🚀 HSC Chemistry 1st Paper | Chapter 2 : Qualitative Chemistry – Previous Board Exam Comprehension Questions & Answers

October 18, 2026 9:03 pm No Comments

Qualitative Chemistry Comprehension Question Answer

Welcome to qChem+! Chapter 2 of HSC Chemistry 1st Paper — Qualitative Chemistry — carries a significant weight in both HSC Board Examinations and University Admission Tests (Engineering, Medical, and Public Universities).

In Creative Questions (CQ), Comprehension Questions (Question ‘b’ / 2-marks) require a clear understanding of core scientific laws, electron configuration rules, quantum mechanics, analytical reactions, and mathematical reasoning. Writing logical, precise, and well-structured answers with correct chemical formulas is essential for securing full marks.

To help NCTB HSC English Version students achieve top grades, we have compiled and solved all 53 authentic comprehension questions from past Board Exams (2015–2026) with detailed, step-by-step explanations.

To get these board exam questions together along with their correct answers, see the post below:

🚀 HSC Chemistry 1st Paper | Chapter 2 : Qualitative Chemistry – Past Board Exam Comprehension-based Questions Bank

📌 1. Quantum Numbers, Sub-shells & Electron Capacity

1. Why is a 2d-orbital not possible?

[Md(I) B 26; MB, CB 22; MB, RB 21; JB 19; CB 17; BB 16; CB 15]

Answer: For a 2d-orbital, the principal quantum number n = 2. We know that the azimuthal quantum number l can have integer values ranging from 0 to (n − 1). For n = 2, l = 0, 1 (which correspond to s and p subshells). For a d-subshell, l must equal 2. Since l = 2 is not permissible when n = 2, a 2d-orbital cannot exist.

2. Why is a 3f-orbital not possible?

[CtgB 22; DB, BB 21; CtgB 19]

Answer: For a 3f-orbital, the principal quantum number n = 3. The permissible values of the azimuthal quantum number l are 0, 1, 2 (representing s, p, and d subshells respectively). For an f-subshell, l must equal 3. Since l = 3 is impossible when n = 3, the 3f-orbital is not possible.

3. Fe²⁺ and Co³⁺ are isoelectronic with each other — Explain.

[DinB 19]

Answer: Chemical species (atoms or ions) having the same total number of electrons are called isoelectronic. Neutral Fe (atomic number Z = 26) loses 2 electrons to form Fe²⁺: Total electrons = 26 − 2 = 24. Neutral Co (atomic number Z = 27) loses 3 electrons to form Co³⁺: Total electrons = 27 − 3 = 24. Since both Fe²⁺ and Co³⁺ contain 24 electrons, they are isoelectronic with each other.

4. The f-subshell contains 7 orbitals — Explain.

[MB, CB 25]

Answer: For the f-subshell, the azimuthal quantum number l = 3. The magnetic quantum number m can take (2l + 1) values ranging from −l through 0 to +l. For l = 3, m = −3, −2, −1, 0, +1, +2, +3 (a total of 7 distinct values). Since each unique value of m represents one spatial orientation (orbital), the f-subshell contains 7 degenerate orbitals.

📌 2. Aufbau Principle, Pauli’s Exclusion Principle & Hund’s Rule

5. Explain the Aufbau Principle.

[JB 21; CB 19]

Answer: According to the Aufbau Principle, in the ground state of an atom, electrons sequentially fill atomic orbitals in order of increasing orbital energy. An electron enters the lowest energy orbital first before occupying higher energy orbitals. The energy of an orbital is determined by its (n + l) value. An orbital with a lower (n + l) value possesses lower energy and gets filled first.

6. Between 3d and 4p orbitals, in which one does an electron enter first?

[DB 16]

Answer: For the 3d-orbital: (n + l) = 3 + 2 = 5. For the 4p-orbital: (n + l) = 4 + 1 = 5. When two orbitals have equal (n + l) values, the orbital with the lower principal quantum number n has lower energy. Since n = 3 for 3d and n = 4 for 4p, 3d has lower energy than 4p. Therefore, an electron enters the 3d orbital first.

7. Between 4d and 4f, in which orbital does an electron enter first and why?

[CB 21]

Answer: For the 4d-orbital: (n + l) = 4 + 2 = 6. For the 4f-orbital: (n + l) = 4 + 3 = 7. According to the Aufbau Principle, the orbital with the smaller (n + l) value possesses lower energy. Since 6 is less than 7, the 4d orbital has lower energy than 4f, so electrons enter 4d first.

8. Why does the 19th electron of K enter the 4s orbital instead of the 3d orbital?

[DB, RB, DinB, SB, JB 18; RB 15]

Answer: For Potassium (K, Z = 19), the energy levels of 3d and 4s are compared using the (n + l) rule: For 3d, (n + l) = 3 + 2 = 5; for 4s, (n + l) = 4 + 0 = 4. Since 4 is less than 5, the 4s orbital has lower energy than the 3d orbital. By the Aufbau Principle, the 19th electron enters the lower energy 4s orbital instead of 3d.

9. Why does the 37th electron of Rb enter the 5s orbital instead of the 4d orbital?

[DinB 21]

Answer: For Rubidium (Rb, Z = 37): For 4d, (n + l) = 4 + 2 = 6; for 5s, (n + l) = 5 + 0 = 5. Since the 5s orbital has a smaller (n + l) value (5 is less than 6), it possesses lower energy than 4d. Following the Aufbau principle, the 37th electron occupies the 5s orbital before 4d.

10. Explain Pauli’s Exclusion Principle.

[DinB 22; DB, RB, DinB, SB, JB 18]

Answer: Pauli’s Exclusion Principle states that “No two electrons in the same atom can have the same set of all four quantum numbers (n, l, m, s).” If two electrons occupy the same orbital and thus share identical values for n, l, and m, they must have opposite spin quantum numbers (s = +½ and s = −½).

11. Why is a half-filled ‘p’ orbital more stable?

[CtgB 21]

Answer: A p-subshell contains three degenerate orbitals (px, py, pz). According to Hund’s rule, in a half-filled state (p³), each orbital is singly occupied by an electron with parallel spin. This symmetric electron distribution provides maximum exchange energy and spatial symmetry, resulting in lower potential energy and enhanced thermodynamic stability.

12. “A p-subshell can hold a maximum of 6 electrons” — Explain according to Pauli’s Exclusion Principle.

[SB 19]

Answer: The p-subshell has l = 1, yielding three magnetic quantum values (m = −1, 0, +1), which correspond to 3 orbitals (px, py, pz). By Pauli’s Exclusion Principle, each orbital can accommodate a maximum of 2 electrons with opposite spins. Therefore, the maximum capacity of a p-subshell is 3 orbitals × 2 e⁻/orbital = 6 electrons.

13. Explain how the electronic configuration of He obeys Pauli’s Exclusion Principle.

Answer: Helium (He, Z = 2) has the configuration 1s². The quantum numbers for its two electrons are:

1st electron: n = 1, l = 0, m = 0, s = +½
2nd electron: n = 1, l = 0, m = 0, s = −½

Although both electrons share the same values for n, l, and m, their spin quantum numbers (s) are opposite. Thus, it strictly obeys Pauli’s Exclusion Principle.

14. Why does the 19th electron of a calcium atom enter the 4s orbital instead of the 3d orbital? Explain.

[Md(I) B 26; CtgB 21]

Answer: For Calcium (Ca, Z = 20), the (n + l) value for 4s is 4 + 0 = 4, whereas for 3d it is 3 + 2 = 5. Because the 4s orbital has lower energy than 3d, the 19th and 20th electrons fill the 4s orbital first before any electron can enter 3d, according to the Aufbau Principle.

15. Explain Hund’s Rule along with electron configuration.

[RB 22]

Answer: Hund’s Rule states that “In degenerate orbitals (orbitals of equal energy), electron pairing will not occur until each orbital is singly occupied by an electron with parallel spin.” For example, the electronic configuration of Nitrogen (N, Z = 7) is 1s² 2s² 2px¹ 2py¹ 2pz¹. The three 2p electrons singly occupy the three degenerate 2p orbitals with identical spin directions.

16. Why is Hund’s rule of maximum multiplicity applicable to the electron configuration of N?

[All(I) B. 26]

Answer: The electronic configuration of Nitrogen (N, Z = 7) is 1s² 2s² 2p³. Its 2p subshell consists of three degenerate orbitals (2px, 2py, 2pz) of equal energy containing 3 unpaired electrons. Because Hund’s rule governs how electrons are distributed among degenerate orbitals—requiring each orbital of equal energy to be singly occupied with parallel spins (1s² 2s² 2px¹ 2py¹ 2pz¹) to achieve maximum spin multiplicity before pairing begins—Hund’s rule is directly applicable to Nitrogen.

17. Explain the electronic configuration of phosphorus according to Hund’s Rule.

[DB 19]

Answer: Phosphorus (P, Z = 15) has a ground-state configuration of 1s² 2s² 2p⁶ 3s² 3p³. Applying Hund’s rule, the 3 electrons in the 3p subshell distribute singly into the three degenerate orbitals with parallel spins: P(15): 1s² 2s² 2p⁶ 3s² 3px¹ 3py¹ 3pz¹.

18. Between Fe²⁺ and Fe³⁺ ions, which one is more stable? Explain.

[Md(I) B 26; DB 21]

Answer: The ground state configurations are:

Fe²⁺ (24 e⁻): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
Fe³⁺ (23 e⁻): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

The Fe³⁺ ion possesses a symmetrically half-filled 3d⁵ subshell, which offers extra stability due to high exchange energy. Therefore, Fe³⁺ is more stable than Fe²⁺.

19. Why does the normal electronic configuration of Cu show an exception?

[CB 19; RB, CB, CtgB, BB 18]

Answer: According to the Aufbau principle, Copper (Cu, Z = 29) should have the configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s². However, fully filled (d¹⁰) or half-filled (d⁵) subshells exhibit extra thermodynamic stability and electronic symmetry. Thus, one electron shifts from 4s to 3d to yield the stable configuration: Cu(29): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹.

20. Why is the electronic configuration of Cr (24) an exception to the general rule?

[CB 25; JB 23; MB 22; MB, CB 21; CtgB, SB 15]

Answer: The expected configuration of Chromium (Cr, Z = 24) is [Ar] 3d⁴ 4s². Since a half-filled 3d⁵ subshell is exceptionally stable due to maximum exchange energy and spatial symmetry, one electron jumps from 4s to 3d, giving the true configuration: Cr(24): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹.

📌 3. Electromagnetic Spectrum

21. Calculate the energy of a radiation having a wavelength of 675 nm.

[BB 17]

Answer: We know the energy equation: E = hc / λ

Wavelength, λ = 675 nm = 675 × 10⁻⁹ m
Planck’s constant, h = 6.626 × 10⁻³⁴ J·s
Speed of light, c = 3 × 10⁸ m/s
Calculation: E = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (675 × 10⁻⁹) = 2.944 × 10⁻¹⁹ Joules

📌 4. Bohr Atom Model & Hydrogen Spectrum

22. Bohr’s theory is applicable for He⁺ — Explain.

[CtgB 17]

Answer: The Bohr atomic model is applicable only to single-electron species (like H, He⁺, Li²⁺). Helium (He) has 2 electrons; removing one electron forms He⁺ (2 − 1 = 1 electron). Since He⁺ possesses a single electron, Bohr’s theory successfully explains its spectral transitions. (Note: Bohr’s model does not apply to He²⁺ as it has 0 electrons).

23. Why are many spectral lines observed in the atomic spectrum of hydrogen despite having only one electron?

[RB 16]

Answer: Although a single hydrogen atom contains only 1 electron, a sample of hydrogen gas contains a vast number of hydrogen molecules/atoms. Upon absorbing energy, electrons in different atoms get excited to various higher energy levels (n = 2, 3, 4…). When they de-excite back to lower energy levels (n = 1, 2, 3…), they emit photons of different frequencies, generating numerous spectral lines across Lyman, Balmer, Paschen, etc., series.

24. How is the hydrogen spectrum produced? Explain.

[DB 25; BB 23]

Answer: Ground-state hydrogen electrons absorb specific quanta of energy and jump to higher energy orbits (n ≥ 2). In the excited state, the electron is unstable and returns to lower energy levels within 10⁻⁸ seconds by emitting radiation. The emitted photons correspond to specific wavelengths, appearing as distinct lines in the atomic emission spectrum.

📌 5. Applications of UV & IR Rays

25. Why are UV rays used in detecting fake passports or currency notes?

[All B. 26; CB 19; BB 17; DB 16 / DinB, CtgB, JB 25; DinB, SB 23; JB 22]

Answer: Genuine currency notes and passports are printed with invisible security watermarks containing fluorescent chemical dyes (fluorophores). When exposed to Ultraviolet (UV) light, these fluorophores absorb UV radiation and emit characteristic visible glowing light (luminescence). Fake notes lack these special fluorophores and fail to glow.

26. How is fluorescence / phosphorescence created? Explain.

[DB 15 / JB 17]

Answer:

Fluorescence: Occurs when molecules absorb high-energy radiation and instantly (within 10⁻⁸ s) re-emit light of longer wavelength. The emission stops immediately when the excitation source is removed.
Phosphorescence: Involves intersystem crossing to a metastable triplet state. The stored energy is re-emitted slowly over time, continuing long after the light source is turned off.

27. Write down the uses of IR rays in medical science.

[DinB 23; SB 22; SB 19]

Answer: Infrared (IR) radiation is widely used in medicine:

FIR (Far Infrared) Therapy: Relieves muscle pain, improves blood circulation, and reduces joint inflammation.
Thermography: Detects abnormal body heat patterns for early diagnosis of tumors and localized infections.

28. How are IR rays related to global warming?

Answer: Solar radiation warms the Earth’s surface, which re-radiates energy as long-wavelength Infrared (IR) thermal radiation. Greenhouse gases (CO₂, CH₄, H₂O vapor) in the atmosphere absorb this outgoing IR radiation and trap heat, raising atmospheric temperatures and driving global warming.

📌 6. Qualitative Chemical Analysis (Wet & Flame Tests)

29. How will you detect the presence of Na⁺ ion in a solution?

[RB, SB 21]

Answer: Adding Potassium Pyroantimonate (K₂H₂Sb₂O₇) solution to the test sample forms a white crystalline precipitate of Sodium Pyroantimonate: 2Na⁺ + K₂H₂Sb₂O₇ → Na₂H₂Sb₂O₇↓ (White ppt) + 2K⁺

30. How will you detect Al³⁺ ion in a solution?

[DinB 17; CB 16]

Answer: Adding a few drops of NaOH solution yields a gelatinous white precipitate of Aluminium Hydroxide, which dissolves in excess NaOH: Al³⁺ + 3OH⁻ → Al(OH)₃↓ (Gelatinous white ppt)

31. Why is concentrated HCl acid used in flame tests?

[All B. 26; DB, BB 19; DB, CB, SB 17; CtgB 16; DB, CtgB 15]

Answer: Metal salts like nitrates or carbonates are non-volatile. Concentrated HCl converts metal salts into volatile metal chlorides. Metal chlorides readily vaporize in the non-luminous Bunsen flame, imparting characteristic flame colors.

32. How can NH₄⁺ ion be detected in a solution?

[BB 16]

Answer: Adding Nessler’s reagent (K₂[HgI₄] in alkaline KOH) to the solution yields a reddish-brown precipitate of iodide of Millon’s base (HgO·Hg(NH₂)I), confirming NH₄⁺ ions.

33. How will you detect Fe³⁺ ion in a solution?

[JB 16]

Answer: Adding Potassium Thiocyanate (KSCN) solution to the sample produces a blood-red coloration of Ferric Thiocyanate complex: Fe³⁺ + 3SCN⁻ → [Fe(SCN)]²⁺ / Fe(SCN)₃ (Blood Red Solution)

34. How will you detect SO₄²⁻ ion in a solution? Write down the reaction.

[All(I) B. 26; DB 25; BB 21; SB 17]

Answer: Adding Barium Chloride (BaCl₂) solution yields a heavy white precipitate of Barium Sulphate, which is insoluble in dilute HCl: SO₄²⁻ + BaCl₂ → BaSO₄↓ (Heavy White ppt) + 2Cl⁻

📌 7. Solubility, Solubility Product & Common Ion Effect

35. What is meant by “Solubility of a salt is 30”?

[JB 21]

Answer: It means that at a specified temperature, a maximum of 30 grams of the salt can be dissolved in 100 grams of solvent to form a saturated solution.

36. The solubility of NaCl at 25°C is 36 — Explain.

[MB 21; DB 19]

Answer: It signifies that at 25°C, a maximum of 36 grams of NaCl dissolves in 100 grams of water to form a saturated solution.

37. What is meant by the solubility product of CaCO₃ being 8.7 × 10⁻⁹?

[RB 21; DinB 16]

Answer: It means that in a saturated aqueous solution of CaCO₃ at a specified temperature, the product of the molar concentrations of its ions is [Ca²⁺][CO₃²⁻] = 8.7 × 10⁻⁹ mol²·L⁻².

38. What do you understand by solubility product?

[All(I) B. 26; DB 21; DinB 17]

Answer: At a constant temperature, the solubility product (Kₛₚ) of a sparingly soluble salt is the product of the molar concentrations of its constituent ions in a saturated solution, raised to powers equal to their stoichiometric coefficients.

39. What is meant by the solubility product of Zn(OH)₂ at 25°C being 1 × 10⁻¹⁷?

[SB 19]

Answer: In a saturated solution of Zn(OH)₂ at 25°C, the ionic product under equilibrium is [Zn²⁺][OH⁻]² = 1 × 10⁻¹⁷ mol³·L⁻³.

40. The solubility product of Al(OH)₃ is 3.7 × 10⁻¹⁵ — Explain the meaning of this statement.

[JB 21]

Answer: It means that in a saturated solution of Al(OH)₃ at a given temperature, [Al³⁺][OH⁻]³ = 3.7 × 10⁻¹⁵ mol⁴·L⁻⁴.

41. What is meant by the solubility product of BaSO₄ at room temperature being 1.1 × 10⁻¹⁰?

[RB 17]

Answer: In a saturated BaSO₄ solution at room temperature, the product of barium and sulphate ion concentrations is [Ba²⁺][SO₄²⁻] = 1.1 × 10⁻¹⁰ mol²·L⁻².

42. What is meant by the solubility product of M(OH)₃ at 25°C being 2.5 × 10⁻³⁹?

[SB 16]

Answer: In a saturated solution of M(OH)₃ at 25°C, the ion concentration product [M³⁺][OH⁻]³ = 2.5 × 10⁻³⁹ mol⁴·L⁻⁴.

43. Explain the effect of temperature on solubility.

[JB 22; RB 17]

Answer:

For endothermic dissolution (ΔH is positive, e.g., KNO₃), solubility increases with an increase in temperature.
For exothermic dissolution (ΔH is negative, e.g., Ca(OH)₂), solubility decreases as temperature increases.

44. Why does the solubility product increase with an increase in temperature?

[MB 21]

Answer: Dissolution of sparingly soluble salts is generally endothermic. Raising the temperature increases the molar solubility (S), producing a higher concentration of ions in solution, which directly increases Kₛₚ.

45. Why does the ionic product of water change with temperature?

[DB 22; DB 19]

Answer: The auto-ionization of water (H₂O ⇌ H⁺ + OH⁻) is an endothermic process. According to Le Chatelier’s principle, increasing temperature shifts the equilibrium forward, raising [H⁺] and [OH⁻], thereby increasing K𝒘.

46. Why does solubility decrease due to the common ion effect?

[RB 22; RB 19; SB 17]

Answer: Adding a strong electrolyte with a common ion to a solution of a sparingly soluble salt increases the common ion concentration. By Le Chatelier’s principle, the equilibrium shifts backward to precipitate the salt, thereby reducing solubility.

47. “The solubility of hydrated salts increases up to a certain temperature and then decreases with further temperature rise” — Explain.

[RB 19]

Answer: Glauber’s salt (Na₂SO₄·10H₂O) loses hydration water endothermically up to 32.4°C, so solubility increases. Above 32.4°C, it becomes anhydrous Na₂SO₄, whose dissolution is exothermic, causing solubility to decrease with further heating.

48. What is meant by the solubility of KNO₃ at 25°C being 31.6?

[CB 17]

49. Why does the solubility of metal chlorides decrease in an HCl solution? Explain.

[MdB 26]

📌 8. Crystallization, Distillation & Chromatography

50. Why is HCl used in the crystallization of pure NaCl?

[DinB 25; DB 16]

51. Why is HCl used for purifying impure NaCl in solution? Explain.

[MdB 18]

52. Why is ether not rendered completely anhydrous during distillation?

[JB 17]

53. Write down the working principle of chromatography.

[DinB 16]

54. Why is the value of R𝒇 less than 1?

[BB 19; CB, CtgB, BB 18; RB 17; SB 15]

55. Why does R𝒇 have no unit?

[JB 19]

56. Why is the value of R𝒇 affected by the solvent in paper chromatography?