Chemistry 9701/23 — October/November 2025

Background Concept

In nuclide notation $^{A}_{Z}\text{X}$, $Z$ is the proton (atomic) number and $A$ is the nucleon (mass) number. Chromium's proton number is 24. In a neutral atom, number of electrons = number of protons, and number of neutrons = $A - Z$.

Understanding the Question

Chromium-52 means $A = 52$ for Cr, whose $Z = 24$. State the counts of the three sub-atomic particles.

Approach

Use neutrons = mass number − proton number; electrons = protons for a neutral atom.

Step-by-Step Reasoning

• Protons: $Z = 24$.
• Neutrons: $52 - 24 = 28$.
• Electrons: neutral atom, so $24$.

Key Takeaways

Isotope notation instantly gives all three particle counts for a neutral atom.

Common Mistakes

Subtracting the wrong way (24 − 52), or giving 28 electrons by confusing mass number with electron count.

Things to Be Careful About

The atom is neutral here — only for ions would electron count differ from proton number.

Background Concept

Isotopes have the same proton number but different numbers of neutrons, hence different mass numbers.

Understanding the Question

Compare $^{54}\text{Cr}$ with $^{52}\text{Cr}$, quoting particle numbers.

Approach

Protons and electrons are identical (both 24); only neutrons differ: $54 - 52 = 2$ more neutrons.

Step-by-Step Reasoning

$^{54}\text{Cr}$: 24p, 30n, 24e. $^{52}\text{Cr}$: 24p, 28n, 24e. Difference: two extra neutrons.

Key Takeaways

Isotopes differ only in neutron number.

Common Mistakes

Saying 'different mass' without referring to particle numbers, as the command word requires.

Things to Be Careful About

The question demands reference to numbers of particles — a vague 'heavier atom' scores no mark.

Background Concept

$A_r = \frac{\sum (\text{RIM} \times \text{abundance})}{100}$ — a weighted mean requiring both isotopic masses and their abundances.

Understanding the Question

Mass spectrometry gives relative isotopic masses; what else is needed to compute $A_r$?

Approach

Recall the weighted-average formula and identify the missing quantity.

Step-by-Step Reasoning

Without abundances, a simple average of isotopic masses would be meaningless; the natural abundance of each isotope weights the mean.

Key Takeaways

$A_r$ is an abundance-weighted average of isotopic masses.

Common Mistakes

Answering 'the number of isotopes' — that alone does not allow a weighted mean.

Things to Be Careful About

'Abundance' must be relative abundance (percentage or fraction), not just counts.

Background Concept

$A_r$ lies between the lightest and heaviest isotopic masses, weighted by abundance. If three isotopes of mass ≥52 dominate (>95%) and the mean is 51.996 — just below 52 — the remaining small contribution must pull the average down.

Understanding the Question

Deduce what the two statements imply about the unseen fourth isotope's relative isotopic mass (RIM).

Approach

Note that 51.996 < 52, the lowest RIM of the three named isotopes. A weighted mean below the minimum of the dominant contributors requires the minor contributor to be lighter.

Step-by-Step Reasoning

If all four isotopes had RIM ≥ 52, $A_r$ could not fall below 52. Since $A_r = 51.996 < 52$, the fourth isotope must have RIM < 51.996 (in fact $^{50}\text{Cr}$, ~4.3% abundance).

Key Takeaways

A weighted mean constrains the range of possible component values.

Common Mistakes

Claiming the fourth isotope must be exactly chromium-50, or saying 'heavier' — the direction of the deviation from 52 is the key deduction.

Things to Be Careful About

The mark scheme accepts any value of RIM less than 51.996; you need only the inequality, not the exact isotope.

Background Concept

$[\text{Ar}]$ denotes the electron configuration of argon, $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6$ (18 electrons).

Understanding the Question

Expand the shorthand $[\text{Ar}]\,3\text{d}^5 4\text{s}^1$ into the full configuration.

Approach

Replace $[\text{Ar}]$ with its full configuration and append the valence subshells.

Step-by-Step Reasoning

Argon's 18 electrons fill $1\text{s}$ through $3\text{p}$; chromium then has $3\text{d}^5 4\text{s}^1$ (the anomalous half-filled d subshell arrangement).

Key Takeaways

Know the $[\text{Ar}]$ core = $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6$.

Common Mistakes

Writing $3\text{p}^6 4\text{s}^2$ (calcium-like) or omitting a subshell.

Things to Be Careful About

Keep the given $3\text{d}^5 4\text{s}^1$ exactly as printed — do not 'correct' it to $3\text{d}^4 4\text{s}^2$.

Background Concept

By Hund's rule, electrons occupy degenerate orbitals singly before pairing. A half-filled $3\text{d}^5$ has five singly occupied d orbitals (5 unpaired); $4\text{s}^1$ has one unpaired electron. Paired subshells ($\text{s}^2$, $\text{p}^6$) contribute none.

Understanding the Question

Count total unpaired electrons in Cr: $3\text{d}^5 4\text{s}^1$.

Approach

$5 + 1 = 6$.

Step-by-Step Reasoning

Each of the five 3d orbitals holds one electron (all spin-parallel), and the single 4s electron is also unpaired: total 6.

Key Takeaways

Half-filled d subshells maximise unpaired electrons; s¹ adds one more.

Common Mistakes

Answering 5 (forgetting the 4s electron) or 4 (assuming $3\text{d}^4 4\text{s}^2$).

Things to Be Careful About

Use the anomalous configuration given in the stem, not the 'expected' filling order.

Background Concept

Reduction = gain of electrons (OIL RIG). In a half-equation, electrons appearing on the reactant side indicate reduction.

Understanding the Question

Half-equation 2 shows electrons on the left; name the reduced species and explain.

Approach

Electrons are reactants → the species consuming them is reduced: dichromate(VI) ions (Cr goes from +6 to +3).

Step-by-Step Reasoning

$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$: Cr oxidation state falls from +6 to +3, consistent with electron gain.

Key Takeaways

Electrons on the left of a half-equation = reduction.

Common Mistakes

Naming $\text{H}^+$ or $\text{H}_2\text{O}$, or saying 'oxidised'.

Things to Be Careful About

The explanation must state electron gain, not just 'oxidation number decreases' alone — electron gain is the credited reason.

Background Concept

Oxygen is −2 in oxides; the sum of oxidation states in a neutral molecule is zero.

Understanding the Question

Find carbon's oxidation state in $\text{CO}_2$, given it is 0 in methanal.

Approach

Set up $x + 2(-2) = 0$.

Step-by-Step Reasoning

$x = +4$. Carbon is oxidised from 0 to +4, losing 4 electrons — matching half-equation 1.

Key Takeaways

Oxidation-state arithmetic: assign O as −2, solve for the unknown.

