(a) The elements of Group 17 are called halogens. Complete Table 1.1. Table 1.1 halogen colour at 293 K --- --- chlorine bromine iodine (b) State the trend in volatility of the halogens chlorine, bromine and iodine. Explain your answer. (c) Iodine is made by reacting bromine with sodium iodide. (i) Construct an ionic equation for the reaction of bromine with sodium iodide. (ii) State the role of bromine in the reaction. Explain your answer. (d) Concentrated sulfuric acid is…

(a) Chlorine: green; bromine: brown; iodine: dark grey · (b) Volatility decreases down the group because the molecules have more electrons, giving stronger instantaneous dipole–induced dipole (van der Waals) forces. · (c)(i) Br2 + 2I^- -> 2Br^- + I2 · (c)(ii) Bromine is the oxidising agent because it removes electrons from iodide ions (oxidises I^- to I2). · (d)(i) NaBr + H2SO4 -> HBr + NaHSO4 · (d)(ii) NaCl — Cl^- is too weak a reducing agent to reduce H2SO4, whereas Br^-…

Hydrocarbon molecules contain covalent bonds. (a) Define covalent bond. (b) A C=C bond in an alkene is made from a $\sigma$ bond and a $\pi$ bond. (i) Identify the hybridisation of the carbon atoms in a C=C bond in an alkene. (ii) Draw labelled diagrams to show, in terms of orbital overlap, how the $\sigma$ and $\pi$ bonds are made in a C=C bond. $\sigma$ bond $\pi$ bond (c) In electrophilic reactions involving alkenes the $\pi$ bond of C=C is broken. (i) Suggest one…

(a) (electrostatic) attraction between the nuclei of two atoms and a shared pair of electrons. · (b)(i) sp^2 · (b)(ii) Diagram showing head-on overlap of sp2 orbitals for sigma bond; diagram showing sideways overlap of p orbitals for pi bond. · (c)(i) The electrons in the pi bond are further from the nuclei, resulting in weaker attraction. · (c)(ii) See mechanism diagram: HBr adds to 2-methylpropene via tertiary carbocation intermediate to form 2-bromo-2-methylpropane.

(a) V shows stereoisomerism. (i) Explain what is meant by stereoisomerism. (ii) Deduce the number of stereoisomers of V. Explain your reasoning. (iii) Deduce the molecular formula of V. (iv) Name all the functional groups present in V. (b) Fig. 6.2 shows two reactions involving V. (i) Identify the role of reagent T for each functional group that reacts in reaction 1. (ii) Suggest the identity of reagent U in reaction 2. (c) Both functional groups in one molecule of Y react…

(a)(i) Same structural formula but different arrangement of atoms/groups in space · (a)(ii) 8 · (a)(iii) C13H20O3 · (a)(iv) ester, ketone (carbonyl), alkene (C=C) · (b)(i) reducing agent for C=O and C=C · (b)(ii) NaBH4 (sodium borohydride) · (c)(i) 13.2 · (c)(ii) Y: orange precipitate; Q: no precipitate · (c)(iii) 2 · (c)(iv) Y: C=O peak 1670-1740 cm-1 (Q none); Q: O-H peak 3200-3650 cm-1 (Y none)

9701/21 May/June 2024 — Chemistry Solved Past Paper

Q1MediumGroup 17Chemical BondingAtoms, Molecules and StoichiometryElectrochemistryChemical Periodicity

(a)

The elements of Group 17 are called halogens.

Complete Table 1.1.

Table 1.1

halogen	colour at 293 K
chlorine
bromine
iodine

1M Similar

DifficultyEasy

Worked solution

Answer

halogen	colour at 293 K
chlorine	(pale) green
bromine	brown (red-brown)
iodine	(dark) grey (solid)

Final answer

Chlorine: green; bromine: brown; iodine: dark grey

Detailed explanation

Background Concept

The halogens are diatomic molecular elements, $\text{Cl}_2$ , $\text{Br}_2$ and $\text{I}_2$ , whose physical appearance at room temperature (293 K) is standard recall material for the 9701 syllabus. Their colours arise from the absorption of certain wavelengths of visible light by the molecules.

Understanding the Question

The command word is 'Complete Table 1.1' — a pure recall task: state the colour of each halogen at 293 K.

Approach

Recall the three standard colours in order down the group.

Step-by-Step Reasoning

Chlorine is a pale green (yellow-green) gas.
Bromine is a brown (red-brown), volatile liquid — its vapour is also brown.
Iodine is a dark grey (black-grey) crystalline solid that sublimes to give a purple vapour.

All three colours must be correct for the single mark.

Key Takeaways

Learn the halogen colours as a fixed set: green → brown → grey down the group.

Common Mistakes

Writing 'purple' for iodine: that is the colour of its vapour; the solid is dark grey.
Writing 'red' or 'orange' for bromine — the accepted colour is brown (red-brown).
Giving only two of the three colours correctly: the mark requires all three.

Things to Be Careful About

The question specifies 293 K (room temperature), so describe the physical state as it exists at that temperature — chlorine gas, bromine liquid, iodine solid.

Techniques used

recall the colours of the halogens at room temperature

(b)

State the trend in volatility of the halogens chlorine, bromine and iodine. Explain your answer.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Volatility decreases down the group from chlorine to iodine.
The halogen molecules contain more electrons going down the group.
Therefore the strength of the instantaneous dipole–induced dipole (van der Waals') forces of attraction between molecules increases, so more energy is needed to separate the molecules.

Final answer

Volatility decreases down the group because the molecules have more electrons, giving stronger instantaneous dipole–induced dipole (van der Waals) forces.

Detailed explanation

Background Concept

Volatility is the ease with which a substance evaporates. For molecular elements like the halogens, volatility depends entirely on the strength of the intermolecular forces between molecules — not on the covalent bonds within them. The relevant intermolecular force here is the instantaneous dipole–induced dipole force (London/dispersion force), which arises when momentary uneven electron distributions in one molecule induce dipoles in neighbouring molecules. The more electrons a molecule has, the more polarisable its electron cloud is, and the stronger these forces become.

Understanding the Question

The command word is 'State ... Explain your answer' — you must give the trend (1 mark) and a full causal explanation (2 marks) linking electron number to intermolecular force strength.

Approach

State the trend, then build the explanation in two logical steps: electron count → force strength → volatility.

Step-by-Step Reasoning

Trend (M1): Volatility decreases down the group: $\text{Cl}_2$ (gas) > $\text{Br}_2$ (volatile liquid) > $\text{I}_2$ (solid).
Electron count (M2): Going from $\text{Cl}_2$ to $\text{Br}_2$ to $\text{I}_2$ , each molecule contains more electrons (larger atoms, more shells).
Force strength (M3): More electrons mean a more polarisable electron cloud, so the instantaneous dipole–induced dipole forces between molecules are stronger. Stronger intermolecular attraction means more energy is needed to vaporise the substance, so volatility falls.

Key Takeaways

For halogens (and any non-polar molecular substance), volatility, melting point and boiling point all increase as molecular size/electron number increases, because van der Waals forces strengthen.

Common Mistakes

Saying 'stronger van der Waals forces' without the causal chain through electron number — M2 and M3 are separate marks.
Talking about covalent bond strength: the X–X bond is irrelevant to volatility; the question concerns forces between molecules.
Saying 'mass increases' without linking mass to electron number and polarisability.
Calling the forces 'permanent dipole' forces — the halogen molecules are non-polar.

Things to Be Careful About

Use the exact phrase 'instantaneous dipole–induced dipole forces' (or 'van der Waals' forces') — vague terms like 'intermolecular attraction' alone may not earn M3. Ensure the direction of the trend is explicit: volatility decreases down the group.

Techniques used

state the volatility trend down Group 17relate number of electrons to instantaneous dipole–induced dipole forces

(c)

Iodine is made by reacting bromine with sodium iodide.

2M

(i)

Construct an ionic equation for the reaction of bromine with sodium iodide.

1M Similar

(ii)

State the role of bromine in the reaction. Explain your answer.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Bromine is the oxidising agent, because it removes electron(s) from the iodide ions, $\text{I}^-$ (equivalently, it increases the oxidation number of iodine from −1 in $\text{I}^-$ to 0 in $\text{I}_2$ ).

Final answer

Bromine is the oxidising agent because it removes electrons from iodide ions (oxidises I^- to I2).

Detailed explanation

Background Concept

An oxidising agent is a species that oxidises another species by accepting electrons from it; the oxidising agent is itself reduced. Equivalently, it causes the oxidation number of the other species to increase.

Understanding the Question

'State the role of bromine ... Explain your answer' — name the role (1 mark) AND give the electron or oxidation-number justification; the mark requires both parts together.

Approach

Identify which species gains electrons (bromine), then justify using either electron transfer or oxidation number change.

Step-by-Step Reasoning

In the equation $\text{Br}_2 + 2\text{I}^- \rightarrow 2\text{Br}^- + \text{I}_2$ , bromine gains electrons: $\text{Br}_2 + 2\text{e}^- \rightarrow 2\text{Br}^-$ . A species that accepts electrons is an oxidising agent.
The iodide ion's oxidation number rises from −1 to 0 in $\text{I}_2$ — it has been oxidised by the bromine.

Key Takeaways

Oxidising agent = electron acceptor = causes oxidation number of the other species to increase; it is itself reduced.

Common Mistakes

Saying 'bromine is oxidised' — bromine is reduced; it causes oxidation.
Giving the role without the explanation (or vice versa) — the mark requires both.
Confusing oxidising agent with reducing agent.

Things to Be Careful About

The explanation must mention electron removal from iodide OR the increase in oxidation number of iodide — a vague 'it takes electrons' without naming from what may not score.

Techniques used

identify the oxidising agentexplain in terms of electron removal or oxidation number change

(d)

Concentrated sulfuric acid is added to separate samples containing equal amounts of NaCl, NaBr and NaI. All three samples initially react to produce the hydrogen halide.

4M

(i)

Write an equation to describe the acid–base reaction that occurs when concentrated sulfuric acid reacts with NaBr.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{HBr} + \text{NaHSO}_4

(or, alternatively)

2\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow 2\text{HBr} + \text{Na}_2\text{SO}_4

Final answer

NaBr + H2SO4 -> HBr + NaHSO4

Detailed explanation

Background Concept

Concentrated sulfuric acid is a strong acid and its first proton transfer to a solid halide salt is a straightforward acid–base (proton transfer) reaction: the halide ion acts as a Brønsted–Lowry base, accepting H⁺ to form the hydrogen halide. The acid's conjugate base is $\text{HSO}_4^-$ , giving sodium hydrogensulfate, $\text{NaHSO}_4$ , at room temperature.

Understanding the Question

'Write an equation to describe the acid–base reaction' — only the initial proton-transfer step is required, not the subsequent redox reactions that concentrated $\text{H}_2\text{SO}_4$ can also undergo with Br⁻ and I⁻.

Approach

Swap the acid's H⁺ for the metal: NaBr + H₂SO₄ → HBr + NaHSO₄. Check atom balance.

