NSEC 2024 previous year paper · official paper and PYQ-pattern practice

NSEC 2024 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 · Structure of Atom

   An electron in a hydrogen atom undergoes a transition from n = 5 to n = 2. If the Rydberg constant is 1.097 × 10⁷ m⁻¹, the wavelength of the emitted photon (in nm) is closest to:

   • A. 434
   • B. 486
   • C. 656
   • D. 410
   Answer and solution

   Answer: A

   Using 1/λ = R(1/n₁² - 1/n₂²) = 1.097×10⁷(1/4 - 1/25) = 1.097×10⁷(0.21) ≈ 2.30×10⁶ m⁻¹. Therefore λ ≈ 434 nm, which corresponds to the blue-green line in the Balmer series.

2. Q2 · Chemical Kinetics

   For a first-order reaction, the rate constant at 300 K is 2.5 × 10⁻³ s⁻¹ and at 320 K is 8.0 × 10⁻³ s⁻¹. What is the activation energy (Ea) for this reaction? (Given: R = 8.314 J mol⁻¹ K⁻¹, ln 3.2 ≈ 1.16)

   • A. 45.2 kJ mol⁻¹
   • B. 58.1 kJ mol⁻¹
   • C. 72.4 kJ mol⁻¹
   • D. 89.6 kJ mol⁻¹
   Answer and solution

   Answer: A

   Using Arrhenius: ln(k2/k1)=Ea/R(1/T1-1/T2). Ratio=3.2, ln=1.16. (1/300-1/320)=2.083×10⁻⁴. Ea=1.16×8.314/2.083×10⁻⁴≈46.3 kJ/mol, closest to option A.

3. Q3 · Coordination Compounds

   What is the IUPAC name of the coordination compound [Co(NH₃)₄Cl₂]Cl?

   • A. Tetraamminedichloridocobalt(III) chloride
   • B. Dichloridotetraamminecobaltic chloride
   • C. Tetraamminedichloridocobalt(II) chloride
   • D. Dichloridotetraamminecobalt(III) chloride
   Answer and solution

   Answer: A

   According to IUPAC nomenclature, ligands are named alphabetically (ammine before chlorido), followed by the metal name and its oxidation state. Co has oxidation state +3 (4×0 + 2×(-1) + x = +1, x = +3). The correct name is tetraamminedichloridocobalt(III) chloride.

4. Q4 · Hydrocarbons

   When propyne is treated with dil. H₂SO₄ in the presence of HgSO₄, the major product formed is:

   • A. Propan-1-ol
   • B. Propan-2-ol
   • C. Propanone
   • D. Propanal
   Answer and solution

   Answer: C

   Hydration of propyne under acidic conditions with HgSO₄ catalyst follows Markovnikov's rule, forming an enol intermediate that tautomerizes to propanone (acetone).

5. Q5 · Biomolecules

   A disaccharide X on hydrolysis gives two molecules of D-glucose. It does not reduce Fehling's solution. The glycosidic linkage in X is:

   • A. α(1→4)
   • B. β(1→4)
   • C. α(1→2)
   • D. α,α(1→1)
   Answer and solution

   Answer: D

   The disaccharide is non-reducing (no free anomeric carbon), formed from two glucose units. This indicates α,α(1→1) linkage as in trehalose, where both anomeric carbons are involved.

6. Q6 · Environmental Chemistry

   A water sample from an industrial area has a BOD value of 45 mg/L. If the dissolved oxygen content is 3.5 mg/L and the saturation DO at that temperature is 9.0 mg/L, what is the approximate oxygen deficit in this water sample?

   • A. 2.5 mg/L
   • B. 5.5 mg/L
   • C. 12.5 mg/L
   • D. 41.5 mg/L
   Answer and solution

   Answer: B

   Oxygen deficit = Saturation DO - Actual DO = 9.0 - 3.5 = 5.5 mg/L. The BOD value indicates future oxygen demand but doesn't directly affect current deficit.

7. Q7 · Olympiad-style Problem Solving

   An organic compound A (C₄H₈O) reacts with hydroxylamine to form an oxime and with Grignard reagent followed by hydrolysis to give a tertiary alcohol. Compound A does not show geometrical isomerism. Which of the following is compound A?

   • A. Butanal
   • B. Butanone
   • C. 2-Methylpropanal
   • D. Cyclobutanol
   Answer and solution

   Answer: B

   Oxime formation indicates carbonyl group. Tertiary alcohol from Grignard reagent indicates a ketone (not aldehyde). Butanone (CH₃COCH₂CH₃) is the only ketone among options.

8. Q8 · Equilibrium

   A gaseous equilibrium 2X(g) + Y(g) ⇌ 2Z(g) has Kp = 0.25 at 400 K. If initially 3.0 mol of X and 1.5 mol of Y are placed in a 5.0 L container and the system reaches equilibrium, which of the following statements is necessarily true?

   • A. Partial pressure of Z equals partial pressure of Y at equilibrium
   • B. Moles of Z formed equal moles of Y consumed
   • C. Total moles at equilibrium exceed initial moles
   • D. Partial pressure of X is always greater than that of Z
   Answer and solution

   Answer: D

   Stoichiometry gives Z formed = 2x, Y consumed = x, so B is false; total moles = 4.5 - x decreases, so C is false; A depends on specific values. With small Kp = 0.25, the reaction lies far toward reactants, keeping x small so X remains in excess over Z, making partial pressure of X greater than Z the most defensible statement.

9. Q9 · s-Block Elements

   Among the alkaline earth metals, which element exhibits anomalous behavior due to its small size and high charge density, forming covalent compounds and showing diagonal relationship with aluminum?

   • A. Magnesium
   • B. Calcium
   • C. Beryllium
   • D. Strontium
   Answer and solution

   Answer: C

   Beryllium shows anomalous behavior among alkaline earth metals due to its smallest size and highest charge density, forming predominantly covalent compounds and exhibiting diagonal relationship with aluminum.

10. Q10 · Coordination Compounds

    A coordination complex of formula [Co(NH₃)₄Br₂]Cl gives one mole of AgCl precipitate upon treatment with excess AgNO₃ solution. The most likely geometry of the complex cation is:

    • A. Tetrahedral with two Br ligands inside the coordination sphere
    • B. Octahedral with both Br ligands inside the coordination sphere
    • C. Square planar with two Br ligands inside the coordination sphere
    • D. Octahedral with one Br outside the coordination sphere
    Answer and solution

    Answer: B

    Only the chloride outside the coordination sphere precipitates with AgNO₃. The formula indicates 6 ligands (4 NH₃ and 2 Br inside), hence octahedral geometry.