

AgCN ⇌ Ag⁺ + CN^–K_sp = 2.2 × 10^–16

x x

H⁺ + CN^– ⇌ HCN

K_sp × 1/k_a = [Ag⁺] [CN^–] × [HCN] / [H⁺][CN^–]

2.2 × 10^-16 × 1 / 6.2 × 10^-10 = [S][S] / 10^-3

S² = 2.2 / 6.2 × 10^-9

S² = 3.55 × 10^–10

S = √3.55 × 10^–10

S = 1.88 × 10^–5 ⇒ 1.9 × 10^–5

SECTION B

Question 1. The reaction of cyanamide, NH₂CN_(s) with oxygen was run in a bomb calorimeter and ΔU was found to be –742.24 kJ mol^–1. The magnitude of ΔH₂₉₈ for the reaction NH₂CN (s) + 3/2 O₂ (g) → N₂ (g) + O₂ (g) + H_2O(l) is______kJ. (Rounded off to the nearest integer). [Assume ideal gases and R = 8.314 J mol^–1 K^–1]

Answer: 741 kJ/mol

NH₂CN(s) + 3/2 O₂(g) → N₂(g) + CO₂(g) + H₂O(λ)

Δn_g = (1 + 1) – 3/2 = ½

ΔH = ΔU + Δn_g RT

= -742.24 + ½ × 8.314 × 298 / 1000

= -742.24 + 1.24

= 741 kJ/mol

Question 2. In basic medium CrO₄^2– oxidizes S₂O₃^2– to form SO²₄ and itself changes into Cr(OH)₄^–. The volume of 0.154 M CrO₄^2– required to react with 40 mL of 0.25 M S₂O₃^2– is ______ mL. (Rounded-off to the nearest integer)

Answer: 173 mL

17H₂O + 8CrO₄ + 3S₂O₃ → 6SO₄ + 8Cr(OH)₄^– + 2OH^–

Applying mole-mole analysis

0.154 × v / 8 = 40 × 0.25 / 3

V = 173 mL

Question 3. For the reaction, aA + bB → cC + dD, the plot of log k vs 1/T is given below:

The temperature at which the rate constant of the reaction is 10^–4s^–1 is ________ K. [Rounded off to the nearest integer)

[Given: The rate constant of the reaction is 10^–5 s^–1 at 500 K]

Answer: 526 K

T = 10000 / 19

T= 526 K

Question 4. 0.4g mixture of NaOH, Na₂CO₃ and some inert impurities was first titrated with N / 10HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrated. 1.5 mL of the same HCl was required for the next end point. The weight percentage of Na₂CO₃ in the mixture is ______. (Rounded-off to the nearest integer)

Answer: 3%

1^st end point reaction

NaOH + HCl → NaCl + H₂O

n_f = 1

NaCO₃ + HCl → NaHCO₃

n_f = 1

Eq of HCl used = n_NaOH × 1 + n_Na2CO3 × 1

17.5 × 1/10 × 10^-3 = n_NaOH + n_Na2CO3

2^nd end point

NaHCO₃ + HCl → H₂CO₃

1.5 × 1/10 × 10^-3 = n_NaHCO3 × 1 = n_NaHCO3

0.15 mmol = n_Na2CO3

0.15 = n_Na2CO3

= 3 × 106 × 10^–2

= 3 × 1.06 = 3.18%

Question 5. The ionization enthalpy of Na⁺ formation from Na_(g) is 495.8 kJ mol^–1, while the electron gain enthalpy of Br is –325.0 kJ mol^–1. Given the lattice enthalpy of NaBr is –728.4 kJ mol^–1. The energy for the formation of NaBr ionic solid is (–)_____ × 10^–1 kJ mol^–1.

Answer: 5576 kJ

Na(s) → Na⁺(g) ΔH = 495.8

½ Br₂(λ) + e^– → Br^–(g) ΔH = 325

Na⁺(g) + Br^–(g) → NaBr(s) ΔH = –728.4

Na(s) + ½ Βr₂(λ) → NaBr(s). ΔH =?

ΔH = 495.8 – 325 – 728.4

–557.6 kJ = –5576 × 10^–1 kJ

Question 6. Consider the following chemical reaction.

The number of sp² hybridized carbon atom(s) present in the product is _______.

Answer: 7

All carbon atoms in benzaldehyde are sp2 hybridized.

Question 7. A car tire is filled with nitrogen gas at 35 psi at 27°C. It will burst if pressure exceeds 40 psi. The temperature in °C at which the car tyre will burst is ______. (Rounded-off to the nearest integer)

Answer: 69.85°C ≃ 70°C

P₁ / T₁ = P₂ / T₂

35 / 300 = 40 / T₂

T₂ = 40 × 300 / 35

= 342.86 K

= 69.85°C ≃ 70°C

Question 8. Among the following, the number of halide(s) which is/are inert to hydrolysis is ______.

a. BF₃
b. SiCl₄
c. PCl₅
d. SF₆

Answer: 1

Due to crowding, SF₆ is not hydrolyzed.

Question 9. 1 molal aqueous solution of an electrolyte A₂B₃ is 60% ionised. The boiling point of the solution at 1 atm is _____ K. (Rounded-off to the nearest integer). [Given K_b for (H₂O) = 0.52 K kg mol^–1]

Answer: 375 K

A₂B₃ → 2A⁺³ + 3B^–2

No. of ions = 2 + 3 = 5

wi = 1 + (n – 1) ∝

= 1 + (5 – 1) × 0.6

= 1 + 4 × 0.6 = 1 + 2.4 = 3.4

ΔT_b = K_b × m × i

= 0.52 × 1 × 3.4 = 1.768°C

ΔT_b = (T_b)_solution – [(T_bH2O)]_solution

1.768 = (T_b)_solution – 100

(T_b)_solution = 101.768 °C

= 375 K

Question 10. Using the provided information in the following paper chromatogram:

The calculated R_f value of A ______ × 10^–1.

Answer: 4

On chromatogram distance travelled by compound is → 2 cm

Distance travelled by solvent = 5 cm

So R_f = 2 / 5 = 4 × 10^–1 = 0.4