JEE Main 2021 August 27 Shift 1 Chemistry Question Paper

Question 1. In the following equation, find B.

Answer: (a)

Hence, option (a) is the correct answer.

Question 2. A metal having a much low melting point is refined by:

a. Liquation
b. Electrolytic refining
c. Zone Refining
d. None of these

Answer: (a)

Liquation: In this method, a low melting metal like a tin can be made to flow on a sloping surface. In this way, it is separated from higher melting impurities. This process is used for the purification of Sn and Zn, and for removing Pb from Zn-Ag alloy.

Hence, option (a) is the correct answer.

Question 3. From 0.2 gm of a compound, 0.188 gm of AgBr is formed by Carius Method. Find % of Br?

a. 80%
b. 20%
c. 40%
d. 10%

Answer: (c)

Molar mass of AgBr = (108 + 80) g/mol = 188 g/mol

188g of AgBr contains 80 g of Br

0.188 g of AgBr will contain 80 x 0.188 / 188 g Br

In 0.2 g of compound % of Br = 80 x 0.188 / 188 x 0.2 x 100% = 40%

Hence, option (c) is the correct answer.

Question 4. In which form uracil is present in DNA?

Answer: (a)

Fact-based.

Question 5. Which of the following shows the Tyndall effect?

a. True solution
b. Lyphobic solution
c. Lyophilic solution
d. Suspension

Answer: (b)

Hence, option (b) is the correct answer.

Question 6. When the intensity of radiation incident on a photographic plate is increased keeping the frequency constant, then the number and K.E of photoelectrons emitted?

a. Remains same and increases
b. Increases and remains the same
c. Decreases and remains the same
d. Decreases and decreases

Answer: (b)

As the intensity increases the number of photons striking on the surface increases hence the number of photoelectrons emitted increases whereas the energy of incident photons remains the same and hence the kinetic energy of emitted photoelectrons remains the same.

Hence, option (b) is the correct answer.

Question 7. 0.075 molal solution of sucrose solution containing 1 kg water is at the temperature of – 0.4⁰C, If (Kf)H₂O= 1.86 K/m. How much ice will freeze out?

a. 350 g
b. 100 g
c. 180 g
d. 650 g

Answer: (d)

On freezing moles of sucrose does not change.

ΔT_f = i × K_f × m

0.4 = 1 x 1.86 x 0.075 / W_H2O x 1000 = 348.75 g

Amount of ice separated= (1000 – 348.75) g

Amount of ice separated = 651.25 ≈ 650 g

Hence, option (d) is the correct answer.

Question 8. One mole of octahedral compound MCl3E2 is reacted with AgNO3 and 1 mol AgCl is formed. Find denticity of E:

Answer: (2)

One mole of an octahedral compound [MCl₃E₂] is reacted with AgNO₃ to give 1 mol AgCl which means the complex exists as [MCl₂E₂]Cl. The formation of AgCl is due to the reaction of AgNO₃ with Cl⁻ giving AgCl. One mole of AgCl is obtained, which means only one chlorine is free. Hence, the complex has to be [MCl₂E₂]Cl. For an octahedral complex, the coordination number must be 6. Now the denticity of Cl⁻ is 1 and two Cl⁻ are present so two valencies are satisfied by chlorine. To make the coordination number 6 ligand E has to be bidentate.

Therefore, the denticity of E must be 2.

Hence, 2 is the correct answer.

Question 9. Find the strength of KMnO4 in gm/L. If 10mL of KMnO4 requires 0.1 M FeSO4 of the same value for decolourisation.

Answer: (3.16)

MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺

Vf = 5 Vf = 1

Milliequivalents of MnO₄⁻ = Milliequivalents of Fe²⁺

5 × [M × 10] = 1 × [0.1 × 10]

M = 0.02 x mol / L

Strength of KMnO₄ = Molarity × Molar mass

= (0.02 ×158) g/L = 3.16 g/L

Hence, 3.16 is the correct answer.

Question 10. The number of electrons in the outermost f orbital in the ground state of neptunium (Z = 93).

a. 2
b. 3
c. 4
d. 5

Answer: (c)

Electronic configuration of Neptunium: [Rn]5f⁴ 6d¹ 7s²

Thus, the number of electrons in the outermost f orbital is 4.

Hence, option (b) is the correct answer.

Question 11. Which of the following is correct with respect to ‘H’ and ‘D’?

a. D produces β⁺
b. D is more reactive than H
c. H is more reactive than D
d. Equal reactivity

Answer: (c)

Reactivity of Deuterium is less than hydrogen due to the high bond dissociation energy of D₂

Hence, option (c) is the correct answer.

Question 12. Nature of oxides of V₂O₃ and CrO?

a. Amphoteric and Basic
b. Basic and Basic
c. Basic and Amphoteric
d. Acidic and Acidic

Answer: (b)

In the case of transition metal oxides, the oxides with metals in lower oxidation states are basic in nature. Since, in V₂O₃ and CrO, oxidation states of V and Cr are in lower oxidation states i.e., +3 and +2 respectively. Hence both are basic oxides.

