I85PREP
⌘ K
Galicia PAU Chemistry / Química 2º Bachillerato CIUG study materials
Galicia PAU Chemistry / Química 2º Bachillerato CIUG study resources

Galicia PAU Chemistry / Química 2º Bachillerato CIUG practice questions

One question at a time. Build your Galicia PAU Chemistry / Química 2º Bachillerato CIUG understanding.

First 50 of 190 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Physical chemistry, amounts and reactions

In the reaction NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), what are the Brønsted-Lowry acid and its conjugate base?

  1. H₂O is the acid; OH⁻ is its conjugate base. Correct answer.
  2. NH₃ is the acid; NH₄⁺ is its conjugate base.
  3. H₂O is the acid; NH₄⁺ is its conjugate base.
  4. NH₃ is the base; H₂O is its conjugate acid.

Explanation

According to Brønsted-Lowry theory, an acid is a proton (H⁺) donor. Here, H₂O donates a proton to NH₃, acting as an acid. After donating H⁺, H₂O becomes OH⁻, which is its conjugate base.

Question 2Physical chemistry, amounts and reactions

What is the pH of a 0.0020 M aqueous solution of nitric acid, HNO₃(aq), at 25 °C?

  1. 2.70 Correct answer.
  2. 3.00
  3. 2.00
  4. 11.30

Explanation

HNO₃ is a strong monoprotic acid that dissociates completely in water: HNO₃ → H⁺ + NO₃⁻. Thus, [H⁺] = 0.0020 M = 2.0 × 10⁻³ M. pH = -log₁₀[H⁺] = -log₁₀(2.0 × 10⁻³) = 3 - log₁₀(2) = 3 - 0.301 = 2.70.

Question 3Physical chemistry, amounts and reactions

What is the pH of a 0.0050 M aqueous solution of barium hydroxide, Ba(OH)₂(aq), at 25 °C?

  1. 12.00 Correct answer.
  2. 11.70
  3. 2.30
  4. 2.00

Explanation

Ba(OH)₂ is a strong diprotic base: Ba(OH)₂ → Ba²⁺ + 2 OH⁻. [OH⁻] = 2 × 0.0050 M = 0.010 M = 1.0 × 10⁻² M. pOH = -log₁₀(1.0 × 10⁻²) = 2.00. pH = 14.00 - pOH = 14.00 - 2.00 = 12.00.

Question 4Physical chemistry, amounts and reactions

Calculate the pH of a 0.10 M aqueous solution of acetic acid, CH₃COOH, given Ka = 1.8 × 10⁻⁵ at 25 °C.

  1. 2.87 Correct answer.
  2. 1.00
  3. 4.74
  4. 3.37

Explanation

CH₃COOH ⇌ H⁺ + CH₃COO⁻. Ka = [H⁺][CH₃COO⁻] / [CH₃COOH] ≈ x² / c 1.8 × 10⁻⁵ = x² / 0.10 ⇒ x² = 1.8 × 10⁻⁶ x = [H⁺] = √(1.8 × 10⁻⁶) = 1.342 × 10⁻³ M. pH = -log₁₀(1.342 × 10⁻³) = 2.87.

Question 5Physical chemistry, amounts and reactions

Calculate the pH of a 0.20 M aqueous solution of ammonia, NH₃, given Kb = 1.8 × 10⁻⁵ at 25 °C.

  1. 11.28 Correct answer.
  2. 2.72
  3. 4.74
  4. 9.26

Explanation

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. Kb = [NH₄⁺][OH⁻] / [NH₃] ≈ x² / 0.20 1.8 × 10⁻⁵ = x² / 0.20 ⇒ x² = 3.6 × 10⁻⁶ x = [OH⁻] = √(3.6 × 10⁻⁶) = 1.897 × 10⁻³ M. pOH = -log₁₀(1.897 × 10⁻³) = 2.72. pH = 14.00 - 2.72 = 11.28.

Question 6Physical chemistry, amounts and reactions

What is the degree of ionization α of a 0.010 M weak acid HA solution with Ka = 1.0 × 10⁻⁵?

  1. 0.0316 (or 3.16%) Correct answer.
  2. 0.0010 (or 0.10%)
  3. 0.100 (or 10.0%)
  4. 0.00316 (or 0.316%)

Explanation

For weak acid ionization: Ka = c α² / (1 - α) ≈ c α². α = √(Ka / c) = √(1.0 × 10⁻⁵ / 0.010) = √(1.0 × 10⁻³) = √(10 × 10⁻⁴) = 0.0316 = 3.16%.

Question 7Physical chemistry, amounts and reactions

What is the pH of a 0.10 M aqueous solution of ammonium chloride, NH₄Cl? (Ka for NH₄⁺ = 5.6 × 10⁻¹⁰ at 25 °C)

  1. 5.13 Correct answer.
  2. 8.87
  3. 7.00
  4. 4.75

Explanation

NH₄Cl dissociates completely to NH₄⁺ and Cl⁻. Cl⁻ does not hydrolyze. NH₄⁺ undergoes acid hydrolysis: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. Ka = x² / 0.10 ⇒ 5.6 × 10⁻¹⁰ = x² / 0.10 ⇒ x² = 5.6 × 10⁻¹¹ x = [H⁺] = √(5.6 × 10⁻¹¹) = 7.483 × 10⁻⁶ M. pH = -log₁₀(7.483 × 10⁻⁶) = 5.13.

