NSEC 2021 previous year paper · official paper and PYQ-pattern practice

NSEC 2021 PYQ-pattern questions

Showing 10 of the 10 PYQ-pattern questions in this set — freshly worded in that year's pattern, not the original questions. Answers and solutions open under each question.

1. Q1 · States of Matter

   A gas occupies 5.0 L at 27°C and 2 atm pressure. What volume will it occupy at 127°C and 4 atm pressure?

   • A. 2.5 L
   • B. 3.3 L
   • C. 5.0 L
   • D. 6.7 L
   Answer and solution

   Answer: B

   Using combined gas law: P₁V₁/T₁ = P₂V₂/T₂. Converting to Kelvin: T₁ = 300 K, T₂ = 400 K. V₂ = (2×5.0×400)/(4×300) = 4000/1200 = 3.33 L ≈ 3.3 L.

2. Q2 · Chemical Bonding and Molecular Structure

   According to Molecular Orbital Theory, which of the following species has the highest bond order?

   • A. O₂
   • B. O₂⁺
   • C. O₂⁻
   • D. O₂²⁻
   Answer and solution

   Answer: B

   O₂ has 16 electrons, bond order = (10-6)/2 = 2. O₂⁺ has 15 electrons, bond order = (10-5)/2 = 2.5. O₂⁻ has 17 electrons, bond order = (10-7)/2 = 1.5. O₂²⁻ has bond order = 1. Thus O₂⁺ has the highest bond order.

3. Q3 · p-Block Elements

   Xenon hexafluoride (XeF₆) undergoes partial hydrolysis to form compound X and complete hydrolysis to form compound Y. What are X and Y respectively?

   • A. XeOF₄ and XeO₃
   • B. XeO₂F₂ and XeO₄
   • C. XeF₄ and XeO₂
   • D. XeO₃F₂ and Xe
   Answer and solution

   Answer: A

   Partial hydrolysis of XeF₆ gives XeOF₄ (xenon oxotetrafluoride), while complete hydrolysis produces XeO₃ (xenon trioxide) along with HF. The reaction sequence is: XeF₆ + H₂O → XeOF₄ + 2HF (partial); XeF₆ + 3H₂O → XeO₃ + 6HF (complete).

4. Q4 · Electrochemistry

   During the electrolysis of aqueous copper sulfate solution using copper electrodes, if 0.25 F of electricity is passed through the solution, what mass of copper will be deposited at the cathode? (Atomic mass of Cu = 63.5 g mol⁻¹)

   • A. 7.94 g
   • B. 15.88 g
   • C. 31.75 g
   • D. 63.5 g
   Answer and solution

   Answer: A

   The cathode reaction is Cu²⁺ + 2e⁻ → Cu, requiring 2F for 1 mole of Cu. Using Faraday's law: mass = (0.25 F × 63.5 g mol⁻¹) / 2 = 7.94 g of copper deposited.

5. Q5 · Aldehydes, Ketones and Carboxylic Acids

   A carboxylic acid A (C₅H₁₀O₂) on heating with soda lime gives a hydrocarbon B which on monochlorination gives only one product C. Compound A is:

   • A. Pentanoic acid
   • B. 2,2-Dimethylpropanoic acid
   • C. 3-Methylbutanoic acid
   • D. 2-Methylbutanoic acid
   Answer and solution

   Answer: B

   2,2-Dimethylpropanoic acid on decarboxylation with soda lime gives neopentane (2,2-dimethylpropane) which has all equivalent hydrogen atoms, giving only one monochlorination product.

6. Q6 · Chemistry in Everyday Life

   Aspartame, an artificial sweetener, is a methyl ester of a dipeptide. Upon hydrolysis, it yields aspartic acid, phenylalanine, and methanol. Consumption of aspartame should be avoided by individuals with which genetic disorder?

   • A. Galactosemia
   • B. Phenylketonuria
   • C. Cystic fibrosis
   • D. Sickle cell anemia
   Answer and solution

   Answer: B

   Aspartame releases phenylalanine upon hydrolysis. Individuals with phenylketonuria (PKU) cannot metabolize phenylalanine properly, leading to toxic accumulation.

7. Q7 · Chemical Kinetics

   A first-order reaction has a rate constant of 0.035 min⁻¹ at 25°C. If the activation energy is 52 kJ/mol, what will be the rate constant at 45°C? (R = 8.314 J mol⁻¹ K⁻¹)

   • A. 0.078 min⁻¹
   • B. 0.098 min⁻¹
   • C. 0.115 min⁻¹
   • D. 0.142 min⁻¹
   Answer and solution

   Answer: D

   Applying Arrhenius: ln(k2/k1) = -(Ea/R)(1/T2 - 1/T1) = (52000/8.314)(1/298 - 1/318) ≈ 1.32, giving k2/k1 ≈ 3.74. Thus k2 ≈ 0.035 × 3.74 ≈ 0.131 min⁻¹, closest to option D (0.142).

8. Q8 · Chemical Kinetics

   A reaction follows the mechanism: Step 1 (slow): A + B → C; Step 2 (fast): C + B → D. What is the overall rate law for the formation of product D?

   • A. Rate = k[A][B]²
   • B. Rate = k[A][B]
   • C. Rate = k[C][B]
   • D. Rate = k[A]²[B]
   Answer and solution

   Answer: B

   The rate-determining step (slow step) is Step 1: A + B → C. Therefore, the overall rate law is determined by this step: Rate = k[A][B].

9. Q9 · Hydrocarbons

   An alkyne with molecular formula C₅H₈ on catalytic hydrogenation gives 2-methylbutane. The alkyne is:

   • A. Pent-1-yne
   • B. Pent-2-yne
   • C. 3-Methylbut-1-yne
   • D. 3,3-Dimethylprop-1-yne
   Answer and solution

   Answer: C

   3-Methylbut-1-yne upon catalytic hydrogenation adds H₂ to the triple bond, yielding 2-methylbutane with the same carbon skeleton.

10. Q10 · Coordination Compounds

    A square planar complex [Pt(NH₃)₂Cl₂] shows geometrical isomerism. The total number of isomers (including both geometrical and any other type) is:

    • A. 1
    • B. 2
    • C. 3
    • D. 4
    Answer and solution

    Answer: B

    Square planar [Pt(NH₃)₂Cl₂] exhibits cis-trans (geometrical) isomerism only; the complex has 2 isomers: cis (both Cl adjacent) and trans (Cl opposite).