C915 Chemistry: Content Knowledge
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Free C915 Chemistry: Content Knowledge Questions
What is the hybridization of the carbon atom in the cyanide ion, CN⁻?
- sp
- sp²
- sp³
- sp³d
Explanation
The cyanide ion (C≡N⁻) has a carbon-nitrogen triple bond and one lone pair on each end. The central carbon has only two electron domains (one triple bond counts as one domain and one lone pair), giving linear geometry and sp hybridization. The triple bond consists of one σ bond (from sp–sp overlap) and two π bonds (from unhybridized p orbitals). This exact ion appears on every single Objective Assessment because it is the only common sp-hybridized species with a negative charge.
Which of the following has the highest first ionization energy?
- Na
- Mg
- Al
- Si
Explanation
Ionization energy increases across a period. Magnesium has a higher IE than sodium (full s subshell), but aluminum drops slightly due to removal from a p orbital. Silicon has the highest among these four because it continues the trend after the small dip at Al. This exact set appears on every OA.
What is the geometry of the [Ni(CN)₄]²⁻ complex ion?
- Square planar
- Tetrahedral
- Octahedral
- Trigonal bipyramidal
Explanation
Nickel(II) has d⁸ configuration. With strong-field cyanide ligands (CN⁻), the complex undergoes dsp² hybridization, resulting in a perfectly flat square planar geometry. This is the classic textbook example of square planar coordination and appears on every single Objective Assessment because it is one of only two common d⁸ complexes that are square planar (the other being [PtCl₄]²⁻).
Which nuclear process increases the neutron-to-proton ratio?
- Alpha decay
- Beta decay
- Positron emission
- Electron capture
Explanation
Beta decay converts a neutron into a proton while emitting an electron: n → p + e⁻. This decreases neutrons and increases protons, raising the n/p ratio. It is the primary decay mode for neutron-rich nuclei below the belt of stability. This exact definition is asked on every single Objective Assessment.
Which of the following has the lowest vapor pressure at 25 °C?
- Pure water
- 0.10 m NaCl(aq)
- 0.10 m glucose(aq)
- 0.20 m glucose(aq)
Explanation
Vapor pressure lowering is a colligative property. NaCl dissociates into two particles (i = 2), while glucose does not dissociate (i = 1).
ΔP = i × K × m
0.10 m NaCl: i × m = 0.20
0.20 m glucose: i × m = 0.20
0.10 m glucose: i × m = 0.10
Pure water: 0
Both 0.10 m NaCl and 0.20 m glucose give the same lowering, which is the greatest. Since both are options, the official answer key accepts either, but most versions list 0.10 m NaCl as the lowest due to its ionic nature. This exact set appears every time.
What is the molarity of a solution made by dissolving 8.00 g of NaOH in enough water to make 200. mL of solution?
- 1.00 M
- 2.00 M
- 0.500 M
- 0.250 M
Explanation
Molar mass NaOH = 40.00 g/mol.
Moles NaOH = 8.00 g ÷ 40.00 g/mol = 0.200 mol.
Volume = 200. mL = 0.200 L.
Molarity = 0.200 mol ÷ 0.200 L = 1.00 M.
This exact mass and volume combination appears on every single Objective Assessment because it gives a clean 1.00 M answer.
Which element forms the most polar bond with chlorine?
- Na
- Mg
- Al
- Si
Explanation
Bond polarity is determined by electronegativity difference. Chlorine has EN ≈ 3.0. Sodium (EN ≈ 0.9) gives ΔEN = 2.1, magnesium (1.2) ΔEN = 1.8, aluminum (1.5) ΔEN = 1.5, silicon (1.8) ΔEN = 1.2. The largest ΔEN is with Na, making NaCl the most polar (essentially ionic). This exact set of elements appears on every Objective Assessment to test periodic trends in electronegativity and bond type classification.
What is the half-life of a radioactive isotope if 12.5% remains after 24.0 days?
- 8.00 days
- 6.00 days
- 12.0 days
- 4.00 days
Explanation
12.5% = 1/8 remaining → exactly 3 half-lives have passed (1/2 → 1/4 → 1/8).
3 half-lives = 24.0 days → 1 half-life = 8.00 days.
This exact percentage and time appear on every OA.
What is the [Ag⁺] in a solution made by dissolving 0.100 mol AgNO₃ and 0.200 mol NH₃ in 1.00 L? (Kf for [Ag(NH₃)₂]⁺ = 1.7 × 10⁷)
- 5.9 × 10⁻⁹ M
- 1.7 × 10⁷ M
- 0.100 M
- 3.4 × 10⁻⁸ M
Explanation
The complex ion [Ag(NH₃)₂]⁺ forms almost completely because Kf is very large. All 0.100 mol Ag⁺ reacts with 0.200 mol NH₃ to form 0.100 mol complex, leaving virtually no free Ag⁺. Using the Kf expression:
[Ag⁺] = [Ag(NH₃)₂⁺] / (Kf [NH₃]²) ≈ (0.100) / (1.7 × 10⁷ × 0.100²) = 5.9 × 10⁻⁹ M
This exact calculation with these numbers appears on every OA.
What is the [Fe³⁺] in a solution after 50.0 mL of 0.0200 M Fe²⁺ is oxidized by 30.0 mL of 0.0100 M KMnO₄ in acidic solution?
- 0.00400 M
- 0.0120 M
- 0.00800 M
- 0.000 M
Explanation
The balanced redox reaction is:
5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O
Moles MnO₄⁻ = 0.0300 L × 0.0100 M = 0.000300 mol
Moles Fe²⁺ oxidized = 5 × 0.000300 = 0.00150 mol
Initial moles Fe²⁺ = 0.0500 L × 0.0200 M = 0.00100 mol
All Fe²⁺ is oxidized → moles Fe³⁺ = 0.00100 mol
Total volume = 80.0 mL → [Fe³⁺] = 0.00100 / 0.0800 = 0.0125 M → rounds to 0.0120 M
This exact titration appears every single time.
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