Elements in Period 3 of the Periodic Table show trends in their properties. (a) Complete Table 1.1 by identifying the lattice structures of the crystalline solids of Mg, Si and P. Table 1.1 element Mg Si P --- --- --- --- lattice structure in crystalline solid (b) The relative electrical conductivities of the Period 3 elements are shown in Fig. 1.1. Explain why there is an increase in conductivity from Na to Al and why P, S and Cl are non-conductors of electricity. (c) The…

(a) Mg: giant metallic; Si: giant covalent; P: simple molecular · (b) Na to Al have delocalised electrons that can move; P, S, Cl have no delocalised electrons. Increasing number of delocalised electrons from Na to Al increases conductivity. · (c)(i) Ar²⁺(g) → Ar³⁺(g) + e⁻ · (c)(ii) 1. Nuclear charge; 2. Distance of outer electron from nucleus (or shielding) · (c)(iii) Distance from nucleus / shielding; Na and Mg 3rd IE much greater than Al-Ar because electron removed from…

The skeletal formula of E is shown in Fig. 4.1. (a) State the molecular formula of E . (b) Name E . (c) E is made when a carboxylic acid and an alcohol react together. (i) Name the type of reaction that takes place during this esterification reaction. (ii) Write an equation for the formation of E from a carboxylic acid and an alcohol. Use structural formulae to represent the organic species. (iii) State the systematic name of the alcohol used to make E . (d) F is an isomer…

(a) C5H10O2 · (b) propyl ethanoate · (c)(i) condensation · (c)(ii) CH3COOH + CH3CH2CH2OH -> CH3COOCH2CH2CH3 + H2O · (c)(iii) propan-1-ol · (d)(i) CH3CN + HCl + 2H2O -> CH3COOH + NH4Cl · (d)(ii) Displayed formula of propan-2-ol (see diagram) · (d)(iii) Step 1: PCl3 and heat; Step 2: KCN in ethanol and heat

The skeletal formula of W is shown in Fig. 5.1. (a) W has two positional isomers. Only one shows stereoisomerism. Draw the structures of the two positional isomers of W in the boxes. (i) Describe the origin of stereoisomerism in W . (ii) State why W shows stereoisomerism but one of the positional isomers in (a) does not. (c) Draw the skeletal structure of the stereoisomer of W . (d) W reacts with reagent X to produce ethanoic acid and pentanoic acid. (i) State the role of X…

(a) hept-1-ene and hept-3-ene · (b)(i) Restricted rotation around the C=C double bond · (b)(ii) W has different groups on each C of the C=C; hept-1-ene has two H's on C1 · (c) Skeletal structure of (E)-hept-2-ene (or (Z)-hept-2-ene, opposite to W) · (d)(i) Oxidising agent · (d)(ii) hot concentrated acidified KMnO4

P , Q and R are three different hydrocarbon molecules. (i) Define empirical formula. (ii) Hydrocarbon P contains 85.7% by mass of carbon. Calculate the empirical formula of P . Show your working. empirical formula of P = .............................. (iii) Molecules of P have straight chains. Name the homologous series to which P belongs. (b) Q is a volatile hydrocarbon. $0.194\text{ g}$ of gaseous Q occupies a volume of $71.5\text{ cm}^3$ at $100\text{ }^\circ\text{C}$ and…

(a)(i) simplest whole-number ratio of atoms of each element in a compound · (a)(ii) CH2 · (a)(iii) alkenes · (b) 84.1 · (c)(i) 10 · (c)(ii) C10H18 · (c)(iii) (C4H9)+

9701/24 May/June 2025 — Chemistry Solved Past Paper

Q1MediumChemical PeriodicityStates of MatterAtomic Structure

Elements in Period 3 of the Periodic Table show trends in their properties.

(a)

Complete Table 1.1 by identifying the lattice structures of the crystalline solids of Mg, Si and P.

Table 1.1

element	Mg	Si	P
lattice structure in crystalline solid

3M Similar

DifficultyEasy

Worked solution

Answer

Element	Lattice structure
Mg	Giant metallic
Si	Giant covalent
P	Simple molecular

Final answer

Mg: giant metallic; Si: giant covalent; P: simple molecular

Detailed explanation

Background Concept

The Period 3 elements display a progression in bonding and structure across the period. On the left, the metals (Na, Mg, Al) have giant metallic lattices consisting of positive metal ions in a sea of delocalised electrons. Silicon in the middle is a metalloid with a giant covalent (diamond-like) structure where every atom is bonded to four others by strong covalent bonds. Moving right, phosphorus, sulfur, chlorine, and argon exist as simple molecular substances — discrete molecules held together by weak van der Waals forces between them.

Understanding the Question

The question asks you to identify the lattice structure of the crystalline solid form of three Period 3 elements: magnesium, silicon, and phosphorus. The command word is "complete" — you simply need to name the correct structure type for each.

Approach

Recall the position of each element in Period 3 and the type of bonding it exhibits. Metals form giant metallic lattices, metalloids like silicon form giant covalent lattices, and non-metals on the right form simple molecular lattices.

Step-by-Step Reasoning

Magnesium (Mg): A Group 2 metal. In its crystalline solid form, it consists of Mg²⁺ ions arranged in a regular lattice surrounded by a sea of delocalised electrons. This is a giant metallic structure.
Silicon (Si): A Group 14 element and metalloid. Each silicon atom forms four covalent bonds with neighbouring atoms in a tetrahedral arrangement, creating a continuous three-dimensional network identical in type to diamond. This is a giant covalent (giant molecular) structure.
Phosphorus (P): A Group 15 non-metal. In its crystalline solid form (white phosphorus), it exists as discrete P₄ molecules held together by weak intermolecular van der Waals forces. This is a simple molecular structure.

Key Takeaways

The Period 3 elements show a clear progression from metallic bonding (left) through giant covalent (middle) to simple molecular (right), reflecting the change from metallic to non-metallic character across the period.

Common Mistakes

Calling silicon's structure "giant molecular" when the mark scheme accepts this but "giant covalent" is the more precise and commonly expected term. Note: the mark scheme uses "giant molecular" for Si, which is acceptable.
Writing "simple covalent" for phosphorus instead of "simple molecular" — while sometimes accepted, "simple molecular" is the standard term.
Confusing the lattice structure with the bonding type. The question asks for the lattice structure, not the type of bond within it.

Things to Be Careful About

Use the exact terminology expected: "giant metallic", "giant covalent" (or "giant molecular"), and "simple molecular". Avoid vague terms like "covalent" or "molecular" alone.

Techniques used

identify the type of bonding in each elementclassify the lattice structure from bonding type

(b)

The relative electrical conductivities of the Period 3 elements are shown in Fig. 1.1.

Fig. 1.1

Explain why there is an increase in conductivity from Na to Al and why P, S and Cl are non-conductors of electricity.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Na to Al are metals with delocalised electrons that can move through the lattice and carry charge; P, S, and Cl are simple molecular substances with no delocalised electrons (no mobile charge carriers), so they cannot conduct.

From Na to Al, the number of delocalised electrons per atom increases, so conductivity increases.

Final answer

Na to Al have delocalised electrons that can move; P, S, Cl have no delocalised electrons. Increasing number of delocalised electrons from Na to Al increases conductivity.

Detailed explanation

Background Concept

Electrical conductivity in solids requires mobile charge carriers. In metallic lattices, the charge carriers are delocalised electrons that are free to move through the structure under an applied potential difference. In simple molecular substances, all electrons are localised in covalent bonds within molecules, and there are no free-moving charges, so these substances are electrical insulators. The number of delocalised electrons per atom in a metal depends on how many electrons are contributed to the sea — Na contributes 1, Mg contributes 2, and Al contributes 3.

Understanding the Question

The question asks you to explain two features visible in Fig. 1.1: (1) why conductivity increases from Na to Al, and (2) why P, S, and Cl are non-conductors. The command word is "explain", so you must give reasons, not just state observations.

Approach

Identify the structural difference between the metallic elements (Na, Mg, Al) and the non-metallic elements (P, S, Cl). For the increase from Na to Al, focus on the number of delocalised electrons. For the non-conductors, focus on the absence of mobile charge carriers.

Step-by-Step Reasoning

Na to Al (increase in conductivity): All three are metals with giant metallic lattices containing delocalised electrons. Sodium has 1 valence electron per atom delocalised, magnesium has 2, and aluminium has 3. More delocalised electrons means more charge carriers available to move through the lattice, hence higher conductivity. This explains the steady rise from Na to Al in Fig. 1.1.
P, S, Cl (non-conductors): These are simple molecular substances. Their electrons are all held in localised covalent bonds within discrete molecules. There are no free electrons and no mobile ions to carry charge, so they cannot conduct electricity regardless of the applied voltage.
Note that Si shows very low conductivity (it is a semiconductor), which is consistent with its giant covalent structure having no free electrons under normal conditions.

Key Takeaways

Electrical conductivity requires mobile charge carriers. Metals conduct because of delocalised electrons; the more delocalised electrons per atom, the higher the conductivity. Simple molecular substances lack mobile charges and are insulators.

Common Mistakes

Saying "metals conduct because they have free electrons" without specifying that these are delocalised electrons that can move through the lattice.
Forgetting to explain why P, S, Cl do NOT conduct (the mark scheme requires both the positive and negative statements in M1).
Saying "they don't have electrons" — all substances have electrons; the point is that the electrons are not free to move.
Not mentioning the increasing number of delocalised electrons for the Na-to-Al trend.

Things to Be Careful About

The mark scheme combines both the presence of delocalised electrons (Na-Al) and their absence (P-S-Cl) into one mark (M1), so both must appear together. The second mark (M2) specifically requires the point about increasing number of delocalised electrons.

Techniques used

relate delocalised electrons to electrical conductivitycompare metallic and molecular structures for charge carriers

(c)

The third ionisation energies of the Period 3 elements are shown in Fig. 1.2.

Fig. 1.2

7M

(i)

Write an equation, including state symbols, to represent the third ionisation energy of argon.

2M Similar

DifficultyEasy

Worked solution

Answer

\text{Ar}^{2+}(\text{g}) \rightarrow \text{Ar}^{3+}(\text{g}) + \text{e}^-

Final answer

Ar²⁺(g) → Ar³⁺(g) + e⁻

Detailed explanation

Background Concept

Ionisation energy is the energy required to remove an electron from a gaseous atom or ion. The first ionisation energy removes an electron from a neutral gaseous atom; the second removes an electron from the resulting 1+ gaseous ion; the third removes an electron from the 2+ gaseous ion. Each successive ionisation energy is larger because the remaining electrons experience greater effective nuclear attraction. State symbols (g) are essential because ionisation energy is defined for gaseous species only.

Understanding the Question

The question asks for an equation representing the third ionisation energy of argon. The command word is "write an equation, including state symbols". You must show the removal of one electron from Ar²⁺(g) to form Ar³⁺(g).

Approach

Apply the general form: Xⁿ⁺(g) → X⁽ⁿ⁺¹⁾⁺(g) + e⁻, with n = 2 for the third ionisation energy.

Step-by-Step Reasoning

The third ionisation energy involves removing the third electron, so we start with the 2+ ion: Ar²⁺(g)
After removal, we get the 3+ ion: Ar³⁺(g)
The removed particle is an electron: e⁻
All species must be gaseous: include (g) after each
The equation is: Ar²⁺(g) → Ar³⁺(g) + e⁻

Key Takeaways

The nth ionisation energy always involves removing an electron from the (n-1)+ gaseous ion. State symbols are mandatory for full marks.

Common Mistakes

Writing Ar(g) → Ar⁺(g) + e⁻ (this is the first ionisation energy, not the third).
Omitting state symbols (g) — this loses a mark.
Writing the electron as e without the minus sign.
Using (aq) instead of (g).

Things to Be Careful About

The mark scheme awards M1 for a correct ionisation equation format (Xⁿ⁺ → X⁽ⁿ⁺¹⁾⁺ + e) and M2 specifically for correct state symbols with argon. Both must be present for full marks.

Techniques used

write an ionisation energy equation with correct charge and state symbols

(ii)

The differences in the values for third ionisation energy shown in Fig. 1.2 are due to differences in the strength of attraction between the nucleus and the outer electron of each ion.

State two factors that affect the strength of attraction between the nucleus and the outer electron.

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

2M Similar

DifficultyEasy

Worked solution

Answer

Nuclear charge (number of protons)
Distance of the outer electron from the nucleus (or shielding by inner electrons)

Final answer

Nuclear charge; 2. Distance of outer electron from nucleus (or shielding)

Detailed explanation

Background Concept

The strength of attraction between the nucleus and an outer electron is governed by Coulomb's law: the force is proportional to the product of the charges and inversely proportional to the square of the distance between them. In multi-electron atoms, the effective nuclear charge felt by an outer electron is reduced by shielding from inner electrons. The two key factors are therefore (1) the nuclear charge (number of protons) and (2) the distance of the electron from the nucleus, which is influenced by the number of inner shells (shielding).

Understanding the Question

The question asks you to state two factors that affect the strength of attraction between the nucleus and the outer electron. The command word is "state", so brief answers are sufficient. The mark scheme accepts nuclear charge as one factor and distance/shielding/ionic radius as the other.

Approach

Recall the factors that determine effective nuclear attraction from the study of periodic trends in ionisation energy and atomic radius.

Step-by-Step Reasoning

Factor 1 — Nuclear charge: The more protons in the nucleus, the stronger the electrostatic attraction on the outer electron (assuming other factors are constant).
Factor 2 — Distance from nucleus / Shielding: Electrons in shells further from the nucleus experience weaker attraction because the distance is greater and because inner electrons shield the nuclear charge. The mark scheme accepts "distance of outer electron(s) from nucleus", "ionic radius", or "shielding by inner shells and/or sub-shells" as the second factor.

Key Takeaways

Effective nuclear attraction depends on nuclear charge and the balance between distance and shielding. These are the two factors that explain all periodic trends in ionisation energy.

Common Mistakes

Writing "number of electrons" as a factor — it is the nuclear charge (protons) that matters, not the total electron count.
Giving "electron-electron repulsion" as a factor when the question asks about nucleus-electron attraction specifically.
Writing only one factor and leaving the second blank.

Things to Be Careful About

The mark scheme is specific: M1 must be "nuclear charge" and M2 must be distance/shielding/ionic radius. Do not confuse these with "number of shells" alone, which is really a proxy for distance.

Techniques used

identify factors affecting electrostatic attraction in atoms

(iii)

Use Fig. 1.2 to suggest the most significant factor that determines the size of attraction between the nucleus and the outer electron. Explain your answer.

2M Similar

DifficultyMedium

Worked solution

Answer

The most significant factor is the distance of the outer electron from the nucleus (or shielding).

The third ionisation energies of Na and Mg are much greater than those of Al to Ar because the electron removed from Na⁺ and Mg²⁺ comes from the n = 2 shell (closer to the nucleus, less shielding), whereas for Al to Ar the electron is removed from the n = 3 shell.

Final answer

Distance from nucleus / shielding; Na and Mg 3rd IE much greater than Al-Ar because electron removed from n=2 shell (closer to nucleus, less shielding)

Detailed explanation

Background Concept

When examining successive ionisation energies, a large jump occurs when an electron is removed from an inner shell (closer to the nucleus, experiencing much less shielding). This is the basis for deducing the number of electrons in the outer shell and hence the group of an element. For the third ionisation energy across Period 3, the key insight is that Na and Mg have their third electron removed from the n = 2 shell (inner shell), while Al through Ar have it removed from the n = 3 shell (outer shell).

Understanding the Question

The question asks you to use Fig. 1.2 to identify the most significant factor determining the attraction between nucleus and outer electron, and to explain your choice. The graph shows a dramatic drop from Mg to Al in the third ionisation energy, which is the key feature to interpret.

Approach

Compare the electronic configurations of the ions from which the electron is being removed:

Na²⁺: 1s² 2s² 2p⁵ (electron removed from n=2)
Mg²⁺: 1s² 2s² 2p⁶ (electron removed from n=2)
Al²⁺: 1s² 2s² 2p⁶ 3s¹ (electron removed from n=3)
Si²⁺: 1s² 2s² 2p⁶ 3s² (electron removed from n=3)

The huge drop from Mg to Al indicates a change in shell, pointing to distance/shielding as the dominant factor.

Step-by-Step Reasoning

From Fig. 1.2, Na and Mg have very high third ionisation energies, while Al to Ar have much lower values.
For Na: the electron configuration is 1s² 2s² 2p⁶ 3s¹. The third electron removed comes from the 2p subshell (n = 2 shell), which is much closer to the nucleus.
For Mg: the electron configuration is 1s² 2s² 2p⁶ 3s². The third electron removed comes from the 2p subshell (n = 2 shell).
For Al: the electron configuration is 1s² 2s² 2p⁶ 3s² 3p¹. The third electron removed comes from the 3p subshell (n = 3 shell), which is further from the nucleus and experiences more shielding.
The dramatic drop from Mg to Al in the graph confirms that the distance from the nucleus (and associated shielding) is the most significant factor, not nuclear charge alone.

Key Takeaways

A large jump in successive ionisation energies indicates removal of an electron from an inner shell. The distance from the nucleus (number of shells) is the dominant factor in determining attraction strength, as evidenced by the sharp drop between Mg and Al.

Common Mistakes

Saying "nuclear charge" is the most significant factor — it increases steadily across the period but does not explain the dramatic drop from Mg to Al.
Not explaining WHY Na and Mg have high values (electron from inner shell) versus Al-Ar (electron from outer shell).
Simply restating the graph without identifying the shell change.

Things to Be Careful About

The mark scheme requires both the identification of the factor (M1) and the explanation linking it to the data (M2). The explanation must reference the shell difference between Na/Mg and Al-Ar.

Techniques used

interpret ionisation energy data to identify dominant factorcompare electronic configurations across ions

(iv)

In the third ionisation energy of argon, the ion produced has the electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^3$ .

Complete Fig. 1.3 to show the arrangement of electrons in the orbitals of this ion.

Fig. 1.3

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The completed orbital diagram for $\text{Ar}^{3+}$ ( $1s^2 2s^2 2p^6 3s^2 3p^3$ ):

The 2s box contains two opposite arrows (↑↓), the three 2p boxes each contain two opposite arrows (↑↓ ↑↓ ↑↓), the 3s box contains two opposite arrows (↑↓), and the three 3p boxes each contain one upward arrow (↑ ↑ ↑), following Hund's rule.

Final answer

2s: ↑↓; 2p: ↑↓ ↑↓ ↑↓; 3s: ↑↓; 3p: ↑ ↑ ↑ (three unpaired electrons)

Detailed explanation

Background Concept

The electrons-in-boxes (orbital) diagram represents the arrangement of electrons in individual orbitals within subshells. Each box represents one orbital and can hold a maximum of two electrons with opposite spins (↑↓). The s subshell has 1 orbital, the p subshell has 3 orbitals. Hund's rule states that when electrons occupy degenerate (equal-energy) orbitals, they fill singly with parallel spins before pairing up. The Pauli exclusion principle states that no two electrons in the same orbital can have the same spin.

Understanding the Question

The question gives the electronic configuration of the ion produced after the third ionisation of argon: $1s^2 2s^2 2p^6 3s^2 3p^3$ . Fig. 1.3 shows the orbital boxes with only the 1s box already filled. You must complete the diagram by filling in the remaining boxes according to the configuration.

Approach

Work through each subshell in order, placing the correct number of electrons in each box, applying Hund's rule for the 3p subshell which has 3 electrons in 3 orbitals.

Step-by-Step Reasoning

1s: 2 electrons → one box with ↑↓ (already shown in Fig. 1.3)
2s: 2 electrons → one box with ↑↓
2p: 6 electrons → three boxes, each with ↑↓ (all paired since 6 electrons fill 3 orbitals completely)
3s: 2 electrons → one box with ↑↓
3p: 3 electrons → three boxes, each with ↑ (one electron per orbital, all with the same spin, following Hund's rule — they do NOT pair up)

The key point is the 3p³ configuration: three electrons in three p orbitals means each orbital gets one electron with parallel spin. This is analogous to the ground-state configuration of nitrogen.

Key Takeaways

Hund's rule is essential when filling degenerate orbitals. Three electrons in three p orbitals gives three unpaired electrons with parallel spins, not one paired and one unpaired.

Common Mistakes

Pairing electrons in the 3p subshell (e.g., ↑↓ ↑ ↑ or ↑↓ ↑↓) instead of placing one in each orbital.
Drawing opposite spins for the unpaired 3p electrons (should all be the same direction, typically up).
Forgetting to fill the 2s box (it is empty in Fig. 1.3 and needs ↑↓ added).
Drawing only 2 electrons in 3p instead of 3.

Things to Be Careful About

Arrows must be clearly drawn as up or down. All three 3p electrons should have the same spin direction. The 2p boxes must all be filled with pairs (↑↓) since there are 6 electrons.

Techniques used

apply Hund's rule to fill degenerate orbitalscomplete an electrons-in-boxes diagram

(d)

Most Period 3 elements react with oxygen to form oxides.

2M

(i)

Write an equation for the reaction of phosphorus with oxygen.

1M Similar

(ii)

Write an equation for the reaction of aluminium oxide with an excess of aqueous sodium hydroxide.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

\text{Al}_2\text{O}_3 + 2\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 2\text{NaAl}(\text{OH})_4

Final answer

Al₂O₃ + 2NaOH + 3H₂O → 2NaAl(OH)₄

Detailed explanation

Background Concept

Aluminium oxide is amphoteric — it reacts with both acids and bases. With excess aqueous sodium hydroxide, it acts as an acid and forms the tetrahydroxoaluminate(III) ion, [Al(OH)₄]⁻. The sodium salt is written as NaAl(OH)₄ (sodium tetrahydroxoaluminate). Water is a reactant in this equation because the oxide ion (O²⁻) must be hydrated to form the hydroxo complex.

Understanding the Question

The command word is "write an equation". You need the balanced equation for Al₂O₃ reacting with excess aqueous NaOH. The key challenge is knowing the correct product formula and that water is needed as a reactant.

Approach

Al₂O₃ is amphoteric and dissolves in excess NaOH to form the soluble complex NaAl(OH)₄. Balance the equation by ensuring Al, Na, O, and H atoms are conserved.

Step-by-Step Reasoning

Al₂O₃ contains 2 aluminium atoms, so we need 2 NaAl(OH)₄ on the product side.
Each NaAl(OH)₄ contains 1 Na, so we need 2 NaOH.
Count oxygen: left has 3 (from Al₂O₃) + 2 (from NaOH) = 5 from those two, plus 3 from water = 8. Right has 2 × 4 = 8 from NaAl(OH)₄. ✓
Count hydrogen: left has 2 (from NaOH) + 6 (from 3H₂O) = 8. Right has 2 × 4 = 8. ✓
Balanced: Al₂O₃ + 2NaOH + 3H₂O → 2NaAl(OH)₄

Key Takeaways

Al₂O₃ is amphoteric. With excess NaOH, it forms NaAl(OH)₄ (sodium tetrahydroxoaluminate), and water must be included as a reactant. This is a common exam equation that requires memorisation of the product formula.

Common Mistakes

Writing Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O (this is the equation with molten NaOH or at high temperature, not aqueous excess NaOH).
Omitting water as a reactant.
Writing the product as Na₃AlO or Al(OH)₃ instead of NaAl(OH)₄.
Not balancing correctly.

Things to Be Careful About

The question specifies "excess aqueous sodium hydroxide", which means the product is the soluble complex NaAl(OH)₄, not a precipitate of Al(OH)₃. The ionic equation is Al₂O₃ + 2OH⁻ + 3H₂O → 2[Al(OH)₄]⁻.

Techniques used

write the equation for an amphoteric oxide reacting with excess alkali

Q2Medium-EasyChemical BondingStates of MatterElectrochemistry

(a)

Fig. 2.1 shows the covalent bonds and lone pairs of electrons in a molecule of $\text{SCl}_2$ .

Fig. 2.1

2M

(i)

Use VSEPR theory to predict the shape of a molecule of $\text{SCl}_2$ .

1M Similar

DifficultyEasy

Worked solution

Answer

non-linear (or bent / angular)

Final answer

non-linear

Detailed explanation

Background Concept

VSEPR (Valence Shell Electron Pair Repulsion) theory predicts the three-dimensional shape of molecules based on the repulsion between electron pairs (both bonding and lone pairs) in the valence shell of the central atom. Electron pairs arrange themselves as far apart as possible to minimise repulsion. Lone pairs occupy more space than bonding pairs, so lone pair–lone pair repulsion > lone pair–bonding pair repulsion > bonding pair–bonding pair repulsion.

Understanding the Question

The question asks for the molecular shape of $\text{SCl}_2$ using VSEPR theory. Fig. 2.1 shows the central sulfur atom bonded to two chlorine atoms via single covalent bonds, with two lone pairs of electrons on the sulfur atom.

Approach

Count the total number of electron domains (bonding pairs + lone pairs) around the central sulfur atom to determine the electron geometry, then ignore the lone pairs to name the molecular shape.

Step-by-Step Reasoning

The central sulfur atom has 6 valence electrons.
It forms 2 single covalent bonds with two chlorine atoms, using 2 electrons.
The remaining 4 electrons form 2 lone pairs on the sulfur atom.
Total electron domains around S = 2 bonding pairs + 2 lone pairs = 4 domains.
Four electron domains arrange themselves in a tetrahedral electron geometry.
Because two of these domains are lone pairs, the molecular shape (considering only the positions of the atoms) is non-linear (also accepted as bent or angular).

Key Takeaways

When a central atom has 4 electron domains with 2 lone pairs and 2 bonding pairs, the electron geometry is tetrahedral but the molecular shape is non-linear (bent).

Common Mistakes

Stating the shape is "tetrahedral". Tetrahedral is the electron geometry; the question asks for the molecular shape, which is non-linear.
Stating "trigonal planar" or "linear" without correctly counting the lone pairs.

Things to Be Careful About

Use precise terminology: "non-linear", "bent", or "angular" are all acceptable for the molecular shape. "Tetrahedral" alone is incorrect for the molecular shape.

Techniques used

apply VSEPR theory to predict molecular shapecount bonding and lone pairs around the central atom

(ii)

Predict the bond angle in a molecule of $\text{SCl}_2$ .

.................... °

1M Similar

DifficultyEasy

Worked solution

Answer

104.5

Final answer

104.5

Detailed explanation

Background Concept

In a perfect tetrahedral arrangement of four electron domains, the bond angles are $109.5^\circ$ . However, lone pairs repel bonding pairs more strongly than bonding pairs repel each other. This extra repulsion compresses the bonding pair–bonding pair angle to less than $109.5^\circ$ . For a molecule with two lone pairs and two bonding pairs (an $\text{AX}_2\text{E}_2$ system, analogous to water $\text{H}_2\text{O}$ ), the bond angle is typically compressed to approximately $104.5^\circ$ .

Understanding the Question

Predict the bond angle in the $\text{SCl}_2$ molecule. The central sulfur has two bonding pairs and two lone pairs.

Approach

Use the $\text{AX}_2\text{E}_2$ VSEPR model. The electron geometry is tetrahedral ( $109.5^\circ$ ), but the two lone pairs on sulfur compress the $\text{Cl}-\text{S}-\text{Cl}$ bond angle. The standard expected value for this configuration is $104.5^\circ$ .

Step-by-Step Reasoning

Electron domains around S: 4 (tetrahedral arrangement, ideal angle $109.5^\circ$ ).
Lone pair–lone pair and lone pair–bonding pair repulsions are stronger than bonding pair–bonding pair repulsions.
This compresses the $\text{Cl}-\text{S}-\text{Cl}$ bond angle below $109.5^\circ$ .
The accepted value for a molecule with this electron configuration (similar to water) is $104.5^\circ$ .

Key Takeaways

Molecules with 4 electron domains and 2 lone pairs have bond angles less than $109.5^\circ$ , typically around $104.5^\circ$ .

Common Mistakes

Writing $109.5^\circ$ : this ignores the compressive effect of the lone pairs.
Writing $120^\circ$ : this would be for a trigonal planar or trigonal pyramidal geometry, which is incorrect here.

Things to Be Careful About

Ensure the numerical value is correct ( $104.5$ ). The degree symbol ( $^\circ$ ) is already provided in the question prompt, so only the number is required in the answer line.

Techniques used

predict bond angle using VSEPR theoryaccount for lone pair repulsion on bond angles

(b)

$\text{SCl}_2$ is a red liquid that reacts with water.

2\text{SCl}_2(\text{l}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_2\text{SO}_3(\text{aq}) + \text{S}(\text{s}) + 4\text{HCl}(\text{aq})

5M

(i)

Describe two observations, other than temperature change, that are made when an excess of water is added to $\text{SCl}_2$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

the (red) liquid/solution becomes colourless
a yellow solid appears

Final answer

red liquid to colourless; yellow solid appears

Detailed explanation

Background Concept

When observing a chemical reaction, physical observations include colour changes, state changes (formation of a precipitate or gas), and temperature changes. The balanced chemical equation provides the identities and physical states of the reactants and products, which allow us to predict these observations.

Understanding the Question

The reaction is $2\text{SCl}_2(\text{l}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_2\text{SO}_3(\text{aq}) + \text{S}(\text{s}) + 4\text{HCl}(\text{aq})$ . We are told $\text{SCl}_2$ is a red liquid. We need to describe two observations other than temperature change when excess water is added.

Approach

Compare the initial physical states and colours with the final products. $\text{SCl}_2$ is a red liquid; water is colourless. The products are aqueous $\text{H}_2\text{SO}_3$ , solid sulfur $\text{S}$ , and aqueous $\text{HCl}$ . Aqueous solutions of these products are colourless, and elemental sulfur is a yellow solid.

Step-by-Step Reasoning

Initial state: $\text{SCl}_2$ is a red liquid mixed with colourless water.
Final state: The products $\text{H}_2\text{SO}_3(\text{aq})$ and $\text{HCl}(\text{aq})$ are dissolved in water, forming a colourless aqueous solution. Thus, the red liquid/solution becomes colourless.
Solid sulfur $\text{S}(\text{s})$ is produced. Elemental sulfur is a yellow solid. Thus, a yellow solid (precipitate) appears.
These two observations (colour change from red to colourless, and appearance of a yellow solid) are the expected visual changes.

Key Takeaways

Always check the physical states and known colours of reactants and products in a balanced equation to predict observable changes in a reaction.

Common Mistakes

Describing the disappearance of the red liquid without mentioning the resulting colourless solution.
Not specifying the colour of the solid (sulfur is yellow, not just "a solid appears").
Mentioning temperature change, which the question explicitly excludes.

Things to Be Careful About

The question asks for observations "other than temperature change". Do not include exothermic/endothermic observations.
Ensure both observations are distinct: one relates to the liquid/solution (red to colourless), the other to the solid (yellow solid appears).

Techniques used

predict physical observations from reaction products and statesidentify colour changes and precipitate formation

(ii)

When $\text{SCl}_2$ reacts with water, the $\text{SCl}_2$ is broken down and a disproportionation reaction occurs.

State the oxidation numbers of sulfur in $\text{SCl}_2$ and $\text{H}_2\text{SO}_3$ .

oxidation number of S in $\text{SCl}_2$ ...........................................................................................

oxidation number of S in $\text{H}_2\text{SO}_3$ ...........................................................................................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

oxidation number of S in $\text{SCl}_2$ : +2
oxidation number of S in $\text{H}_2\text{SO}_3$ : +4

Final answer

+2 in SCl2; +4 in H2SO3

Detailed explanation

Background Concept

Oxidation numbers are assigned using a set of rules: fluorine is always -1, oxygen is usually -2 (except in peroxides), hydrogen is +1 (except in metal hydrides), and the sum of oxidation numbers in a neutral molecule is zero. In $\text{SCl}_2$ , chlorine is more electronegative than sulfur, so chlorine takes the -1 oxidation state.

Understanding the Question

State the oxidation numbers of sulfur in $\text{SCl}_2$ and in $\text{H}_2\text{SO}_3$ . The reaction is a disproportionation, meaning sulfur is both oxidised and reduced.

Approach

Apply standard oxidation number rules to each compound. For $\text{SCl}_2$ , Cl is -1. For $\text{H}_2\text{SO}_3$ , H is +1 and O is -2.

Step-by-Step Reasoning

In $\text{SCl}_2$ : Let the oxidation number of S be $x$ . Chlorine is more electronegative, so each Cl is -1. The molecule is neutral: $x + 2(-1) = 0 \Rightarrow x = +2$ . Oxidation number of S is +2.
In $\text{H}_2\text{SO}_3$ : Let the oxidation number of S be $y$ . Hydrogen is +1, oxygen is -2. The molecule is neutral: $2(+1) + y + 3(-2) = 0 \Rightarrow 2 + y - 6 = 0 \Rightarrow y = +4$ . Oxidation number of S is +4.
(Note: In the products, elemental sulfur $\text{S}(\text{s})$ has an oxidation number of 0, confirming disproportionation: S(+2) is oxidised to S(+4) and reduced to S(0)).

Key Takeaways

Oxidation numbers are calculated using standard rules: H = +1, O = -2, halogens = -1 (when bonded to less electronegative elements). The sum must equal the overall charge of the species.

Common Mistakes

Assigning chlorine an oxidation number other than -1 in $\text{SCl}_2$ .
Forgetting that hydrogen is +1 in $\text{H}_2\text{SO}_3$ (it is not a hydride here).
Calculating the sum incorrectly (e.g., $2 + y - 6 = 0 \Rightarrow y = 4$ , not -4).

Things to Be Careful About

Use the correct sign convention: write "+2" or "+4", not "2-" or "4-". Roman numerals (+II, +IV) are also acceptable but must have the correct sign.

Techniques used

calculate oxidation number of an element in a compoundapply oxidation number rules for halogens and oxygen

(iii)

State the general name to describe the type of reaction that occurs when a substance is broken down by water.

1M Similar

DifficultyEasy

Worked solution

Answer

hydrolysis

Final answer

hydrolysis

Detailed explanation

Background Concept

When a chemical compound is broken down by reaction with water, the process is called hydrolysis. This typically involves the cleavage of covalent bonds in the molecule, with the $\text{H}^+$ and $\text{OH}^-$ components of water adding to the fragments. Common examples include the hydrolysis of esters, halogenoalkanes, and covalent non-metal halides like $\text{SCl}_2$ or $\text{SiCl}_4$ .

Understanding the Question

The question asks for the general name of the reaction type where a substance is broken down by water.

Approach

Recall the standard terminology for reactions involving water as a reactant that breaks chemical bonds.

Step-by-Step Reasoning

The reaction is $2\text{SCl}_2(\text{l}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_2\text{SO}_3(\text{aq}) + \text{S}(\text{s}) + 4\text{HCl}(\text{aq})$ .
Water is reacting with $\text{SCl}_2$ to break it down into new products.
The general term for a reaction in which a substance is decomposed by water is hydrolysis.

Key Takeaways

Reactions where water breaks down a compound are classified as hydrolysis reactions.

Common Mistakes

Writing "hydration": hydration is the addition of water molecules to form a hydrated compound without breaking the main covalent framework (e.g., $\text{CuSO}_4(\text{s}) + 5\text{H}_2\text{O}(\text{l}) \rightarrow \text{CuSO}_4 \cdot 5\text{H}_2\text{O}(\text{s})$ ).
Writing "dissolution" or "dissolving": these refer to the physical process of a substance mixing with water, not a chemical reaction breaking bonds.

Things to Be Careful About

Use the precise term "hydrolysis". Ensure correct spelling: h-y-d-r-o-l-y-s-i-s.

Techniques used

identify reaction type from descriptionrecognise hydrolysis as bond cleavage by water

Q3MediumAtoms, Molecules and StoichiometryEquilibriaChemical BondingGroup 17

$\text{H}_2(\text{g})$ and $\text{I}_2(\text{g})$ react to form $\text{HI}(\text{g})$ in a reversible reaction, as shown in equation 1.

\text{equation 1} \quad \text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g})

In three separate experiments, a student combines different amounts of two or more of the gases from equation 1. In each experiment, the gases are left to reach equilibrium at a given temperature.

(a)

In experiment 1, $\text{H}_2(\text{g})$ and $\text{I}_2(\text{g})$ are combined.

Water is then added to the equilibrium mixture to produce $1.00\text{ dm}^3$ of solution A.

The amount of $\text{I}_2(\text{aq})$ present in $1.00\text{ dm}^3$ of solution A is found by titration with $\text{Na}_2\text{S}_2\text{O}_3(\text{aq})$ .

Exactly $32.90\text{ cm}^3$ of $0.200\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3$ reacts with all of the $\text{I}_2$ in a $25.0\text{ cm}^3$ sample of solution A.

\text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq})

3M

(i)

Calculate the amount, in mol, of $\text{S}_2\text{O}_3^{2-}$ that reacts in the titration.

1M Similar

DifficultyEasy

Worked solution

Working

Convert the titre volume to $\text{dm}^3$ :

V = 32.90 \text{ cm}^3 = \frac{32.90}{1000} \text{ dm}^3 = 0.03290 \text{ dm}^3

Amount of thiosulfate:

n(\text{S}_2\text{O}_3^{2-}) = cV = 0.200 \times 0.03290 = 6.58 \times 10^{-3} \text{ mol}

Answer

$0.00658 \text{ mol}$

Final answer

0.00658 mol

Detailed explanation

Background Concept

In a titration, the amount of a reagent that reacts is found from its concentration and the volume used: $n = cV$ , where $V$ must be in $\text{dm}^3$ when $c$ is in $\text{mol dm}^{-3}$ . Because $1\text{ dm}^3 = 1000\text{ cm}^3$ , a volume in $\text{cm}^3$ is converted to $\text{dm}^3$ by dividing by $1000$ .

Understanding the Question

The question gives the titre volume $32.90\text{ cm}^3$ and the concentration $0.200\text{ mol dm}^{-3}$ of $\text{Na}_2\text{S}_2\text{O}_3$ . It asks only for the amount of thiosulfate reacting. The balanced equation with iodine is not needed for this part; it is needed in the next part.

Approach

Apply $n = cV$ directly, first converting the titre from $\text{cm}^3$ to $\text{dm}^3$ .

Step-by-Step Reasoning

Convert $32.90\text{ cm}^3$ to $\text{dm}^3$ : $32.90/1000 = 0.03290\text{ dm}^3$ .
Substitute into $n = cV$ : $n = 0.200 \times 0.03290 = 0.00658\text{ mol}$ .
The answer is $6.58 \times 10^{-3}\text{ mol}$ , or $0.00658\text{ mol}$ .

Key Takeaways

Always express volume in $\text{dm}^3$ when using $c$ in $\text{mol dm}^{-3}$ . The product $cV$ gives amount in mol directly.

Common Mistakes

Leaving the volume in $\text{cm}^3$ and multiplying, which gives an answer 1000 times too large. Also misreading the value as $65.8$ instead of $0.00658$ .

Things to Be Careful About

The mark scheme accepts $0.00658(0)\text{ mol}$ . Use consistent significant figures; here three are appropriate.

Techniques used

convert volume from cm3 to dm3apply n = cV

(ii)

Use your answer to (a)(i) to calculate the amount, in mol, of $\text{I}_2$ present in $1.00\text{ dm}^3$ of solution A.

(If you were unable to obtain an answer in (a)(i), then use $9.42 \times 10^{-3}\text{ mol}$ . This is not the correct answer.)

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From the equation, $1\text{ mol}$ of $\text{I}_2$ reacts with $2\text{ mol}$ of $\text{S}_2\text{O}_3^{2-}$ :

n(\text{I}_2 \text{ in }25.0\text{ cm}^3) = \frac{0.00658}{2} = 0.00329\text{ mol}

The $25.0\text{ cm}^3$ sample is a fraction of $1.00\text{ dm}^3$ :

n(\text{I}_2 \text{ in }1.00\text{ dm}^3) = 0.00329 \times \frac{1000}{25.0} = 0.132\text{ mol}

Answer

$0.132\text{ mol}$

Final answer

0.132 mol

Detailed explanation

Background Concept

The titration reaction is $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$ . The ratio shows that twice the amount of thiosulfate is needed as iodine. An aliquot is a portion of the whole solution; any amount in the aliquot is proportional to the amount in the whole solution by volume.

Understanding the Question

We already know the amount of thiosulfate from part (i) is $0.00658\text{ mol}$ . That thiosulfate reacted with all the iodine in only $25.0\text{ cm}^3$ of solution A. We need the total iodine in the full $1.00\text{ dm}^3$ of solution A.

Approach

First use the mole ratio to find iodine in the $25.0\text{ cm}^3$ sample, then scale up by the ratio of total volume to sample volume.

Step-by-Step Reasoning

From $\text{I}_2 : \text{S}_2\text{O}_3^{2-} = 1:2$ , moles of iodine in the sample $= 0.00658/2 = 0.00329\text{ mol}$ .
$1.00\text{ dm}^3 = 1000\text{ cm}^3$ , so the sample is $25.0/1000$ of the total solution.
Total iodine $= 0.00329 \times 1000/25.0 = 0.00329 \times 40 = 0.1316 \approx 0.132\text{ mol}$ .

If the alternative fallback value $9.42 \times 10^{-3}\text{ mol}$ were used, the corresponding answers would be $0.00471\text{ mol}$ in the sample and $0.1884\text{ mol}$ in $1.00\text{ dm}^3$ .

Key Takeaways

Balanced equations give the mole ratio for stoichiometry. Dilution or sampling means the amount scales with volume.

Common Mistakes

Using the ratio the wrong way, i.e. multiplying $0.00658$ by $2$ , or forgetting to scale up from the aliquot to the whole $1.00\text{ dm}^3$ .

Things to Be Careful About

The mark scheme awards M1 for the half of thiosulfate and M2 for the volume scaling. Keep your working visible so the method marks can be awarded even if the final value slips.

Techniques used

use the stoichiometric ratio from the balanced equationscale an aliquot amount to the full solution volume

(b)

In experiment 2, $0.100\text{ mol}$ of $\text{H}_2(\text{g})$ , $0.200\text{ mol}$ of $\text{I}_2(\text{g})$ and $0.300\text{ mol}$ of $\text{HI}(\text{g})$ are combined.

At equilibrium $0.154\text{ mol}$ of $\text{I}_2$ is present.

Calculate the amounts, in mol, of $\text{H}_2$ and $\text{HI}$ in the equilibrium mixture produced in experiment 2.

amount of $\text{H}_2$ in equilibrium mixture = .............................. mol

amount of $\text{HI}$ in equilibrium mixture = .............................. mol

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Equation: $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$ .

The amount of $\text{I}_2$ that reacts is:

0.200 - 0.154 = 0.046\text{ mol}

So $0.046\text{ mol}$ of $\text{H}_2$ also reacts, and $2 \times 0.046 = 0.092\text{ mol}$ of $\text{HI}$ is formed.

n(\text{H}_2)_{\text{eq}} = 0.100 - 0.046 = 0.054\text{ mol}

n(\text{HI})_{\text{eq}} = 0.300 + 0.092 = 0.392\text{ mol}

Answer

$\text{H}_2 = 0.054\text{ mol}$ ; $\text{HI} = 0.392\text{ mol}$

Final answer

0.054 mol; 0.392 mol

Detailed explanation

Background Concept

At equilibrium, the amounts of reactants and products are constant, but reaction has occurred from the starting amounts. For the reaction $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$ , any amount $x$ of $\text{H}_2$ that is consumed is accompanied by the same amount $x$ of $\text{I}_2$ consumed and the production of $2x$ of $\text{HI}$ .

Understanding the Question

Initial amounts are $0.100\text{ mol }\text{H}_2$ , $0.200\text{ mol }\text{I}_2$ and $0.300\text{ mol }\text{HI}$ . At equilibrium the amount of $\text{I}_2$ is $0.154\text{ mol}$ , which is less than the initial $0.200\text{ mol}$ , so the forward reaction has taken place. We need the equilibrium amounts of $\text{H}_2$ and $\text{HI}$ .

Approach

Subtract the equilibrium iodine amount from its initial amount to find the change $x$ . Then apply the same change to $\text{H}_2$ and twice the change to $\text{HI}$ .

Step-by-Step Reasoning

Change in $\text{I}_2$ : $0.200 - 0.154 = 0.046\text{ mol}$ consumed.
$\text{H}_2$ is consumed in the same mole ratio $1:1$ , so $n(\text{H}_2) = 0.100 - 0.046 = 0.054\text{ mol}$ .
$\text{HI}$ is produced with coefficient $2$ : $n(\text{HI}) = 0.300 + 2 \times 0.046 = 0.300 + 0.092 = 0.392\text{ mol}$ .

Key Takeaways

The reaction stoichiometry ties all equilibrium amounts together. The change in one species fixes the change in the others.

Common Mistakes

Forgetting that $\text{HI}$ increases by twice the decrease in $\text{I}_2$ ; or adding to $\text{H}_2$ instead of subtracting.

Things to Be Careful About

State symbols are not needed in the calculation. The mark scheme uses error carried forward, so if you misread one change but use it correctly afterwards you can still gain later marks.

Techniques used

deduce the extent of reaction from an equilibrium amountapply stoichiometric changes to calculate other equilibrium amounts

(c)

In experiment 3, $0.0772\text{ mol}$ of $\text{H}_2$ and $0.0986\text{ mol}$ of $\text{I}_2$ are present in an equilibrium mixture at $298\text{ K}$ .

\text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g}) \quad K_c = 7.94 \times 10^2

Use the equilibrium constant, $K_c$ , to calculate the amount, in mol, of $\text{HI}$ present in the equilibrium mixture in experiment 3.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

For a fixed volume $V$ , the volumes cancel in the $K_c$ expression:

K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{n_{\text{HI}}^2}{n_{\text{H}_2}n_{\text{I}_2}}

n_{\text{HI}}^2 = 7.94 \times 10^2 \times 0.0772 \times 0.0986 = 6.0438...

n_{\text{HI}} = \sqrt{6.0438...} = 2.46\text{ mol}

Answer

$2.46\text{ mol}$

Final answer

2.46 mol

Detailed explanation

Background Concept

For $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$ , the equilibrium constant is $K_c = [\text{HI}]^2/([\text{H}_2][\text{I}_2])$ . Concentrations are amounts divided by volume. If all gases share the same volume $V$ , the $V$ terms cancel, leaving $K_c = n_{\text{HI}}^2/(n_{\text{H}_2}n_{\text{I}_2})$ . This is an approximation that works for a constant volume, as here.

Understanding the Question

At equilibrium there are $0.0772\text{ mol}$ of $\text{H}_2$ and $0.0986\text{ mol}$ of $\text{I}_2$ ; $K_c = 7.94 \times 10^2$ . We need the amount of $\text{HI}$ .

Approach

Set the simplified amount expression equal to the given $K_c$ , substitute the known amounts, and solve for $n_{\text{HI}}$ .

Step-by-Step Reasoning

Write $K_c = n_{\text{HI}}^2/(n_{\text{H}_2}n_{\text{I}_2})$ .
Substitute: $7.94 \times 10^2 = n_{\text{HI}}^2/(0.0772 \times 0.0986)$ .
Multiply both sides: $n_{\text{HI}}^2 = 7.94 \times 10^2 \times 0.0772 \times 0.0986 = 6.0438$ .
Square root: $n_{\text{HI}} = \sqrt{6.0438} = 2.458 \approx 2.46\text{ mol}$ .

Key Takeaways

When all species are in the same container, the volume cancels out of the $K_c$ expression, so amounts can be used directly. Check the power of each concentration when writing the expression.

Common Mistakes

Forgetting the coefficient 2 as a square on $[\text{HI}]$ , or taking the square root of the wrong side.

Things to Be Careful About

The answer is $2.46\text{ mol}$ ; the mark scheme quotes $2.46$ . Round appropriately and include mol.

Techniques used

write the Kc expressionsubstitute equilibrium amounts into Kcsolve for the unknown amount using a square root

(d)

The value of $K_c$ for the dissociation of $\text{HCl}$ at $298\text{ K}$ is $5.50 \times 10^{-34}$ .

2\text{HCl}(\text{g}) \rightleftharpoons \text{H}_2(\text{g}) + \text{Cl}_2(\text{g}) \quad K_c = 5.50 \times 10^{-34}

4M

(i)

Define covalent bond.

1M Similar

(ii)

Describe and explain the relative thermal stabilities of the hydrogen halides $\text{HCl}$ and $\text{HI}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\text{HCl}$ is more thermally stable than $\text{HI}$ .

The Cl atom is smaller than the I atom, so the H–Cl bond is shorter and the shared electron pair is closer to the nucleus. There is also less shielding in HCl, so the attraction between the nucleus and the shared pair is stronger than in HI. More energy is therefore needed to break the H–Cl bond.

Thus $\text{HCl}$ decomposes only at higher temperatures than $\text{HI}$ .

Final answer

HCl is more thermally stable than HI.

Detailed explanation

Background Concept

Thermal stability of a hydrogen halide means how readily it remains undecomposed when heated; decomposition breaks the H–X bond. Down Group 17 the halogen atom becomes larger and has more electron shells. A longer bond and greater shielding weaken the attraction between the nucleus and the shared electron pair, lowering the bond dissociation enthalpy. Thus H–F is most stable and H–I is least stable.

Understanding the Question

Compare only HCl and HI for relative thermal stability. The explanation must use atomic size and bond strength, not electronegativity alone. The command is explain, so each observation needs a reason.

Approach

State which is more stable, then give two linked reasons: (1) Cl is smaller than I, giving a shorter bond and a closer electron pair; (2) there is less shielding in HCl, giving a stronger nucleus–shared-pair attraction.

Step-by-Step Reasoning

Stability is determined by the strength of the H–X bond.
Cl has fewer electron shells than I, so the Cl atom is smaller.
The H–Cl bond is shorter than H–I, so the shared electron pair is closer to the Cl nucleus.
There is less shielding by inner electrons in HCl.
Therefore the nucleus attracts the shared pair more strongly in HCl; more energy is needed to break H–Cl than H–I.
So HCl decomposes at a higher temperature and is more thermally stable.

Key Takeaways

The trend down Group 17: larger atom, longer and weaker bond, lower thermal stability. Use core concepts of atomic radius, shielding, and bond energy.

Common Mistakes

Citing only "HCl is more stable" without a reason, or giving electronegativity as the sole explanation. Also saying H–I is more stable because I is bigger is wrong.

Things to Be Careful About

The mark scheme awards one mark for HCl more stable plus a correct size/bond-length reason, and one for stronger attraction/less shielding/greater orbital overlap. Include both ideas.

Techniques used

compare atomic size and shielding down Group 17relate bond length and strength to thermal stability

(iii)

Use the data given in (c) and (d) to suggest a value for the equilibrium constant for the dissociation of $\text{HBr}(\text{g})$ at $298\text{ K}$ .

1M Similar

DifficultyMedium

Worked solution

Working

For the reverse of the reaction in (c):

K_c(2\text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2) = \frac{1}{7.94 \times 10^2} = 1.26 \times 10^{-3}

The dissociation of $\text{HBr}$ is intermediate in extent between $\text{HCl}$ and $\text{HI}$ , so its $K_c$ lies between:

5.50 \times 10^{-34} < K_c < 1.26 \times 10^{-3}

Answer

Any value in this range, e.g. $1.00 \times 10^{-20}$ .

Final answer

1.00 × 10^-20 (any value between 5.50×10^-34 and 1.26×10^-3)

Detailed explanation

Background Concept

For a reversible reaction, $K_c$ for the reverse reaction is the reciprocal of $K_c$ for the forward reaction. A small $K_c$ means very little product at equilibrium; for dissociation of a hydrogen halide, a smaller $K_c$ means the hydrogen halide is more stable. The hydrogen halides become less stable down the group, so the order of dissociation constants is $K_c(\text{HCl}) < K_c(\text{HBr}) < K_c(\text{HI})$ .

Understanding the Question

Part (c) gives $K_c = 7.94 \times 10^2$ for $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$ . Part (d) gives $K_c = 5.50 \times 10^{-34}$ for $2\text{HCl} \rightleftharpoons \text{H}_2 + \text{Cl}_2$ . We are to suggest a $K_c$ for the dissociation of $\text{HBr}$ , which lies between HCl and HI in stability.

Approach

Convert the (c) constant to the reverse reaction, $2\text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2$ , by taking the reciprocal. Then, using the trend that HBr is intermediate between HCl and HI, choose a value between the two dissociation constants.

Step-by-Step Reasoning

The reaction in (c) written in the forward direction is the reverse of HI dissociation. Therefore $K_c(2\text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2) = 1/(7.94 \times 10^2) = 1.26 \times 10^{-3}$ .
The given $K_c$ for $\text{HCl}$ dissociation is $5.50 \times 10^{-34}$ , extremely small because HCl is very stable.
HBr is less stable than HCl but more stable than HI, so its dissociation constant should be larger than that of HCl but smaller than that of HI.
Hence $5.50 \times 10^{-34} < K_c(\text{HBr}) < 1.26 \times 10^{-3}$ . Any value in this range, e.g. $1.00 \times 10^{-20}$ , is acceptable.

Key Takeaways

The relationship between forward and reverse equilibrium constants is reciprocal. Trends in bond stability translate into trends in $K_c$ : a more stable compound means a smaller dissociation constant.

Common Mistakes

Using $7.94 \times 10^2$ directly as the HI dissociation constant instead of taking its reciprocal, or suggesting a value outside the allowed range.

Things to Be Careful About

The mark scheme accepts any value strictly between the two limits. Do not include the endpoints.

Techniques used

take the reciprocal of Kc for the reverse reactionapply the trend in hydrogen halide stability to estimate an intermediate Kc

Q4MediumIntroduction to Organic ChemistryCarboxylic Acids and DerivativesHydroxy CompoundsNitrogen CompoundsHalogen Compounds

The skeletal formula of E is shown in Fig. 4.1.

Fig. 4.1

(a)

State the molecular formula of E.

1M Similar

DifficultyEasy

Worked solution

Answer

C $_{5}$ H $_{10}$ O $_{2}$

Final answer

C5H10O2

Detailed explanation

Background Concept

A skeletal formula represents organic molecules where carbon atoms are implied at the vertices and ends of lines, and hydrogen atoms attached to carbons are omitted. Heteroatoms (like O) and hydrogens attached to heteroatoms are shown explicitly. To find the molecular formula, one must count all carbon, hydrogen, and oxygen atoms implied by the structure.

Understanding the Question

The question asks for the molecular formula of compound E, whose skeletal formula is given in Fig. 4.1. The image shows a carbonyl group (C=O) bonded to a methyl group on one side and an oxygen atom on the other, which is bonded to a three-carbon chain (propyl group). This is an ester.

Approach

Count the carbon atoms at each vertex and end of the chain. Count the oxygen atoms explicitly shown. Calculate the number of hydrogen atoms by ensuring each carbon has four bonds (octet rule for carbon).

Step-by-Step Reasoning

Count Carbons: The left side has a methyl group (1 C) and a carbonyl carbon (1 C). The right side has a propyl chain (3 C). Total C = 1 + 1 + 3 = 5.
Count Oxygens: There is a C=O oxygen and an -O- oxygen. Total O = 2.
Count Hydrogens:
- Left methyl group (CH $_3$ -): 3 H.
- Carbonyl carbon (C=O): 0 H (4 bonds used: 2 to O, 1 to C, 1 to O).
- Oxygen atom: 0 H.
- Propyl chain (-CH $_2$ -CH $_2$ -CH $_3$ ): 2 + 2 + 3 = 7 H.
- Total H = 3 + 7 = 10.
Assemble Formula: C $_5$ H $_{10}$ O $_2$ .

Key Takeaways

Skeletal formulas hide C-H bonds. Always verify carbon valency (4 bonds) to determine hidden hydrogens.

Common Mistakes

Forgetting to count hydrogens on the terminal methyl group or miscounting the propyl chain length.

Things to Be Careful About

Ensure the formula is empirical or molecular as requested. Here, molecular formula is required, so C $_5$ H $_{10}$ O $_2$ is correct (not simplified to C $_{2.5}$ H $_5$ O $_1$ ).

Techniques used

count atoms in skeletal formula

(b)

Name E.

1M Similar

(c)

E is made when a carboxylic acid and an alcohol react together.

3M

(i)

Name the type of reaction that takes place during this esterification reaction.

1M Similar

(ii)

Write an equation for the formation of E from a carboxylic acid and an alcohol. Use structural formulae to represent the organic species.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3 + \text{H}_2\text{O}$

Final answer

CH3COOH + CH3CH2CH2OH -> CH3COOCH2CH2CH3 + H2O

Detailed explanation

Background Concept

Esterification involves the carboxyl group (-COOH) of a carboxylic acid reacting with the hydroxyl group (-OH) of an alcohol. The -OH from the acid and the -H from the alcohol combine to form water. The remaining parts join to form the ester linkage (-COO-).

Understanding the Question

Write an equation for the formation of E (propyl ethanoate) from a carboxylic acid and an alcohol. Structural formulae must be used for organic species.

Approach

Identify the carboxylic acid (ethanoic acid) and alcohol (propan-1-ol) that form propyl ethanoate. Write the balanced equation.

Step-by-Step Reasoning

Identify reactants: To get propyl ethanoate (CH $_3$ COOCH $_2$ CH $_2$ CH $_3$ ), we need ethanoic acid (CH $_3$ COOH) and propan-1-ol (CH $_3$ CH $_2$ CH $_2$ OH).
Write equation: CH $_3$ COOH + CH $_3$ CH $_2$ CH $_2$ OH $ightarrow$ CH $_3$ COOCH $_2$ CH $_2$ CH $_3$ + H $_2$ O.
Check structural formulae: CH $_3$ COOH is structural for ethanoic acid. CH $_3$ CH $_2$ CH $_2$ OH is structural for propan-1-ol. The ester is written as CH $_3$ COOCH $_2$ CH $_2$ CH $_3$ .

Key Takeaways

Esterification equation: RCOOH + R'OH $ightarrow$ RCOOR' + H $_2$ O. Ensure structural formulae are used, not molecular.

Common Mistakes

Using molecular formulae (C $_2$ H $_4$ O $_2$ ) instead of structural. Forgetting water as a product. Incorrectly linking the propyl group (e.g., writing CH $_3$ COOCH(CH $_3$ ) $_2$ which is isopropyl ethanoate).

Things to Be Careful About

The question asks for 'structural formulae'. CH $_3$ COOH is acceptable. Displayed formulae are not required unless specified. Balance the equation (it is already balanced 1:1:1:1).

Techniques used

write balanced esterification equation with structural formulae

(iii)

State the systematic name of the alcohol used to make E.

1M Similar

(d)

F is an isomer of E.

F reacts with $\text{HCl}(\text{aq})$ to produce G and H.

G is a secondary alcohol.

H is also produced when $\text{HCl}(\text{aq})$ is added to $\text{CH}_3\text{CN}$ .

6M

(i)

Construct an equation for the acid hydrolysis of $\text{CH}_3\text{CN}$ with $\text{HCl}(\text{aq})$ .

1M Similar

DifficultyMedium

Worked solution

Answer

$\text{CH}_3\text{CN} + \text{HCl} + 2\text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + \text{NH}_4\text{Cl}$

Final answer

CH3CN + HCl + 2H2O -> CH3COOH + NH4Cl

Detailed explanation

Background Concept

Nitriles (R-CN) undergo acid hydrolysis to form carboxylic acids (R-COOH) and ammonium salts (NH $_4^+$ X $^-$ ). The reaction requires water and an acid catalyst (like HCl). The nitrile carbon is hydrolyzed to a carboxyl group, and the nitrogen becomes ammonia, which immediately reacts with the acid to form an ammonium salt.

Understanding the Question

Construct an equation for the acid hydrolysis of CH $_3$ CN with HCl(aq). This produces H (identified later as ethanoic acid).

Approach

Write the hydrolysis reaction. R-CN + 2H $_2$ O $ightarrow$ R-COOH + NH $_3$ . In acidic conditions (HCl), NH $_3$ + HCl $ightarrow$ NH $_4$ Cl. Combine these.

Step-by-Step Reasoning

Hydrolysis: CH $_3$ CN + 2H $_2$ O $ightarrow$ CH $_3$ COOH + NH $_3$ .
Neutralization: The NH $_3$ produced is basic and reacts with HCl: NH $_3$ + HCl $ightarrow$ NH $_4$ Cl.
Overall equation: CH $_3$ CN + 2H $_2$ O + HCl $ightarrow$ CH $_3$ COOH + NH $_4$ Cl.
Check atoms:
- Left: C=2, H=3+4+1=8, N=1, O=2, Cl=1.
- Right: C=2, H=4+4=8, N=1, O=2, Cl=1.
- Balanced.

Key Takeaways

Acid hydrolysis of nitriles yields carboxylic acid and ammonium salt (not free ammonia). Requires 2 water molecules.

Common Mistakes

Writing NH $_3$ as a product instead of NH $_4$ Cl. Forgetting the HCl in the reactants. Incorrect number of water molecules (need 2 for full hydrolysis to acid, not amide).

Things to Be Careful About

The mark scheme specifically requires NH $_4$ Cl. If written as NH $_3$ + HCl separately, it might not score depending on strictness, but usually combined is best. Ensure state symbols are not required (mark scheme doesn't show them, but good practice is (aq) or (l)).

Techniques used

balance nitrile hydrolysis equation

(ii)

Draw the displayed formula of G.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The displayed formula is propan-2-ol:

      H   H   H
      |   |   |
  H - C - C - C - H
      |   |   |
      H   O   H
          |
          H

(Note: The central carbon is bonded to H, OH, CH $_3$ , and CH $_3$ ).

Final answer

Displayed formula of propan-2-ol (see diagram)

Detailed explanation

Background Concept

G is a secondary alcohol produced from the acid hydrolysis of F. F is an isomer of E (propyl ethanoate, C $_5$ H $_{10}$ O $_2$ ). Acid hydrolysis of an ester RCOOR' gives RCOOH and R'OH. We established H is ethanoic acid (CH $_3$ COOH) from part (d)(i). Therefore, F must be an ester of ethanoic acid with a 3-carbon alcohol part. Since G is a secondary alcohol, the alcohol part cannot be propan-1-ol (primary). It must be propan-2-ol (isopropanol). Thus, F is isopropyl ethanoate (propan-2-yl ethanoate).

Understanding the Question

Draw the displayed formula of G, which is a secondary alcohol. Based on deduction, G is propan-2-ol.

Approach

Deduce that G is propan-2-ol. Draw its displayed formula showing all atoms and bonds.

Step-by-Step Reasoning

Identify G: Hydrolysis of ester F (isomer of propyl ethanoate) gives acid H (ethanoic acid) and alcohol G (secondary). The only 3-carbon secondary alcohol is propan-2-ol.
Structure: CH $_3$ -CH(OH)-CH $_3$ .
Displayed formula: Show all C-C, C-H, C-O, O-H bonds.
- Central C bonded to: H (top), OH (bottom), CH $_3$ (left), CH $_3$ (right).
- Left C bonded to 3 H.
- Right C bonded to 3 H.
- O bonded to H.

Key Takeaways

Secondary alcohols have the OH group on a carbon bonded to two other carbons. Propan-2-ol is the only secondary C $_3$ alcohol.

Common Mistakes

Drawing propan-1-ol (primary). Missing any hydrogen atoms in the displayed formula. Incorrect connectivity.

Things to Be Careful About

'Displayed formula' means showing all bonds. Skeletal or condensed formulae will not score. Ensure the OH is on the middle carbon.

Techniques used

draw displayed formula of secondary alcohol

(iii)

$\text{CH}_3\text{OH}$ is used to make $\text{CH}_3\text{CN}$ in a two-step process, as shown in Fig. 4.2.

Fig. 4.2

State the reagents and conditions required for steps 1 and 2.

step 1 ................................................................................................................................

step 2 ................................................................................................................................

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

step 1: PCl $_3$ and heat (or HCl(g) / PCl $_5$ / SOCl $_2$ )
step 2: KCN dissolved in ethanol, and heat

Final answer

Step 1: PCl3 and heat; Step 2: KCN in ethanol and heat

Detailed explanation

Background Concept

Step 1: Conversion of alcohol (CH $_3$ OH) to halogenoalkane (CH $_3$ Cl). Reagents for this include phosphorus(III) chloride (PCl $_3$ ), phosphorus(V) chloride (PCl $_5$ ), thionyl chloride (SOCl $_2$ ), or concentrated HCl with a catalyst (though for methanol, HCl gas or ZnCl $_2$ catalyst is needed, but PCl $_3$ /PCl $_5$ are standard). For methanol specifically, PCl $_3$ or PCl $_5$ or SOCl $_2$ are common. NaCl/KCl with conc. H $_2$ SO $_4$ can also work but is less standard for primary alcohols in some syllabuses, though marked acceptable here.

Step 2: Conversion of halogenoalkane (CH $_3$ Cl) to nitrile (CH $_3$ CN). This is a nucleophilic substitution reaction using potassium cyanide (KCN) or sodium cyanide (NaCN). The solvent is typically ethanol (to dissolve both organic halide and ionic cyanide), and heat is required.

Understanding the Question

Fig 4.2 shows CH $_3$ OH $ightarrow$ CH $_3$ Cl $ightarrow$ CH $_3$ CN. State reagents and conditions for steps 1 and 2.

Approach

Identify the reaction type for each step and recall standard reagents/conditions from the syllabus.

Step-by-Step Reasoning

Step 1: Alcohol to chloroalkane.

Reagent: PCl $_3$ (phosphorus(III) chloride) or PCl $_5$ or SOCl $_2$ or HCl(g).
Condition: Heat (or reflux).
Mark scheme accepts: PCl $_3$ and heat; OR HCl(g); OR PCl $_5$ ; OR SOCl $_2$ ; OR NaCl/KCl and conc. H $_2$ SO $_4$ .

Step 2: Chloroalkane to nitrile.

Reagent: KCN (potassium cyanide) or NaCN.
Solvent/Condition: Dissolved in ethanol (aqueous ethanol is often used, but 'dissolved in ethanol' is key to avoid elimination/substitution with OH $^-$ ). Heat (reflux).
Mark scheme requires: reagent KCN AND conditions dissolved in ethanol AND heat.

Key Takeaways

Alcohol $ightarrow$ RCl: PCl $_3$ /PCl $_5$ /SOCl $_2$ + heat.
RCl $ightarrow$ RCN: KCN in ethanol + heat.

Common Mistakes

Using aqueous KCN (leads to alcohol product via hydrolysis instead of nitrile). Forgetting 'heat' or 'ethanol' for step 2. Using PCl $_5$ for step 1 is fine, but condition 'heat' is often needed.

Things to Be Careful About

For step 2, the solvent is crucial. KCN(aq) gives alcohol. KCN(ethanol) gives nitrile. Must specify 'ethanol'. Heat is also required.

Techniques used

identify reagents for alcohol to halide conversionidentify reagents for halide to nitrile conversion

Q5Medium-EasyIntroduction to Organic ChemistryHydrocarbons

The skeletal formula of W is shown in Fig. 5.1.

Fig. 5.1

(a)

W has two positional isomers. Only one shows stereoisomerism.

Draw the structures of the two positional isomers of W in the boxes.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Box 1:

Box 2:

Final answer

hept-1-ene and hept-3-ene

Detailed explanation

Background Concept

Structural isomerism occurs when compounds have the same molecular formula but different structural arrangements of atoms. For alkenes, positional isomerism arises when the carbon-carbon double bond is located at different positions along the carbon chain. A skeletal formula represents the carbon backbone as a zigzag line, where each vertex and endpoint is a carbon atom, and hydrogen atoms attached to carbons are implied. Double bonds are shown as two parallel lines.

Understanding the Question

Compound W is given as hept-2-ene, a seven-carbon alkene with the double bond between C2 and C3. The question asks for the two positional isomers of W to be drawn in skeletal form. Positional isomers of hept-2-ene must have the same molecular formula ( $C_7H_{14}$ ) but the double bond at a different position. Moving the double bond along the chain gives hept-1-ene (double bond at C1-C2) and hept-3-ene (double bond at C3-C4). Note that hept-4-ene is identical to hept-3-ene when numbered from the other end, and hept-5-ene is identical to hept-2-ene.

Approach

Identify the molecular formula and carbon chain length of W (7 carbons, one double bond).
Shift the double bond to all unique positions: C1-C2 (hept-1-ene) and C3-C4 (hept-3-ene).
Draw the skeletal structures for hept-1-ene and hept-3-ene, ensuring the double bond is clearly represented.

Step-by-Step Reasoning

Hept-1-ene: The double bond is at the end of the chain. The skeletal structure is a 7-carbon chain (6 line segments) where the first segment is a double line. This is drawn as a zigzag starting with a double bond at the terminal end.
Hept-3-ene: The double bond is in the middle of the chain, between the third and fourth carbons. The skeletal structure is a 7-carbon chain where the third segment is a double line. The chain can be drawn in a standard zigzag conformation.

Key Takeaways

Positional isomers of alkenes differ only in the location of the double bond along the carbon chain.
When numbering the chain, always start from the end that gives the double bond the lowest possible number, so hept-4-ene is incorrectly named and is actually hept-3-ene.

Common Mistakes

Drawing isomers with a different carbon chain length (e.g., branching), which would be chain isomers, not positional isomers.
Forgetting to draw the double bond explicitly in the skeletal formula.
Numbering the chain incorrectly and drawing hept-4-ene instead of hept-3-ene.

Things to Be Careful About

Ensure the skeletal formula has exactly 7 carbon atoms (7 vertices/ends).
The double bond must be clearly drawn as two parallel lines.
Do not add hydrogens to the skeletal formula; they are implied.

Techniques used

draw positional isomers of an alkeneinterpret skeletal formulae

(b)

2M

(i)

Describe the origin of stereoisomerism in W.

1M Similar

(ii)

State why W shows stereoisomerism but one of the positional isomers in (a) does not.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

W has two different groups bonded to each carbon atom of the C=C double bond, whereas hept-1-ene has two identical hydrogen atoms bonded to one of the carbon atoms of the double bond.

Final answer

W has different groups on each C of the C=C; hept-1-ene has two H's on C1

Detailed explanation

Background Concept

For an alkene to exhibit cis-trans stereoisomerism, each carbon atom of the double bond must be bonded to two different groups. If either carbon of the C=C bond has two identical groups (usually two hydrogen atoms), rotation (if it were possible) or spatial arrangement does not create a distinct isomer because the molecule is superimposable on its mirror image or alternative orientation.

Understanding the Question

The question asks why W (hept-2-ene) shows stereoisomerism, but one of its positional isomers (hept-1-ene) does not. We need to compare the groups attached to the double-bonded carbons in both molecules.

Approach

Analyze the groups on the C=C carbons in W (hept-2-ene).
Analyze the groups on the C=C carbons in hept-1-ene.
Show that W meets the condition (different groups on each C) while hept-1-ene fails it (two H's on C1).

Step-by-Step Reasoning

In W (hept-2-ene), C2 is bonded to a methyl group ( $CH_3$ ) and a hydrogen atom ( $H$ ). C3 is bonded to a butyl group ( $C_4H_9$ ) and a hydrogen atom ( $H$ ). Since both carbons have two different groups, stereoisomerism is possible.
In hept-1-ene, C1 is bonded to two hydrogen atoms ( $H$ ) and C2 is bonded to a pentyl group and a hydrogen. Because C1 has two identical groups (two H's), no stereoisomerism exists.

Key Takeaways

The condition for cis-trans isomerism in alkenes is that each carbon of the double bond must have two different substituents.
Terminal alkenes (like hept-1-ene) generally do not show cis-trans isomerism because the terminal carbon always has two hydrogen atoms.

Common Mistakes

Simply stating "hept-1-ene has two hydrogens" without specifying that they are on the same carbon of the double bond.
Failing to mention that W has different groups on each carbon.

Things to Be Careful About

Be precise: say "two identical groups (e.g., two hydrogen atoms) bonded to one carbon of the double bond".
Explicitly name the isomer that lacks stereoisomerism (hept-1-ene) or refer to it as "the other positional isomer".

Techniques used

compare structural features of stereoisomers and non-stereoisomers

(c)

Draw the skeletal structure of the stereoisomer of W.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Final answer

Skeletal structure of (E)-hept-2-ene (or (Z)-hept-2-ene, opposite to W)

Detailed explanation

Background Concept

Geometric (cis-trans or E-Z) isomers of alkenes differ in the spatial arrangement of groups around the C=C double bond. In the cis (or Z) isomer, the higher priority groups (or main chain continuation) are on the same side of the double bond. In the trans (or E) isomer, they are on opposite sides. When drawing skeletal structures, the cis isomer often looks like a 'cup' or 'C' shape around the double bond, while the trans isomer has a zigzag continuation through the double bond.

Understanding the Question

The question asks to draw the stereoisomer of W. Since W is hept-2-ene and shows stereoisomerism, it must be either (E)-hept-2-ene or (Z)-hept-2-ene. The drawing must show the opposite geometric arrangement to W.

Approach

Identify the main chain continuation across the double bond in W.
Draw the same carbon chain (7 carbons) but with the double bond geometry reversed (trans if W is cis, or cis if W is trans).

Step-by-Step Reasoning

W is shown in Fig 5.1. Assuming W is the cis (Z) isomer (where the chain continues on the same side), the stereoisomer is the trans (E) isomer.
Draw a 7-carbon chain. The double bond is between C2 and C3.
For the trans isomer, the C1 methyl group and the C4-C7 butyl group are on opposite sides of the double bond. This is drawn as a zigzag line where the double bond segment is parallel to the adjacent single bond segments but on opposite sides.

Key Takeaways

Stereoisomers of alkenes have the same connectivity but different spatial arrangements around the double bond.
Trans isomers typically have the main chain continuing in a zigzag pattern across the double bond.

Common Mistakes

Drawing a structural isomer instead of a stereoisomer (e.g., changing the double bond position).
Drawing the same isomer as W (must be the opposite geometry).
Incorrectly numbering or drawing the carbon chain length.

Things to Be Careful About

Ensure the skeletal structure has exactly 7 carbons.
The double bond must be clearly shown.
The geometry must be clearly trans (or cis, opposite to W). Avoid ambiguous drawings where it's unclear if it's cis or trans.

Techniques used

draw stereoisomer of an alkene

(d)

W reacts with reagent X to produce ethanoic acid and pentanoic acid.

2M

(i)

State the role of X in this reaction.

1M Similar

(ii)

Identify X and state the conditions used for this reaction.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

X is $\text{KMnO}_4$ (potassium manganate(VII)).
Conditions: hot, concentrated, acidified (with dilute $\text{H}_2\text{SO}_4$ ).

Final answer

hot concentrated acidified KMnO4

Detailed explanation

Background Concept

The oxidative cleavage of alkenes is typically carried out using acidified potassium manganate(VII) ( $KMnO_4$ ) or ozone ( $O_3$ ) followed by hydrolysis. For A-Level Chemistry, the standard reagent taught is hot, concentrated, acidified $KMnO_4$ . The acid is usually dilute sulfuric acid ( $H_2SO_4$ ). Under these vigorous conditions, the double bond is completely broken, and each carbon of the double bond is oxidised to a carbonyl group. If the carbon is monosubstituted (has one H), it becomes a carboxylic acid. If disubstituted, it becomes a ketone. If it has two H's (terminal alkene), it becomes $CO_2$ .

Understanding the Question

The question asks to identify reagent X and state the conditions. We know X cleaves hept-2-ene into ethanoic acid and pentanoic acid. This matches the oxidative cleavage using $KMnO_4$ .

Approach

Identify the reagent: potassium manganate(VII) ( $KMnO_4$ ).
State the required conditions: hot, concentrated, and acidified.

Step-by-Step Reasoning

Reagent X must be $KMnO_4$ to perform oxidative cleavage.
The conditions must be hot and concentrated to ensure complete cleavage to carboxylic acids (dilute, cold $KMnO_4$ would only form diols).
The solution must be acidified, typically with dilute $H_2SO_4$ , to provide the acidic medium required for the reaction.

Key Takeaways

Hot, concentrated, acidified $KMnO_4$ cleaves alkenes to carboxylic acids/ketones.
Cold, dilute, acidified $KMnO_4$ converts alkenes to diols (addition reaction, not cleavage).
Conditions are critical: temperature, concentration, and pH must be specified.

Common Mistakes

Saying "cold dilute $KMnO_4$ " (this gives a diol, not cleavage products).
Forgetting to specify "acidified" or "hot" and "concentrated".
Writing $MnO_4^-$ instead of the full reagent $KMnO_4$ (though $MnO_4^-$ is often accepted, $KMnO_4$ is safer).

Things to Be Careful About

Always include all three conditions: hot, concentrated, acidified.
Do not say "boiling"; "hot" or "warm" is the standard terminology, though boiling is implied by concentrated hot conditions. Stick to the mark scheme wording: "hot concentrated acidified $KMnO_4$ ".

Techniques used

identify reagent and conditions for alkene cleavage

Q6MediumAtoms, Molecules and StoichiometryIntroduction to Organic ChemistryStates of MatterAnalytical Techniques

P, Q and R are three different hydrocarbon molecules.

(a)

4M

(i)

Define empirical formula.

1M Similar

DifficultyEasy

Worked solution

Answer

The simplest whole-number ratio of atoms of each element present in a compound.

Final answer

simplest whole-number ratio of atoms of each element in a compound

Detailed explanation

Background Concept

An empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. It does not necessarily represent the actual number of atoms in a molecule. For example, hydrogen peroxide has the molecular formula $\text{H}_2\text{O}_2$ but its empirical formula is $\text{HO}$ ; glucose has the molecular formula $\text{C}_6\text{H}_{12}\text{O}_6$ but its empirical formula is $\text{CH}_2\text{O}$ .

The empirical formula is derived from percentage composition or reacting masses by converting to moles and then dividing by the smallest number of moles to obtain a ratio, which is then scaled to whole numbers if necessary.

Understanding the Question

This part asks for the definition of empirical formula. No calculation is required — it is a straightforward recall question worth one mark.

Approach

Recall the standard Cambridge definition: simplest whole-number ratio of atoms of the elements in a compound.

Step-by-Step Reasoning

The mark scheme awards one mark for stating that the empirical formula is the simplest whole-number ratio of atoms of each element in a compound. Key words that must appear: simplest, whole number ratio, atoms, and elements. Omitting any of these can cost the mark.

Key Takeaways

Be able to distinguish between molecular formula (actual numbers of atoms) and empirical formula (simplest ratio). Both describe the same compound but give different information.

Common Mistakes

Saying "the formula with the smallest numbers" without mentioning it is a ratio of atoms — this is vague and may not score.
Writing "the formula of the simplest compound" — this is incorrect; empirical formula applies to any compound, not just simple ones.
Forgetting to say whole number — fractional ratios are not empirical formulas.

Things to Be Careful About

Use the exact phrasing: "simplest whole-number ratio of atoms of each element in a compound." Mark scheme examiners look for these specific keywords.

Techniques used

define empirical formula

(ii)

Hydrocarbon P contains 85.7% by mass of carbon.

Calculate the empirical formula of P.

Show your working.

empirical formula of P = ..............................

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Assume 100 g of P:

\text{moles of C} = \frac{85.7}{12} = 7.14

\text{mass of H} = 100 - 85.7 = 14.3 \text{ g}

\text{moles of H} = \frac{14.3}{1} = 14.3

Ratio C : H = $7.14 : 14.3 = 1 : 2$

Answer

empirical formula of P = $\text{CH}_2$

Final answer

CH2

Detailed explanation

Background Concept

To find an empirical formula from percentage composition:

Assume a 100 g sample, so percentages become grams directly.
Divide each mass by the atomic mass (or molar mass) of the element to get moles.
Divide all mole values by the smallest mole value to obtain a ratio.
If the ratio contains fractions (e.g., 0.5, 1.33, 1.5), multiply through by the appropriate integer (2, 3, 2) to get whole numbers.

For hydrocarbons, only carbon and hydrogen are present, so the hydrogen percentage is $100\% - \%\text{C}$ .

Understanding the Question

Hydrocarbon P is 85.7% carbon by mass. Since it is a hydrocarbon, the remainder (14.3%) must be hydrogen. We need to find the simplest whole-number ratio of C to H atoms.

Approach

Convert the mass percentages to moles using atomic masses ( $A_r(\text{C}) = 12$ , $A_r(\text{H}) = 1$ ), then divide by the smallest value to get the ratio.

Step-by-Step Reasoning

Step 1: Assume 100 g of P. Then mass of C = 85.7 g, mass of H = 100 - 85.7 = 14.3 g.

Step 2: Calculate moles:

n_\text{C} = \frac{85.7}{12} = 7.14 \text{ mol}

n_\text{H} = \frac{14.3}{1} = 14.3 \text{ mol}

Step 3: Divide by the smallest (7.14):

\text{C} : \text{H} = \frac{7.14}{7.14} : \frac{14.3}{7.14} = 1 : 2

Step 4: The empirical formula is $\text{CH}_2$ .

The mark scheme awards M1 for showing the mole calculation ( $85.7/12 = 7.14$ and $14.3/1 = 14.3$ ) and M2 for the final empirical formula $\text{CH}_2$ .

Key Takeaways

Always assume 100 g when given percentages — this makes the conversion from percentage to grams trivial. The ratio step (dividing by the smallest value) is where most errors occur; check your arithmetic carefully.

Common Mistakes

Forgetting to calculate the hydrogen percentage: if C is 85.7%, H must be 14.3%, not 100%.
Using wrong atomic masses (e.g., using 13.0 for carbon instead of 12.0).
Writing the empirical formula as $\text{C}_7\text{H}_{14}$ instead of reducing to $\text{CH}_2$ .
Not showing the division step — the mark scheme requires seeing the ratio calculation.

Things to Be Careful About

Show all working: the mark scheme gives one method mark (M1) for the mole calculation and one accuracy mark (M2) for the final formula.
Use correct significant figures in intermediate steps (at least 3 sf is acceptable here).
The final answer must be written as $\text{CH}_2$ , not $\text{C}_1\text{H}_2$ .

Techniques used

convert percentage composition to molesdivide by smallest mole value to find ratiowrite empirical formula from ratio

(iii)

Molecules of P have straight chains.

Name the homologous series to which P belongs.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

alkenes

Final answer

alkenes

Detailed explanation

Background Concept

A homologous series is a family of organic compounds with the same functional group, similar chemical properties, and a general formula. Each member differs from the next by a $\text{CH}_2$ unit.

Key homologous series and their general formulas:

Alkanes: $\text{C}_n\text{H}_{2n+2}$
Alkenes: $\text{C}_n\text{H}_{2n}$
Alkynes: $\text{C}_n\text{H}_{2n-2}$
Cycloalkanes: $\text{C}_n\text{H}_{2n}$

The empirical formula $\text{CH}_2$ corresponds to a general formula of $\text{C}_n\text{H}_{2n}$ , which fits both alkenes and cycloalkanes.

Understanding the Question

Hydrocarbon P has empirical formula $\text{CH}_2$ (from part (ii)) and its molecules have straight chains. We need to name the homologous series.

Approach

The empirical formula $\text{CH}_2$ suggests a general formula of $\text{C}_n\text{H}_{2n}$ . This matches both alkenes and cycloalkanes. The clue "straight chains" rules out cycloalkanes (which are cyclic, not straight-chain). Therefore, P belongs to the alkenes.

Step-by-Step Reasoning

Empirical formula $\text{CH}_2$ → general formula $\text{C}_n\text{H}_{2n}$ .
$\text{C}_n\text{H}_{2n}$ corresponds to alkenes (containing one C=C double bond) or cycloalkanes (ring structures with only C-C single bonds).
The question states P has straight chains, which means it is not cyclic. Cycloalkanes are excluded.
Therefore, P is an alkene.

Note: Alkynes ( $\text{C}_n\text{H}_{2n-2}$ ) have a different hydrogen count and are ruled out.

Key Takeaways

When deducing a homologous series from an empirical formula, consider both the general formula and any structural information given (straight chain, cyclic, etc.). The combination of $\text{CH}_2$ empirical formula + straight chain uniquely identifies alkenes.

Common Mistakes

Answering "alkanes" — alkanes have general formula $\text{C}_n\text{H}_{2n+2}$ , empirical formula would be close to $\text{CH}_3$ for large n, not $\text{CH}_2$ .
Answering "alkynes" — these have empirical formula closer to $\text{CH}$ for large n.
Answering "cycloalkanes" — the question specifies straight chains, ruling out rings.
Writing "alkene" in singular — the mark scheme accepts "alkene(s)" but the series name is "alkenes".

Things to Be Careful About

The command word is "name" — give the name of the series, not a description.
"Alkenes" is the correct term for the series; "alkene" (singular) may or may not score depending on examiner discretion, but "alkenes" is safer.

Techniques used

identify homologous series from empirical formula and structural information

(b)

Q is a volatile hydrocarbon.

$0.194\text{ g}$ of gaseous Q occupies a volume of $71.5\text{ cm}^3$ at $100\text{ }^\circ\text{C}$ and $100\text{ kPa}$ .

Use the ideal gas equation to calculate the $M_r$ of Q.

$M_r$ of Q = ..............................

2M Similar

DifficultyMedium

Worked solution

Working

pV = nRT

Convert units:

$T = 100 \text{ }^\circ\text{C} = 373 \text{ K}$
$V = 71.5 \text{ cm}^3 = 71.5 \times 10^{-6} \text{ m}^3$
$p = 100 \text{ kPa} = 100 \times 10^3 \text{ Pa}$
$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$

n = \frac{pV}{RT} = \frac{100 \times 10^3 \times 71.5 \times 10^{-6}}{8.314 \times 373} = \frac{7.15}{3101.1} = 2.306 \times 10^{-3} \text{ mol}

M_r = \frac{\text{mass}}{n} = \frac{0.194}{2.306 \times 10^{-3}} = 84.1

Answer

$M_r$ of Q = 84.1

Final answer

84.1

Detailed explanation

Background Concept

The ideal gas equation relates pressure, volume, temperature, and amount of gas:

pV = nRT

where:

$p$ = pressure in pascals (Pa)
$V$ = volume in cubic metres (m³)
$n$ = amount of substance in moles (mol)
$R$ = gas constant = $8.314 \text{ J K}^{-1} \text{ mol}^{-1}$
$T$ = temperature in kelvin (K)

Critical unit requirements:

Pressure must be in Pa (not kPa, atm, or mmHg)
Volume must be in m³ (not cm³ or dm³)
Temperature must be in K (not °C)

To convert:

$T(\text{K}) = T(\text{°C}) + 273$ (or 273.15, but 273 is acceptable at A-Level)
$1 \text{ cm}^3 = 10^{-6} \text{ m}^3$
$1 \text{ kPa} = 10^3 \text{ Pa}$

Once $n$ is found, molar mass is: $M_r = \text{mass (g)} / n$ .

Understanding the Question

We are given:

Mass of gaseous Q: $0.194 \text{ g}$
Volume: $71.5 \text{ cm}^3$
Temperature: $100 \text{ °C}$
Pressure: $100 \text{ kPa}$

We need to calculate $M_r$ using the ideal gas equation.

Approach

Convert all quantities to SI units (Pa, m³, K).
Rearrange $pV = nRT$ to find $n$ .
Calculate $M_r = \text{mass} / n$ .

Step-by-Step Reasoning

Step 1: Convert units

T = 100 + 273 = 373 \text{ K}

V = 71.5 \text{ cm}^3 = 71.5 \times 10^{-6} \text{ m}^3 = 7.15 \times 10^{-5} \text{ m}^3

p = 100 \text{ kPa} = 100 \times 10^3 \text{ Pa} = 1.00 \times 10^5 \text{ Pa}

Step 2: Calculate moles

n = \frac{pV}{RT} = \frac{(1.00 \times 10^5) \times (7.15 \times 10^{-5})}{8.314 \times 373}

n = \frac{7.15}{3101.1} = 2.306 \times 10^{-3} \text{ mol}

Step 3: Calculate $M_r$

M_r = \frac{0.194}{2.306 \times 10^{-3}} = 84.1

The mark scheme awards M1 for converting temperature to 373 K and using it correctly, and M2 for the correct rearrangement and final answer of 84.1.

Key Takeaways

Unit conversion is the most common source of error in ideal gas calculations. Always check:

Temperature in K (add 273 to °C)
Volume in m³ (multiply cm³ by $10^{-6}$ )
Pressure in Pa (multiply kPa by $10^3$ )

Common Mistakes

Forgetting to convert °C to K: using $T = 100$ instead of $T = 373$ gives a wildly wrong answer.
Forgetting to convert cm³ to m³: using $V = 71.5$ directly gives $n$ that is $10^6$ times too large.
Using $R = 8.314$ with pressure in kPa and volume in cm³ without converting — the units must be consistent with $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$ .
Rounding too early: keep at least 4 significant figures in intermediate calculations.
Writing $M_r = 84$ instead of 84.1: the mark scheme accepts 84.1 (3 sf is appropriate given the data).

Things to Be Careful About

The gas constant $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$ requires SI units throughout.
If you use $R = 0.0821 \text{ dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1}$ , you must use atm and dm³ instead. Both approaches are valid but must be internally consistent.
$M_r$ is dimensionless (relative molecular mass), so no unit is needed in the final answer.

Techniques used

convert temperature to kelvinconvert volume to cubic metresapply ideal gas equation to find molescalculate molar mass from moles and mass

(c)

The mass spectrum of hydrocarbon R is recorded.

Information about the two peaks with $m/e$ greater than 135 is shown in Fig. 6.1.

Fig. 6.1

4M

(i)

Use Fig. 6.1 to deduce the number of carbon atoms in a molecule of R.

Show your working.

number of carbon atoms = ..............................

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DifficultyMedium

Worked solution

Working

The peak at $m/e = 138$ is the molecular ion $\text{M}^+$ (relative abundance = 100).
The peak at $m/e = 139$ is the $[\text{M}+1]^+$ peak (relative abundance = 11.0).

The $[\text{M}+1]^+$ peak arises from molecules containing one $^{13}\text{C}$ atom.

Natural abundance of $^{13}\text{C} \approx 1.1\%$ .

\text{Number of C atoms} = \frac{\text{abundance of } [\text{M}+1]^+}{\text{abundance of M}^+} \times \frac{100}{1.1}

= \frac{11.0}{100} \times \frac{100}{1.1} = 10

Answer

number of carbon atoms = 10

Final answer

10

Detailed explanation

Background Concept

In mass spectrometry, molecules are ionised (typically by electron impact) to form a molecular ion $\text{M}^+$ . The mass spectrum shows peaks at various $m/e$ values corresponding to the molecular ion and fragment ions.

The molecular ion peak ( $\text{M}^+$ ) appears at the highest $m/e$ value (excluding isotope peaks) and gives the relative molecular mass.

The $[\text{M}+1]^+$ peak appears at $m/e = M^+ + 1$ and is primarily due to molecules containing one $^{13}\text{C}$ atom instead of $^{12}\text{C}$ . The natural abundance of $^{13}\text{C}$ is approximately 1.1% (more precisely 1.07-1.11%, but 1.1% is used at A-Level).

The relationship is:

\frac{\text{abundance of } [\text{M}+1]^+}{\text{abundance of M}^+} \times 100 = n \times 1.1\%

where $n$ is the number of carbon atoms. Rearranging:

n = \frac{\text{abundance of } [\text{M}+1]^+}{\text{abundance of M}^+} \times \frac{100}{1.1}

Note: Hydrogen contributes negligibly to the $[\text{M}+1]^+$ peak (natural abundance of $^2\text{H}$ is only 0.015%), so we can ignore it.

Understanding the Question

Fig. 6.1 shows a mass spectrum with two peaks:

$m/e = 138$ , relative abundance = 100 (this is $\text{M}^+$ )
$m/e = 139$ , relative abundance = 11.0 (this is $[\text{M}+1]^+$ )

We need to deduce the number of carbon atoms in a molecule of R.

Approach

Use the ratio of the $[\text{M}+1]^+$ peak to the $\text{M}^+$ peak and the known natural abundance of $^{13}\text{C}$ (1.1%) to calculate the number of carbon atoms.

Step-by-Step Reasoning

Step 1: Identify the peaks.

$m/e = 138$ with abundance 100 → molecular ion $\text{M}^+$ , so $M_r = 138$ .
$m/e = 139$ with abundance 11.0 → $[\text{M}+1]^+$ peak.

Step 2: Apply the carbon counting formula.

\text{Number of C atoms} = \frac{11.0}{100} \times \frac{100}{1.1} = \frac{11.0}{1.1} = 10

The mark scheme shows this as $\frac{100}{1.1} \times \frac{11}{100} = 10$ .

Step 3: Verify.

With 10 carbon atoms, the expected $[\text{M}+1]^+$ abundance relative to $\text{M}^+$ is:

10 \times 1.1\% = 11\%

This matches the observed 11.0%, confirming 10 carbon atoms.

Key Takeaways

The $[\text{M}+1]^+$ peak in a mass spectrum is a powerful tool for determining the number of carbon atoms. The formula is:

n_\text{C} = \frac{\text{height of } [\text{M}+1]^+}{\text{height of M}^+} \times \frac{100}{1.1}

This works because each carbon atom has a 1.1% chance of being $^{13}\text{C}$ , and for a molecule with $n$ carbons, the probability of exactly one being $^{13}\text{C}$ is approximately $n \times 1.1\%$ (valid for small percentages).

Common Mistakes

Confusing $m/e = 138$ and $m/e = 139$ : the molecular ion is at the lower $m/e$ (138), not 139.
Using the wrong isotope abundance: using 1.0% or 1.5% instead of 1.1% gives the wrong number of carbons.
Forgetting to divide by 100 or multiply incorrectly: the ratio 11.0/100 must be multiplied by 100/1.1, not just 100/1.1.
Assuming the $[\text{M}+1]^+$ peak is due to $^2\text{H}$ : hydrogen's contribution is negligible at A-Level.

Things to Be Careful About

The mark scheme accepts the calculation as $\frac{100}{1.1} \times \frac{11}{100}$ or equivalently $\frac{11.0}{1.0} \div 1.1$ . Show the working clearly.
The answer must be a whole number (10). If you get a non-integer, recheck your calculation.
This method assumes only $^{13}\text{C}$ contributes significantly to $[\text{M}+1]^+$ . For molecules with many hydrogen atoms, a small correction for deuterium ( $^2\text{H}$ ) would be needed, but this is beyond A-Level.

Techniques used

identify molecular ion peak from mass spectrumuse M+1 peak abundance to count carbon atoms

(ii)

Use Fig. 6.1 and your answer to (c)(i) to deduce the molecular formula of R.

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DifficultyMedium-Easy

Worked solution

Answer

$M_r = 138$ (from $\text{M}^+$ peak at $m/e = 138$ )

Mass of 10 carbon atoms = $10 \times 12 = 120$

Mass of hydrogen = $138 - 120 = 18$

Number of hydrogen atoms = 18

molecular formula of R = $\text{C}_{10}\text{H}_{18}$

Final answer

C10H18

Detailed explanation

Background Concept

From the mass spectrum:

The $\text{M}^+$ peak at $m/e = 138$ gives $M_r = 138$ .
Part (c)(i) established 10 carbon atoms.

For a hydrocarbon $\text{C}_x\text{H}_y$ :

M_r = 12x + y

With $x = 10$ and $M_r = 138$ :

138 = 120 + y \implies y = 18

So the molecular formula is $\text{C}_{10}\text{H}_{18}$ .

Note: $\text{C}_{10}\text{H}_{18}$ has a degree of unsaturation (index of hydrogen deficiency) of:

\text{IHD} = \frac{2 \times 10 + 2 - 18}{2} = \frac{4}{2} = 2

This means R has either two double bonds, one triple bond, or two rings (or a combination). This is consistent with it being a hydrocarbon that is not fully saturated.

Understanding the Question

We know:

$M_r = 138$ (from the molecular ion peak)
Number of carbon atoms = 10 (from part (c)(i))

We need to deduce the molecular formula.

Approach

Calculate the mass contributed by carbon, subtract from $M_r$ to get the mass of hydrogen, then divide by 1 to get the number of hydrogen atoms.

Step-by-Step Reasoning

Step 1: $M_r = 138$ from the $\text{M}^+$ peak.

Step 2: Mass of 10 C atoms = $10 \times 12 = 120$ .

Step 3: Mass of H = $138 - 120 = 18$ .

Step 4: Number of H atoms = $18 / 1 = 18$ .

Step 5: Molecular formula = $\text{C}_{10}\text{H}_{18}$ .

Key Takeaways

Once you know $M_r$ and the number of carbon atoms in a hydrocarbon, finding the molecular formula is straightforward subtraction. Always verify: $12 \times 10 + 1 \times 18 = 138$ ✓.

Common Mistakes

Using the wrong $M_r$ : the molecular ion peak is at $m/e = 138$ , not 139.
Arithmetic errors: $138 - 120 = 18$ , not 16 or 20.
Writing $\text{C}_{10}\text{H}_{18}$ as $\text{C}_{10}\text{H}_{18}$ with incorrect subscripts.

Things to Be Careful About

The molecular formula must be written with correct subscripts: $\text{C}_{10}\text{H}_{18}$ .
Check your answer: $12(10) + 1(18) = 138 = M_r$ ✓.

Techniques used

determine molecular formula from Mr and number of carbon atoms

(iii)

Suggest the molecular formula of the fragment of R with $m/e = 57$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$M_r = 57$ corresponds to $\text{C}_4\text{H}_9^+$ :

(4 \times 12) + (9 \times 1) = 48 + 9 = 57

fragment = $(\text{C}_4\text{H}_9)^+$

Final answer

(C4H9)+

Detailed explanation

Background Concept

In mass spectrometry, the molecular ion $\text{M}^+$ fragments into smaller ions and neutral radicals. The fragment ions appear at lower $m/e$ values.

A peak at $m/e = 57$ in a hydrocarbon mass spectrum commonly corresponds to the butyl cation $\text{C}_4\text{H}_9^+$ :

M_r(\text{C}_4\text{H}_9^+) = (4 \times 12) + (9 \times 1) = 48 + 9 = 57

This is a very common fragment in the mass spectra of alkanes and alkyl-substituted compounds. The butyl cation can arise from cleavage of a C-C bond in a molecule containing a butyl group ( $\text{-C}_4\text{H}_9$ ).

For $\text{C}_{10}\text{H}_{18}$ , losing a $\text{C}_6\text{H}_{13}$ fragment (Mr = 85) from the molecular ion gives:

138 - 85 = 53 \text{ (not 57)}

Actually, let's reconsider: $138 - 57 = 81$ . The neutral fragment lost has $M_r = 81$ , which could be $\text{C}_6\text{H}_9$ (81) — but that's an unusual radical. More likely, the $m/e = 57$ peak is simply the $\text{C}_4\text{H}_9^+$ ion formed by fragmentation.

The mark scheme accepts $(\text{C}_4\text{H}_9)^+$ as the answer.

Understanding the Question

Given molecular formula $\text{C}_{10}\text{H}_{18}$ ( $M_r = 138$ ) and a peak at $m/e = 57$ , we need to suggest the molecular formula of this fragment ion.

Approach

Find a combination of C and H atoms that gives $M_r = 57$ . The most common hydrocarbon fragment with $M_r = 57$ is $\text{C}_4\text{H}_9^+$ (butyl cation).

Step-by-Step Reasoning

Step 1: $m/e = 57$ means the fragment ion has $M_r = 57$ .

Step 2: Try $\text{C}_4\text{H}_9^+$ :

M_r = (4 \times 12) + (9 \times 1) = 48 + 9 = 57 \quad \checkmark

Step 3: This is a well-known fragment ion (butyl cation) commonly seen in mass spectra of compounds containing butyl groups.

Step 4: The fragment is written as $(\text{C}_4\text{H}_9)^+$ to indicate it is a cation.

Key Takeaways

Common fragment ions to remember:

$m/e = 15$ : $\text{CH}_3^+$
$m/e = 29$ : $\text{C}_2\text{H}_5^+$
$m/e = 43$ : $\text{C}_3\text{H}_7^+$
$m/e = 57$ : $\text{C}_4\text{H}_9^+$
$m/e = 71$ : $\text{C}_5\text{H}_{11}^+$

These are alkyl cations with the general formula $\text{C}_n\text{H}_{2n+1}^+$ .

Common Mistakes

Writing $\text{C}_4\text{H}_9$ without the charge: the fragment is an ion, so $(\text{C}_4\text{H}_9)^+$ is required.
Trying $\text{C}_3\text{H}_{15}$ : this doesn't make chemical sense (too many hydrogens for 3 carbons).
Not showing any working or justification for the choice.

Things to Be Careful About

The answer must include the positive charge: $(\text{C}_4\text{H}_9)^+$ .
The mark scheme accepts $(\text{C}_4\text{H}_9)^+$ specifically. Other valid fragments with $M_r = 57$ are not commonly expected at this level.

Techniques used

identify fragment ion from m/e value and molecular formula

Chemistry 9701/24 — May/June 2025

Answer

Background Concept

Understanding the Question

Approach

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

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Understanding the Question

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Step-by-Step Reasoning

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Understanding the Question

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Step-by-Step Reasoning

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Understanding the Question

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Step-by-Step Reasoning

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

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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

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

Common Mistakes