Question 8Physical chemistry, amounts and reactions

Calculate the pH of a 0.20 M aqueous solution of sodium acetate, CH₃COONa, at 25 °C. (Ka for CH₃COOH = 1.8 × 10⁻⁵, Kw = 1.0 × 10⁻¹⁴)

  1. 9.02 Correct answer.
  2. 4.98
  3. 7.00
  4. 11.30

Explanation

CH₃COONa dissociates completely to Na⁺ and CH₃COO⁻. Acetate hydrolyzes: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.56 × 10⁻¹⁰. Kb = x² / 0.20 ⇒ 5.56 × 10⁻¹⁰ = x² / 0.20 ⇒ x² = 1.111 × 10⁻¹⁰ x = [OH⁻] = √(1.111 × 10⁻¹⁰) = 1.054 × 10⁻⁵ M. pOH = -log₁₀(1.054 × 10⁻⁵) = 4.98. pH = 14.00 - 4.98 = 9.02.

Question 9Physical chemistry, amounts and reactions

Calculate the pH of a buffer solution containing 0.30 M acetic acid (CH₃COOH) and 0.20 M sodium acetate (CH₃COONa). (Ka for CH₃COOH = 1.8 × 10⁻⁵, pKa = 4.74)

  1. 4.57 Correct answer.
  2. 4.92
  3. 4.74
  4. 5.06

Explanation

Use Henderson-Hasselbalch equation: pH = pKa + log₁₀([Base] / [Acid]) pH = 4.74 + log₁₀(0.20 / 0.30) = 4.74 + log₁₀(0.6667) = 4.74 - 0.176 = 4.564 ≈ 4.57.

Question 10Physical chemistry, amounts and reactions

What is the pH of a buffer solution composed of 0.15 M NH₃ and 0.25 M NH₄Cl? (Kb for NH₃ = 1.8 × 10⁻⁵, pKb = 4.74, pKa for NH₄⁺ = 9.26)

  1. 9.04 Correct answer.
  2. 9.48
  3. 4.96
  4. 9.26

Explanation

Use Henderson-Hasselbalch equation for conjugate acid-base pair NH₄⁺ / NH₃: pH = pKa + log₁₀([NH₃] / [NH₄⁺]) = 9.26 + log₁₀(0.15 / 0.25) = 9.26 + log₁₀(0.60) = 9.26 - 0.22 = 9.04.

Question 11Physical chemistry, amounts and reactions

What volume of 0.200 M NaOH is required to neutralize completely 25.0 mL of 0.120 M H₂SO₄ aqueous solution?

  1. 30.0 mL Correct answer.
  2. 15.0 mL
  3. 60.0 mL
  4. 25.0 mL

Explanation

Balanced equation: H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O. Moles of H₂SO₄ = 0.0250 L × 0.120 mol/L = 0.00300 mol. Stoichiometric ratio: 1 mol H₂SO₄ reacts with 2 mol NaOH. Moles of NaOH required = 2 × 0.00300 = 0.00600 mol. Volume of NaOH = 0.00600 mol / 0.200 mol/L = 0.0300 L = 30.0 mL.

Question 12Physical chemistry, amounts and reactions

During the titration of 25.0 mL of 0.10 M weak acid HA (Ka = 1.0 × 10⁻⁵) with 0.10 M NaOH, what is the pH of the solution after adding EXACTLY 12.5 mL of NaOH?

  1. 5.00 Correct answer.
  2. 7.00
  3. 2.87
  4. 8.72

Explanation

Adding 12.5 mL of NaOH neutralizes exactly half of the initial 25.0 mL of HA, reaching the HALF-EQUIVALENCE POINT. At this point, [HA] = [A⁻]. From Henderson-Hasselbalch: pH = pKa + log₁₀(1) = pKa = -log₁₀(1.0 × 10⁻⁵) = 5.00.

Question 13Physical chemistry, amounts and reactions

Why is the pH at the equivalence point of a weak acid (such as CH₃COOH) titrated with a strong base (such as NaOH) GREATER THAN 7.0 at 25 °C?

  1. At the equivalence point, all acid is converted to its conjugate base (CH₃COO⁻), which undergoes basic hydrolysis with water to produce OH⁻ ions. Correct answer.
  2. Excess NaOH remains unreacted in the solution.
  3. Strong base titrations always have pH = 14 at equivalence.
  4. Water autoionization increases significantly at the equivalence point.

Explanation

At the equivalence point of a weak acid-strong base titration, exact stoichiometric amounts have reacted, forming the salt CH₃COONa. The conjugate base CH₃COO⁻ hydrolyzes: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, generating OH⁻ ions and making the solution basic (pH > 7.0).

Question 14Physical chemistry, amounts and reactions

Which indicator is most suitable for detecting the equivalence point in the titration of 0.10 M CH₃COOH (weak acid) with 0.10 M NaOH (strong base)?