Common Mistakes

Writing −4 or 4 without the sign; the oxidation state is $+4$.

Things to Be Careful About

Include the positive sign explicitly.

Background Concept

To combine half-equations, multiply each so the electrons lost equal electrons gained (LCM of 4 and 6 is 12), then cancel electrons and any species appearing on both sides.

Understanding the Question

The published paper withdrew this part due to a printing issue, so no marks were awarded; the method is shown for learning purposes.

Approach

Multiply half-equation 1 by 3 (12 e⁻ lost) and half-equation 2 by 2 (12 e⁻ gained); add and cancel.

Step-by-Step Reasoning

$3\text{CH}_2\text{O} + 3\text{H}_2\text{O} \rightarrow 3\text{CO}_2 + 12\text{H}^+ + 12\text{e}^-$ and $2\text{Cr}_2\text{O}_7^{2-} + 28\text{H}^+ + 12\text{e}^- \rightarrow 4\text{Cr}^{3+} + 14\text{H}_2\text{O}$. Adding and cancelling 12 e⁻, 12 H⁺ and 3 H₂O gives $3\text{CH}_2\text{O} + 2\text{Cr}_2\text{O}_7^{2-} + 16\text{H}^+ \rightarrow 3\text{CO}_2 + 4\text{Cr}^{3+} + 11\text{H}_2\text{O}$. (Alternatively, dividing through by 2 gives the simplest whole-number ratio shown in the solution.)

Key Takeaways

Electron balance dictates the multipliers; cancel species common to both sides.

Common Mistakes

Forgetting to cancel H⁺ and H₂O after cancelling electrons.

Things to Be Careful About

Check atoms and charges balance as a final step.

Background Concept

Physical state at a given temperature depends on the strength of intermolecular forces (IMF), not covalent bond strength. $\text{CO}_2$ is linear and symmetrical ($\text{O}=\text{C}=\text{O}$), so bond dipoles cancel — the molecule is non-polar overall and only London dispersion forces act between molecules. Methanal ($\text{H}_2\text{C}=\text{O}$) is polar (the C=O bond dipole does not cancel), giving permanent dipole–permanent dipole forces in addition to dispersion forces.

Understanding the Question

Explain why two small covalent molecules differ in state at $-30\ ^\circ\text{C}$ — the answer lies in comparing IMFs, with two marks: identify the force types, then compare strengths.

Approach

Step 1: classify polarity of each molecule. Step 2: name the IMF each exhibits. Step 3: state which IMFs are stronger and link to the physical state.

Step-by-Step Reasoning

M1: $\text{CO}_2$ — only instantaneous dipole–induced dipole forces; methanal — permanent dipole–permanent dipole (plus id-id). M2: methanal's IMFs are stronger, so its molecules need more energy to separate; at $-30\ ^\circ\text{C}$ methanal remains liquid while $\text{CO}_2$ molecules easily escape to the gas phase.

Key Takeaways

Molecular polarity determines whether permanent dipole forces exist; stronger IMFs mean higher boiling point and a liquid/solid state at higher temperatures.

Common Mistakes

Talking about covalent C=O bonds being broken (bonds are not broken on boiling); calling $\text{CO}_2$'s forces 'weak van der Waals' without naming the specific type; omitting the comparison of strength (M2).

Things to Be Careful About

Use the precise terms 'instantaneous dipole–induced dipole' and 'permanent dipole–permanent dipole'; both the force identification AND the strength comparison are needed for the two marks.

Background Concept

'Dynamic' means both reactions continue to occur; 'equilibrium' means no net change. Rates are equal and macroscopic properties (concentrations) are constant, not necessarily equal.

Understanding the Question

State the meaning of dynamic equilibrium for this $\text{CrO}_4^{2-}/\text{Cr}_2\text{O}_7^{2-}$ system.

Approach

Give either the rate equality or the constant-concentrations statement (mark scheme accepts either).

Step-by-Step Reasoning

Forward ($2\text{CrO}_4^{2-} + 2\text{H}^+ \rightarrow \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}$) and backward reactions occur at the same rate, so the yellow/orange composition stays constant.

Key Takeaways

Equal rates, constant (not equal) concentrations.

Common Mistakes

Saying 'concentrations of reactants and products are equal' — that is wrong.

Things to Be Careful About

'Constant' not 'equal' — a classic trap.

Background Concept

Dynamic equilibrium can only be established in a closed system, where no reactants or products escape or enter.

Understanding the Question

Name the condition needed for the equilibrium to be established.

Approach

Recall: closed system.

Step-by-Step Reasoning

In an open system, species leaving would prevent the rates from balancing; here all species remain in solution.

Key Takeaways

Closed system is the necessary condition.

Common Mistakes

Answering 'constant temperature' — temperature affects position, but the necessary condition for establishment is a closed system.

Things to Be Careful About

The question asks for the condition necessary to establish equilibrium — closed system.

Background Concept

Le Chatelier's principle: when a change is made to a system in dynamic equilibrium, the equilibrium shifts to oppose (minimise) the change. Here, $2\text{CrO}_4^{2-} + 2\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}$, $\Delta H > 0$ (forward endothermic). Yellow = $\text{CrO}_4^{2-}$; orange = $\text{Cr}_2\text{O}_7^{2-}$.

Understanding the Question

For each perturbation, state the colour observed and explain it via the direction of shift.

Approach

Temperature increase → shift in the endothermic direction (forward, right). Adding H⁺ → shift to consume H⁺ (forward, right). Both produce more orange dichromate.

Step-by-Step Reasoning

• Warming: extra heat is 'on the reactant side' of the endothermic forward reaction; the system shifts forward to absorb it → more $\text{Cr}_2\text{O}_7^{2-}$ → more orange (M1 observation, M2 explanation).
• HCl: raises $[\text{H}^+]$; equilibrium shifts forward to remove the added acid → more orange (M3 observation, M4 explanation).

Key Takeaways

Endothermic forward reactions shift right on heating; concentration additions shift the equilibrium away from the added species.

Common Mistakes

Saying the mixture turns yellow (wrong direction); explaining the HCl effect as 'Cl⁻ reacts' rather than via added H⁺; omitting 'to oppose the change' in the explanation.

Things to Be Careful About

Both the observation AND the explanation carry marks — a bare colour without the Le Chatelier reasoning loses half the marks.

Background Concept

Atomic radius is the distance from the nucleus to the outermost electrons. Across a period in the periodic table, each successive element has one more proton in the nucleus and one more electron in the same principal energy level. The inner-shell electrons provide shielding that screens the outer electrons from the full nuclear charge. Because the additional electrons are added to the same shell, the shielding effect increases only slightly, while the nuclear charge increases by one unit for each element.

Understanding the Question

The question asks for an explanation of the observed decrease in atomic radii from sodium (Na) to chlorine (Cl) across Period 3. The bar chart in Fig. 2.1 shows atomic radii falling from approximately 155 pm for Na to about 100 pm for Cl. The command word "explain" requires giving the cause, not merely describing the trend.

Approach

To explain a periodic trend in atomic radius, identify the two competing factors: (1) the increasing nuclear charge, which pulls electrons inward, and (2) the shielding effect from inner electrons, which opposes this pull. Show that the increase in nuclear charge dominates because shielding remains roughly constant across a period.

Step-by-Step Reasoning

Point 1 (M1): Across Period 3, each element has one more proton than the previous one, so the nuclear charge increases. For example, Na has nuclear charge +11 and Cl has nuclear charge +17. Simultaneously, the outer electrons are all added to the n = 3 shell, so the number of inner (shielding) electrons remains the same at 10 (1s² 2s² 2p⁶). Therefore, the shielding effect is similar across the period.

Point 2 (M2): Because the nuclear charge increases while shielding is similar, the effective nuclear charge felt by the outer electrons increases. This greater positive charge exerts a stronger electrostatic attraction on the outer-shell electrons, drawing them closer to the nucleus and reducing the atomic radius.

Key Takeaways

When explaining atomic radius trends across a period, always mention both (a) the increase in nuclear charge and (b) the similar (or constant) shielding by inner electrons. Omitting either point typically costs a mark.

Common Mistakes

• Saying "there are more electrons" without specifying they are in the same shell — this is irrelevant to the trend.
• Saying "shielding decreases" — shielding is roughly constant, not decreasing.
• Describing the trend without giving the cause (e.g., "the radius decreases" without explaining why).

Things to Be Careful About

The mark scheme requires both the increase in nuclear charge AND similar shielding as part of M1. A single point alone will not earn full credit. Use precise language: "nuclear charge increases" not "more protons pull harder", and "similar shielding" not "no shielding change".

Background Concept

When atoms form ions, their radius changes dramatically. Metals like aluminium lose electrons to form cations, which are smaller than the parent atom because electrons are removed from the outermost shell, often leaving the ion with one fewer complete shell. Non-metals like phosphorus gain electrons to form anions, which are larger than the parent atom because the added electrons increase electron-electron repulsion and the ion retains the same outer shell.

Understanding the Question

The bar chart shows Al³⁺ has an ionic radius of about 50 pm while P³⁻ has an ionic radius of about 210 pm — a factor of four difference. The question asks why this large difference exists.

Approach

Compare the electron configurations and number of electron shells in Al³⁺ and P³⁻. Aluminium is a metal that forms a cation by losing electrons; phosphorus is a non-metal that forms an anion by gaining electrons. The key difference is the number of electron shells in the resulting ions.

Step-by-Step Reasoning

Point 1 (M1): Aluminium (electron configuration 1s² 2s² 2p⁶ 3s² 3p¹) loses its 3 outer-shell electrons to form Al³⁺ with configuration 1s² 2s² 2p⁶. Phosphorus (1s² 2s² 2p⁶ 3s² 3p³) gains 3 electrons to form P³⁻ with configuration 1s² 2s² 2p⁶ 3s² 3p⁶.

Point 2 (M2): Al³⁺ has electrons only in the n = 1 and n = 2 shells (2 shells total). P³⁻ has electrons in the n = 1, n = 2, and n = 3 shells (3 shells total). The Al³⁺ ion has one fewer electron shell than P³⁻, which accounts for the large difference in ionic radii.

Key Takeaways

Cations are smaller than their parent atoms (often losing an outer shell); anions are larger. When comparing ions of different elements, the number of electron shells is the dominant factor affecting ionic radius.

Common Mistakes

• Saying "Al has a higher charge" without explaining the shell difference — this alone is not sufficient.
• Confusing ionic radius with atomic radius trends.
• Not specifying that Al loses electrons and P gains electrons.

Things to Be Careful About

The mark scheme awards one mark for identifying that Al loses 3 electrons (forming Al³⁺) and P gains 3 electrons (forming P³⁻), and the second mark for stating that Al³⁺ has one fewer shell than P³⁻. Both points are required for full credit.

Background Concept

Period 3 oxides show a clear trend from basic (ionic) on the left to acidic (covalent) on the right, with aluminium oxide and silicon dioxide being amphoterically or inert. The oxidation number of the Period 3 element in its oxide equals the number of valence electrons it effectively donates or shares with oxygen. When these oxides react with water, basic oxides form alkaline solutions (high pH), acidic oxides form acidic solutions (low pH), and amphoteric/inert oxides do not dissolve.

Understanding the Question

Table 2.1 has blank cells for oxidation numbers (row A) and pH values (row B) for Na₂O, MgO, SiO₂, P₄O₁₀, and SO₃. The oxidation number +3 is given for Al₂O₃ and dashes for Al₂O₃ and SiO₂ in row B. The task is to fill in the blanks.

Approach

For oxidation numbers: use the rule that oxygen is –2 and the sum of oxidation numbers in a neutral compound is zero. For pH values: recall that Na₂O and MgO are basic oxides (high pH), P₄O₁₀ and SO₃ are acidic oxides (low pH), and Al₂O₃ and SiO₂ are insoluble (no pH data).

Step-by-Step Reasoning

Oxidation numbers (M1):

• Na₂O: 2x + (–2) = 0 → x = +1
• MgO: x + (–2) = 0 → x = +2
• SiO₂: x + 2(–2) = 0 → x = +4
• P₄O₁₀: 4x + 10(–2) = 0 → 4x = +20 → x = +5
• SO₃: x + 3(–2) = 0 → x = +6

pH values (M2):

• Na₂O reacts with water to form NaOH (strong base): pH 12–14
• MgO reacts with water to form Mg(OH)₂ (weak base): pH 8–10
• P₄O₁₀ reacts with water to form H₃PO₄ (strong acid): pH 0–4
• SO₃ reacts with water to form H₂SO₄ (strong acid): pH 0–4

Key Takeaways

Oxidation numbers in Period 3 oxides increase from +1 to +6, matching the group number. The pH of oxide-water solutions reflects the acid-base character: basic oxides (Na, Mg) give high pH, acidic oxides (P, S) give low pH.

Common Mistakes

• Calculating wrong oxidation numbers (e.g., saying P is +4 in P₄O₁₀ instead of +5).
• Giving a single pH value instead of a range — the mark scheme accepts ranges.
• Forgetting that SiO₂ does not react with water (it is covalent and insoluble).

Things to Be Careful About

The mark scheme gives acceptable ranges: pH 12–14 for Na₂O, 8–10 for MgO, and 0–4 for both P₄O₁₀ and SO₃. Values outside these ranges may not score. Oxidation numbers must be exact integers.

Background Concept

Aluminium oxide (Al₂O₃) is amphoteric — it reacts with both acids and bases — but it does not dissolve in water. Silicon dioxide (SiO₂) is a giant covalent structure (like quartz) and is completely insoluble in water. Neither oxide reacts with water to produce a solution with a measurable pH.

Understanding the Question

Row B of Table 2.1 has dashes (—) for Al₂O₃ and SiO₂, indicating no pH data is given. The question asks why.

Approach

The pH of a solution requires a dissolved substance. If the oxide does not dissolve in water, no solution forms and no pH can be recorded.

Step-by-Step Reasoning

Al₂O₃ is amphoteric but insoluble in water. SiO₂ has a giant covalent structure and is also insoluble in water. Since neither oxide dissolves to form an aqueous solution, there is no solution whose pH can be measured. Hence the dashes in row B.

Key Takeaways

Not all Period 3 oxides react with water. Al₂O₃ and SiO₂ are exceptions — they are insoluble and do not produce a solution with a measurable pH.

Common Mistakes

• Saying "they are neutral" — Al₂O₃ is amphoteric, not neutral, and SiO₂ is acidic (just insoluble).
• Saying "they don't react with water" — while technically true, the key reason is insolubility.

Things to Be Careful About

The mark scheme specifically accepts "insoluble" as the answer. Other wordings like "do not dissolve" or "not soluble" also score.

Background Concept

Sodium oxide (Na₂O) is a basic oxide. Basic oxides react with acids in a neutralisation reaction to form a salt and water. This is analogous to the reaction of a metal oxide with an acid, similar to how NaOH (a base) reacts with HCl.

Understanding the Question

Write a balanced equation for the reaction of Na₂O with dilute hydrochloric acid (HCl).

Approach

Na₂O is a basic oxide; HCl is an acid. The products are the salt (NaCl) and water (H₂O). Balance the equation.

Step-by-Step Reasoning

Na₂O + HCl → NaCl + H₂O

Balancing: Na₂O has 2 Na atoms, so we need 2 NaCl on the right. This requires 2 HCl on the left. The 2 H from HCl and the O from Na₂O give H₂O.

Balanced equation: Na₂O + 2HCl → 2NaCl + H₂O

Key Takeaways

Basic oxides react with acids to give salt + water. Always balance the equation and include correct state symbols if required (though not always required in this question).

Common Mistakes

• Writing Na₂O + HCl → NaCl + H₂O without balancing (missing the coefficient 2 for HCl and NaCl).
• Including state symbols incorrectly if the mark scheme doesn't require them.

Things to Be Careful About

The mark scheme gives the equation without state symbols. If state symbols are required, Na₂O is (s), HCl is (aq), NaCl is (aq), H₂O is (l). The equation must be fully balanced.

Background Concept

Aluminium oxide (Al₂O₃) is amphoteric, meaning it reacts with both acids and bases. When it reacts with a base like sodium hydroxide (NaOH), it forms a sodium aluminate salt. The formula NaAlO₂ (sodium aluminate or sodium meta-aluminate) is given in the question. The reaction is analogous to how amphoteric hydroxides like Al(OH)₃ react with NaOH.

Understanding the Question

Construct a balanced equation for the reaction of Al₂O₃ with a base (NaOH) to form NaAlO₂.

Approach

Reactants: Al₂O₃ and NaOH. Product given: NaAlO₂. Also produce H₂O (from the oxide ion combining with H⁺ from NaOH). Balance the equation.

Step-by-Step Reasoning

Al₂O₃ + NaOH → NaAlO₂ + H₂O

Balancing Al: Al₂O₃ has 2 Al, so need 2 NaAlO₂ on the right.
Al₂O₃ + NaOH → 2NaAlO₂ + H₂O

Balancing Na: 2 NaAlO₂ requires 2 NaOH on the left.
Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

Check O: Left: 3 (from Al₂O₃) + 2 (from 2NaOH) = 5. Right: 4 (from 2NaAlO₂) + 1 (from H₂O) = 5. ✓
Check H: Left: 2 (from 2NaOH). Right: 2 (from H₂O). ✓

Balanced equation: Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

Key Takeaways

Amphoteric oxides react with bases to form salts. When given the product formula, use it to construct and balance the equation.

Common Mistakes

• Writing the wrong product formula (e.g., NaAl(OH)₄ instead of NaAlO₂ — the question specifies NaAlO₂).
• Not balancing the equation correctly.
• Forgetting water as a product.

Things to Be Careful About

The mark scheme gives the equation as Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O. This must be exactly balanced. State symbols are not required here unless specified.

Background Concept

Group 2 nitrates decompose on heating to give the metal oxide, nitrogen dioxide, and oxygen. The thermal stability of Group 2 nitrates increases down the group: Be(NO₃)₂ decomposes most easily, while Ba(NO₃)₂ requires the highest temperature. This is because the larger cations (like Ba²⁺) have lower charge density and polarise the nitrate ion less, making it more stable to heat.

Understanding the Question

State the trend in thermal stability of Group 2 nitrates as you go down the group from Be to Ba.

Approach

Recall the standard trend: thermal stability increases down Group 2 for both nitrates and carbonates.

Step-by-Step Reasoning

As you go down Group 2, the cation size increases and charge density decreases. The larger cation polarises the nitrate ion less effectively, so the nitrate is more stable and requires more heat to decompose. Therefore, thermal stability increases down the group.

Key Takeaways

Thermal stability of Group 2 nitrates (and carbonates) increases down the group. This is a standard trend that should be recalled.

Common Mistakes

• Saying "stability decreases down the group" — this is the opposite of the correct trend.
• Giving the trend for Group 1 instead (Group 1 nitrates decompose to nitrites + O₂, not oxides + NO₂ + O₂).

Things to Be Careful About

The mark scheme simply accepts "increases down the group". Be precise: it is the thermal stability that increases, not the decomposition temperature (though they are related).

Background Concept

Group 2 nitrates decompose on heating to produce the metal oxide, nitrogen dioxide gas, and oxygen gas. The general equation is:

$2\text{M(NO}_3)_2(\text{s}) \rightarrow 2\text{MO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})$

This is different from Group 1 nitrates, which decompose to give the metal nitrite and oxygen only (no NO₂).

Understanding the Question

The question states that Group 2 nitrates decompose to form oxides. Identify the other products besides the oxide.

Approach

Recall the full decomposition equation for Group 2 nitrates. The products are: metal oxide (given), nitrogen dioxide (NO₂), and oxygen (O₂).

Step-by-Step Reasoning

For example, with magnesium nitrate:
$2\text{Mg(NO}_3)_2(\text{s}) \rightarrow 2\text{MgO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})$

The other products besides the oxide are nitrogen dioxide (NO₂) and oxygen (O₂). Both must be stated for full credit.

Key Takeaways

Group 2 nitrate decomposition produces: metal oxide + NO₂ + O₂. Group 1 nitrate decomposition produces: metal nitrite + O₂ (no NO₂). Don't confuse the two.

Common Mistakes

• Only stating one product (e.g., just NO₂ or just O₂) — both are required.
• Saying CO₂ or H₂O — these are not products of nitrate decomposition.
• Confusing with Group 1 nitrate decomposition products.

Things to Be Careful About

The mark scheme requires both NO₂ AND O₂. Stating only one will not earn full credit. Also, write the formulas correctly: NO₂ (not NO, not N₂O₄), O₂ (not O).

Background Concept

Bond fission can be homolytic (each atom takes one electron from the shared pair, producing radicals) or heterolytic (one atom takes both electrons, producing ions). In free-radical substitution, UV light provides enough energy to break the Cl-Cl bond homolytically, generating two chlorine radicals.

Understanding the Question

The question states the reaction is initiated by bond fission of Cl₂ under UV light. The candidate must name the type of fission.

Approach

UV-initiated reactions that produce radicals involve homolytic fission. The word 'homolytic' is the required answer.

Step-by-Step Reasoning

Cl₂ + UV light → 2Cl•. Each chlorine atom receives one electron from the bond pair, producing two neutral radicals. This is the definition of homolytic fission.

Key Takeaways

Homolytic fission produces radicals; heterolytic fission produces ions. UV light and radical chain reactions are always associated with homolytic fission.

Common Mistakes

• Writing 'heterolytic' (wrong — that produces ions, not radicals).
• Writing 'covalent' (that describes the bond type, not the fission type).

Things to Be Careful About

The mark scheme requires the exact word 'homolytic'. 'Homolytic bond fission' or 'homolysis' are also acceptable.

Background Concept

In free-radical substitution, the propagation steps form a chain: a radical reacts with a stable molecule to produce a new radical, which then reacts with another stable molecule to produce the product and regenerate the original radical. This cycle sustains the chain reaction.

Understanding the Question

The initiation step produces Cl• radicals. The candidate must write the two propagation steps that convert cyclopentane to chlorocyclopentane while regenerating the chlorine radical.

Approach

Step 1: Cl• abstracts a hydrogen from cyclopentane, forming HCl and a cyclopentyl radical (C₅H₉•). Step 2: The cyclopentyl radical attacks Cl₂, forming chlorocyclopentane (C₅H₉Cl) and regenerating Cl•.

Step-by-Step Reasoning

• Propagation 1: The chlorine radical (electron-deficient) attacks a C-H bond in cyclopentane. One electron from the C-H bond pairs with the unpaired electron on Cl to form H-Cl; the other electron remains on carbon, giving C₅H₉•. M1 is awarded for correctly writing this step.
• Propagation 2: The cyclopentyl radical attacks Cl₂. One electron from the Cl-Cl bond pairs with the unpaired electron on carbon to form C-Cl; the other electron goes to the second Cl, regenerating Cl•. M2 is awarded for correctly writing this step.
• Note: The Cl• produced in propagation 2 can re-enter propagation 1, sustaining the chain.

Key Takeaways

The two propagation steps always involve: (1) radical + stable molecule → new radical + stable product, and (2) new radical + stable molecule → final product + regenerated radical. The net result is substitution of H by Cl.

Common Mistakes

• Writing Cl₂ instead of Cl• in propagation 1.
• Forgetting the radical dot on C₅H₉•.
• Writing the steps in the wrong order.
• Including the initiation or termination step instead.

Things to Be Careful About

Radical dots (•) must be shown on Cl• and C₅H₉•. The mark scheme awards M1 and M2 separately, so both steps must be correct.

Background Concept

A free-radical chain reaction has three stages: initiation (radicals are created), propagation (radicals are consumed and regenerated in a cycle), and termination (two radicals combine to form a stable molecule, ending the chain).

Understanding the Question

The equation C₅H₉• + Cl• → C₅H₉Cl shows two radicals combining to form a stable product. The candidate must name this stage.

Approach

Two radicals combining with no new radical produced = termination.

Step-by-Step Reasoning

In this step, the unpaired electron on C₅H₉• pairs with the unpaired electron on Cl• to form a covalent C-Cl bond. No radicals remain, so the chain is terminated. This is the definition of a termination step.

Key Takeaways

Termination steps always involve radical + radical → stable molecule. They reduce the number of radicals in the system.

Common Mistakes

• Calling it a 'propagation' step (propagation regenerates a radical).
• Calling it 'initiation' (initiation creates radicals from a non-radical).

Things to Be Careful About

The answer must be 'termination' — not 'terminating step' or any other phrasing that might be ambiguous.

Background Concept

Alkenes undergo electrophilic addition reactions. The C=C double bond acts as a nucleophile (electron source) and attacks an electrophile. HCl adds across the double bond: H⁺ attacks first to form a carbocation, then Cl⁻ completes the addition, giving a chloroalkane.

Understanding the Question

Reaction 3 converts cyclopentene to chlorocyclopentane. The candidate must identify a suitable reagent. Since the product has one H and one Cl added across the former double bond, the reagent is HCl.

Approach

Compare the structures: cyclopentene (C₅H₈) → chlorocyclopentane (C₅H₉Cl). The difference is +H +Cl = HCl addition.

Step-by-Step Reasoning

The molecular formula changes from C₅H₈ to C₅H₉Cl, a net addition of HCl. The mark scheme specifies HCl(g) — the gaseous form is used because aqueous HCl would give a different product (or no clean addition). The state symbol (g) is important.

Key Takeaways

Addition of HX to an alkene gives a haloalkane. The reagent must be specified with the correct state.

Common Mistakes

• Writing 'HCl(aq)' — aqueous HCl is not the correct reagent for this addition.
• Writing 'Cl₂' — that would give a dichloro product.

Things to Be Careful About

The mark scheme requires HCl(g). The state symbol matters.

Background Concept

Hess's law states that the total enthalpy change for a reaction is independent of the pathway taken, provided the initial and final states are the same. This allows unknown enthalpy changes to be calculated from known ones by constructing a cycle. Combustion data can be used to find the enthalpy change of a reaction via: ΔH_reaction = ΣΔHc(reactants) − ΣΔHc(products).

Understanding the Question

The cycle shows three reactions:

• Reaction 1: cyclopentene + H₂ → cyclopentane (ΔH₁ = unknown)
• Reaction 2: cyclopentane + Cl₂ → chlorocyclopentane + HCl (ΔH₂ = unknown, to find)
• Reaction 3: cyclopentene + HCl → chlorocyclopentane (ΔH₃ = −53 kJ mol⁻¹)

From the cycle geometry: going from cyclopentene to chlorocyclopentane via cyclopentane (reactions 1 then 2) equals going directly (reaction 3). So ΔH₁ + ΔH₂ = ΔH₃.

Approach

Step 1: Calculate ΔH₁ from combustion data using ΔH = ΣΔHc(reactants) − ΣΔHc(products).
Step 2: Rearrange the cycle equation to find ΔH₂ = ΔH₃ − ΔH₁.

Step-by-Step Reasoning

Finding ΔH₁:
Reactants of reaction 1: cyclopentene + H₂
Product of reaction 1: cyclopentane

Finding ΔH₂:
From the cycle: ΔH₁ + ΔH₂ = ΔH₃

The positive sign indicates reaction 2 is endothermic.

Key Takeaways

Hess's law cycles can be solved by writing the algebraic relationship between the reactions (pathway equivalence) and then substituting known values. Combustion data gives reaction enthalpies via reactants minus products.

Common Mistakes

• Getting the sign wrong in the combustion formula (should be reactants − products, not products − reactants).
• Forgetting that ΔHc of H₂ is −286 kJ mol⁻¹ (it is the combustion of hydrogen to water).
• Writing ΔH₂ = ΔH₃ + ΔH₁ instead of ΔH₃ − ΔH₁ (sign error in rearranging the cycle).
• Omitting the positive sign in the final answer.

Things to Be Careful About

M1 is awarded for correctly calculating ΔH₁ = −109. M2 is awarded for the final ΔH₂ = +56. If ΔH₁ is wrong but used correctly to find ΔH₂, M2 may still be awarded via ecf. The sign of ΔH₂ is important — it is positive (endothermic).

Background Concept

Hot concentrated acidified potassium manganate(VII) (KMnO₄) is a strong oxidising agent that cleaves carbon-carbon double bonds. In a cyclic alkene, the ring opens at the double bond, and both carbons of the former C=C are oxidised to carboxylic acid groups (−COOH) if they each bear at least one hydrogen. For cyclopentene, both alkene carbons have one H each, so both become −COOH groups, giving a dicarboxylic acid.

Understanding the Question

Cyclopentene (C₅H₈) is a five-membered ring with one C=C. Oxidative cleavage opens the ring at the double bond, producing a straight-chain compound with two −COOH ends. The molecular formula C₅H₈O₄ confirms a dicarboxylic acid (two COOH groups = 2×CO₂H, leaving a 3-carbon chain between them).

Approach

Open the ring at the C=C, add an oxygen to each former alkene carbon to form C=O, and add OH to each to form −COOH. The remaining three CH₂ groups form the chain between them.

Step-by-Step Reasoning

Cyclopentene: five carbons in a ring, one C=C between C1 and C2.
After oxidative cleavage:
C1 becomes −COOH, C2 becomes −COOH, C3-C4-C5 remain as −CH₂−CH₂−CH₂−.
Product: HOOC−CH₂−CH₂−CH₂−COOH = pentanedioic acid.
Check formula: 5C, 8H (2 from COOH + 6 from CH₂ groups), 4O (2 from each COOH). ✓

Key Takeaways

Hot concentrated KMnO₄ cleaves C=C to give carboxylic acids (or ketones if the carbon has no H). In cyclic alkenes, the ring opens to give a dicarboxylic acid (or keto-acid).

Common Mistakes

• Drawing a diol (that would be cold dilute KMnO₄, not hot concentrated).
• Drawing only one COOH group (forgetting both carbons of the double bond are oxidised).
• Incorrect chain length (must have 3 CH₂ groups between the two COOH).

Things to Be Careful About

The structure must show all atoms and bonds (displayed formula) or at least a clear skeletal structure showing both COOH groups and the three-carbon chain between them.

Background Concept

IR spectroscopy identifies functional groups by their characteristic bond stretching absorptions. A carboxylic acid shows two key absorptions not found in alkenes: a very broad O-H stretch (2500–3000 cm⁻¹ for carboxyl O-H, distinct from the sharper alcohol O-H at 3200–3650 cm⁻¹) and a strong C=O stretch (1670–1740 cm⁻¹ for carboxyl). Cyclopentene shows C=C (1500–1680 cm⁻¹) and C-H (2850–2950 cm⁻¹) but no O-H or C=O.

Understanding the Question

The candidate must identify two absorptions in the IR spectrum of W (pentanedioic acid) that would NOT appear in the spectrum of cyclopentene. From Table 3.3, the relevant new bonds in W are O-H (carboxyl) and C=O (carboxyl). The candidate labels these on Fig. 3.3 and writes the bond in Table 3.2.

Approach

Compare functional groups: cyclopentene has C=C and C-H only. W has C=O, O-H (carboxyl), and C-H. The two absorptions unique to W are O-H and C=O.

Step-by-Step Reasoning

• O-H (carboxyl): The broad absorption centred around 2500–3000 cm⁻¹ (labelled '1' on the spectrum) corresponds to the O-H stretch of the carboxylic acid group. Cyclopentene has no O-H bond, so this absorption is absent from its spectrum. M1 awarded.
• C=O (carboxyl): The sharp, strong absorption near 1710 cm⁻¹ (labelled '2' on the spectrum) corresponds to the C=O stretch. Cyclopentene has C=C (which absorbs at 1500–1680 cm⁻¹) but no C=O, so this is a new absorption. M2 awarded.

Note: The C=C absorption of cyclopentene (1500–1680 cm⁻¹) is NOT present in W, but the question asks for absorptions in W that are NOT in cyclopentene, so C=C is irrelevant here.

Key Takeaways

Carboxylic acids are identified in IR by the combination of a very broad O-H (2500–3000 cm⁻¹) and a strong C=O (1670–1740 cm⁻¹). These are absent in hydrocarbons.

Common Mistakes

• Identifying C-H as a new absorption (both compounds have C-H bonds).
• Confusing the carboxyl O-H range (2500–3000) with the alcohol O-H range (3200–3650).
• Labeling the C=C absorption (which is in cyclopentene, not W).

Things to Be Careful About

The question asks for absorptions in W that are NOT in cyclopentene. The two answers must be O-H and C=O. The wavenumber positions must match the spectrum (approximately 2500–3000 for O-H and approximately 1710 for C=O).

Background Concept

Carboxylic acids can be reduced to primary alcohols using strong reducing agents such as lithium aluminium hydride (LiAlH4) followed by acid workup, or catalytic hydrogenation under high pressure. In this process, the oxidation state of the carbonyl carbon decreases, which defines a reduction reaction.

Understanding the Question

The question asks for the type of reaction occurring in reaction 1, where propanoic acid ($\text{C}_2\text{H}_5\text{COOH}$) is converted to propan-1-ol ($\text{C}_2\text{H}_5\text{CH}_2\text{OH}$). This involves the addition of hydrogen (or removal of oxygen) to the carboxyl group.

Approach

Compare the functional groups of the reactant and product. A carboxylic acid ($-\text{COOH}$) becoming a primary alcohol ($-\text{CH}_2\text{OH}$) is a classic reduction.

Step-by-Step Reasoning

The functional group changes from $-\text{COOH}$ to $-\text{CH}_2\text{OH}$. This requires the addition of hydrogen atoms and the removal of an oxygen atom. In organic chemistry, an increase in the number of C-H bonds or a decrease in C-O bonds is classified as reduction. Therefore, reaction 1 is a reduction.

Key Takeaways

Converting a carboxylic acid to a primary alcohol is a reduction reaction. This is a fundamental transformation in organic synthesis.

Common Mistakes

Students often confuse reduction with oxidation or simply write 'hydrogenation' without considering the specific functional group change. While hydrogenation is a type of reduction, 'reduction' is the broader and more accurate term here, especially if LiAlH4 is the reagent (which doesn't involve $\text{H}_2$ gas).

Things to Be Careful About

Ensure the term used matches the mark scheme. 'Reduction' is the precise term for this functional group transformation.

Background Concept

Primary alcohols can be converted to chloroalkanes using several reagents. The most common and cleanest methods involve phosphorus chlorides or thionyl chloride, which produce gaseous by-products that are easy to separate. Alternatively, a halide salt can be used with a strong acid to generate HCl in situ.

Understanding the Question

Reaction 2 converts propan-1-ol ($\text{C}_2\text{H}_5\text{CH}_2\text{OH}$) to 1-chloropropane ($\text{C}_2\text{H}_5\text{CH}_2\text{Cl}$). We need to suggest a suitable reagent for this substitution.

Approach

Recall the standard reagents for converting $-\text{OH}$ to $-\text{Cl}$ in alcohols.

Step-by-Step Reasoning

• $\text{PCl}_5$: Reacts vigorously at room temperature. $\text{C}_2\text{H}_5\text{CH}_2\text{OH} + \text{PCl}_5 \rightarrow \text{C}_2\text{H}_5\text{CH}_2\text{Cl} + \text{POCl}_3 + \text{HCl}$.
• $\text{PCl}_3$: Requires gentle warming. $3\text{C}_2\text{H}_5\text{CH}_2\text{OH} + \text{PCl}_3 \rightarrow 3\text{C}_2\text{H}_5\text{CH}_2\text{Cl} + \text{H}_3\text{PO}_3$.
• $\text{SOCl}_2$ (thionyl chloride): Produces gaseous $\text{SO}_2$ and $\text{HCl}$, making purification easy. $\text{C}_2\text{H}_5\text{CH}_2\text{OH} + \text{SOCl}_2 \rightarrow \text{C}_2\text{H}_5\text{CH}_2\text{Cl} + \text{SO}_2 + \text{HCl}$.
• $\text{NaCl}$ / $\text{KCl}$ + conc. $\text{H}_2\text{SO}_4$ / $\text{H}_3\text{PO}_4$: The acid reacts with the chloride salt to produce HCl, which then substitutes the OH group. Note: conc. $\text{H}_2\text{SO}_4$ with $\text{NaCl}$ can sometimes cause elimination or oxidation side reactions, but it is accepted in the mark scheme. $\text{H}_3\text{PO}_4$ is often preferred to avoid oxidation.

Key Takeaways

Know the three main phosphorus/chlorine reagents ($\text{PCl}_5$, $\text{PCl}_3$, $\text{SOCl}_2$) and the salt + acid method for converting alcohols to chloroalkanes.

Common Mistakes

Writing $\text{HCl}$ alone: while possible, it is slow and reversible for primary alcohols; the mark scheme specifically expects the phosphorus chlorides or the salt/acid combination. Writing $\text{Cl}_2$ is incorrect as this would lead to free-radical substitution or addition, not simple substitution of the OH group.

Things to Be Careful About

If using $\text{NaCl}$ or $\text{KCl}$, you must state the acid (conc. $\text{H}_2\text{SO}_4$ or conc. $\text{H}_3\text{PO}_4$) as well. Just writing 'HCl' or 'chloride ions' is insufficient.

Background Concept

Halogenoalkanes can undergo two competing reactions with hydroxide ions ($\text{OH}^-$): nucleophilic substitution (forming alcohols) and elimination (forming alkenes). The outcome depends heavily on the solvent and temperature.

• Aqueous NaOH / cold: Favors substitution ($\text{S}_\text{N}2$ for primary halogenoalkanes), producing an alcohol.
• Ethanolic (alcoholic) NaOH / hot (reflux): Favors elimination (E2 mechanism), producing an alkene. The ethanolic solvent makes the $\text{OH}^-$ a stronger base rather than a nucleophile.

Understanding the Question

Reaction 3 converts 1-chloropropane to propene using NaOH. The question states this is an elimination reaction and asks for the equation, solvent, and conditions.

Approach

Write the balanced chemical equation for the dehydrohalogenation of 1-chloropropane. Then, specify the conditions that favor elimination: a non-aqueous solvent (ethanol) and heat.

Step-by-Step Reasoning

1. Equation: The reactant is $\text{C}_2\text{H}_5\text{CH}_2\text{Cl}$ (or $\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}$). It reacts with $\text{NaOH}$. The products are propene ($\text{C}_3\text{H}_6$), water ($\text{H}_2\text{O}$), and sodium chloride ($\text{NaCl}$).
   $\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{H}_2\text{O} + \text{NaCl}$
   (Using the condensed formula $\text{C}_2\text{H}_5\text{CH}_2\text{Cl}$ as given in the question is also perfectly acceptable and often preferred to match the scheme).
2. Solvent: Must be ethanol (or 'alcoholic'). Aqueous conditions would lead to substitution (propan-1-ol).
3. Conditions: Must include heat or reflux. Heating provides the activation energy required for the elimination pathway.

Key Takeaways

The solvent is the key differentiator between substitution and elimination of halogenoalkanes with $\text{OH}^-$. Ethanol + heat = elimination; water + cold = substitution.

Common Mistakes

• Writing $\text{KOH}$ instead of $\text{NaOH}$: while chemically valid, stick to the reagent given in the question ($\text{NaOH}$) unless asked otherwise.
• Forgetting the by-products ($\text{H}_2\text{O}$ and $\text{NaCl}$). The equation must be balanced.
• Writing 'water' as the solvent: this is the classic trap. Water favors substitution.
• Forgetting 'heat': elimination has a higher activation energy than substitution, so heat is required to favor it.

Things to Be Careful About

Ensure the equation is balanced. The mark scheme awards one mark for the correct organic/inorganic products and one for the solvent and conditions. Do not write 'alcohol' generally; specify 'ethanol'.

Background Concept

Alcohols can be dehydrated to form alkenes. This is an elimination reaction (specifically, an acid-catalyzed dehydration or catalytic dehydration).

• Acid-catalyzed: Uses concentrated sulfuric acid ($\text{H}_2\text{SO}_4$) or concentrated phosphoric acid ($\text{H}_3\text{PO}_4$) with heating (typically around $170^\circ\text{C}$ for $\text{H}_2\text{SO}_4$).
• Catalytic: Passes alcohol vapor over a heated catalyst like aluminum oxide ($\text{Al}_2\text{O}_3$) at around $300^\circ\text{C}$.

Understanding the Question

The question asks for a direct conversion of propan-1-ol ($\text{C}_2\text{H}_5\text{CH}_2\text{OH}$) to propene ($\text{C}_3\text{H}_6$). This is a dehydration reaction. We need to suggest the reagent and conditions.

Approach

Recall the standard reagents for dehydrating alcohols to alkenes.

Step-by-Step Reasoning

• Reagent: Concentrated sulfuric acid ($\text{conc. H}_2\text{SO}_4$) is the most common. Concentrated phosphoric acid ($\text{conc. H}_3\text{PO}_4$) is also acceptable and often preferred in industry as it is less oxidizing. Alternatively, aluminum oxide ($\text{Al}_2\text{O}_3$) can be used as a solid catalyst.
• Conditions: Heat is required. For acids, this is typically 'heat' or 'reflux' or a specific temperature like $170^\circ\text{C}$. For $\text{Al}_2\text{O}_3$, 'heat' or 'vapor over heated catalyst' is used.

Key Takeaways

Dehydration of alcohols to alkenes requires an acid catalyst (conc. $\text{H}_2\text{SO}_4$ or $\text{H}_3\text{PO}_4$) and heat, or a solid catalyst ($\text{Al}_2\text{O}_3$) and heat.

Common Mistakes

Writing 'dilute' acid: dilute acid favors hydration of alkenes (the reverse reaction) or substitution to form alcohols, not dehydration.
Writing just 'heat': heat alone is insufficient; a catalyst or reagent is needed to protonate the OH group and make it a good leaving group ($\text{H}_2\text{O}$).

Things to Be Careful About

Always specify concentrated for the acids. 'Conc. $\text{H}_2\text{SO}_4$' is the safest, most standard answer. Don't forget to include 'heat' or 'high temperature'.

Background Concept

Alkenes contain a carbon-carbon double bond ($\text{C}=\text{C}$), which consists of a sigma bond and a pi bond. The pi bond is relatively weak and can be broken to form new sigma bonds with other monomer molecules. This process, where monomers add together without the loss of any small molecules, is called addition polymerisation.

Understanding the Question

Propene ($\text{CH}_3\text{CH}=\text{CH}_2$) is an alkene. The question asks for the type of polymerisation that forms poly(propene) from propene.

Approach

Identify the functional group in the monomer (alkene) and match it to the polymerisation type.

Step-by-Step Reasoning

Since the monomer is an alkene and the polymer chain contains only single bonds (the double bond opens up to link monomers), the reaction is addition polymerisation. No by-products (like water in condensation polymerisation) are formed.

Key Takeaways

Monomers with $\text{C}=\text{C}$ double bonds (alkenes) undergo addition polymerisation.

Common Mistakes

Writing 'condensation': this is for monomers with two different functional groups (e.g., diols and dicarboxylic acids) that lose a small molecule like water.

Things to Be Careful About

The answer is simply 'addition'. Do not overcomplicate it.

Background Concept

Stereoisomers are molecules that have the same molecular formula and the same connectivity of atoms (same structural formula), but differ in the way their atoms are arranged in space. This includes:

• Geometric (cis/trans or E/Z) isomerism: Due to restricted rotation around a $\text{C}=\text{C}$ double bond or a ring.
• Optical isomerism: Due to the presence of a chiral center (a carbon atom bonded to four different groups).

Understanding the Question

The question asks for a definition of stereoisomerism. This is a standard recall question.

Approach

Provide the two-part definition: same connectivity, different 3D arrangement.

Step-by-Step Reasoning

1. M1: Must state they have the same structural formula (and/or molecular formula). This distinguishes them from structural isomers, which have different connectivity.
2. M2: Must state they have a different 3D or spatial arrangement of atoms or groups. This distinguishes them from the same molecule drawn differently.

Key Takeaways

Stereoisomerism = same structural formula + different spatial arrangement.

Common Mistakes

Saying 'same molecular formula but different structural formula': this is the definition of structural isomerism.
Saying 'different shape': 'shape' is vague; 'spatial arrangement' or '3D arrangement' is the precise terminology.

Things to Be Careful About

Both parts of the definition are usually required for the 2 marks. Ensure 'structural formula' is used, not just 'formula'.

Background Concept

Poly(propene) is formed from propene ($\text{CH}_2=\text{CHCH}_3$) via addition polymerisation. The repeat unit is $-\text{CH}_2-\text{CH}(\text{CH}_3)-$.
In the polymer chain, every other carbon atom (the one bearing the methyl group) is bonded to:

1. A hydrogen atom ($-\text{H}$)
2. A methyl group ($-\text{CH}_3$)
3. A $-\text{CH}_2-$ group (part of the chain)
4. Another carbon group from the rest of the chain (which is effectively different from $-\text{CH}_2-$ due to length/continuation)

Actually, strictly speaking, in a polymer chain, the two 'chain' directions are considered different if the chain is long enough, making the carbon a chiral center. This leads to optical isomerism (specifically, isotactic, syndiotactic, or atactic arrangements, which are forms of stereoisomerism related to the chiral centers). For A-Level purposes, identifying the chiral center and stating 'optical isomerism' is the key.

Understanding the Question

We need to:

1. Draw a section of poly(propene) showing two repeat units (displayed formula, showing all atoms and bonds).
2. Identify the type of stereoisomerism.
3. Explain the answer using the diagram.

Approach

Draw the displayed formula for $-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}(\text{CH}_3)-$. Identify the carbon with the methyl group. Check if it has 4 different groups. Conclude the type of stereoisomerism.

Step-by-Step Reasoning

1. Drawing: Poly(propene) repeat unit is $-\text{CH}_2-\text{CH}(\text{CH}_3)-$. Two repeat units: $-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}(\text{CH}_3)-$. Must show all C-H bonds and C-C bonds. The mark scheme shows a displayed formula with H and CH3 on alternating carbons.
2. Type: The carbon atom bonded to the $\text{CH}_3$ group is bonded to: $-\text{H}$, $-\text{CH}_3$, $-\text{CH}_2-\text{CH}(\text{CH}_3)-...$ (left side), and $-\text{CH}_2-\text{CH}(\text{CH}_3)-...$ (right side). In a polymer, these chain ends are different, making it a chiral center. Thus, optical isomerism.
3. Explanation: The carbon atom carrying the methyl group is bonded to four different groups (it is a chiral centre / asymmetric carbon).