Step-by-Step Reasoning

The Br⁻ ion is protonated: $\text{Br}^- + \text{H}^+ \rightarrow \text{HBr}$ .
The remaining $\text{HSO}_4^-$ pairs with $\text{Na}^+$ to give $\text{NaHSO}_4$ .
The equation $\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{HBr} + \text{NaHSO}_4$ is balanced as written (1 Na, 1 Br, 2 H, 1 S, 4 O each side).
Alternatively, with excess/heat the full neutralisation gives $2\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow 2\text{HBr} + \text{Na}_2\text{SO}_4$ — both forms are credited.

Key Takeaways

The first reaction of any solid halide with concentrated $\text{H}_2\text{SO}_4$ is acid–base, producing the hydrogen halide and $\text{NaHSO}_4$ .

Common Mistakes

Writing $\text{Na}_2\text{SO}_4$ with only 1 NaBr (unbalanced).
Writing $\text{NaSO}_4$ (incorrect formula for sodium sulfate/hydrogensulfate).
Including the redox products ( $\text{SO}_2$ , $\text{Br}_2$ ) — the question asks only for the acid–base step.

Things to Be Careful About

Either $\text{NaHSO}_4$ (1:1) or $\text{Na}_2\text{SO}_4$ (2:1) is accepted, but the equation must be internally balanced with the salt you choose.

Techniques used

write the acid–base (proton transfer) equation for a halide salt with concentrated sulfuric acid

(ii)

Deduce which sodium halide, NaCl, NaBr or NaI, produces the largest percentage yield of hydrogen halide when concentrated sulfuric acid is added. Explain your answer by considering the relative reactivity of the halide ions as reducing agents.

identity of sodium halide:

explanation:

3M Similar

DifficultyMedium

Worked solution

Answer

Identity of sodium halide: NaCl (sodium chloride)

Explanation:

$\text{Cl}^-$ is not a strong enough reducing agent to reduce the sulfur in $\text{H}_2\text{SO}_4$ , so all the chloride is converted to HCl, which is not consumed in any further reaction.
$\text{Br}^-$ and $\text{I}^-$ are strong enough reducing agents to react further with (the S in) concentrated $\text{H}_2\text{SO}_4$ , so some of the HBr/HI produced is oxidised (to $\text{Br}_2$ / $\text{I}_2$ ), lowering the yield of hydrogen halide.

Final answer

NaCl — Cl^- is too weak a reducing agent to reduce H2SO4, whereas Br^- and I^- reduce the sulfuric acid further, consuming some of the HBr/HI formed.

Detailed explanation

Background Concept

Reducing power of the halide ions increases down Group 17: $\text{F}^- < \text{Cl}^- < \text{Br}^- < \text{I}^-$ . This is because the halide ion gets larger, so the outer electrons are further from the nucleus and more shielded, and are lost more easily. Concentrated sulfuric acid contains sulfur in the +6 oxidation state and can itself be reduced (to $\text{SO}_2$ , S or $\text{H}_2\text{S}$ ) by a sufficiently strong reducing agent. So while all three halides initially give the hydrogen halide by acid–base reaction, $\text{Br}^-$ and especially $\text{I}^-$ then reduce the acid further, and the hydrogen halide is simultaneously oxidised to the free halogen — consuming some of the HBr/HI product.

Understanding the Question

'Deduce which sodium halide produces the largest percentage yield of hydrogen halide' — you must pick one (1 mark) and explain using the relative reducing-agent strength of the halide ions (2 marks). The key insight is that yield is reduced when the hydrogen halide is destroyed in a secondary redox reaction.

Approach

Ask: for which halide does no further (redox) reaction occur after the acid–base step? That halide gives 100% conversion to hydrogen halide and hence the largest yield.

Step-by-Step Reasoning

Identity (M1): NaCl. Chloride is the weakest reducing agent of the three.
Why Cl⁻ gives full yield (M2): $\text{Cl}^-$ is not a strong enough reducing agent to reduce the S in $\text{H}_2\text{SO}_4$ , so no secondary redox reaction occurs — every mole of NaCl ends up as HCl.
Why Br⁻ and I⁻ lose yield (M3): $\text{Br}^-$ and $\text{I}^-$ are strong enough reducing agents to react further with the sulfur in $\text{H}_2\text{SO}_4$ ; e.g. $2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}$ . This consumes some of the hydrogen halide, so the percentage yield of HBr/HI is lower. (For iodide the reduction goes further, even to $\text{H}_2\text{S}$ .)

The alternative acceptable reasoning is the general trend: reducing ability increases down the group, so Cl⁻ cannot reduce the acid while Br⁻/I⁻ can.

Key Takeaways

Halide reducing power increases down Group 17 (larger ion, more shielding, electrons lost more easily).
Concentrated $\text{H}_2\text{SO}_4$ is an oxidising agent; only sufficiently strong reducing halides (Br⁻, I⁻) reduce it.
Percentage yield of hydrogen halide is highest for the halide that undergoes no further redox reaction.

Common Mistakes

Choosing NaI because 'I⁻ is the best reducing agent' — being a good reducing agent means the acid gets reduced more, destroying more HI, so the yield is lowest, not highest.
Explaining only in terms of the halide ions without linking to the yield of the hydrogen halide product.
Saying 'Cl⁻ is less reactive' without specifying 'less strong a reducing agent / cannot reduce $\text{H}_2\text{SO}_4$ '.

Things to Be Careful About

The explanation must explicitly compare: Cl⁻ not strong enough to reduce $\text{H}_2\text{SO}_4$ , while Br⁻ and I⁻ are strong enough to react further with it. Mentioning the sulfur in $\text{H}_2\text{SO}_4$ being reduced makes the redox chemistry explicit and secures the marks.

Techniques used

compare reducing power of halide ions down the groupdeduce which hydrogen halide is not further oxidised by concentrated sulfuric acid

Q2MediumChemical PeriodicityChemical BondingElectrochemistryAtoms, Molecules and StoichiometryEquilibriaNitrogen and SulfurStates of Matter

(a)

Sulfur chloride, $\text{SCl}_2$ , is a liquid at room temperature. When $\text{SCl}_2$ is added to water, misty fumes are seen and a solution is made that turns universal indicator red.

2M

(i)

Identify the type of reaction that occurs when $\text{SCl}_2$ is added to water.

1M Similar

(ii)

Name a chloride of a different Period 3 element that is also a liquid at room temperature and produces misty fumes when added to water.

1M Similar

(b)

A molecule of $\text{SCl}_2$ contains two S–Cl covalent bonds.

4M

(i)

Complete the dot-and-cross diagram in Fig. 2.1 to show the arrangement of the outer electrons in a molecule of $\text{SCl}_2$ .

Use $\times$ to show electrons from the chlorine atoms.
Use $\bullet$ to show electrons from the sulfur atom.

Fig. 2.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Central S atom with 4 non-bonding electrons (dots, •) arranged as two lone pairs.
Two Cl atoms, each with 6 non-bonding electrons (crosses, ×) arranged as three lone pairs.
One shared pair (•×) between S and each Cl in the overlapping regions.

Final answer

Dot-and-cross diagram: S has 2 lone pairs (4 dots), each Cl has 3 lone pairs (6 crosses), and each S-Cl bond has 1 dot and 1 cross.

Detailed explanation

Background Concept

A dot-and-cross diagram shows the arrangement of outer (valence) electrons in a covalent molecule. Electrons from one atom are typically shown as dots (•) and from the other as crosses (×). Shared pairs (bonds) sit in the overlap between atoms. Lone pairs (non-bonding pairs) sit outside the overlap on the respective atoms.

Sulfur is in Group 16, so it has 6 outer electrons. Chlorine is in Group 17, so it has 7 outer electrons.

Understanding the Question

We must complete the diagram for SCl₂, using × for Cl electrons and • for S electrons. The molecule has two S–Cl single bonds.

Approach

Determine outer electrons: S (6), Cl (7 each). Total = 20.
Form two S–Cl bonds: uses 4 electrons (2 from S, 2 from Cl).
Remaining electrons: S has 4 left (2 lone pairs), each Cl has 6 left (3 lone pairs).
Place them in the diagram according to the symbols specified.

Step-by-Step Reasoning

Bonding pairs: In each overlap (S–Cl region), place one • (from S) and one × (from Cl). This represents the shared pair in the covalent bond.
Lone pairs on S: S has 6 outer electrons; 2 are used in bonding. 4 remain. Place them as two lone pairs (4 •) around the central S atom, outside the overlaps.
Lone pairs on Cl: Each Cl has 7 outer electrons; 1 is used in bonding. 6 remain. Place them as three lone pairs (6 ×) around each Cl atom, outside the overlap.
Check total: 2 bonds (4 e⁻) + 2 lone pairs on S (4 e⁻) + 6 lone pairs on Cls (12 e⁻) = 20 e⁻. Correct.

Key Takeaways

Always count total valence electrons first. Ensure the correct symbol (• or ×) is used for each atom's electrons, and that lone pairs are placed on the correct atom.

Common Mistakes

Using the same symbol for both atoms' electrons (e.g., all dots).
Forgetting lone pairs on sulfur (S needs an octet, so 2 lone pairs).
Placing too many or too few electrons in the bonds or lone pairs.

Things to Be Careful About

The question explicitly states: 'Use × to show electrons from the chlorine atoms. Use • to show electrons from the sulfur atom.' Mixing these up loses marks. Also, ensure lone pairs are clearly shown as separate pairs, not just 4 or 6 dots scattered randomly.

Techniques used

draw dot-and-cross diagram for covalent molecule

(ii)

Predict the shape of, and bond angle in, a molecule of $\text{SCl}_2$ by using VSEPR theory.

shape:

bond angle:

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

shape: non-linear (or bent)
bond angle: 103–105°

Final answer

shape: non-linear; bond angle: 103–105°

Detailed explanation

Background Concept

VSEPR (Valence Shell Electron Pair Repulsion) theory states that electron pairs (bonding and lone pairs) around a central atom arrange themselves to minimise repulsion. Lone pairs repel more strongly than bonding pairs, compressing bond angles.

For a central atom with 4 electron domains (2 bonding pairs + 2 lone pairs), the basic arrangement is tetrahedral (109.5°). With 2 lone pairs, the molecular shape is 'bent' or 'non-linear', and the bond angle is less than 109.5° (typically around 104.5° for water, and 103–105° for SCl₂).

Understanding the Question

Predict the shape and bond angle of SCl₂ using VSEPR theory. S is the central atom.

Approach

Count electron domains around S: 2 bonding pairs (to Cl) + 2 lone pairs = 4 domains.
Basic geometry: tetrahedral.
Molecular shape (ignoring lone pairs for name): non-linear / bent.
Bond angle: <109.5° due to lone pair repulsion. For SCl₂, it is 103–105°.

Step-by-Step Reasoning

Sulfur has 6 outer electrons. It forms 2 single bonds with Cl atoms, using 2 electrons. 4 electrons remain as 2 lone pairs.
Total electron domains = 4 (2 bonding + 2 lone pair).
Electron geometry is tetrahedral.
Molecular shape is determined by the positions of atoms only: non-linear (or bent / angular).
Lone pairs repel bonding pairs more strongly, reducing the Cl–S–Cl bond angle from the ideal 109.5°. In SCl₂, this angle is 103–105°.

Key Takeaways

4 electron domains with 2 lone pairs → non-linear shape, bond angle <109.5°. Compare with water (H₂O) which has the same arrangement.

Common Mistakes

Saying 'tetrahedral' for the shape: tetrahedral is the electron geometry, not the molecular shape.
Giving the bond angle as exactly 109.5°: lone pairs compress the angle.
Saying 'linear': this would be for 2 domains with 0 lone pairs (e.g., BeCl₂ gas).

Things to Be Careful About

The mark scheme accepts 'non-linear' or 'bent'. For the angle, give a value in the range 103–105°. Do not just say '<109.5°' without a specific value if the mark scheme asks for a prediction (though '<109.5' is often accepted, 103-105 is precise for SCl₂). Actually, CIE often accepts <109.5, but 103-105 is the specific value for SCl₂/Water-like molecules. I will provide 103-105° as per mark scheme.

Techniques used

apply VSEPR theory to predict molecular shape and bond angle

(c)

Solid magnesium nitride, $\text{Mg}_3\text{N}_2$ , is a crystalline solid.

4M

(i)

Deduce the oxidation numbers of magnesium and nitrogen in magnesium nitride to complete Table 2.1.

Table 2.1

	oxidation number in $\text{Mg}_3\text{N}_2$
magnesium
nitrogen

1M Similar

DifficultyEasy

Worked solution

Answer

	oxidation number in Mg₃N₂
magnesium	+2 (or 2+)
nitrogen	–3 (or 3–)

Final answer

magnesium: +2; nitrogen: -3

Detailed explanation

Background Concept

Oxidation numbers in binary ionic compounds follow group trends. Group 2 metals (like Mg) always have an oxidation number of +2. Nitrogen, being in Group 15 and more electronegative than magnesium, takes an oxidation number of -3 in nitrides (N³⁻).

Understanding the Question

Deduce the oxidation numbers of Mg and N in Mg₃N₂.

Approach

Mg is Group 2 → +2.
N is Group 15, forms N³⁻ → -3.
Check: 3(+2) + 2(-3) = 0. Correct.

Step-by-Step Reasoning

Magnesium is an alkaline earth metal (Group 2), so it loses 2 electrons to form Mg²⁺. Oxidation number = +2.
Nitrogen is a non-metal (Group 15) and is more electronegative than Mg. It gains 3 electrons to form N³⁻. Oxidation number = -3.
The formula Mg₃N₂ balances the charges: 3 × (+2) + 2 × (-3) = 0.

Key Takeaways

Group 2 elements are always +2. In binary compounds with metals, Group 15 elements are -3.

Common Mistakes

Giving nitrogen as +5 or -5: nitrogen is -3 in nitrides.
Forgetting the sign: oxidation numbers must include + or -.

Things to Be Careful About

Write '+2' and '-3' (or '2+' and '3-'). The mark scheme accepts both formats.

Techniques used

deduce oxidation numbers from formula and group trends

(ii)

Magnesium nitride reacts with an excess of water to produce ammonia and magnesium hydroxide only. Construct an equation to describe this reaction.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

\text{Mg}_3\text{N}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{Mg(OH)}_2 + 2\text{NH}_3

Final answer

Mg3N2 + 6H2O -> 3Mg(OH)2 + 2NH3

Detailed explanation

Background Concept

Writing and balancing chemical equations from word descriptions requires identifying reactants and products, writing their correct formulae, and then balancing atoms on both sides.

Understanding the Question

Reactants: magnesium nitride (Mg₃N₂) and water (H₂O, excess).
Products: ammonia (NH₃) and magnesium hydroxide (Mg(OH)₂) only.

Approach

Write unbalanced equation: Mg₃N₂ + H₂O → Mg(OH)₂ + NH₃
Balance Mg: 3 on left → 3Mg(OH)₂ on right.
Balance N: 2 on left → 2NH₃ on right.
Balance O: 3 × 2 = 6 O on right → 6H₂O on left.
Check H: 6 × 2 = 12 H on left. Right: 3 × 2 (from OH) + 2 × 3 (from NH₃) = 6 + 6 = 12 H. Balanced.

Step-by-Step Reasoning

Unbalanced: $\text{Mg}_3\text{N}_2 + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2 + \text{NH}_3$
Mg: 3 on left, so need 3 Mg(OH)₂.
N: 2 on left, so need 2 NH₃.
Now right side has 3×2 = 6 O atoms and (3×2)+(2×3) = 12 H atoms.
Left side needs 6 H₂O to provide 6 O and 12 H.
Final balanced equation: $\text{Mg}_3\text{N}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{Mg(OH)}_2 + 2\text{NH}_3$
State symbols are not explicitly required by the mark scheme for this part, but if included: Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g). The mark scheme does not show state symbols in the answer line, so they are optional or ignored.

Key Takeaways

Always balance metals first, then non-metals, then H and O. Check your work by counting all atoms.

Common Mistakes

Forgetting to multiply NH₃ by 2 (gives 1 N on right instead of 2).
Balancing H before O, leading to fractional coefficients for water.
Writing MgOH instead of Mg(OH)₂.

Things to Be Careful About

Ensure the formula for magnesium hydroxide is Mg(OH)₂, not MgOH. Ammonia is NH₃, not NH₄ (that's ammonium).

Techniques used

construct and balance a chemical equation from word description

(iii)

Explain why the solution produced in the reaction in (c)(ii) has a pH greater than 7. Refer to the products of the reaction in your answer.

2M Similar

DifficultyMedium

Worked solution

Answer

The solution contains OH⁻(aq) ions (from Mg(OH)₂), making it alkaline.
Ammonia (NH₃) is a weak base that reacts with water (acid-base reaction) to produce OH⁻(aq) ions: $\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$ .
Alternatively, Mg(OH)₂ is slightly soluble and dissociates to give OH⁻(aq).
The presence of OH⁻(aq) means pH > 7.

Final answer

The products are Mg(OH)₂ (source of OH⁻) and NH₃ (weak base that reacts with water to form OH⁻), both making the solution alkaline (pH > 7).

Detailed explanation

Background Concept

A pH greater than 7 indicates an alkaline solution, which has a higher concentration of hydroxide ions (OH⁻) than hydrogen ions (H⁺). This can arise from:

A soluble or slightly soluble base (like Mg(OH)₂) dissociating to release OH⁻.
A weak base (like NH₃) reacting with water to accept a proton, leaving excess OH⁻.

Understanding the Question

Explain why the solution from (c)(ii) has pH > 7, referring to the products (Mg(OH)₂ and NH₃).

Approach

Identify the basic nature of both products and how they generate OH⁻(aq) in water.

Mg(OH)₂ is a base (metal hydroxide). Even though it is sparingly soluble, the dissolved part dissociates: Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq).
NH₃ is a weak base. In water: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq).
Both processes increase [OH⁻], so pH > 7.

Step-by-Step Reasoning

The reaction produces Mg(OH)₂ and NH₃.
Mg(OH)₂ is a metal hydroxide (base). It is slightly soluble in water and dissociates to give OH⁻(aq) ions: $\text{Mg(OH)}_2 \rightarrow \text{Mg}^{2+} + 2\text{OH}^-$ .
NH₃ is a weak base (non-metal hydride). It reacts with water in an acid-base reaction: $\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$ .
Both reactions produce OH⁻(aq) ions in solution.
An increase in [OH⁻(aq)] means the solution is alkaline, so pH > 7.

Key Takeaways

Both metal hydroxides (even slightly soluble ones) and ammonia produce OH⁻ ions in water, leading to alkaline solutions.

Common Mistakes

Saying 'Mg(OH)₂ is soluble': it is only slightly soluble/sparingly soluble, but enough to give pH > 7.
Forgetting to mention NH₃'s basicity: the question says 'refer to the products', so both must be addressed or at least the main source of OH⁻.
Saying 'NH₃ is a base' without explaining how (i.e., it produces OH⁻ with water).

Things to Be Careful About

The mark scheme gives 2 marks: M1 for OH⁻(aq) from Mg(OH)₂, M2 for either the acid-base reaction of NH₃ with water OR the dissociation of Mg(OH)₂. Ensure you mention OH⁻(aq) explicitly. State symbols are not strictly required but good practice.

Techniques used

explain pH of solution using properties of reaction products

(d)

Boron nitride is a white solid that melts above 2900 °C.

Fig. 2.2 shows part of the lattice structure of a crystal of boron nitride.

Fig. 2.2

2M

(i)

Use Fig. 2.2 to deduce the empirical formula of boron nitride.

1M Similar

(ii)

Suggest the identity of another crystalline solid that has atoms arranged in layers similar to that of solid boron nitride.

1M Similar

Q3MediumEquilibria

(a)

Define Le Chatelier’s principle.

2M Similar

DifficultyEasy

Worked solution

Answer

If a change in conditions is applied to a system at dynamic equilibrium, the equilibrium shifts in the direction that minimises (opposes) the change in conditions.

Final answer

When a change in conditions is applied to a system at equilibrium, the equilibrium shifts to minimise the change in conditions.

Detailed explanation

Background Concept

Le Chatelier's principle describes how a reversible reaction at dynamic equilibrium responds when conditions (concentration, pressure, or temperature) are changed. At equilibrium the forward and reverse rates are equal, so concentrations are constant. A change in conditions disturbs this balance; the system responds by shifting the position of equilibrium — favouring either the forward or reverse reaction — in the direction that reduces (minimises) the effect of the change.

Understanding the Question

The command word is 'define', so a precise two-part statement is required: (1) the equilibrium shifts, and (2) it shifts so as to minimise/oppose the change. Each half earns one mark.

Approach

Learn the definition in two clauses. The mark scheme explicitly credits M1 'the equilibrium shifts' and M2 'to minimise the change in conditions'. Both ideas must be present.

Step-by-Step Reasoning

M1: A change in conditions causes the equilibrium to shift — i.e. the system is no longer at the original position of equilibrium and the forward/reverse rates become temporarily unequal until a new equilibrium is reached.
M2: The direction of that shift is such that the change in conditions is minimised (opposed) — the system counteracts the imposed change as far as it can, though it cannot undo it completely.

Key Takeaways

Le Chatelier's principle is a qualitative rule, not a calculation. Always phrase it as 'shifts to minimise/oppose the change' — vague answers like 'moves to the right' without context do not define the principle.

Common Mistakes

Writing only 'the equilibrium shifts' without saying why or in what manner — this loses M2.
Saying the shift 'reverses' or 'undoes' the change completely — it only minimises it.
Confusing the principle with statements about rate (a change in conditions also changes rate, but that is not what the definition asks).

Things to Be Careful About

Use the words 'minimise' or 'oppose' exactly; examiners look for them. Do not mention specific conditions (temperature, pressure) in the definition unless asked — keep it general.

Techniques used

state Le Chatelier's principle

(b)

Reaction 1 describes the reversible reaction between yellow $\text{Fe}^{3+}(\text{aq})$ and colourless $\text{SCN}^-(\text{aq})$ to produce red $\text{FeSCN}^{2+}(\text{aq})$ .

\begin{aligned} \text{reaction 1} \quad \text{Fe}^{3+}(\text{aq}) + \text{SCN}^-(\text{aq}) &\rightleftharpoons \text{FeSCN}^{2+}(\text{aq}) \quad \Delta H = -x \text{ kJ mol}^{-1} \\ \text{yellow} \quad \text{colourless} &\quad \text{red} \end{aligned}

A mixture of $\text{Fe}^{3+}(\text{aq})$ , $\text{SCN}^-(\text{aq})$ and $\text{FeSCN}^{2+}(\text{aq})$ is at equilibrium at 20 °C.