In intermediate oxidation states, they are amphoteric in nature. For example, V₂O₄, CrO₂ MnO₂ etc. In higher oxidation states, they are acidic in nature. for example, V₂O₅, CrO₃, Mn₂O₇ etc.

Question 13. Find the oxidation number of ‘S’ in H₂S_(x+2)O₆

a. 0 and +5 only
b. +5 only
c. 0 and +3 only
d. +3 only

Answer: (a)

When x = 2,

The compound is H₂S₄O₆:

Structure:

So, the oxidation states can be 0 and +5.

Hence, option (c) is the correct answer.

Question 14. Match the following:

a. 1-(i), 2-(iv), 3-(ii), 4-(iii)
b. 1-(iv), 2-(ii), 3-(i), 4-(iii)
c. 1-(iv), 2-(ii), 3-(iii), 4-(i)
d. 1-(iv), 2-(iii), 3-(i), 4-(ii)

Answer: (b)

Hence, option (b) is the correct answer.

Question 15. Find the incorrect statement regarding primary aliphatic amines:

a. They can be produced by Gabriel phthalimide process
b. Its solubility is greater than 2⁰ amines
c. Can be distinguished by carbylamine test
d. It is less basic than aromatic amines

Answer: (d)

Gabriel Phthalimide Synthesis is used to get primary amines from primary alkyl halides. Since primary amines have a higher number of hydrogens than secondary amines, they show more H-bonding and hence are more soluble than secondary amines. Primary amines can be distinguished from secondary and tertiary amines by reacting with. Primary amine reacts with CHCl₃ and alc. KOH to form isocyanide while secondary and tertiary amines do not react.

In aromatic amines, lone pair of nitrogen is involved in resonance and hence less available for donation. So, aromatic amines are less basic than primary amines.

Hence, option (d) is the correct answer.

Question 16. A reacts with B to form a Prussian blue compound Fe₄[Fe(CN)₆]₃. A and B are respectively:

a. FeCl₃ and K₃[Fe(CN)₆]
b. FeCl₃ and K₄[Fe(CN)₆]
c. K₃[Fe(CN)₆] and K₄[Fe(CN)₆]
d. FeCl₃ and KCNS

Answer: (b)

FeCl₃ + K₄[Fe(CN)₆] → Fe₄[Fe(CN)₆]₃

A B Blue colour

Hence, option (b) is the correct answer.

Question 17. Find the order of Lewis acid strength of BI₃, BCl₃, BF₃, BBr₃.

a. BF3 < BCl3 < BBr3 < BI3
b. BF3 < BBr3 < BI3 < BCl3
c. BCl3 < BF3 < BBr3 < BI3
d. BBr3 < BF3 < BI3 < BCl3

Answer: (a)

On the basis of back bonding, the smaller the halide atom, the more effective be the back bonding in BX₃.

More effective back bonding shows less tendency to accept a pair of electrons. Hence, correct order of Lewis acid for boron halide is BF₃ < BCl₃ < BBr₃ < BI₃.

Question 18. If we have two particles in a system i.e. A & B where particle ‘A’ has 4 amu mass and 2- unit charge and another particle B has 9 amu mass and 3-unit charge. Find the correct option regarding A and B:

a. Deviation of A is more than B
b. Deviation of B is more than A
c. None of them will deviate
d. Same deviation

Answer: (a)

The deviation of the particle depends on its mass to charge ratio. In general, higher mass to charge ratio particles will deviate less. Here, the mass to charge ratio of A is 2 and the mass to charge ratio of B is 3. Hence, A will deviate more than B.

Question 19. The van der Waal equation is given as:

a. [atm][m⁶][mol]^-2
b. [atm]²[m²][mol]²
c. [atm]³[m³][mol]¹
d. [atm]⁰

Answer: (a)

Since,

Hence, dimension of ‘a’ has to be [atm][m⁶][mol]^-2.

Question 20. Find the number of water molecules in gypsum, dead burnt plaster, and Plaster of Paris.

a. 3, 0, 1
b. 1, 0, 1
c. 2, 0, 0.5
d. 5, 0, 0.5

Answer: (c)

Gypsum = CaSO₄. 2H₂O

Plaster of Paris = CaSO₄.½ H₂O

Dead burnt plaster = CaSO₄

Hence, option (c) is correct.

Question 21. Given below are two statements:

Statement 1: Ethyl phenyl ether can be prepared by Williamson ether synthesis.

Statement 2: Bromobenzene on reacting with sodium ethoxide gives ethyl phenyl ether.

a. Statement I is correct and Statement II is correct
b. Statement I is correct and Statement II is incorrect
c. Statement I is incorrect and Statement II is incorrect
d. Statement I is incorrect and Statement II is correct

Answer: (b)

Due to the double bond character in the C-Br bond in bromobenzene, a reaction with sodium ethoxide is not possible.

Bromobenzene does not undergo nucleophilic aromatic substitution unless they have an electron-withdrawing group at ortho or para position.

Preparation of Ethyl phenyl ether by Williamson ether synthesis: