The 15 one-mark questions from the March 2025 12th Standard Chemistry public exam, in paper order, with the correct option marked and a short explanation.
Q1
During the decomposition of \(H_2O_2\) to give dioxygen, 48 g \(O_2\) is formed per minute at certain point of time. The rate of formation of water at this point is :
- A. \(2.25\ mol\ min^{-1}\)
- B. \(0.75\ mol\ min^{-1}\)
- C. \(3.0\ mol\ min^{-1}\)Correct
- D. \(1.5\ mol\ min^{-1}\)
Explanation. 2H₂O₂ → 2H₂O + O₂. 48 g O₂ = 48/32 = 1.5 mol per minute. Two moles of water form for every mole of O₂, so the rate of formation of water = 2 × 1.5 = 3.0 mol min⁻¹.
Q2
How many moles of \(I_2\) are liberated when 1 mole of potassium dichromate react with potassium iodide ?
- A. 3Correct
- B. 1
- C. 4
- D. 2
Explanation. K₂Cr₂O₇ + 7H₂SO₄ + 6KI → 4K₂SO₄ + Cr₂(SO₄)₃ + 3I₂ + 7H₂O. One Cr₂O₇²⁻ gains 6 electrons, which oxidises 6 I⁻ to 3 I₂, so 3 moles of I₂ are liberated.
Q3
Non-stick cookwares generally have a coating of a polymer, whose monomer is :
- A. chloroethene
- B. ethane
- C. 1,1,2,2-tetrafluoroethaneCorrect
- D. prop-2-enenitrile
Explanation. Non-stick coating is Teflon (PTFE), made by polymerising tetrafluoroethylene, CF₂=CF₂. The paper prints "1,1,2,2-tetrafluoroethane" (as in the textbook); the intended monomer is tetrafluoroethene, so option (c) is the answer.
Q4
The compound that reacts with nitrous acid to give yellow oily liquid is ________.
- A. N-methylanilineCorrect
- B. Nitro benzene
- C. N,N-dimethyl aniline
- D. Aniline
Explanation. Secondary amines react with nitrous acid to form N-nitrosoamines, which are yellow oily liquids. N-methylaniline is a secondary amine: C₆H₅NHCH₃ + HNO₂ → C₆H₅N(CH₃)–N=O + H₂O.
Q5
Boric acid is an acid because its molecule :
- A. combines with proton to form water molecule.
- B. contains replaceable \(H^+\) ion.
- C. accepts \(OH^-\) from water, releasing proton.Correct
- D. gives up a proton.
Explanation. Boric acid is a weak monobasic Lewis acid. It does not donate its own proton; the electron-deficient boron accepts OH⁻ from water: B(OH)₃ + H₂O → [B(OH)₄]⁻ + H⁺.
Q6
In an electrical field, the particles of a Colloidal system move towards cathode. The coagulation of the same sol is studied using (i) \(K_2SO_4\), (ii) \(Na_3PO_4\), (iii) \(K_4[Fe(CN)_6]\) and (iv) NaCl. Their coagulating power should be :
- A. (iii) > (ii) > (i) > (iv)Correct
- B. (i) > (ii) > (iii) > (iv)
- C. (ii) > (i) > (iv) > (iii)
- D. None of these
Explanation. Particles moving to the cathode are positively charged, so the anions cause coagulation. By the Hardy–Schulze rule, higher anion charge means greater coagulating power: [Fe(CN)₆]⁴⁻ > PO₄³⁻ > SO₄²⁻ > Cl⁻, i.e. (iii) > (ii) > (i) > (iv).
Q7
Assertion : Bond dissociation energy of Fluorine is greater than Chlorine gas.
Reason : Chlorine has more electronic repulsion than Fluorine.
- A. Assertion is true but Reason is false.
- B. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
- C. Both Assertion and Reason are false.Correct
- D. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Explanation. The F–F bond is weaker than the Cl–Cl bond (about 158 vs 242 kJ mol⁻¹), so the assertion is false. The reason is also false: the small fluorine atom has greater repulsion between its lone pairs, which is why its bond is weak.
Q8
In calcium fluoride, having the flurite structure, the coordination number of \(Ca^{2+}\) ion and \(F^-\) ion are :
- A. 8 and 4Correct
- B. 4 and 2
- C. 4 and 8
- D. 6 and 6
Explanation. In fluorite (CaF₂), Ca²⁺ ions form an fcc lattice and F⁻ ions occupy all tetrahedral voids. Each Ca²⁺ is surrounded by 8 F⁻ and each F⁻ by 4 Ca²⁺, giving 8 and 4.
Q9
The secondary structure of a protein refers to ________.
- A. sequence of \(\alpha\)-amino acids
- B. fixed configuration of the polypeptide backboneCorrect
- C. \(\alpha\)-helical backbone
- D. hydrophobic interaction
Explanation. The sequence of α-amino acids is the primary structure. The secondary structure is the regular, hydrogen-bonded folding (fixed configuration) of the polypeptide backbone; the α-helix is only one example of it (β-sheet is another), so (b) is the complete answer.
Q10
At 25°C, ionic product constant \(K_w\) of water is \(1.00 \times 10^{-14}\). Its value at 40°C is ________.
- A. \(1.00 \times 10^{-14}\)
- B. \(1.14 \times 10^{-15}\)
- C. \(2.71 \times 10^{-14}\)Correct
- D. \(2.95 \times 10^{-15}\)
Explanation. Ionisation of water is endothermic, so Kw increases as temperature rises. Only option (c), 2.71 × 10⁻¹⁴, is larger than 1.00 × 10⁻¹⁴.
Q11
What is the oxidation number of the central metal ion in the complex, \([Pt(NO_2)(H_2O)(NH_3)_2]Br\) ?
- A. +4
- B. +2Correct
- C. +6
- D. +3
Explanation. Br⁻ outside the bracket makes the complex ion +1. NO₂⁻ is −1, H₂O and NH₃ are neutral: x + (−1) = +1, so x = +2.
Q12
The number of electrons that have a total charge of 9650 coulombs is :
- A. \(6.022 \times 10^{22}\)Correct
- B. \(6.22 \times 10^{23}\)
- C. \(6.022 \times 10^{-34}\)
- D. \(6.022 \times 10^{24}\)
Explanation. 1 F = 96500 C is the charge of 1 mole (6.022 × 10²³) of electrons. 9650 C = 0.1 F, so number of electrons = 0.1 × 6.022 × 10²³ = 6.022 × 10²².
Q13
Which one of the following is the Strongest acid ?
- A. 4-nitrophenolCorrect
- B. 2-nitrophenol
- C. 3-nitrophenol
- D. 4-chlorophenol
Explanation. The –NO₂ group at the para position stabilises the phenoxide ion by both −I and −R effects, so 4-nitrophenol is the most acidic. In 2-nitrophenol intramolecular hydrogen bonding makes the proton harder to lose, 3-nitrophenol has only the −I effect, and Cl is a much weaker electron-withdrawing group.
Q14
\(CH_3Br \xrightarrow{KCN} (A) \xrightarrow{H_3O^+} (B) \xrightarrow{PCl_5} (C)\)
Product (C) is :
- A. chloro acetic acid
- B. \(\alpha\)-chlorocyano ethanoic acid
- C. acetylchlorideCorrect
- D. none of these
Explanation. CH₃Br + KCN → CH₃CN (A); acid hydrolysis gives CH₃COOH (B); PCl₅ converts the –COOH group to –COCl, giving CH₃COCl, acetyl chloride (C).
Q15
Extraction of gold and silver involves leaching with cyanide ion. Silver is later recovered by :
- A. Displacement with ZincCorrect
- B. Distillation
- C. Liquation
- D. Zone refining
Explanation. Leaching gives the soluble complex Na[Ag(CN)₂]. Zinc, being more electropositive, displaces silver: 2Na[Ag(CN)₂] + Zn → Na₂[Zn(CN)₄] + 2Ag.