The temperature of this mixture is then increased to 50 °C and allowed to reach equilibrium.

Deduce the changes that occur, if any, in the equilibrium mixture at 50 °C compared to the equilibrium mixture at 20 °C.

change in appearance:
change in relative concentration of $\text{FeSCN}^{2+}(\text{aq})$ :
change in value of the equilibrium constant, $K_c$ :

3M Similar

DifficultyMedium

Worked solution

Answer

Change in appearance: the mixture becomes paler red / more yellow.
Change in relative concentration of $\text{FeSCN}^{2+}(\text{aq})$ : lower.
Change in value of $K_c$ : lower.

The forward reaction is exothermic ( $\Delta H$ negative), so raising the temperature shifts the equilibrium in the endothermic (reverse) direction to minimise the temperature rise. Less $\text{FeSCN}^{2+}$ (red) is formed, so the red colour fades and the yellow of $\text{Fe}^{3+}$ becomes more apparent. Since the reverse shift lowers $[\text{FeSCN}^{2+}]$ relative to $[\text{Fe}^{3+}]$ and $[\text{SCN}^-]$ , $K_c$ decreases.

Final answer

Paler red / more yellow; [FeSCN2+] lower; Kc lower.

Detailed explanation

Background Concept

For an exothermic forward reaction ( $\Delta H$ negative), increasing the temperature adds heat to the system. By Le Chatelier's principle, the equilibrium shifts in the endothermic direction — here the reverse reaction — to absorb the added heat. Crucially, a temperature change is the only condition change that alters the value of $K_c$ : for an exothermic reaction, $K_c$ decreases as temperature rises (and increases as temperature falls). Concentration and pressure changes shift the position of equilibrium but leave $K_c$ unchanged.

Understanding the Question

Reaction 1 is exothermic in the forward direction and produces the red complex $\text{FeSCN}^{2+}$ from yellow $\text{Fe}^{3+}$ and colourless $\text{SCN}^-$ . The mixture is heated from 20 °C to 50 °C. You must deduce three things: the visible change, the change in $[\text{FeSCN}^{2+}]$ , and the change in $K_c$ .

Approach

First identify the direction of the shift using the sign of $\Delta H$ . Then translate that shift into (a) the colour observed, (b) the concentration of the red species, and (c) the effect on $K_c$ using the rule that temperature is the only factor that changes $K_c$ .

Step-by-Step Reasoning

The forward reaction is exothermic. Heating the mixture is 'adding heat', so the equilibrium shifts in the reverse (endothermic) direction to minimise the change.
Reverse shift means $\text{FeSCN}^{2+}$ decomposes back to $\text{Fe}^{3+}$ and $\text{SCN}^-$ . The concentration of the red complex falls (M2), so the red colour fades — the mixture appears paler red or more yellow (M1).
Because the equilibrium position has moved against the product, the ratio $[\text{FeSCN}^{2+}]/([\text{Fe}^{3+}][\text{SCN}^-])$ is smaller at the new equilibrium. Since $K_c$ equals exactly this ratio, $K_c$ must be lower at 50 °C than at 20 °C (M3). This is consistent with the general rule: for an exothermic reaction, $K$ decreases with increasing temperature.

Key Takeaways

Temperature increase shifts an exothermic equilibrium backwards and decreases $K_c$ ; the reverse holds for endothermic reactions.
Colour observations must be linked to species concentrations: red fades because the red complex concentration falls.
Only temperature changes the value of $K_c$ — a very common exam discriminator.

Common Mistakes

Saying the mixture becomes 'redder' — students often assume heating increases reaction rate and hence product; rate is not position of equilibrium.
Stating $K_c$ is unchanged 'because $K_c$ only depends on temperature' — the logic is right but the conclusion is backwards: temperature HAS changed, so $K_c$ changes.
Describing the colour as 'less red' without an acceptable alternative like 'paler red' or 'more yellow' — be specific.

Things to Be Careful About

Quote the sign of $\Delta H$ explicitly in your reasoning ('forward reaction is exothermic') — it anchors the whole argument. All three parts here are consequences of one shift; if you get the direction wrong, all three marks are lost, so check the sign of $\Delta H$ first.

Techniques used

apply Le Chatelier's principle to a temperature changepredict the direction of equilibrium shift for an exothermic reactiondeduce the effect of temperature on Kc

(c)

In another experiment, equimolar amounts of $\text{Fe}^{3+}(\text{aq})$ and $\text{SCN}^-(\text{aq})$ are mixed together and allowed to reach equilibrium. The total volume of the mixture is $25.0 \text{ cm}^3$ .

\text{reaction 1} \quad \text{Fe}^{3+}(\text{aq}) + \text{SCN}^-(\text{aq}) \rightleftharpoons \text{FeSCN}^{2+}(\text{aq})

At equilibrium the mixture contains:

$[\text{SCN}^-] = 1.30 \times 10^{-3} \text{ mol dm}^{-3}$
$[\text{FeSCN}^{2+}] = 0.300 \times 10^{-3} \text{ mol dm}^{-3}$

4M

(i)

Calculate the initial amount, in mol, of $\text{Fe}^{3+}(\text{aq})$ added to $\text{SCN}^-(\text{aq})$ to produce this mixture.

initial amount of $\text{Fe}^{3+}(\text{aq})$ = .............................. mol

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Since $\text{Fe}^{3+}$ and $\text{SCN}^-$ react 1:1 and equimolar amounts were mixed, at equilibrium:

[\text{Fe}^{3+}]_{\text{eqm}} = [\text{SCN}^-]_{\text{eqm}} = 1.30 \times 10^{-3} \text{ mol dm}^{-3}

Initial $[\text{SCN}^-]$ (and $[\text{Fe}^{3+}]$ ) = equilibrium value + amount converted to $\text{FeSCN}^{2+}$ :

[\text{Fe}^{3+}]_{\text{initial}} = 1.30 \times 10^{-3} + 0.300 \times 10^{-3} = 1.60 \times 10^{-3} \text{ mol dm}^{-3}

n = cV = 1.60 \times 10^{-3} \times \frac{25.0}{1000} = 4.00 \times 10^{-5} \text{ mol}

Answer

Initial amount of $\text{Fe}^{3+}(\text{aq})$ = $4.00 \times 10^{-5}$ mol

Final answer

4.00 x 10^-5 mol

Detailed explanation

Background Concept

In an equilibrium mixture, the amount of product formed equals the amount of reactants consumed (for a 1:1 stoichiometry). So the initial concentration of a reactant = its equilibrium concentration + the concentration of product formed. To convert concentration to amount, use $n = c \times V$ , with $V$ in dm³ (divide cm³ by 1000).

Understanding the Question

Equimolar $\text{Fe}^{3+}$ and $\text{SCN}^-$ are mixed in a total volume of 25.0 cm³. At equilibrium, $[\text{SCN}^-] = 1.30 \times 10^{-3}$ and $[\text{FeSCN}^{2+}] = 0.300 \times 10^{-3}$ mol dm⁻³. You must work backwards to find how much $\text{Fe}^{3+}$ was added initially.

Approach

Build an ICE-style bookkeeping argument. Because the reaction is 1:1:1 and the starting amounts of $\text{Fe}^{3+}$ and $\text{SCN}^-$ were equal, their equilibrium concentrations are equal. The amount of $\text{Fe}^{3+}$ that reacted equals the amount of $\text{FeSCN}^{2+}$ formed. Add that back to the equilibrium concentration to get the initial concentration, then multiply by the volume.

Step-by-Step Reasoning

M1: The $\text{FeSCN}^{2+}$ present at equilibrium ( $0.300 \times 10^{-3}$ mol dm⁻³) was formed from equal concentrations of $\text{Fe}^{3+}$ and $\text{SCN}^-$ . So the initial concentration of each reactant was $1.30 \times 10^{-3} + 0.300 \times 10^{-3} = 1.60 \times 10^{-3}$ mol dm⁻³.
M2: Convert to moles using the total volume of 25.0 cm³ = 0.0250 dm³:

n = 1.60 \times 10^{-3} \times 0.0250 = 4.00 \times 10^{-5} \text{ mol}

Because equimolar amounts were added, this is also the initial amount of $\text{SCN}^-$ .

Key Takeaways

For a 1:1 equilibrium with equimolar starting reactants, equilibrium concentrations of the reactants are equal.
Initial = equilibrium + amount reacted (for a reactant).
$n = cV$ with $V$ in dm³ is the bridge between concentration and moles.

Common Mistakes

Forgetting to add the $\text{FeSCN}^{2+}$ concentration back, giving $1.30 \times 10^{-3}$ mol dm⁻³ and hence $3.25 \times 10^{-5}$ mol.
Using the volume as 25.0 instead of 0.0250 dm³, giving an answer 1000× too large.
Assuming $[\text{Fe}^{3+}]$ at equilibrium is unknown and getting stuck — the equimolar condition makes it equal to $[\text{SCN}^-]$ .

Things to Be Careful About

Keep powers of ten tidy: $1.30 \times 10^{-3} + 0.300 \times 10^{-3} = 1.60 \times 10^{-3}$ , not $1.630 \times 10^{-3}$ . Give the final answer to 3 significant figures ( $4.00 \times 10^{-5}$ ) to match the data's precision.

Techniques used

use stoichiometry of the equilibrium to find initial concentrationconvert concentration and volume to moles

(ii)

Calculate $K_c$ for reaction 1 and state its units.

Show your working.

$K_c$ = ..............................

units: ..............................

2M Similar

DifficultyMedium-Easy

Worked solution

Working

At equilibrium, $[\text{Fe}^{3+}] = [\text{SCN}^-] = 1.30 \times 10^{-3}$ mol dm⁻³ and $[\text{FeSCN}^{2+}] = 0.300 \times 10^{-3}$ mol dm⁻³.

K_c = \frac{[\text{FeSCN}^{2+}]}{[\text{Fe}^{3+}][\text{SCN}^-]} = \frac{0.300 \times 10^{-3}}{(1.30 \times 10^{-3}) \times (1.30 \times 10^{-3})}

K_c = \frac{0.300 \times 10^{-3}}{1.69 \times 10^{-6}} = 178

Units: $\dfrac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})^2} = \text{mol}^{-1} \text{dm}^{3}$

Answer

$K_c = 178$ ; units = $\text{mol}^{-1} \text{ dm}^{3}$

Final answer

Kc = 178, units mol^-1 dm^3

Detailed explanation

Background Concept

For a general equilibrium $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$ , $K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$ . The units of $K_c$ follow from the powers in this expression: units = $(\text{mol dm}^{-3})^{(c+d)-(a+b)}$ . Here there is one product and two reactants, each to the power 1, so the units are $(\text{mol dm}^{-3})^{-1} = \text{mol}^{-1}\text{dm}^{3}$ .

Understanding the Question

You are given two equilibrium concentrations and must calculate $K_c$ for reaction 1, showing working, and state its units. The third equilibrium concentration, $[\text{Fe}^{3+}]$ , comes from the equimolar argument in part (i): it equals $[\text{SCN}^-] = 1.30 \times 10^{-3}$ mol dm⁻³.