  1. Phenolphthalein (transition pH range: 8.2 – 10.0) Correct answer.
  2. Methyl orange (transition pH range: 3.1 – 4.4)
  3. Bromophenol blue (transition pH range: 3.0 – 4.6)
  4. Methyl red (transition pH range: 4.4 – 6.2)

Explanation

The equivalence point of a weak acid titrated with a strong base occurs in the basic region (pH ~8.7). An appropriate indicator must undergo color change within this pH range. Phenolphthalein (pH 8.2 - 10.0) matches this equivalence point perfectly.

Question 15Physical chemistry, amounts and reactions

According to Lewis acid-base theory, what occurs when ammonia (NH₃) reacts with boron trifluoride (BF₃) to form H₃N:BF₃?

  1. NH₃ acts as a Lewis base by donating an electron pair, and BF₃ acts as a Lewis acid by accepting an electron pair. Correct answer.
  2. BF₃ acts as a Lewis base by donating a proton to NH₃.
  3. NH₃ acts as a Lewis acid by accepting an electron pair from BF₃.
  4. Both species act as Brønsted-Lowry acids.

Explanation

A Lewis base is an electron-pair donor; NH₃ has a lone pair on nitrogen. A Lewis acid is an electron-pair acceptor; BF₃ has an incomplete octet (6 valence electrons on B) with a vacant p orbital. NH₃ donates its lone pair to B to form a coordinate covalent bond (H₃N→BF₃).

Question 16Physical chemistry, amounts and reactions

At 60 °C, the ion product constant of pure water is Kw = 1.0 × 10⁻¹³. What is the pH of pure, neutral water at 60 °C?

  1. 6.50 Correct answer.
  2. 7.00
  3. 7.50
  4. 13.00

Explanation

Autoionization of water: 2 H₂O ⇌ H₃O⁺ + OH⁻ is endothermic (ΔH > 0). Increasing temperature increases Kw. In neutral water, [H₃O⁺] = [OH⁻] = √Kw = √(1.0 × 10⁻¹³) = 3.162 × 10⁻⁷ M. pH = -log₁₀(3.162 × 10⁻⁷) = 6.50. Neutral water at 60 °C has pH = 6.50 (it is neutral because [H⁺] = [OH⁻]).

Question 17Physical chemistry, amounts and reactions

For phosphoric acid, H₃PO₄, step ionization constants are Ka₁ = 7.5 × 10⁻³, Ka₂ = 6.2 × 10⁻⁸, Ka₃ = 4.8 × 10⁻¹³. In a 0.10 M H₃PO₄ solution, which phosphate species is present in the HIGHEST concentration at equilibrium?

  1. H₃PO₄ Correct answer.
  2. H₂PO₄⁻
  3. HPO₄²⁻
  4. PO₄³⁻

Explanation

Because Ka₁ >> Ka₂ >> Ka₃ (differing by ~5 orders of magnitude), first-stage ionization occurs to a small extent (~25%), while second and third stage ionizations are negligible. Un-ionized H₃PO₄ remains the predominant species in solution (~0.075 M remaining).

Question 18Physical chemistry, amounts and reactions

What happens when 1.0 mL of 1.0 M HCl is added to 100 mL of a buffer solution containing 0.10 M CH₃COOH and 0.10 M CH₃COONa?

  1. The added H⁺ ions react with CH₃COO⁻ conjugate base to form CH₃COOH, maintaining a nearly constant pH. Correct answer.
  2. The added H⁺ ions react with CH₃COOH acid, causing pH to drop drastically to 1.0.
  3. The added H⁺ ions react with Na⁺ ions to precipitate NaCl.
  4. The buffer capacity is completely exhausted, and pH rises above 12.

Explanation

A buffer resists pH change because added strong acid (H⁺) is consumed by the weak conjugate base component of the buffer: H⁺ + CH₃COO⁻ → CH₃COOH. This converts a strong acid into a weak acid, keeping the [Base]/[Acid] ratio and pH nearly constant.

Question 19Physical chemistry, amounts and reactions

What is the pH of a solution prepared by mixing 50.0 mL of 0.10 M HCl with 50.0 mL of 0.10 M NaOH?

  1. 7.00 Correct answer.
  2. 0.00
  3. 14.00
  4. 1.00

Explanation

Moles of H⁺ = 0.050 L × 0.10 M = 0.0050 mol. Moles of OH⁻ = 0.050 L × 0.10 M = 0.0050 mol. Exact neutralization occurs: H⁺ + OH⁻ → H₂O. The resulting solution contains only NaCl(aq) and H₂O. Since NaCl is the salt of a strong acid and strong base, neither ion hydrolyzes, giving pH = 7.00 at 25 °C.

Question 20Physical chemistry, amounts and reactions

Given that the acid dissociation constant of nitrous acid, HNO₂, is Ka = 4.5 × 10⁻⁴ at 25 °C, what is the base dissociation constant Kb of its conjugate base, the nitrite ion NO₂⁻?

  1. 2.22 × 10⁻¹¹ Correct answer.
  2. 4.5 × 10⁻¹⁰
  3. 2.22 × 10⁻³
  4. 1.0 × 10⁻¹⁴

Explanation

For any conjugate acid-base pair in aqueous solution: Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25 °C. Kb(NO₂⁻) = Kw / Ka = (1.0 × 10⁻¹⁴) / (4.5 × 10⁻⁴) = 2.222 × 10⁻¹¹.

Question 21Structure, bonding and inorganic chemistry

Which set of quantum numbers (n, l, ml, ms) is allowed for a valence electron in a ground-state sulfur atom (Z = 16)?

  1. n = 3, l = 1, ml = -1, ms = -1/2 Correct answer.
  2. n = 3, l = 2, ml = 0, ms = +1/2
  3. n = 2, l = 1, ml = +1, ms = -1/2
  4. n = 3, l = 1, ml = +2, ms = +1/2

Explanation

Sulfur (Z = 16) has the ground-state electron configuration 1s² 2s² 2p⁶ 3s² 3p⁴. The valence electrons occupy the n = 3 shell. For 3p electrons, n = 3 and l = 1. Permissible values for ml range from -l to +l (-1, 0, +1), and ms can be +1/2 or -1/2. Therefore, n = 3, l = 1, ml = -1, ms = -1/2 is a valid set.

Question 22Structure, bonding and inorganic chemistry

What is the ground-state electron configuration of the iron(III) ion, Fe³⁺ (Z = 26)?

  1. [Ar] 3d⁵ Correct answer.
  2. [Ar] 4s² 3d³
  3. [Ar] 4s¹ 3d⁴
  4. [Ar] 3d⁶

Explanation

Neutral iron (Z = 26) has the electron configuration [Ar] 4s² 3d⁶. When transition metals ionize, electrons are removed first from the outermost 4s subshell, then from the 3d subshell. Removing three electrons yields Fe³⁺ with configuration [Ar] 3d⁵, representing a stable half-filled d subshell.

Question 23Structure, bonding and inorganic chemistry

Why is the first ionization energy of phosphorus (Z = 15) higher than that of sulfur (Z = 16)?

  1. Phosphorus has a stable half-filled 3p³ subshell, whereas sulfur has paired electrons in one 3p orbital experiencing electron-electron repulsion. Correct answer.
  2. Sulfur has a smaller nuclear charge than phosphorus, resulting in weaker attraction for valence electrons.
  3. Phosphorus has a significantly smaller atomic radius than sulfur.
  4. Sulfur's valence electrons are shielded by an extra inner electron shell.

Explanation

Phosphorus ([Ar] 3s² 3p³) has three unpaired electrons in three degenerate 3p orbitals, giving an extra exchange energy stability due to a half-filled subshell. Sulfur ([Ar] 3s² 3p⁴) has two paired electrons in one 3p orbital. The interelectronic repulsion between these paired electrons lowers the energy required to remove one electron from sulfur.

Question 24Structure, bonding and inorganic chemistry

Which list arranges the isoelectronic species N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺ in order of DECREASING ionic radius?

  1. N³⁻ > O²⁻ > F⁻ > Na⁺ > Mg²⁺ Correct answer.
  2. Mg²⁺ > Na⁺ > F⁻ > O²⁻ > N³⁻
  3. F⁻ > O²⁻ > N³⁻ > Mg²⁺ > Na⁺
  4. N³⁻ > F⁻ > O²⁻ > Na⁺ > Mg²⁺

Explanation

All five species are isoelectronic with 10 electrons (configuration 1s² 2s² 2p⁶). The ionic radius in an isoelectronic series is determined by nuclear charge (Z). As Z increases from N (Z=7) to Mg (Z=12), the greater nuclear pull contracts the electron cloud. Thus, N³⁻ (Z=7) is largest and Mg²⁺ (Z=12) is smallest.

Question 25Structure, bonding and inorganic chemistry

Given standard enthalpies for NaCl formation: Sublimation of Na(s) = +107 kJ/mol, 1st Ionization Energy of Na(g) = +496 kJ/mol, Bond dissociation of Cl₂(g) = +244 kJ/mol, Electron affinity of Cl(g) = -349 kJ/mol, and Standard enthalpy of formation ΔH°f(NaCl, s) = -411 kJ/mol. Calculate the lattice enthalpy (ΔH_lattice) for NaCl(s) → Na⁺(g) + Cl⁻(g).

  1. +787 kJ/mol Correct answer.
  2. +665 kJ/mol
  3. +388 kJ/mol
  4. +909 kJ/mol

Explanation

According to Hess's Law applied to the Born-Haber cycle: ΔH°f = ΔH_sub(Na) + IE₁(Na) + 1/2 ΔH_diss(Cl₂) + EA₁(Cl) - ΔH_lattice -411 = +107 + 496 + 1/2(244) + (-349) - ΔH_lattice -411 = 107 + 496 + 122 - 349 - ΔH_lattice = +376 - ΔH_lattice ΔH_lattice = +376 - (-411) = +787 kJ/mol.

Question 26Structure, bonding and inorganic chemistry

In the preferred Lewis structure of the sulfate ion, SO₄²⁻, with expanded octet to minimize formal charges, what are the formal charges on the sulfur atom and the double-bonded oxygen atoms?

  1. Formal charge on S = 0; formal charge on double-bonded O = 0 Correct answer.
  2. Formal charge on S = +2; formal charge on double-bonded O = -1
  3. Formal charge on S = +1; formal charge on double-bonded O = 0
  4. Formal charge on S = 0; formal charge on double-bonded O = -1

Explanation

In SO₄²⁻, sulfur has 6 valence electrons. In the structure with two S=O double bonds and two S-O⁻ single bonds, sulfur forms 6 covalent bonds (12 shared electrons) and has no lone pairs: FC(S) = 6 - 0 - 6 = 0. The double-bonded oxygens have 2 lone pairs and 2 bonds: FC(O=) = 6 - 4 - 2 = 0. The single-bonded oxygens have 3 lone pairs and 1 bond: FC(O⁻) = 6 - 6 - 1 = -1.

Question 27Structure, bonding and inorganic chemistry

According to VSEPR theory, what are the electron domain geometry and molecular geometry of sulfur tetrafluoride, SF₄?

  1. Trigonal bipyramidal electron domain geometry; seesaw molecular geometry Correct answer.
  2. Tetrahedral electron domain geometry; tetrahedral molecular geometry
  3. Square planar electron domain geometry; square planar molecular geometry
  4. Octahedral electron domain geometry; square pyramidal molecular geometry

Explanation

Sulfur in SF₄ has 6 valence electrons + 4 single bonds = 10 electrons around S (5 electron pairs: 4 bonding pairs and 1 lone pair). 5 electron domains adopt a trigonal bipyramidal domain geometry. To minimize lone pair-bonding pair repulsions, the lone pair occupies an equatorial position, giving a seesaw molecular shape.

Question 28Structure, bonding and inorganic chemistry

What are the hybridizations of the carbon atoms in ethyne (acetylene, C₂H₂), ethene (ethylene, C₂H₄), and ethane (C₂H₆), respectively?

  1. C₂H₂: sp; C₂H₄: sp²; C₂H₆: sp³ Correct answer.
  2. C₂H₂: sp²; C₂H₄: sp³; C₂H₆: sp
  3. C₂H₂: sp³; C₂H₄: sp²; C₂H₆: sp
  4. C₂H₂: sp; C₂H₄: sp³; C₂H₆: sp²

Explanation

In ethyne (H-C≡C-H), each carbon forms 2 σ-bonds (linear shape, 180° bond angle) requiring sp hybridization. In ethene (H₂C=CH₂), each carbon forms 3 σ-bonds (trigonal planar shape, 120° bond angle) requiring sp² hybridization. In ethane (H₃C-CH₃), each carbon forms 4 σ-bonds (tetrahedral shape, 109.5° bond angle) requiring sp³ hybridization.

Question 29Structure, bonding and inorganic chemistry

Which of the following molecules has a non-zero net dipole moment (is polar)?

  1. Sulfur dioxide, SO₂ Correct answer.
  2. Carbon dioxide, CO₂
  3. Boron trifluoride, BF₃
  4. Carbon tetrachloride, CCl₄

Explanation

SO₂ has a bent molecular shape due to 1 lone pair and 2 bonding domains on sulfur (VSEPR angle ~119°). The S-O bond dipoles do not cancel out, resulting in a net dipole moment μ > 0. CO₂ (linear), BF₃ (trigonal planar), and CCl₄ (tetrahedral) are highly symmetric, causing their bond dipoles to cancel out completely (μ = 0).

Question 30Structure, bonding and inorganic chemistry

Which of the following compounds has the highest boiling point, and what is the primary intermolecular force responsible?

  1. Ethanol (CH₃CH₂OH); hydrogen bonding Correct answer.
  2. Dimethyl ether (CH₃OCH₃); dipole-dipole interactions
  3. Ethane (CH₃CH₃); London dispersion forces
  4. Fluoromethane (CH₃F); ion-dipole interactions

Explanation

Ethanol contains an -OH group where hydrogen is covalently bonded to highly electronegative oxygen, allowing strong intermolecular hydrogen bonding. Dimethyl ether (polar, dipole-dipole) and ethane (nonpolar, London dispersion) lack hydrogen bond donors, giving ethanol a significantly higher boiling point (78 °C vs -24 °C for dimethyl ether).

Question 31Structure, bonding and inorganic chemistry

What is the average carbon-oxygen bond order in the carbonate ion, CO₃²⁻?

  1. 1.33 (or 4/3) Correct answer.
  2. 1.50 (or 3/2)
  3. 1.00
  4. 2.00

Explanation

The carbonate ion (CO₃²⁻) has three equivalent resonance structures. Total number of C-O bonds across the three domains is 4 (one C=O double bond and two C-O⁻ single bonds). The average C-O bond order is (2 + 1 + 1) / 3 = 4/3 = 1.33.

Question 32Structure, bonding and inorganic chemistry

Which of the following gaseous transition metal ions is diamagnetic (has zero unpaired d-electrons)?

  1. Zn²⁺ Correct answer.
  2. Fe²⁺
  3. Mn²⁺
  4. Cu²⁺

Explanation

Zinc (Z=30) has the neutral configuration [Ar] 4s² 3d¹⁰. Ionization to Zn²⁺ removes the two 4s electrons, leaving [Ar] 3d¹⁰. Since all five 3d orbitals are completely filled with electron pairs, Zn²⁺ has zero unpaired electrons and is diamagnetic. Fe²⁺ (3d⁶, 4 unpaired), Mn²⁺ (3d⁵, 5 unpaired), and Cu²⁺ (3d⁹, 1 unpaired) are paramagnetic.

Question 33Structure, bonding and inorganic chemistry

According to Molecular Orbital (MO) theory, what are the bond order and magnetic property of the peroxide ion, O₂²⁻?

  1. Bond order = 1; diamagnetic Correct answer.
  2. Bond order = 2; paramagnetic
  3. Bond order = 2.5; paramagnetic
  4. Bond order = 1.5; diamagnetic

Explanation

Neutral O₂ has 12 valence electrons. Peroxide O₂²⁻ has 14 valence electrons (configuration: σ2s² σ*2s² σ2p² π2p⁴ π*2p⁴). Bonding electrons = 8, Antibonding electrons = 6. Bond order = (8 - 6)/2 = 1. All electrons in the MO diagram are paired, making O₂²⁻ diamagnetic.

Question 34Structure, bonding and inorganic chemistry

Which element possesses the highest electronegativity on Pauling's scale, and why?

  1. Fluorine; small atomic radius and high effective nuclear charge without d-orbital expansion Correct answer.
  2. Chlorine; highest electron affinity among all elements
  3. Oxygen; greatest willingness to form double bonds
  4. Helium; highest first ionization energy

Explanation

Fluorine is the most electronegative element (Pauling value 3.98). It has a small 2p valence shell (n = 2) with high effective nuclear charge (Zeff = +7), strongly attracting shared bonding electron pairs toward its nucleus.

Question 35Structure, bonding and inorganic chemistry

Which ionic compound exhibits the HIGHEST lattice energy?

  1. MgO Correct answer.
  2. NaCl
  3. MgCl₂
  4. CaO

Explanation

Lattice energy is proportional to (q₁ · q₂) / r. In MgO, charges are q₁ = +2 and q₂ = -2 (product = 4), compared to NaCl (+1 · -1 = 1). Comparing MgO to CaO (both +2/-2), Mg²⁺ has a smaller ionic radius than Ca²⁺, making the interionic distance r smaller and the lattice energy for MgO the highest (~3795 kJ/mol vs ~3414 kJ/mol for CaO).

Question 36Structure, bonding and inorganic chemistry

Why are metals ductile and good electrical conductors in the solid state, whereas ionic crystals are brittle and non-conducting when solid?

  1. Metals consist of positive metal cations in a delocalized 'sea of electrons' that allows atom layers to slide without breaking bonds; ionic solids consist of rigid alternating charged ions that repel when shifted. Correct answer.
  2. Ionic solids lack electrons, whereas metals contain only negative ions.
  3. Metals have covalent network lattice structures, whereas ionic solids have van der Waals molecular lattices.
  4. Metals expand when deformed, whereas ionic solids melt instantly under physical stress.

Explanation

The metallic bond model features delocalized valence electrons moving freely around fixed metal cations. Shifting layers of cations maintains non-directional metallic bonding. In ionic solids, shifting layers brings ions of like charge adjacent to each other, causing strong electrostatic repulsion and cleaving the crystal.

Question 37Structure, bonding and inorganic chemistry

What is the maximum number of electrons in an atom that can share the quantum numbers n = 4 and l = 2?

  1. 10 Correct answer.
  2. 6
  3. 2
  4. 14

Explanation

The quantum numbers n = 4 and l = 2 specify the 4d subshell. For l = 2, there are 5 degenerate d orbitals (ml = -2, -1, 0, +1, +2). By the Pauli exclusion principle, each orbital holds a maximum of 2 electrons (with opposite spins ms = ±1/2). Thus, 5 × 2 = 10 electrons max.

Question 38Structure, bonding and inorganic chemistry

What are the molecular geometry and hybridization of the central iodine atom in the triiodide ion, I₃⁻?

  1. Linear geometry; sp³d hybridization Correct answer.
  2. Trigonal planar geometry; sp² hybridization
  3. Bent geometry; sp³ hybridization
  4. T-shaped geometry; sp³d hybridization

Explanation

Central iodine has 7 valence electrons + 1 extra negative charge + 2 single bonds to terminal iodines = 10 valence electrons (5 electron pairs: 2 bonding pairs and 3 lone pairs). 5 domains require sp³d hybridization. The 3 lone pairs occupy equatorial positions of a trigonal bipyramid, arranging the 2 bonding pairs axially to form a linear molecular geometry.

Question 39Physical chemistry, amounts and reactions

Calculate the standard reaction enthalpy ΔH°r for the combustion of ethanol: C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l) Given standard enthalpies of formation ΔH°f: C₂H₅OH(l) = -277.7 kJ/mol, CO₂(g) = -393.5 kJ/mol, H₂O(l) = -285.8 kJ/mol.

  1. -1366.7 kJ/mol Correct answer.
  2. -1234.5 kJ/mol
  3. +1366.7 kJ/mol
  4. -401.6 kJ/mol

Explanation

ΔH°r = Σ n ΔH°f(products) - Σ m ΔH°f(reactants) ΔH°r = [2(-393.5) + 3(-285.8)] - [1(-277.7) + 3(0)] ΔH°r = [-787.0 - 857.4] - [-277.7] = -1644.4 + 277.7 = -1366.7 kJ/mol.

Question 40Physical chemistry, amounts and reactions

Given the thermochemical equations: (1) C(s) + O₂(g) → CO₂(g) ΔH₁ = -393.5 kJ (2) 2 CO(g) + O₂(g) → 2 CO₂(g) ΔH₂ = -566.0 kJ Calculate ΔH° for the partial oxidation of carbon to carbon monoxide: 2 C(s) + O₂(g) → 2 CO(g)

  1. -221.0 kJ Correct answer.
  2. -959.5 kJ
  3. +172.5 kJ
  4. -442.0 kJ

Explanation

Target equation: 2 C(s) + O₂(g) → 2 CO(g). Multiply reaction (1) by 2: 2 C(s) + 2 O₂(g) → 2 CO₂(g) ΔH = 2(-393.5) = -787.0 kJ. Reverse reaction (2): 2 CO₂(g) → 2 CO(g) + O₂(g) ΔH = +566.0 kJ. Add equations: 2 C(s) + O₂(g) → 2 CO(g). ΔH° = -787.0 + 566.0 = -221.0 kJ.

Question 41Physical chemistry, amounts and reactions

Estimate ΔH° for the synthesis of ammonia: N₂(g) + 3 H₂(g) → 2 NH₃(g) Using average bond energies: N≡N = 945 kJ/mol, H-H = 436 kJ/mol, N-H = 391 kJ/mol.

  1. -93 kJ/mol Correct answer.
  2. +93 kJ/mol
  3. -105 kJ/mol
  4. -186 kJ/mol

Explanation

ΔH° = Σ BE(bonds broken) - Σ BE(bonds formed). Bonds broken: 1 N≡N (945) + 3 H-H (3 × 436 = 1308) = 2253 kJ. Bonds formed: 2 NH₃ molecules have 2 × 3 = 6 N-H bonds (6 × 391 = 2346 kJ). ΔH° = 2253 - 2346 = -93 kJ/mol.

Question 42Physical chemistry, amounts and reactions

For which of the following processes is the standard entropy change of the system (ΔS°_sys) POSITIVE?

  1. CaCO₃(s) → CaO(s) + CO₂(g) Correct answer.
  2. N₂(g) + 3 H₂(g) → 2 NH₃(g)
  3. 2 SO₂(g) + O₂(g) → 2 SO₃(g)
  4. H₂O(g) → H₂O(l)

Explanation

In CaCO₃(s) → CaO(s) + CO₂(g), 1 mole of solid produces 1 mole of solid and 1 mole of gas. Producing gas from a solid significantly increases molecular disorder, so ΔS° > 0. The other three processes involve a net decrease in moles of gas (3→2, 3→2, or gas→liquid), resulting in ΔS° < 0.

Question 43Physical chemistry, amounts and reactions

A chemical reaction has ΔH° = +178 kJ/mol and ΔS° = +160 J/(mol·K). At what minimum temperature does this reaction become thermodynamic spontaneous under standard conditions?

  1. 1112.5 K (or 839.5 °C) Correct answer.
  2. 890.0 K (or 617.0 °C)
  3. 298.15 K (or 25 °C)
  4. 1450.0 K (or 1177.0 °C)

Explanation

Spontaneity requires ΔG° = ΔH° - T ΔS° < 0. At equilibrium (ΔG° = 0), T = ΔH° / ΔS°. Convert ΔS° to kJ/(mol·K): 160 J/(mol·K) = 0.160 kJ/(mol·K). T = 178 / 0.160 = 1112.5 K. Above 1112.5 K, the -T ΔS° term dominates and ΔG° becomes negative.

Question 44Physical chemistry, amounts and reactions

For the reaction A + 2 B → C, initial rate data are recorded: Exp 1: [A]=0.10 M, [B]=0.10 M, Rate=2.0×10⁻³ M/s Exp 2: [A]=0.20 M, [B]=0.10 M, Rate=4.0×10⁻³ M/s Exp 3: [A]=0.10 M, [B]=0.30 M, Rate=1.8×10⁻² M/s What is the overall rate law for this reaction?

  1. Rate = k [A] [B]² Correct answer.
  2. Rate = k [A]² [B]
  3. Rate = k [A] [B]
  4. Rate = k [A]² [B]²

Explanation

Comparing Exp 1 and Exp 2: [B] is constant, [A] doubles (0.10 → 0.20), Rate doubles (2.0×10⁻³ → 4.0×10⁻³). 2ᵐ = 2 ⇒ m = 1 (1st order in A). Comparing Exp 1 and Exp 3: [A] is constant, [B] triples (0.10 → 0.30), Rate increases by a factor of 9 (2.0×10⁻³ → 1.8×10⁻²). 3ⁿ = 9 ⇒ n = 2 (2nd order in B). Rate = k [A] [B]².

Question 45Physical chemistry, amounts and reactions

The decomposition of hydrogen peroxide in aqueous solution follows first-order kinetics with a rate constant k = 0.0693 min⁻¹. How long will it take for the concentration of H₂O₂ to decrease from 0.80 M to 0.10 M?

  1. 30 minutes Correct answer.
  2. 10 minutes
  3. 20 minutes
  4. 40 minutes

Explanation

Half-life for 1st order: t₁/₂ = ln(2) / k = 0.693 / 0.0693 = 10 min. Decreasing from 0.80 M to 0.10 M represents a reduction to 1/8 of initial concentration (0.80 → 0.40 → 0.20 → 0.10), which corresponds to 3 half-lives. Total time = 3 × 10 min = 30 min.

Question 46Physical chemistry, amounts and reactions

If a reaction's rate constant doubles when the temperature increases from 300 K to 310 K, what is its activation energy Ea? (R = 8.314 J/(mol·K), ln 2 = 0.693)

  1. 53.6 kJ/mol Correct answer.
  2. 26.8 kJ/mol
  3. 107.2 kJ/mol
  4. 12.4 kJ/mol

Explanation

Use Arrhenius equation: ln(k₂/k₁) = (Ea / R) × (1/T₁ - 1/T₂) ln(2) = (Ea / 8.314) × (1/300 - 1/310) 0.693 = (Ea / 8.314) × (10 / 93000) = (Ea / 8.314) × (1.075 × 10⁻⁴) Ea = (0.693 × 8.314) / 1.075×10⁻⁴ = 5.7616 / 1.075×10⁻⁴ = 53584 J/mol = 53.6 kJ/mol.

Question 47Physical chemistry, amounts and reactions

What is the effect of adding a positive catalyst to a chemical reaction at constant temperature?

  1. It lowers the activation energy for both the forward and reverse reactions by providing an alternative mechanism, increasing the rate constant without altering ΔH° or Kc. Correct answer.
  2. It decreases the reaction enthalpy ΔH°, making the reaction more exothermic.
  3. It increases the equilibrium constant Kc, producing a higher yield of products.
  4. It lowers the activation energy of the forward reaction only, shifting equilibrium to the right.

Explanation

A catalyst provides an alternative reaction pathway with lower activation energy (Ea). It speeds up both forward and reverse reaction rates equally. Since ΔH° = Ea(forward) - Ea(reverse) and Kc = k_forward / k_reverse, a catalyst does not change ΔH°, ΔG°, or the equilibrium constant Kc.

Question 48Physical chemistry, amounts and reactions

A 50.0 mL sample of 1.0 M HCl is mixed with 50.0 mL of 1.0 M NaOH in a coffee-cup calorimeter. The temperature rises by 6.7 °C. Assuming the solution density is 1.0 g/mL and specific heat capacity is 4.18 J/(g·°C), calculate the molar enthalpy of neutralization (in kJ/mol of H₂O formed).

  1. -56.0 kJ/mol Correct answer.
  2. -28.0 kJ/mol
  3. -112.0 kJ/mol
  4. -14.0 kJ/mol

Explanation

Total solution volume = 50 + 50 = 100 mL → mass = 100 g. Heat absorbed by solution q_sol = m · c · ΔT = 100 g × 4.18 J/(g·°C) × 6.7 °C = 2800.6 J = 2.80 kJ. Heat released by reaction q_rxn = -2.80 kJ. Moles of H₂O formed = 0.050 L × 1.0 mol/L = 0.050 mol. ΔH_neutralization = -2.80 kJ / 0.050 mol = -56.0 kJ/mol.

Question 49Physical chemistry, amounts and reactions

What are the correct SI units for the rate constant k of a THIRD-ORDER reaction?

  1. L² / (mol² · s) [or M⁻² · s⁻¹] Correct answer.
  2. L / (mol · s) [or M⁻¹ · s⁻¹]
  3. s⁻¹
  4. mol / (L · s) [or M · s⁻¹]

Explanation

Rate law for 3rd order: Rate = k [Conc]³. Units of Rate = mol/(L·s). Units of k = Rate / [Conc]³ = [mol/(L·s)] / [mol³/L³] = L²/(mol²·s) = M⁻²·s⁻¹.

Question 50Physical chemistry, amounts and reactions

A reaction mechanism is proposed: Step 1: NO₂ + F₂ → NO₂F + F (slow) Step 2: NO₂ + F → NO₂F (fast) What is the overall balanced reaction, and what rate law does this mechanism predict?

  1. Overall: 2 NO₂ + F₂ → 2 NO₂F; Rate = k [NO₂] [F₂] Correct answer.
  2. Overall: NO₂ + F₂ → NO₂F + F; Rate = k [NO₂] [F₂]
  3. Overall: 2 NO₂ + F₂ → 2 NO₂F; Rate = k [NO₂]² [F₂]
  4. Overall: 2 NO₂ + F₂ → 2 NO₂F; Rate = k [NO₂] [F]

Explanation

Adding Step 1 and Step 2 cancels the atomic intermediate F, giving the overall reaction 2 NO₂ + F₂ → 2 NO₂F. The rate law is determined by the elementary slow (rate-determining) step 1: Rate = k [NO₂] [F₂].