9701/23

Chemistry 9701/23October/November 2018

Cambridge AS Level · AS Level Structured Questions · worked solutions for every part, with the mark scheme

4
questions
60
marks
75
minutes

Topics Chemical Bonding · Nitrogen and Sulfur · Reaction Kinetics · Atomic Structure · Electrochemistry · Atoms, Molecules and Stoichiometry · +13 more

Q1Atomic StructureElectrochemistryChemical BondingNitrogen and SulfurAtoms, Molecules and StoichiometryFree sample

Iron pyrite, FeS2\text{FeS}_2, has a yellow colour that makes it look like gold metal. The compound contains the ions Fe2+\text{Fe}^{2+} and S22\text{S}_2^{2-}.

(a)
(i)

Give the full electronic configuration of Fe2+\text{Fe}^{2+}.

1s21\text{s}^2 .................................................................................................................................

1M
DifficultyMedium-Easy
Worked solution

Answer

1s22s22p63s23p63d61\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^6\,3\text{d}^6
Final answer

1s2 2s2 2p6 3s2 3p6 3d6

Detailed explanation

Background Concept

Iron has atomic number 26, so neutral Fe is 1s22s22p63s23p63d64s21\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^6\,3\text{d}^6\,4\text{s}^2. Although the 4s subshell fills before 3d, when a transition metal atom loses electrons to form a cation, the 4s electrons are removed first because once occupied, the 3d subshell lies lower in energy than 4s.

Understanding the Question

The command word is 'give', so a single correct configuration earns the mark. The ion is Fe2+\text{Fe}^{2+}, formed from Fe by losing two electrons.

Approach

Write the neutral atom configuration, then remove the two 4s electrons.

Step-by-Step Reasoning

Neutral Fe: 1s22s22p63s23p63d64s21\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^6\,3\text{d}^6\,4\text{s}^2. Removing two 4s electrons gives 1s22s22p63s23p63d61\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^6\,3\text{d}^6 — 24 electrons, correct for Fe2+\text{Fe}^{2+}. The mark scheme shows (4s0)(4\text{s}^0) merely to indicate the empty 4s subshell; writing it is optional.

Key Takeaways

  • 4s fills before 3d but empties before 3d when forming cations.
  • Always count electrons to check the charge.

Common Mistakes

  • Removing the two 3d electrons instead of the 4s electrons, giving 3d44s23\text{d}^4\,4\text{s}^2 — this scores zero.
  • Writing 3d63\text{d}^6 before 4s24\text{s}^2 in the neutral atom (filling order) — not relevant here since 4s is empty, but a common confusion.

Things to Be Careful About

  • The superscripts must sum to 24 for Fe2+\text{Fe}^{2+}; check your arithmetic.
  • Write the full configuration — 'give the full electronic configuration' means no noble-gas shorthand like [Ar]3d⁶.
Techniques used
write the full electron configuration of an ionapply 4s-before-3d filling order and remove 4s electrons first when ionising a transition metal
(ii)

Calculate the oxidation number of sulfur in the S22\text{S}_2^{2-} ion.
Assume that each sulfur atom in the ion has the same oxidation number.

oxidation number of sulfur in the S22\text{S}_2^{2-} ion = ..............................

1M
DifficultyEasy
Worked solution

Answer

The two sulfur atoms share the 2-2 charge of the ion, so each sulfur atom has oxidation number 1-1.

oxidation number of sulfur in S22\text{S}_2^{2-} = 1-1

Final answer

-1

Detailed explanation

Background Concept

The sum of the oxidation numbers of all atoms in an ion equals the charge on that ion. In a polyatomic ion made of identical atoms, the charge is shared equally between them.

Understanding the Question

The S22\text{S}_2^{2-} ion contains two sulfur atoms and carries a total charge of 2-2. The question tells you to assume both sulfur atoms have the same oxidation number.

Approach

Divide the total ionic charge by the number of atoms of the element.

Step-by-Step Reasoning

Total oxidation number for the two S atoms = 2-2. Since each S contributes equally, each sulfur has oxidation number 2÷2=1-2 \div 2 = -1. This is analogous to peroxide, O22\text{O}_2^{2-}, where each oxygen is 1-1 rather than the usual 2-2.

Key Takeaways

  • Oxidation numbers in an ion must sum to the ion's charge.
  • Diatomic ions like S22\text{S}_2^{2-} and O22\text{O}_2^{2-} have per-atom oxidation numbers of half the ionic charge.

Common Mistakes

  • Writing 2-2, assigning the whole ionic charge to each sulfur atom instead of dividing between the two atoms.
  • Assuming sulfur is always 2-2 (its typical state in simple sulfides) without reading the ion formula.

Things to Be Careful About

  • The answer is the oxidation number of each sulfur atom, as the question explicitly assumes equal oxidation numbers.
Techniques used
deduce oxidation number from total ionic chargeapply the rule that the sum of oxidation numbers equals the ion charge
(b)

Describe the metallic bonding in gold.

....................................................................................................................................................

....................................................................................................................................................

..............................................................................................................................................

2M
DifficultyEasy
Worked solution

Answer

  • Positive gold ions (cations) are arranged in a lattice, surrounded by delocalised (mobile) electrons.
  • Electrostatic attraction between the positive ions and the delocalised electrons holds the structure together (this attraction is the metallic bonding).
Final answer

Electrostatic attraction between positive gold ions/cations and delocalised electrons

Detailed explanation

Background Concept

Metallic bonding is the electrostatic attraction between a lattice of positive metal ions and a 'sea' of delocalised electrons. The valence electrons leave individual atoms and become free to move throughout the structure, so they are no longer associated with any particular ion.

Understanding the Question

'Describe' demands a two-mark statement: what is attracted to what (M1: the attraction/holding force; M2: the two species involved — positive ions AND delocalised electrons).

Approach

Name both species present in a metal and state the force of attraction between them.

Step-by-Step Reasoning

In gold, each atom loses its valence electron(s) to form Au+\text{Au}^{+} (or Au^n+) cations packed in a lattice. These electrons become delocalised. The metallic bond is the electrostatic attraction between the cations and this electron sea. Both species must be named for the second mark; the attraction itself is the first mark.

Key Takeaways

  • Metallic bonding = attraction between positive ions and delocalised electrons.
  • Delocalised electrons explain conductivity and malleability of metals.

Common Mistakes

  • Saying 'attraction between metal atoms and electrons' — the species must be positive ions/cations, not neutral atoms.
  • Mentioning only one of the two species (e.g. only 'delocalised electrons') — M2 requires both.
  • Saying electrons are 'shared' (covalent language) rather than delocalised.

Things to Be Careful About

  • A labelled diagram of the lattice of cations with a sea of electrons is accepted, but the attraction must still be stated.
Techniques used
describe metallic bonding in terms of positive ions and delocalised electronsstate the electrostatic attraction holding the structure together
(c)

Iron pyrite is often called fool’s gold because of its appearance. Impure samples of iron pyrite often contain a small amount of gold.

The gold can be obtained from impure iron pyrite. The impure iron pyrite is roasted in oxygen, to produce iron(III) oxide and sulfur dioxide. Gold does not react with oxygen.

(i)

The sulfur dioxide produced during roasting would cause environmental consequences if released into the atmosphere.

State and explain one of these environmental consequences.

.............................................................................................................................................

.............................................................................................................................................

.......................................................................................................................................

2M
DifficultyEasy
Worked solution

Answer

  • Consequence: acid rain.
  • Explanation: SO2\text{SO}_2 dissolves in atmospheric moisture and is oxidised to form sulfuric acid, which falls as acid rain; this damages buildings and statues, and lowers the pH of lakes and soils, harming fish and plants.
Final answer

Acid rain — damages buildings/statues and lowers the pH of lakes/soil, killing fish and plants

Detailed explanation

Background Concept

Sulfur dioxide released into the atmosphere dissolves in water droplets and is slowly oxidised (by O2\text{O}_2 and catalysed by NOx\text{NO}_x) to sulfur trioxide/sulfuric acid. This produces acid rain (rainwater of pH below about 5).

Understanding the Question

The command words are 'state and explain': M1 is naming a consequence (acid rain); M2 is a specific harmful effect of that consequence. One consequence with one explanation is required.

Approach

Name acid rain, then give one concrete damaging effect from the mark scheme list.

Step-by-Step Reasoning

SO2 from roasting sulfide ores enters the air. It dissolves and oxidises to H2SO4, making rain acidic. Credible effects include: erosion of buildings and statues (especially limestone/calcium carbonate), acidification of lakes killing fish, leaching of nutrients or aluminium ions from soil, damage to trees/crops, and breathing difficulties in humans. Any one of these earns M2.

Key Takeaways

  • SO2 → acid rain is a standard AS-level environmental link; always pair the named consequence with a specific effect.

Common Mistakes

  • Stating only 'acid rain' with no explanation — M2 is lost.
  • Vague answers like 'bad for the environment' or 'pollution' — not specific enough.
  • Saying 'global warming' — SO2 is not a greenhouse gas responsible for warming; this scores zero.

Things to Be Careful About

  • The explanation must name something the acid rain actually harms (buildings, fish, plants, soil pH), not just restate that rain becomes acidic.
Techniques used
state an environmental consequence of sulfur dioxide emissionsexplain the consequence with a specific harmful effect
(ii)

Complete the equation to show the roasting of iron pyrite in oxygen.

4FeS2+.......................................2Fe2O3+.......................................4\text{FeS}_2 + \text{.......................................} \rightarrow 2\text{Fe}_2\text{O}_3 + \text{.......................................}
2M
DifficultyMedium-Easy
Worked solution

Answer

4FeS2+11O22Fe2O3+8SO24\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2
Final answer

4FeS2 + 11O2 -> 2Fe2O3 + 8SO2

Detailed explanation

Background Concept

A balanced equation has equal numbers of each atom on both sides. Balancing is done by counting atoms element by element.

Understanding the Question

The Fe and S coefficients are given (4 FeS2 → 2 Fe2O3); you must find the O2 coefficient and the SO2 coefficient.

Approach

Count S atoms first to get the SO2 coefficient, then count O atoms to get O2.

Step-by-Step Reasoning

  • Sulfur: 4 FeS2 contains 8 S atoms, so 8 SO2 is needed (8 S on each side). This is M1's species plus M2's coefficient.
  • Oxygen: right side has 2 × 3 = 6 O in Fe2O3 plus 8 × 2 = 16 O in SO2, total 22 O atoms. So O2 needed = 22 ÷ 2 = 11.
  • Check Fe: 4 on each side. Balanced.

Key Takeaways

  • Balance the element appearing in the fewest species first (S here), then handle the element appearing in several compounds (O) last.

Common Mistakes

  • Writing 8O2 instead of 11O2 by forgetting the oxygen in Fe2O3.
  • Writing SO3 instead of SO2 — the question states sulfur dioxide is produced.
  • Changing the given coefficients (4 and 2) instead of filling the blanks.

Things to Be Careful About

  • Both blanks must be correct for full marks: M1 for identifying O2 and SO2, M2 for the correct numbers 11 and 8.
Techniques used
balance a symbol equation by atom counting
(iii)

A sample of impure iron pyrite was roasted in oxygen. The composition of the mixture of solid products is shown.

solid productmass/g
Fe2O3\text{Fe}_2\text{O}_333.18
Au0.37

Calculate the mass of FeS2\text{FeS}_2 present in the sample of impure iron pyrite.
Assume that all the FeS2\text{FeS}_2 was converted to Fe2O3\text{Fe}_2\text{O}_3 during the roasting process.

(MrM_r: FeS2\text{FeS}_2, 120.0; Fe2O3\text{Fe}_2\text{O}_3, 159.6)

mass of FeS2\text{FeS}_2 = .............................. g

2M
DifficultyMedium-Easy
Worked solution

Working

Moles of Fe2O3\text{Fe}_2\text{O}_3:

n(Fe2O3)=33.18159.6=0.2079 moln(\text{Fe}_2\text{O}_3) = \frac{33.18}{159.6} = 0.2079 \text{ mol}

From the equation, 4FeS22Fe2O34\text{FeS}_2 \rightarrow 2\text{Fe}_2\text{O}_3, so n(FeS2)=0.2079×42=0.4158 moln(\text{FeS}_2) = 0.2079 \times \frac{4}{2} = 0.4158 \text{ mol}.

Mass of FeS2\text{FeS}_2:

m=0.4158×120.0=49.89 gm = 0.4158 \times 120.0 = 49.89 \text{ g}

Answer

mass of FeS2\text{FeS}_2 = 49.89 g

Final answer

49.89 g

Detailed explanation

Background Concept

Stoichiometry links the mass of a product to the mass of the reactant that formed it: mass → moles → mole ratio from the balanced equation → moles of target → mass.

Understanding the Question

All the FeS2 was converted to Fe2O3. We know the mass of Fe2O3 formed (33.18 g) and must work backwards to the mass of FeS2 that produced it, using the ratio 4 FeS2 : 2 Fe2O3 from part (ii).

Approach

Convert 33.18 g of Fe2O3 to moles (M1), apply the 4:2 ratio to get moles of FeS2, then multiply by Mr = 120.0 (M2).

Step-by-Step Reasoning

  1. n(Fe2O3)=33.18÷159.6=0.2079n(\text{Fe}_2\text{O}_3) = 33.18 \div 159.6 = 0.2079 mol.
  2. Equation: 4FeS2+11O22Fe2O3+8SO24\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2. The ratio FeS2 : Fe2O3 is 4 : 2 = 2 : 1, so n(FeS2)=0.2079×2=0.4158n(\text{FeS}_2) = 0.2079 \times 2 = 0.4158 mol.
  3. m(FeS2)=0.4158×120.0=49.89m(\text{FeS}_2) = 0.4158 \times 120.0 = 49.89 g.

Key Takeaways

  • Always go through moles; never convert masses directly without the mole ratio.
  • The ratio comes from the balanced equation you completed in (ii) — an error there is carried forward (ecf).

Common Mistakes

  • Using the ratio 1:1 instead of 4:2, giving 24.95 g.
  • Dividing by 2 instead of multiplying (0.2079 ÷ 2 × 120.0 = 12.47 g).
  • Using Mr of FeS2 as 56 + 32 = 88 (forgetting the second sulfur atom).

Things to Be Careful About

  • M2 depends on correct use of both the stoichiometric ratio and Mr = 120.0 with your M1 value; show the ×4/2 step explicitly so the method mark is visible.
Techniques used
calculate moles from mass and molar massapply stoichiometric ratio from a balanced equationconvert moles back to mass
(iv)

Use your answer to (iii) to calculate the percentage by mass of gold in this sample of impure iron pyrite. Assume that gold is the only impurity in this sample of impure iron pyrite.

Give your answer to two significant figures.

(If you were unable to calculate an answer to (iii), use 55.00 g55.00\text{ g} as the mass of FeS2\text{FeS}_2 in this calculation. This is not the correct answer.)

percentage by mass of gold = .............................. %

1M
DifficultyEasy
Worked solution

Working

Total mass of sample = mass of FeS2\text{FeS}_2 + mass of Au (gold is the only impurity):

%Au=0.370.37+49.89×100=0.3750.26×100=0.736%\%\text{Au} = \frac{0.37}{0.37 + 49.89} \times 100 = \frac{0.37}{50.26} \times 100 = 0.736\%

Answer

percentage by mass of gold = 0.74 % (2 s.f.)

Final answer

0.74 %

Detailed explanation

Background Concept

Percentage by mass of a component = (mass of component ÷ total mass of mixture) × 100. Since gold is stated to be the only impurity, the total sample mass is FeS2 + Au.

Understanding the Question

Use the 49.89 g from (iii) (or the given 55.00 g fallback) with the 0.37 g of gold, and give the answer to exactly two significant figures.

Approach

Add the two masses to get the sample total, divide the gold mass by the total, and round to 2 s.f.

Step-by-Step Reasoning

0.37÷(0.37+49.89)=0.37÷50.26=0.007360.37 \div (0.37 + 49.89) = 0.37 \div 50.26 = 0.00736, i.e. 0.736%0.736\%, which rounds to 0.74%0.74\% (2 s.f.). If the fallback 55.00 g were used: 0.37÷55.37×100=0.668%0.67%0.37 \div 55.37 \times 100 = 0.668\% \approx 0.67\%.

Key Takeaways

  • Percentage by mass needs the total mixture mass in the denominator, not just the FeS2 mass.

Common Mistakes

  • Computing 0.37÷49.89×100=0.74%0.37 \div 49.89 \times 100 = 0.74\% — coincidentally very close here, but the method is wrong; the total mass must include the gold.
  • Giving the answer as 0.736 without rounding to 2 s.f., or rounding to 0.7 (1 s.f.).

Things to Be Careful About

  • The question explicitly demands two significant figures: 0.74 %, not 0.7 % or 0.736 %.
Techniques used
calculate a percentage by massapply correct significant figures

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