Approach

Write the $K_c$ expression for reaction 1, substitute all three equilibrium concentrations, evaluate, then derive the units from the concentration powers in the expression.

Step-by-Step Reasoning

$K_c = \frac{[\text{FeSCN}^{2+}]}{[\text{Fe}^{3+}][\text{SCN}^-]}$ — one species on top, two on the bottom, all to the power 1.
Substitute: numerator $0.300 \times 10^{-3}$ ; denominator $(1.30 \times 10^{-3})^2 = 1.69 \times 10^{-6}$ .
$K_c = \frac{0.300 \times 10^{-3}}{1.69 \times 10^{-6}} = 178$ (3 s.f.).
Units: top has $(\text{mol dm}^{-3})^1$ , bottom has $(\text{mol dm}^{-3})^2$ , so units $= (\text{mol dm}^{-3})^{1-2} = \text{mol}^{-1}\text{dm}^{3}$ .

Key Takeaways

Always write the $K_c$ expression from the balanced equation before substituting.
Units of $K_c$ come from the difference in the number of concentration terms above and below the line.
Equilibrium concentrations of reactants can often be deduced from stoichiometry rather than given directly.

Common Mistakes

Using the initial concentration of $\text{Fe}^{3+}$ ( $1.60 \times 10^{-3}$ ) instead of the equilibrium concentration ( $1.30 \times 10^{-3}$ ) — $K_c$ uses only equilibrium values.
Writing the units as 'mol dm⁻³' by habit — here there is one more concentration term on the bottom, so the units are the inverse.
Arithmetic slips with powers of ten: $(1.30 \times 10^{-3})^2 = 1.69 \times 10^{-6}$ , not $1.69 \times 10^{-5}$ or $2.60 \times 10^{-3}$ .

Things to Be Careful About

Check whether the question expects $K_c$ to a specific number of significant figures — 178 (3 s.f.) matches the data. If a mark scheme shows $K_c$ without units, the units mark is still usually separate and required, so always state them.

Techniques used

write the Kc expressionsubstitute equilibrium concentrationsdetermine the units of Kc

Q4Medium-EasyChemical Energetics

(a)

Define enthalpy change of formation.

2M Similar

DifficultyEasy

Worked solution

Answer

The enthalpy change when one mole of a compound is formed from its elements in their standard states.

Final answer

The enthalpy change when one mole of a compound is formed from its elements in their standard states.

Detailed explanation

Background Concept

Enthalpy changes are measured at constant pressure. The standard enthalpy change of formation, $\Delta H_f^\ominus$ , is the enthalpy change when one mole of a compound is formed from its elements in their standard states, all substances being in their standard states at 298 K and 101 kPa.

Understanding the Question

The question asks for a definition. The mark scheme splits this into two marks: the enthalpy/energy change idea, and the condition that one mole of a compound is formed from its elements in their standard states.

Approach

State the definition directly, making sure both required parts are present: “one mole of a compound” and “from its elements in their standard states”.

Step-by-Step Reasoning

M1: The quantity is an enthalpy or energy change.
M2: It is the change when one mole of a compound (or substance) is formed from its elements, and those elements must be in their standard states.

The phrase “standard states” means the most stable form of each element at 298 K and 101 kPa, for example $\text{Fe(s)}$ , $\text{O}_2\text{(g)}$ and $\text{C(s, graphite)}$ .

Key Takeaways

A formation definition must always include three ideas: one mole of product, formation from elements, and standard states.

Common Mistakes

Missing “one mole” — the definition is per mole of compound formed.
Saying “from its elements” without “standard states” — the standard states are required for the definition.
Using “compound” only when the mark scheme also allows “substance” — either is acceptable.

Things to Be Careful About

Include the standard conditions (298 K and 101 kPa) if you mention them, but the essential wording is “one mole” and “standard states”. Do not confuse this with enthalpy change of combustion or atomisation.

Techniques used

state the definition of standard enthalpy change of formation

(b)

Iron is made when iron(III) oxide is heated with carbon monoxide, as shown by reaction 2.

\text{reaction 2} \quad \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2

Table 4.1 shows enthalpy change of formation data measured at 298 K and 101 kPa.

Table 4.1

substance	equation	value for $\Delta H_f^\ominus$ / $\text{kJ mol}^{-1}$
$\text{Fe}_2\text{O}_3$		$-824.2$
$\text{CO}$		$-110.5$
$\text{CO}_2$	$\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$	$-393.5$

4M

(i)

Complete Table 4.1 by adding equations with relevant state symbols to represent:

standard enthalpy change of formation for $\text{Fe}_2\text{O}_3$
standard enthalpy change of formation for CO.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

For $\text{Fe}_2\text{O}_3$ :

2\text{Fe(s)} + \frac{3}{2}\text{O}_2\text{(g)} \rightarrow \text{Fe}_2\text{O}_3\text{(s)}

For CO:

\text{C(s)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO(g)}

Final answer

2Fe(s) + 3/2O2(g) -> Fe2O3(s); C(s) + 1/2O2(g) -> CO(g)

Detailed explanation

Background Concept

A standard enthalpy change of formation equation always has one mole of the compound as the product, and the reactants are the elements in their standard states. Coefficients may be fractional because the product must be exactly one mole. Standard states at 298 K and 101 kPa are $\text{Fe(s)}$ , $\text{O}_2\text{(g)}$ and $\text{C(s, graphite)}$ .

Understanding the Question

Table 4.1 already gives the equation for $\text{CO}_2$ . We must add the corresponding formation equations for $\text{Fe}_2\text{O}_3$ and CO, with correct state symbols.

Approach

For each compound, write the elements in their standard states on the left, put one mole of the compound on the right, then balance by adjusting coefficients, using fractions if necessary.

Step-by-Step Reasoning

For $\text{Fe}_2\text{O}_3$ : the elements are $\text{Fe(s)}$ and $\text{O}_2\text{(g)}$ . Start with $\text{Fe(s)} + \text{O}_2\text{(g)} \rightarrow \text{Fe}_2\text{O}_3\text{(s)}$ . Balance Fe: $2\text{Fe(s)}$ . Balance O: three O atoms on the right, so $\frac{3}{2}\text{O}_2\text{(g)}$ on the left. The balanced formation equation is $2\text{Fe(s)} + \frac{3}{2}\text{O}_2\text{(g)} \rightarrow \text{Fe}_2\text{O}_3\text{(s)}$ .

For CO: elements are $\text{C(s)}$ and $\text{O}_2\text{(g)}$ . Start with $\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO(g)}$ . Balance O: one O atom on the right, so $\frac{1}{2}\text{O}_2\text{(g)}$ on the left. The equation is $\text{C(s)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO(g)}$ .

The product coefficient is 1 in both cases, as required for a formation equation.

Key Takeaways

Formation equations always have one mole of product; fractional coefficients for reactants are normal. State symbols are essential.

Common Mistakes

Writing $4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3$ : this is a valid reaction but not a formation equation because two moles of $\text{Fe}_2\text{O}_3$ are formed.
Missing state symbols, especially (s) for Fe and C and (g) for $\text{O}_2$ .
Using $\text{C(diamond)}$ instead of $\text{C(s, graphite)}$ — graphite is the standard state of carbon.

Things to Be Careful About

Use a fraction for $\text{O}_2$ when needed. The mark scheme accepts fractional coefficients. Always include state symbols.

Techniques used

write formation equations from elements in standard statesbalance equations with fractional coefficientsinclude state symbols

(ii)

Use the data in Table 4.1 to calculate the enthalpy change of reaction, $\Delta H_r$ , in $\text{kJ mol}^{-1}$ , for reaction 2.

Show your working.

$\Delta H_r$ = .............................. $\text{kJ mol}^{-1}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using $\Delta H_r = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants})$ :

\Delta H_r = [3(-393.5)] - [(-824.2) + 3(-110.5)]

= -1180.5 - (-1155.7)

= -24.8 \text{ kJ mol}^{-1}

Answer

$-24.8 \text{ kJ mol}^{-1}$

Final answer

-24.8 kJ mol^-1

Detailed explanation

Background Concept

Hess's law states that the enthalpy change of a reaction is independent of the route taken. Therefore, using standard enthalpy changes of formation,

\Delta H_r = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants})

Each term is multiplied by the stoichiometric coefficient of the substance in the balanced equation. Elements in their standard states have $\Delta H_f^\ominus = 0$ .

Understanding the Question

Reaction 2 is $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ . Table 4.1 gives formation data for $\text{Fe}_2\text{O}_3$ , CO and $\text{CO}_2$ . We need to calculate $\Delta H_r$ in $\text{kJ mol}^{-1}$ .

Approach

Identify products and reactants, multiply each formation enthalpy by its coefficient, subtract the sum for reactants from the sum for products.

Step-by-Step Reasoning

Products: 3 mol $\text{CO}_2$ , so contribution $= 3(-393.5) = -1180.5$ kJ.

Reactants: 1 mol $\text{Fe}_2\text{O}_3$ and 3 mol CO, so contribution $= (-824.2) + 3(-110.5) = -824.2 - 331.5 = -1155.7$ kJ.

$\Delta H_r = -1180.5 - (-1155.7) = -24.8$ kJ mol $^{-1}$ .

The negative sign shows the reaction is exothermic.

Key Takeaways

Use formation data with Hess's law: products minus reactants, with stoichiometric coefficients. Check signs carefully.

Common Mistakes

Forgetting to multiply by coefficients, especially 3 for CO and $\text{CO}_2$ .
Sign errors when subtracting a negative reactant sum.
Using bond enthalpies instead of formation enthalpies.
Omitting units or writing kJ instead of kJ mol $^{-1}$ .

Things to Be Careful About

The answer is per mole of reaction as written. The mark scheme allows error carried forward if the correct expression is set up with the three values and correct stoichiometry, even if arithmetic is wrong. Include the negative sign and the unit.

Techniques used

apply Hess's law using formation enthalpiesuse stoichiometric coefficientscalculate enthalpy change of reaction

Q5MediumChemical BondingHydrocarbons

Hydrocarbon molecules contain covalent bonds.

(a)

Define covalent bond.

1M Similar

DifficultyEasy

Worked solution

Answer

(electrostatic) attraction between the nuclei of two atoms and a shared pair of electrons.

Final answer

(electrostatic) attraction between the nuclei of two atoms and a shared pair of electrons.

Detailed explanation

Background Concept

A covalent bond is the fundamental bonding interaction in molecular compounds. It arises when two non-metal atoms share one or more pairs of valence electrons to achieve a stable electronic configuration (typically a full outer shell). The bonding force is the electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged shared electron pair(s) located between them.

Understanding the Question

Part (a) asks for a definition of a covalent bond. This is a standard recall question testing precise chemical terminology. The mark scheme specifically looks for the mention of 'attraction', 'nuclei' (or nucleus), and 'shared pair of electrons'.

Approach

Recall the formal definition of a covalent bond. Ensure all key components are present: the particles involved (nuclei, shared electrons) and the nature of the force (electrostatic attraction).

Step-by-Step Reasoning

Identify the particles: two atoms (specifically their nuclei) and the shared electrons.
Identify the force: electrostatic attraction.
Combine: (electrostatic) attraction between nuclei of two atoms and shared pair of electrons.

