Unit 1: Inorganic 1 & Physical 1 Exam Practice

This question is about atomic structure, isotopes, and Time-of-Flight (TOF) mass spectrometry.

(a) State the mass number and the atomic number of the chlorine isotope ³⁷Cl. [1 mark]

(b) Describe how positive ions are formed in a Time-of-Flight (TOF) mass spectrometer using the electron impact ionisation method. [2 marks]

(c) In a TOF mass spectrometer, a singly charged ion of a chlorine isotope, ³⁵Cl⁺, is accelerated to a kinetic energy of 1.15 × 10^-16 J. The drift tube of the spectrometer has a length of 1.50 m. Calculate the time of flight, in seconds, for this ion. Take the mass of one atom of ³⁵Cl as 5.81 × 10^-26 kg. Show your working. (Recall that KE = ½mv² and v = d/t). [4 marks]

(d) A sample of chlorine gas consists of two isotopes, ³⁵Cl and ³⁷Cl. The relative atomic mass (A_r) of this sample is 35.45. Calculate the percentage abundance of each of the two isotopes in this sample of chlorine. [3 marks]

(e) Write the full electron configuration for the chloride ion, Cl^-, showing all sub-shells. [2 marks]

Show Mark Scheme

(a)

Mass number = 37 AND Atomic number = 17 [1 mark] (both required for the mark; accept labeled numbers).

(b)

M1: The sample is vaporised and high-energy electrons are fired at it from an electron gun or hot filament [1 mark].
M2: This knocks off one electron from each atom or molecule to form a 1+ ion: X(g) + e^- → X⁺(g) + 2e^- [1 mark].
Accept/Ignore: Accept X(g) → X⁺(g) + e^-. Do not accept electrospray ionisation mechanisms (e.g. dissolved in solvent, high voltage needle, protonation).

(c)

M1: Rearrange kinetic energy formula: v = √(2KE / m) [1 mark].
M2: Substitute values to find velocity: v = √[(2 × 1.15 × 10^-16) / (5.81 × 10^-26)] = 6.29 × 10⁴ m s^-1 [1 mark].
M3: Rearrange for flight time: t = d / v = 1.50 / (6.29 × 10⁴) [1 mark].
M4: Final evaluation: t = 2.38 × 10^-5 s (accept 2.38 × 10^-5 to 2.39 × 10^-5 s) [1 mark].
ECF: Allow full error carried forward at each stage. Correct numerical answer with no working shown scores maximum 3 marks.

(d)

M1: Set up abundance equation: let abundance of ³⁵Cl = x, so ³⁷Cl = 100 - x; 35.45 = [35x + 37(100 - x)] / 100 [1 mark].
M2: Rearrange and solve for x: 3545 = 35x + 3700 - 37x ⇒ 2x = 155 ⇒ x = 77.5 [1 mark].
M3: State both abundances: ³⁵Cl = 77.5% and ³⁷Cl = 22.5% [1 mark].

(e)

1s² 2s² 2p⁶ 3s² 3p⁶ [2 marks].
(Award 1 mark for correct 3s² 3p⁶ ending if there is a minor slip in earlier sub-shells; do not accept noble gas shorthand [Ar] as the question asks for full configuration).

Examiner tip: In TOF drift calculations, convert mass per mole into mass per individual ion in kg: m = (isotopic mass in grams × 10^-3) / L. Never round intermediate velocity values on your calculator.

This question is about amount of substance, empirical formulas, and the gas syringe method.

(a) Define the term relative molecular mass (M_r). [2 marks]

(b) A gaseous compound A contains carbon, chlorine, and fluorine only. The compound has a relative molecular mass (M_r) of 204.0. Quantitative analysis shows that compound A contains 69.6% chlorine and 18.6% fluorine by mass. Deduce the empirical formula and the molecular formula of compound A. Show your working. [4 marks]

(c) In a separate laboratory experiment, a volatile compound B was studied using a gas syringe. At a temperature of 373 K and a pressure of 102 kPa, a 1.63 g sample of gaseous compound B occupied a volume of 4.85 × 10^-4 m³. Calculate the relative molecular mass (M_r) of compound B. The gas constant R = 8.31 J K^-1 mol^-1. Give your answer to 3 significant figures. [4 marks]

(d) State why it is essential to heat the gas syringe to 373 K rather than conducting this measurement at room temperature (298 K). [3 marks]

Show Mark Scheme

(a)

M1: The weighted average (mean) mass of a molecule of a compound [1 mark].
M2: Compared to (relative to) 1/12th of the mass of an atom of carbon-12 [1 mark].
(Do not accept "mass of a molecule" without mentioning average/mean or relative to 1/12th of carbon-12).

(b)

M1: Calculate % carbon by difference: % C = 100 - 69.6 - 18.6 = 11.8% [1 mark].
M2: Calculate molar ratios in 100 g: C = 11.8 / 12.0 = 0.983 mol; Cl = 69.6 / 35.5 = 1.961 mol; F = 18.6 / 19.0 = 0.979 mol [1 mark].
M3: Divide by smallest (0.979): C : Cl : F = 1 : 2 : 1 ⇒ Empirical formula = CCl₂F [1 mark].
M4: Empirical formula mass = 12.0 + 71.0 + 19.0 = 102.0; 204.0 / 102.0 = 2 ⇒ Molecular formula = C₂Cl₄F₂ [1 mark].
ECF: Award M4 by ECF if an incorrect empirical formula is multiplied correctly to match M_r = 204.0.

(c)

M1: Convert pressure to Pascals: P = 102 kPa = 102,000 Pa (1.02 × 10⁵ Pa) [1 mark].
M2: Rearrange ideal gas equation: n = PV / RT = (102,000 × 4.85 × 10^-4) / (8.31 × 373) [1 mark].
M3: Evaluate amount of substance: n = 49.47 / 3099.63 = 0.01596 mol (allow 0.0160 mol) [1 mark].
M4: Calculate M_r: M_r = mass / n = 1.63 / 0.01596 = 102 (accept 102 or 102.1) [1 mark].
ECF: Full ECF applies for mathematical slips. If temperature is left in Celsius, award max 2 marks.

(d)

M1: Compound B is a liquid at room temperature (298 K) / has a boiling point above room temperature [1 mark].
M2: Heating to 373 K ensures complete vaporisation so that the substance exists entirely in the gas phase [1 mark].
M3: If any liquid remained or condensed in the syringe, the measured gas volume would be smaller than expected, resulting in an erroneously high calculated M_r value [1 mark].

Examiner tip: In PV = nRT calculations, ensure standard units: P in Pa (multiply kPa by 1000), V in m³ (divide cm³ by 10⁶), and T in K (add 273 to °C). Always state final M_r without units.

This question is about chemical bonding, molecular geometry, and intermolecular interactions.

(a) State the meaning of the term dative covalent bond. [2 marks]

(b) Explain the shape of a boron trifluoride (BF₃) molecule. In your answer, state the name of the shape, the theoretical bond angle, and explain why the molecule adopts this shape in terms of electron pair repulsion. [4 marks]

(c) The boiling point of tetrafluoromethane (CF₄) is 145 K, whereas the boiling point of methane (CH₄) is 112 K. Explain why the boiling point of CF₄ is higher than that of CH₄ even though fluorine is significantly more electronegative than carbon, and describe the nature of the intermolecular forces present in both substances. [3 marks]

(d) Fluorine also forms the compound hydrogen fluoride (HF). Explain why the boiling point of HF (293 K) is significantly higher than that of both CH₄ and CF₄. [2 marks]

Show Mark Scheme

(a)

M1: A shared pair of electrons [1 mark].
M2: Where both of the shared electrons are provided (donated) by the same atom [1 mark].
(Reject: "sharing of electrons" without stating that one atom donates both electrons).

(b)

M1: Shape name: Trigonal planar [1 mark].
M2: Bond angle: 120° [1 mark].
M3: Boron has 3 bonding pairs and 0 lone pairs in its outer valence shell [1 mark].
M4: The three electron pairs repel each other equally to achieve maximum separation and minimum electrostatic repulsion [1 mark].

(c)

M1: Both CF₄ and CH₄ are symmetrical tetrahedral molecules, so individual bond dipoles cancel out, meaning neither molecule has a permanent dipole / both are non-polar [1 mark].
M2: The only intermolecular forces acting between molecules in both substances are London dispersion / Van der Waals forces [1 mark].
M3: CF₄ has significantly more electrons than CH₄ (42 electrons in CF₄ vs 10 electrons in CH₄), resulting in stronger temporary induced dipole-dipole attractions that require more energy to overcome [1 mark].
(Do not credit any reference to breaking covalent C-F or C-H bonds).

(d)

M1: HF molecules form intermolecular hydrogen bonds (due to high electronegativity of fluorine and lone pairs on F bonded to H) [1 mark].
M2: Hydrogen bonds are significantly stronger than the London dispersion forces in CH₄ and CF₄ and require considerably more thermal energy to break [1 mark].

Examiner tip: When explaining molecular shapes, always state four elements: (1) count of bonding pairs, (2) count of lone pairs, (3) statement that electron pairs repel to minimise repulsion, and (4) final shape name and bond angle.

This question is about calorimetry, enthalpy of combustion, and thermochemical cycles.

(a) Define the term standard enthalpy of combustion (Δ_cH^θ). [2 marks]

(b) In a laboratory experiment, a student burned 0.800 g of ethanol (C₂H₅OH, M_r = 46.0) in a spirit burner. The heat released was transferred to 150 g of water in a copper calorimeter, causing the water temperature to rise from 20.2 °C to 43.8 °C. The specific heat capacity of water is 4.18 J g^-1 K^-1.

(i) Calculate the heat energy change (q), in kJ, absorbed by the water. [2 marks]

(ii) Calculate the experimental enthalpy of combustion of ethanol (Δ_cH), in kJ mol^-1, from this experiment. Include an algebraic sign in your answer. [3 marks]

(c) The theoretical standard enthalpy of combustion of ethanol is -1367 kJ mol^-1. Suggest two distinct practical reasons why the experimental value obtained in part (b) is significantly less exothermic than this theoretical standard value. [2 marks]

(d) State Hess's Law. [1 mark]

(e) Use the standard enthalpies of formation (Δ_fH^θ) in the table below to calculate the standard enthalpy change (ΔH^θ) for the gas-phase hydration of ethene to form ethanol: C₂H₄(g) + H₂O(g) → C₂H₅OH(g). [2 marks]

Substance	Δ_fH^θ / kJ mol^-1
C₂H₄(g)	+52.2
H₂O(g)	-241.8
C₂H₅OH(g)	-235.1

Show Mark Scheme

(a)

M1: The enthalpy change when one mole of a substance is burned completely in oxygen [1 mark].
M2: Under standard conditions (100 kPa and 298 K) with all reactants and products in their standard states [1 mark].

(b) (i)

M1: Temperature change ΔT = 43.8 - 20.2 = 23.6 °C; q = m * c * ΔT = 150 × 4.18 × 23.6 [1 mark].
M2: q = 14,797 J = 14.80 kJ (accept 14.8 kJ) [1 mark].
(Mass m must be 150 g for water; using 0.800 g scores zero).

(b) (ii)

M1: Amount of ethanol burned: n = 0.800 / 46.0 = 0.01739 mol [1 mark].
M2: Energy per mole: 14.797 / 0.01739 = 850.8 kJ mol^-1 [1 mark].
M3: Correct negative sign: Δ_cH = -851 kJ mol^-1 (accept range -850 to -851 kJ mol^-1) [1 mark].
ECF: Allow full error carried forward from (b)(i). Negative sign is essential for M3.

(c) Any two practical reasons from:

Heat loss from the calorimeter to the surroundings / air / can [1 mark].
Incomplete combustion of ethanol forming carbon monoxide or carbon soot [1 mark].
Evaporation of fuel from the burner wick during weighing [1 mark].
Heat capacity of the copper calorimeter was not included in the calculation [1 mark].
(Do not accept "human error", "parallax error", or "misreading thermometer").

(d)

The enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final states are the same [1 mark].

(e)

M1: ΔH = ∑Δ_fH(products) - ∑Δ_fH(reactants) = (-235.1) - [(+52.2) + (-241.8)] [1 mark].
M2: ΔH = (-235.1) - (-189.6) = -45.5 kJ mol^-1 [1 mark].
(Award 2 marks for correct numerical answer with negative sign; if sign is positive or omitted award 1 mark max).

Examiner tip: Always remember that combustion is an exothermic reaction, so Δ_cH must have a negative sign. In Hess's Law cycles using enthalpies of formation, remember: ΔH = ∑Δ_fH(products) - ∑Δ_fH(reactants).

This question is about redox chemistry, half-equations, and transition metal ion titrations.

(a) State the meaning of the term oxidation in terms of electron transfer. [1 mark]

(b) Deduce the oxidation state of the specified atom in each of the following species: [3 marks]

(i) S in SO₄^2-: .................... [1 mark]

(ii) Cl in NaClO₃: .................... [1 mark]

(iii) Mn in MnO₄^-: .................... [1 mark]

(c) In acidic solution, manganate(VII) ions (MnO₄^-) act as a strong oxidising agent and are reduced to manganese(II) ions (Mn²⁺). Write a balanced ionic half-equation for the reduction of MnO₄^- in acidic conditions. [2 marks]

(d) Acidified manganate(VII) oxidises iron(II) ions (Fe²⁺) to iron(III) ions (Fe³⁺). Write a balanced half-equation for the oxidation of Fe²⁺. [1 mark]

(e) Combine your half-equations from parts (c) and (d) to write the overall balanced ionic equation for the reaction between acidified manganate(VII) ions and iron(II) ions. [2 marks]

(f) State the chemical role of the iron(II) ions in the reaction in part (e). [1 mark]

Show Mark Scheme

(a)

Loss of electrons [1 mark]. (Must refer explicitly to electrons).

(b)

(i) S in SO₄^2-: +6 (accept 6 or VI) [1 mark].
(ii) Cl in NaClO₃: +5 (accept 5 or V) [1 mark].
(iii) Mn in MnO₄^-: +7 (accept 7 or VII) [1 mark].

(c)

MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O [2 marks].
(Award 1 mark for correct formulas of MnO₄^-, H⁺, Mn²⁺, and H₂O; 1 mark for balanced equation with 5e^- on LHS).

(d)

Fe²⁺ → Fe³⁺ + e^- (or Fe²⁺ - e^- → Fe³⁺) [1 mark].

(e)

MnO₄^- + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O [2 marks].
(Award 1 mark for multiplying iron half-equation by 5 to equalise electron transfer; 1 mark for fully correct overall equation with electrons cancelled).
ECF: Allow full error carried forward from incorrect half-equations in (c) and (d).

(f)

Reducing agent / electron donor [1 mark]. (Do not accept simply "it is oxidised" without stating it is a reducing agent or provides electrons).

Examiner tip: When balancing redox half-equations in acidic conditions, follow the systematic rules: (1) balance all atoms other than H and O, (2) balance O by adding H₂O, (3) balance H by adding H⁺, and (4) balance electrical charge by adding electrons.

This question is about Period 3 periodicity trends and Group 2 and Group 7 chemical properties.

(a) State the general trend in first ionisation energy of the elements across Period 3 from Sodium to Argon. [1 mark]

(b) Explain why there is a general increase in first ionisation energy across Period 3 in terms of atomic structure. [3 marks]

(c) The first ionisation energy of sulfur deviates from the general trend. Explain why the first ionisation energy of sulfur is lower than that of phosphorus. [3 marks]

(d) State the trend in the solubility of Group 2 hydroxides down the group from magnesium to barium. [1 mark]

(e) Barium sulfate (BaSO₄) is an important compound in qualitative testing and medicine.

(i) Write an ionic equation, including state symbols, for the precipitation reaction that occurs when aqueous barium chloride is added to an aqueous solution containing sulfate ions (SO₄^2-). [2 marks]

(ii) State why the sample solution must be acidified with hydrochloric acid before aqueous barium chloride is added in the sulfate test. [1 mark]

(iii) Barium ions (Ba²⁺) are highly toxic in aqueous solution. Explain why barium sulfate can be safely swallowed by a patient as a "barium meal" prior to an intestinal X-ray. [1 mark]

Show Mark Scheme

(a)

First ionisation energy increases / general increase [1 mark].

(b)

M1: Nuclear charge increases / number of protons in the nucleus increases [1 mark].
M2: Shielding remains approximately constant / electrons are being added to the same energy shell (third shell) [1 mark].
M3: Stronger electrostatic attraction between nucleus and outer electrons, pulling them closer and requiring more energy to remove [1 mark].

(c)

M1: Phosphorus has outer configuration 3s² 3p³ (unpaired 3p electrons) and sulfur has 3s² 3p⁴ [1 mark].
M2: In sulfur, the fourth 3p electron is paired in an orbital [1 mark].
M3: Spin-pair repulsion between the two electrons in the same 3p orbital makes the electron easier to remove, requiring less energy [1 mark].

(d)

Solubility increases down Group 2 [1 mark]. (Reject: solubility decreases).

(e) (i)

Ba²⁺(aq) + SO₄^2-(aq) → BaSO₄(s) [2 marks].
(Award 1 mark for correct ionic species on both sides; 1 mark for correct state symbols on all species).

(e) (ii)

To react with and remove any carbonate ions (CO₃^2-) or sulfite ions (SO₃^2-) that would also form an insoluble white precipitate with barium ions and give a false-positive result [1 mark].

(e) (iii)

Barium sulfate is extremely insoluble in water and stomach acid, so it cannot be absorbed into the bloodstream or tissues and passes through the digestive tract harmlessly [1 mark].

Examiner tip: Be prepared to explain both Period 3 ionisation energy dips: (1) Al is lower than Mg because the 3p electron is removed from a higher energy orbital that is shielded by 3s electrons; (2) S is lower than P because of electron-pair repulsion in a 3p orbital.

Section B: Structured Written Questions

Question 1: Atomic Structure and TOF Mass Spectrometry

Question 2: Stoichiometry, Empirical Formulas, and Gas Syringe Calculations

Question 3: Chemical Bonding, Molecular Shapes, and Intermolecular Forces

Question 4: Chemical Energetics, Calorimetry, and Hess's Law

Question 5: Oxidation, Reduction, and Redox Equations

Question 6: Periodicity, Group 2 Alkaline Earth Metals, and Group 7 Halogens