This course has 16 chapters with 13,396 multiple-choice
questions. Each question is tied to a topic from the chapter and comes with the correct answer and an
explanation. Open a chapter to see its topics and a free sample of its questions, or read the sample from the whole book below.
10 sample questions with the correct answer and an explanation. Try to answer each one before reading the explanation.
Q1. Which element has the highest affinity for an electron among N, O, F, Ne?
A. Neon
B. Nitrogen
C. Fluorine(correct answer)
D. Oxygen
Answer: C.F (−328 kJ mol⁻¹) is the one nearest to a complete octet with the highest nuclear charge among the elements that can accept an electron favourably (Ne cannot).
Q2. The electron-pair geometry around the central atom of SnCl₂, which has 3 electron pairs including 1 lone pair, is
A. trigonal planar, atoms at the corners of a triangle(correct answer)
B. see-saw, four atoms arranged like a see-saw
C. linear, all atoms in a straight line
D. T-shaped, three atoms arranged like a letter T
Answer: A.3 pairs arrange themselves in a trigonal planar geometry; the lone pairs are then excluded to describe the shape (angular (V-shaped)).
Q3. For BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq), 10.4 g of BaCl₂ is mixed with 5.68 g of Na₂SO₄. Calculate the mass of the excess reactant left unreacted. (Ba = 137, Cl = 35.5, Na = 23, O = 16, S = 32)
A. 10.4 g
B. 8.32 g
C. 4.16 g
D. 2.08 g(correct answer)
Answer: D.n(BaCl₂) = 0.05000 mol, n(Na₂SO₄) = 0.04000 mol; moles per unit of equation: BaCl₂ 0.05000, Na₂SO₄ 0.04000; Na₂SO₄ is limiting. BaCl₂ used = 0.04000 × 1 = 0.04000 mol, so BaCl₂ left = 0.05000 − 0.04000 = 0.01000 mol = 2.08 g.
Q4. The first electron affinity of chlorine is −348.8 kJ mol⁻¹. This is:
A. the same value as for fluorine and for bromine
B. a positive value, as for the noble gases
C. among the most negative values for any element(correct answer)
D. equal to zero, as for the alkaline earths
Answer: C.Chlorine's first electron affinity is greater (more negative) than fluorine's, because fluorine is a very small atom with strong repulsion in its compact shell (standard knowledge).
Q5. Which is the correct equilibrium constant expression (Kc) for the reaction CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)?
A. Kc = ([CH₃COOH] + [C₂H₅OH]) / ([CH₃COOC₂H₅] + [H₂O])
B. Kc = [CH₃COOH][C₂H₅OH] / ([CH₃COOC₂H₅][H₂O])
C. Kc = [CH₃COOC₂H₅][H₂O] / ([CH₃COOH][C₂H₅OH])(correct answer)
D. Kc = ([CH₃COOC₂H₅] + [H₂O]) / ([CH₃COOH] + [C₂H₅OH])
Answer: C.Kc = [products]/[reactants], each concentration raised to its coefficient: Kc = [CH₃COOC₂H₅][H₂O] / ([CH₃COOH][C₂H₅OH]). The liquids take part as one homogeneous liquid mixture, so water is included (as in Sample Problem 8.1).
Q6. Equal volumes of 1 × 10⁻⁶ mol dm⁻³ Ba²⁺ and 1 × 10⁻⁶ mol dm⁻³ CO₃²⁻ solutions are mixed. Ksp of BaCO₃ = 1.1 × 10⁻¹⁰. After mixing the ionic product is 2.5 × 10⁻¹³. Predict:
A. the solution is exactly saturated, with no precipitate
B. no precipitate forms(correct answer)
C. precipitate forms
D. cannot be predicted without the temperature
Answer: B.Concentrations after mixing are halved: [Ba²⁺] = 5 × 10⁻⁷, [CO₃²⁻] = 5 × 10⁻⁷; ionic product = 2.5 × 10⁻¹³ < Ksp = 1.1 × 10⁻¹⁰, so no precipitation occurs.
Q7. What is the major product when hex-2-ene, CH₃CH=CH(CH₂)₂CH₃, reacts with gaseous HBr at room temperature?
A. only 2-bromohexane, since bromine always goes to carbon 2
B. only 1-bromohexane, since bromine goes to the end carbon
C. only 3,4-dibromohexane, since two bromine atoms add
D. a mixture of 2-bromohexane and 3-bromohexane(correct answer)
Answer: D.Electrophilic addition of H−X follows Markovnikov's rule: hydrogen adds to the carbon with more hydrogens and X goes to the more substituted carbon (the more stable carbocation); product: a mixture of 2-bromohexane and 3-bromohexane.
Q8. The progressive change in colour of the halogens is attributed to differences in
A. absorption of light through electron transitions within the molecules(correct answer)
B. the boiling points of the corresponding hydrogen halides
C. the number of neutrons present in each atom of the halogen
D. the number of natural isotopes that each halogen possesses
Answer: A.The book explains that the colour trend arises from changes in the absorption of light as a result of electron transitions within each halogen molecule.
Q9. What is global warming?
A. A change in the Earth's orbit that alters the seasons
B. A long-term rise in the Earth's average surface temperature due to human activities(correct answer)
C. A short-term fall in temperature in winter because of snow
D. A decrease in the Sun's radiation reaching the surface
Answer: B.The chapter defines global warming as the long-term rise in Earth's average surface temperature due to human activities, mainly emission of greenhouse gases.
Q10. Exercise Q.1 I: why is it advisable to read the lower meniscus for colourless liquids and the upper meniscus for coloured liquids?
A. Because lower meniscus does not exist for coloured liquids
B. To avoid parallax error and to read the clearly visible edge(correct answer)
C. Because it is more convenient to read
D. Because colourless liquids have more surface tension
Answer: B.The book gives the answer of parallax; the lower meniscus of colourless liquids is clearly visible, while the dark upper edge of coloured liquids is easier to read.
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