Key Takeaways

Definitions in chemistry must be precise. 'Attraction between atoms and electrons' is not enough; you must specify nuclei and shared pair.

Common Mistakes

Saying 'attraction between two atoms and electrons' (missing 'nuclei' and 'shared pair').
Saying 'sharing of electrons' (this describes the mechanism, not the bond itself, which is the attractive force).

Things to Be Careful About

Always include 'electrostatic' if possible, though 'attraction' alone is often accepted. Ensure 'nuclei' is plural if referring to two atoms.

Techniques used

define covalent bond

(b)

A C=C bond in an alkene is made from a $\sigma$ bond and a $\pi$ bond.

3M

(i)

Identify the hybridisation of the carbon atoms in a C=C bond in an alkene.

1M Similar

(ii)

Draw labelled diagrams to show, in terms of orbital overlap, how the $\sigma$ and $\pi$ bonds are made in a C=C bond.

$\sigma$ bond

$\pi$ bond

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\sigma$ bond:

$\pi$ bond:

Final answer

Diagram showing head-on overlap of sp2 orbitals for sigma bond; diagram showing sideways overlap of p orbitals for pi bond.

Detailed explanation

Background Concept

A carbon-carbon double bond ( $\text{C=C}$ ) consists of one $\sigma$ (sigma) bond and one $\pi$ (pi) bond.

$\sigma$ bond: Formed by the direct (head-on) overlap of two hybrid orbitals (e.g., $\text{sp}^2$ ) along the internuclear axis. This overlap is strong and allows free rotation.
$\pi$ bond: Formed by the sideways (lateral) overlap of two unhybridised $p$ orbitals that are parallel to each other and perpendicular to the internuclear axis. The electron density is concentrated above and below the plane of the atoms.

Understanding the Question

Part (b)(ii) asks for labelled diagrams showing how the $\sigma$ and $\pi$ bonds are formed in terms of orbital overlap in a $\text{C=C}$ bond.

Approach

Draw two separate diagrams:

For the $\sigma$ bond: Show two $\text{sp}^2$ orbitals overlapping end-to-end.
For the $\pi$ bond: Show two $p$ orbitals overlapping side-by-side.
Label the orbitals and the type of overlap.

Step-by-Step Reasoning

$\sigma$ bond diagram:

Draw two $\text{sp}^2$ hybrid orbitals (often drawn as dumbbells with one large lobe).
Position them along a horizontal axis (internuclear axis).
Show them overlapping directly head-on in the region between the nuclei.
Label as ' $\text{sp}^2$ ' or 'hybrid orbital' and indicate 'direct overlap' or 'head-on overlap'.

$\pi$ bond diagram:

Draw two unhybridised $p$ orbitals (dumbbells with two lobes of equal size).
Position them parallel to each other, perpendicular to the internuclear axis (e.g., vertical if the axis is horizontal).
Show them overlapping sideways above and below the internuclear axis.
Label as ' $p$ ' or ' $2p$ ' and indicate 'sideways overlap'.

Key Takeaways

$\sigma$ bonds are always formed by head-on overlap (s-s, s-p, p-p, or hybrid orbitals).
$\pi$ bonds are always formed by sideways overlap of parallel $p$ orbitals.
The $\sigma$ bond is stronger than the $\pi$ bond due to greater orbital overlap.

Common Mistakes

Drawing the $\pi$ bond overlap along the internuclear axis (that would be a $\sigma$ bond).
Forgetting to label the orbitals (e.g., ' $p$ ' orbitals).
Drawing the $\sigma$ bond using $p$ orbitals (while technically possible for $\text{p-p}$ $\sigma$ bonds, in alkenes it's $\text{sp}^2$ - $\text{sp}^2$ ).

Things to Be Careful About

Ensure the diagrams clearly distinguish between head-on and sideways overlap.
Labels are crucial: mark ' $p$ (orbitals)' for the $\pi$ bond diagram.

Techniques used

draw orbital overlap diagrams for sigma and pi bonds

(c)

In electrophilic reactions involving alkenes the $\pi$ bond of C=C is broken.

5M

(i)

Suggest one difference between $\sigma$ and $\pi$ bonds that explains why the $\pi$ bond is broken in electrophilic addition reactions involving alkenes.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The electrons in the $\pi$ bond are further from the nuclei (than those in the $\sigma$ bond), so the electrostatic attraction to the nuclei is weaker (making them more available to be attacked by electrophiles).

Final answer

The electrons in the pi bond are further from the nuclei, resulting in weaker attraction.

Detailed explanation

Background Concept

Electrophilic addition reactions occur at the $\text{C=C}$ double bond because the $\pi$ bond is a region of high electron density. However, not all electrons in the double bond are equally accessible. The $\sigma$ bond electrons are held tightly between the nuclei along the internuclear axis. The $\pi$ bond electrons are located in lobes above and below the plane of the molecule, further away from the direct line between the nuclei.

Understanding the Question

Part (c)(i) asks for a difference between $\sigma$ and $\pi$ bonds that explains why the $\pi$ bond is broken (rather than the $\sigma$ bond) during electrophilic addition.

Approach

Compare the position of the electrons in $\sigma$ and $\pi$ bonds relative to the nuclei. Explain how this position affects the strength of the electrostatic attraction and thus the bond strength/reactivity.

Step-by-Step Reasoning

In a $\pi$ bond, the electron density is concentrated above and below the internuclear axis, further away from the positively charged nuclei.
In a $\sigma$ bond, the electron density is concentrated directly between the nuclei, closer to them.
Coulomb's law states that electrostatic attraction decreases with distance. Therefore, the attraction between the nuclei and the $\pi$ electrons is weaker than for the $\sigma$ electrons.
Because the $\pi$ electrons are less tightly held (weaker attraction), they are more easily donated to an electrophile, leading to the breaking of the $\pi$ bond.

Key Takeaways

$\pi$ bonds are weaker and more reactive than $\sigma$ bonds partly because their electrons are further from the nuclei.
This makes $\pi$ bonds the site of attack in electrophilic addition reactions.

Common Mistakes

Saying ' $\pi$ bonds are weaker' without explaining why (the question asks for the difference that explains why it's broken).
Saying ' $\pi$ bonds have more electrons' (both have 2 electrons).
Forgetting to mention 'distance from nuclei' or 'attraction'.

Things to Be Careful About

The mark scheme accepts either 'pi electrons further away -> weaker attraction' OR 'sigma electrons closer -> stronger attraction'. Ensure the logic is clear.

Techniques used

compare sigma and pi bond electron density/attraction

(ii)

Complete Fig. 5.1 to show the mechanism for the electrophilic addition of hydrogen bromide to 2-methylpropene to produce the major organic product.

Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.

Fig. 5.1

4M Similar

DifficultyMedium

Worked solution

Answer

Step 1:

Step 2:

(See detailed mechanism below for full structure)

Overall Mechanism Description:

Curly arrow from the $\text{C=C}$ double bond to the $\text{H}$ of $\text{H-Br}$ (which has $\delta+$ on $\text{H}$ and $\delta-$ on $\text{Br}$ ).
Curly arrow from the $\text{H-Br}$ bond to the $\text{Br}$ atom.
Intermediate formed: tertiary carbocation ( $2$ -methylpropan- $2$ -yl cation, $(\text{CH}_3)_3\text{C}^+$ ) and bromide ion ( $\text{Br}^-$ with 3 lone pairs).
Curly arrow from a lone pair on $\text{Br}^-$ to the positively charged carbon ( $\text{C}^+$ ) of the carbocation.
Final product: $2$ -bromo- $2$ -methylpropane ( $\text{CH}_3\text{C(Br)(CH}_3\text{)CH}_3$ ).

Final answer

See mechanism diagram: HBr adds to 2-methylpropene via tertiary carbocation intermediate to form 2-bromo-2-methylpropane.

Detailed explanation

Background Concept

Electrophilic addition to alkenes involves the attack of an electrophile (like $\text{H}^+$ from $\text{HBr}$ ) on the electron-rich $\pi$ bond. The mechanism proceeds in two main steps:

Formation of carbocation: The $\pi$ electrons attack the electrophile, breaking the $\text{H-Br}$ bond and forming a carbocation intermediate. Markovnikov's rule dictates that the hydrogen adds to the carbon with more hydrogens, forming the more stable carbocation (tertiary > secondary > primary).
Nucleophilic attack: The nucleophile (bromide ion, $\text{Br}^-$ ) attacks the carbocation to form the final product.

Curly arrows show the movement of electron pairs. They always start from a lone pair or a bond (source of electrons) and point to the atom or bond receiving the electrons (destination). Dipoles ( $\delta+$ , $\delta-$ ) show bond polarity. Charges ( $+$ , $-$ ) show formal charges on ions.

Understanding the Question

Part (c)(ii) asks to complete the mechanism for the electrophilic addition of $\text{HBr}$ to $2$ -methylpropene ( $\text{(CH}_3\text{)}_2\text{C=CH}_2$ ) to produce the major organic product. The starting material is given in Fig 5.1. We need to add dipoles, curly arrows, the intermediate, and the final product.

Approach

Dipoles: $\text{Br}$ is more electronegative than $\text{H}$ , so $\text{H-Br}$ has $\delta+$ on $\text{H}$ and $\delta-$ on $\text{Br}$ .
Step 1 (Electrophilic attack): Draw a curly arrow from the $\text{C=C}$ double bond to the $\text{H}$ of $\text{H-Br}$ . Draw a curly arrow from the $\text{H-Br}$ bond to $\text{Br}$ .
Intermediate: Determine the carbocation. $\text{H}^+$ adds to the $\text{CH}_2$ end (Markovnikov) to form the more stable tertiary carbocation: $\text{(CH}_3\text{)}_3\text{C}^+$ . Draw $\text{Br}^-$ with lone pairs.
Step 2 (Nucleophilic attack): Draw a curly arrow from a lone pair on $\text{Br}^-$ to the $\text{C}^+$ of the carbocation.
Product: $2$ -bromo- $2$ -methylpropane.

Step-by-Step Reasoning

Initial Setup:

Reactant: $2$ -methylpropene ( $\text{H}_2\text{C=C(CH}_3\text{)}_2$ ). In Fig 5.1, it's drawn as $\text{H}_3\text{C-C(CH}_3\text{)=CH}_2$ (actually the image shows $\text{H}_3\text{C-C(CH}_3\text{)=C(H)-H}$ , which is the same).
Reagent: $\text{H-Br}$ . Add dipoles: $\text{H}^{\delta+} - \text{Br}^{\delta-}$ .

Step 1:

Arrow 1: From the center of the $\text{C=C}$ double bond to the $\text{H}$ atom of $\text{H-Br}$ . (This breaks the $\pi$ bond and forms a $\text{C-H}$ bond).
Arrow 2: From the center of the $\text{H-Br}$ bond to the $\text{Br}$ atom. (This breaks the $\text{H-Br}$ bond heterolytically, giving both electrons to $\text{Br}$ ).
Intermediate: The $\text{H}$ adds to the $\text{CH}_2$ carbon (the one with more hydrogens). The other carbon (the one with two methyl groups) becomes positively charged. This is a tertiary carbocation: $\text{(CH}_3\text{)}_3\text{C}^+$ . Draw it with a '+' charge on the central carbon. Draw $\text{Br}^-$ with 3 lone pairs (6 dots) and a '-' charge.

Step 2:

Arrow 3: From a lone pair on $\text{Br}^-$ to the positively charged carbon ( $\text{C}^+$ ) of the carbocation. (This forms the $\text{C-Br}$ bond).
Product: $\text{Br}$ is now bonded to the central carbon. The product is $2$ -bromo- $2$ -methylpropane: $\text{CH}_3\text{-C(Br)(CH}_3\text{)-CH}_3$ .

Key Takeaways

Always show dipoles on polar reagents like $\text{H-Br}$ .
Curly arrows start from electrons (bond or lone pair) and go to the atom/bond being formed.
Follow Markovnikov's rule: H adds to the carbon with more H's to form the most stable carbocation.
Show all lone pairs and charges in intermediates.

Common Mistakes

Drawing the arrow from $\text{H-Br}$ bond to $\text{H}$ (wrong direction).
Forming the wrong carbocation (primary instead of tertiary) - violates Markovnikov's rule.
Forgetting the lone pairs on $\text{Br}^-$ .
Forgetting the '+' charge on the carbocation or '-' charge on bromide.
Drawing the arrow from $\text{C}^+$ to $\text{Br}^-$ (arrows must show electron movement, so from lone pair to $\text{C}^+$ ).

Things to Be Careful About

The question asks for the major organic product. For $2$ -methylpropene + $\text{HBr}$ , the major product is $2$ -bromo- $2$ -methylpropane (tertiary bromide). The minor product would be $1$ -bromo- $2$ -methylpropane.
Ensure the structure of $2$ -methylpropene is correctly interpreted from Fig 5.1. It is $(\text{CH}_3)_2\text{C=CH}_2$ .
In the intermediate, the carbon with the '+' charge should be clearly marked.

Techniques used

draw curly arrow mechanism for electrophilic additionapply Markovnikov's rule

Q6MediumIntroduction to Organic ChemistryHydrocarbonsCarbonyl CompoundsAnalytical TechniquesHydroxy Compounds

(a)

V shows stereoisomerism.

Fig. 6.1

5M

(i)

Explain what is meant by stereoisomerism.

1M Similar

DifficultyEasy

Worked solution

Answer

Stereoisomers have the same structural formula (and same molecular formula) but a different arrangement of atoms or groups in space.

Final answer

Same structural formula but different arrangement of atoms/groups in space

Detailed explanation

Background Concept

Isomerism in organic chemistry is broadly divided into structural (constitutional) isomerism, where atoms are bonded in a different order, and stereoisomerism, where the bonding sequence is identical but the three-dimensional arrangement differs. Stereoisomerism itself has two main types: geometrical (cis/trans or E/Z) arising from restricted rotation around a C=C double bond or in a ring, and optical arising from chiral centres (carbon atoms bonded to four different groups). The defining feature that unites all stereoisomers is that the connectivity is unchanged; only the spatial orientation differs.

Understanding the Question

The command word is 'explain what is meant by', which for a definition question requires a precise, complete statement. The mark scheme awards the mark for stating that the molecules share the same structural (and molecular) formula but differ in the arrangement of atoms/groups in space.

Approach

Recall the standard textbook definition and ensure both halves are present: identical formula/connectivity AND different 3D arrangement. Omitting either half loses the mark.

Step-by-Step Reasoning

The phrase 'same structural formula' establishes that the connectivity is identical (this distinguishes stereoisomerism from structural isomerism). The phrase 'different arrangement of atoms/groups in space' establishes that the molecules differ only in 3D orientation (this captures both optical and geometrical cases). Together these constitute the full definition worth the single mark.

Key Takeaways

A complete definition of stereoisomerism must mention both the identical connectivity and the differing spatial arrangement. The general phrase 'arrangement in space' covers both cis/trans and optical isomerism.

Common Mistakes

Stating only 'different arrangement in space' without noting the same structural formula is incomplete. Confusing stereoisomerism with structural isomerism (different bonding order) is wrong. Saying 'different molecular formula' is incorrect — stereoisomers share the same molecular formula.

Things to Be Careful About

Use the word 'space' (or '3D arrangement'); a vague 'different shape' may not be credited. The mark scheme accepts 'same molecular formula' as an alternative phrasing.

Techniques used

state the definition of stereoisomerism

(ii)

Deduce the number of stereoisomers of V. Explain your reasoning.

2M Similar

DifficultyMedium

Worked solution

Answer

Number of stereoisomers = 8.

Reasoning: V has 2 chiral centres (the two ring carbons bearing the side chains) and 1 C=C double bond capable of cis/trans (E/Z) isomerism. The two chiral centres give $2^2 = 4$ optical arrangements, and the C=C doubles this to $4 \times 2 = 8$ .

Final answer

8

Detailed explanation

Background Concept

The number of stereoisomers a molecule can have is governed by its stereogenic elements. Each chiral centre (a carbon with four different groups) can exist in two configurations (R or S), so n chiral centres give up to $2^n$ optical isomers (assuming no internal symmetry creating meso forms). Independently, each C=C double bond where both carbons carry two different substituents can be either cis (Z) or trans (E), giving a factor of 2 per such bond. When both types are present, the total number of stereoisomers is $2^n \times 2$ (one factor of 2 per geometrical bond).

Understanding the Question

Compound V is a cyclopentanone ring. Looking at Fig. 6.1, two ring carbons are substituted: one bears the $-\text{CH}_2\text{COOCH}_3$ ester side chain and the other bears the $-\text{CH}_2\text{CH=CHCH}_2\text{CH}_3$ alkenyl side chain. Each of these substituted ring carbons is bonded to four different groups (H, the side chain, and two different ring paths), so both are chiral centres. The side chain also contains one C=C double bond with different groups on each carbon, giving geometrical isomerism. The task is to deduce the total number and justify it.

Approach

Count the chiral centres (2) and the geometrical double bonds (1). Apply $2^{\text{chiral}} \times 2^{\text{geometrical}} = 2^2 \times 2^1 = 8$ . The mark scheme wants M1 = 8 and M2 = the reasoning naming 2 chiral centres AND 1 C=C producing cis/trans.

Step-by-Step Reasoning

The ring carbon attached to the ester side chain: its four substituents are H, the $-\text{CH}_2\text{COOCH}_3$ group, and the two ring directions (one leading toward the C=O, the other toward the other chiral centre) which are different — hence chiral. The ring carbon attached to the alkenyl side chain is likewise chiral by the same argument. That is 2 chiral centres → $2^2 = 4$ optical combinations. The internal C=C in the side chain has H and ring-alkyl on one carbon and H and ethyl on the other, so it can be E or Z → factor of 2. Total = $4 \times 2 = 8$ . There is no symmetry that would reduce this to meso forms because the two chiral centres carry different side chains.

Key Takeaways

Combine the optical ( $2^n$ ) and geometrical (×2 per C=C) contributions multiplicatively. Always check each substituted ring carbon for chirality and each double bond for the 'two different groups on each carbon' condition.

Common Mistakes

Forgetting the C=C and answering 4. Forgetting one chiral centre and answering 4 (from 1 chiral + 1 C=C) or 2. Not recognising ring carbons as chiral centres. Assuming meso reduction when the two centres are different.

Things to Be Careful About

Both marks must be present: the correct number AND the reasoning that names both the chiral centres and the C=C. Stating '8' with no explanation scores only M1.

Techniques used

identify chiral centres in a structureidentify geometrical isomerism at a C=C bondcalculate number of stereoisomers from 2^n and a cis/trans pair

(iii)

Deduce the molecular formula of V.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\text{C}_{13}\text{H}_{20}\text{O}_3$

Final answer

C13H20O3

Detailed explanation

Background Concept

In a skeletal (line-angle) formula, carbon atoms sit at each vertex and line-end, and hydrogen atoms attached to carbon are omitted but implied so that each carbon has four bonds. Heteroatoms (O here) and hydrogens on heteroatoms are shown explicitly. The molecular formula is obtained by tallying all C, H, and O atoms.

Understanding the Question

From Fig. 6.1, V is a cyclopentanone ring (5 carbons) bearing a ketone oxygen, an ester side chain $-\text{CH}_2\text{COOCH}_3$ , and an alkenyl side chain $-\text{CH}_2\text{CH=CHCH}_2\text{CH}_3$ . We must count all atoms to give the molecular formula.

Approach

Count carbons by region, then hydrogens by satisfying tetravalency at each carbon, then oxygens by inspection.

Step-by-Step Reasoning

Carbons: ring = 5; ester side chain $-\text{CH}_2\text{COOCH}_3$ = 3; alkenyl side chain $-\text{CH}_2\text{CH=CHCH}_2\text{CH}_3$ = 5. Total C = 5 + 3 + 5 = 13. Oxygens: ketone = 1; ester has 2 (one C=O, one C-O-C) = 2. Total O = 3. Hydrogens: ring CH2 (top) = 2; two ring CH (substituted) = 1 each = 2; ring CH2 (the two remaining ring CH2 are the top one already counted and... recounting the cyclopentanone ring: C1=O, C2(H)(sidechain), C3(H2), C4(H)(sidechain), C5(H2) → ring H = 1+2+1+2 = 6; ester $\text{CH}_2$ = 2, $\text{OCH}_3$ = 3; alkenyl $\text{CH}_2$ = 2, $\text{CH=CH}$ = 2, $\text{CH}_2$ = 2, $\text{CH}_3$ = 3. Total H = 6 + 2 + 3 + 2 + 2 + 2 + 3 = 20. Hence $\text{C}_{13}\text{H}_{20}\text{O}_3$ .

Key Takeaways

Systematic region-by-region counting prevents errors. Remember implicit hydrogens on skeletal carbons and that carbonyl carbons carry no hydrogen.

Common Mistakes

Miscounting ring hydrogens (each substituted ring carbon has only one H). Forgetting the methyl hydrogen on the ester. Writing an incorrect H count like 22 or 18.

Things to Be Careful About

The double bond reduces the hydrogen count by 2 relative to the saturated analogue; ensure the ketone and ester carbons are not given hydrogens.

Techniques used

count carbon, hydrogen and oxygen atoms from a skeletal structureadd implicit hydrogens to satisfy tetravalency

(iv)

Name all the functional groups present in V.

1M Similar

(b)

Fig. 6.2 shows two reactions involving V.

Fig. 6.2

2M

(i)

Identify the role of reagent T for each functional group that reacts in reaction 1.

1M Similar

(ii)

Suggest the identity of reagent U in reaction 2.

1M Similar

(c)

Both functional groups in one molecule of Y react with an inorganic reagent to form one molecule of Q and one molecule of methanol, $\text{CH}_3\text{OH}$ , as shown in Fig. 6.3.

Fig. 6.3

7M

(i)

Part of the mass spectrum for Q is shown in Fig. 6.4. Only peaks with m/e greater than 198 are shown.

Fig. 6.4

Calculate the relative abundance, x, of the peak at m/e = 201.

Show your working.

x = ..............................

2M Similar

DifficultyMedium

Worked solution

Working

The molecular ion of Q is at m/e = 200. Q is formed from Y by hydrolysis of the ester, losing $\text{CH}_3\text{OH}$ (mass 32), so Y has $M_r = 232$ and Q has 12 fewer mass units; the number of carbon atoms in Q is $n = 12$ .

The [M+1] peak arises from $^{13}\text{C}$ , contributing about $1.1\%$ per carbon atom:

\frac{x}{100} = \frac{1.1 \times n}{100} = \frac{1.1 \times 12}{100}

x = 1.1 \times 12 = 13.2

Answer

x = 13.2

Final answer

13.2

Detailed explanation

Background Concept

In a mass spectrum the molecular ion peak (M) corresponds to the molecule containing the most abundant isotopes ( $^{12}\text{C}$ , $^1\text{H}$ , $^{16}\text{O}$ ). A small peak at M+1 comes mainly from molecules containing one $^{13}\text{C}$ atom. Because $^{13}\text{C}$ is about 1.1% abundant relative to $^{12}\text{C}$ , the relative height of the M+1 peak compared with M is approximately $1.1\% \times$ (number of carbon atoms). Thus $x = 1.1n$ when M is set to 100.

Understanding the Question

Fig. 6.4 shows M at m/e = 200 (abundance 100) and the M+1 peak at m/e = 201 with abundance x. We must calculate x. To do so we need the number of carbons in Q. Q is produced from Y by reaction of the ester with an inorganic reagent (aqueous hydroxide/acid hydrolysis) releasing $\text{CH}_3\text{OH}$ ; the mark scheme indicates Q has $n = 12$ carbons (Y has 13 carbons; loss of the methyl carbon of methanol leaves 12 in Q).

Approach

Deduce $n = 12$ for Q, then apply $x = 1.1 \times n$ .

Step-by-Step Reasoning

Y has the same carbon skeleton as V (13 C). Hydrolysis of the ester $-\text{COOCH}_3$ cleaves off the $\text{CH}_3$ of methanol, so Q retains 12 carbons (the carboxylate/acid carbon stays in Q). The M+1/M ratio = $1.1\% \times 12 = 13.2\%$ , so with M = 100, x = 13.2. (Equivalently $(100 \times x)/(1.1 \times 100) = 12$ .)

Key Takeaways

The M+1 peak height relative to M gives the carbon count via the 1.1%-per-carbon factor; conversely, knowing the carbon count lets you predict x. Track how many carbons remain in Q after losing methanol.

Common Mistakes

Using n = 13 (the carbon count of Y/V) instead of 12 for Q, giving x = 14.3. Forgetting that methanol's methyl carbon is removed from Q.

Things to Be Careful About

The 1.1% contribution is per carbon; the formula assumes the dominant M+1 source is $^{13}\text{C}$ (oxygen/hydrogen contributions are negligible here).

Techniques used

deduce number of carbons from the molecular ionapply the 1.1% per carbon rule for the [M+1] peakrearrange to solve for relative abundance x

(ii)

Q contains only hydroxyl functional groups.

Complete Table 6.1 to show the observations that occur when 2,4-dinitrophenylhydrazine (2,4-DNPH reagent) is added to separate samples of Y and Q.

Table 6.1

	observation on addition of 2,4-DNPH reagent
Y
Q

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

	observation on addition of 2,4-DNPH reagent
Y	orange precipitate
Q	no precipitate (no reaction)

Final answer

Y: orange precipitate; Q: no precipitate

Detailed explanation

Background Concept

2,4-Dinitrophenylhydrazine (2,4-DNPH, Brady's reagent) reacts with aldehydes and ketones (the C=O carbonyl) to give an orange/yellow-red precipitate of a hydrazone. It is a positive test for the presence of a carbonyl group. Compounds lacking a C=O (e.g. alcohols, carboxylic acids, esters under these conditions, diols) give no precipitate.

Understanding the Question

Y is the saturated analogue of V that still contains the ketone C=O (Fig. 6.3 shows Y with a ketone and an ester). Q, formed by hydrolysis of the ester with the inorganic reagent, contains only hydroxyl groups (per the question stem) — i.e. the ketone has also been reduced/absent in Q, leaving a diol. We must give the observation for each.

Approach

Y has a C=O → positive (orange precipitate). Q has no C=O (only OH groups) → negative (no precipitate).

Step-by-Step Reasoning

Y retains the ketone carbonyl, so 2,4-DNPH gives an orange precipitate. Q is stated to contain only hydroxyl functional groups, so there is no carbonyl to react; the solution remains clear with no precipitate.

Key Takeaways

2,4-DNPH detects carbonyls (aldehydes/ketones) with an orange precipitate; its absence of reaction indicates no carbonyl is present.

Common Mistakes

Saying Q gives an orange precipitate (Q has no C=O). Confusing 2,4-DNPH with Tollens'/Fehling's (which distinguish aldehyde vs ketone, not presence/absence of carbonyl).

Things to Be Careful About

Both observations must be correct for the single mark; 'orange precipitate' (or yellow/red) for Y and 'no precipitate / no reaction' for Q.

Techniques used

predict the 2,4-DNPH result for a compound containing a carbonylpredict the result for a compound with no carbonyl

(iii)

Under certain conditions, 0.0020 mol of Q reacts with an excess of sodium to produce a total of $44.8 \text{ cm}^3$ of gas at s.t.p.

Calculate the number of hydroxyl groups present in a molecule of Q.

Show your working.

number of hydroxyl groups = ..............................

2M Similar

DifficultyMedium

Worked solution

Working

Each hydroxyl group reacts with sodium as:

2\text{ROH} + 2\text{Na} \rightarrow 2\text{RONa} + \text{H}_2

so $1\ \text{mol OH} : \tfrac{1}{2}\ \text{mol H}_2$ .

Moles of $\text{H}_2 = \dfrac{44.8}{24000} = 0.001867\ \text{mol}$ ... using $22400\ \text{cm}^3$ at s.t.p.:

n(\text{H}_2) = \frac{44.8}{22400} = 0.0020\ \text{mol}

Moles of OH groups = $2 \times n(\text{H}_2) = 2 \times 0.0020 = 0.0040\ \text{mol}$ .

Per molecule: $\dfrac{0.0040}{0.0020} = 2$ .

Answer

number of hydroxyl groups = 2

Final answer

2

Detailed explanation

Background Concept

Sodium metal reacts with each hydroxyl (O-H) group of an alcohol to liberate hydrogen gas: $2\text{ROH} + 2\text{Na} \rightarrow 2\text{RONa} + \text{H}_2$ . Thus two moles of OH groups produce one mole of $\text{H}_2$ , i.e. each OH contributes $\tfrac{1}{2}$ mol $\text{H}_2$ . At s.t.p. one mole of gas occupies $22400\ \text{cm}^3$ (or $24000\ \text{cm}^3$ at room temperature; the question specifies s.t.p., so use 22400).

Understanding the Question

0.0020 mol of Q (a diol candidate) reacts with excess Na to give $44.8\ \text{cm}^3$ of $\text{H}_2$ at s.t.p. We must find how many OH groups each molecule of Q contains.

Approach

Convert the gas volume to moles of $\text{H}_2$ , double it to get moles of OH, then divide by moles of Q to get OH per molecule.

Step-by-Step Reasoning

$n(\text{H}_2) = 44.8/22400 = 0.0020\ \text{mol}$ . Since $\text{H}_2 : \text{OH} = 1 : 2$ , moles of OH = $2 \times 0.0020 = 0.0040\ \text{mol}$ . With 0.0020 mol of Q present, OH per molecule = $0.0040 / 0.0020 = 2$ . Hence Q is a diol with 2 hydroxyl groups, consistent with part (ii) where Q had no carbonyl.

Key Takeaways

The Na–alcohol reaction gives a clean $2\text{OH} : 1\text{H}_2$ ratio; combining it with the ideal-gas volume at s.t.p. lets you count OH groups per molecule.

Common Mistakes

Using 24000 instead of 22400 (the question says s.t.p.). Forgetting the factor of 2 between OH and $\text{H}_2$ , giving 1 OH. Mis-dividing the mole ratio.

Things to Be Careful About

State the $1\text{OH} : \tfrac{1}{2}\text{H}_2$ relationship (M1) and then use the data to reach 2 (M2); both are needed.

Techniques used

write the stoichiometry of alcohol with sodiumcalculate moles of gas from volume at s.t.p.deduce number of OH groups from mole ratio

(iv)

Use Table 6.2 to describe and explain two differences between the infrared spectrum of Y and Q in the region above $1500 \text{ cm}^{-1}$ .

Table 6.2

bond	functional groups containing the bond	characteristic infrared absorption range (in wavenumbers) / $\text{cm}^{-1}$
C–O	hydroxy, ester	1040–1300
C=C	aromatic compound, alkene	1500–1680
C=O	amide carbonyl, carboxyl ester	1640–1690 1670–1740 1710–1750
C $\equiv$ N	nitrile	2200–2250
C–H	alkane	2850–2950
N–H	amine, amide	3300–3500
O–H	carboxyl hydroxy	2500–3000 3200–3650

2M Similar

DifficultyMedium

Worked solution

Answer

Y has a strong absorption (peak/trough) in the range $1670\text{–}1740\ \text{cm}^{-1}$ due to the C=O (ketone) bond; Q has no such absorption because it has no carbonyl group.
Q has a broad absorption in the range $3200\text{–}3650\ \text{cm}^{-1}$ due to the O-H (hydroxyl) bond; Y has no such absorption because it has no hydroxyl group.

Final answer

Y: C=O peak 1670-1740 cm-1 (Q none); Q: O-H peak 3200-3650 cm-1 (Y none)

Detailed explanation

Background Concept

Infrared spectroscopy detects bond vibrations; each bond type absorbs over a characteristic wavenumber range. A C=O stretch appears as a strong peak around 1670–1740 cm⁻¹ (for ketones/carbonyls), while an O-H stretch of an alcohol appears as a broad peak around 3200–3650 cm⁻¹. Comparing two spectra reveals which functional groups are present or absent.

Understanding the Question

Y contains a ketone C=O and an ester but no hydroxyl; Q (the diol product) contains hydroxyl groups but no carbonyl. Using Table 6.2, describe and explain two differences above 1500 cm⁻¹.

Approach

For each difference, name the absorption range, the bond responsible, and which molecule has it and which lacks it, tying it to the functional-group difference.

Step-by-Step Reasoning

Difference 1: Y shows a peak at 1670–1740 cm⁻¹ (C=O of the ketone); Q does not, because Q has no carbonyl (it was reduced/lost). Difference 2: Q shows a broad peak at 3200–3650 cm⁻¹ (O-H of the hydroxyl groups); Y does not, because Y has no hydroxyl. Both the observation (range + bond) and the explanation (group present/absent) are needed for each mark.

Key Takeaways

IR differences between two compounds map directly onto their functional-group differences; quote the wavenumber range, the bond, and the reason.

Common Mistakes

Quoting the C-O range (1040–1300), which is below 1500 and outside the requested region. Saying both have C=O or both have O-H. Failing to explain why the peak is present/absent.

Things to Be Careful About

Restrict to the region above 1500 cm⁻¹ (so C-O is excluded). Use the exact ranges from Table 6.2 and link each to the correct molecule.

Techniques used

compare functional groups present in Y and Qassign characteristic IR absorptions to C=O and O-Hdescribe and explain spectral differences

Chemistry 9701/21 — May/June 2024

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

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Understanding the Question

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Understanding the Question

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Step-by-Step Reasoning

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Things to Be Careful About

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Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes