Chemistry 9701 / 42 — October/November 2024

DifficultyMedium-Easy

Worked solution

Working

The half-life is $t_{1/2} = 900$ s, so over $1800$ s the reaction passes through two half-lives.

After two half-lives:

[X] = 0.180 \times \frac{1}{2} \times \frac{1}{2} = 0.045 \text{ mol dm}^{-3}

Change in $[X]$ :

0.180 - 0.045 = 0.135 \text{ mol dm}^{-3}

Average rate:

\frac{0.135}{1800} = 7.5 \times 10^{-5} \text{ mol dm}^{-3} \text{ s}^{-1}

Answer

$7.5 \times 10^{-5}$ mol dm $^{-3}$ s $^{-1}$

Final answer

7.5 × 10^-5 mol dm^-3 s^-1

Detailed explanation

Background Concept

In first-order kinetics, the rate of reaction is directly proportional to the concentration of one reactant: rate = $k[X]$ . A defining feature of first-order reactions is that the half-life — the time for the concentration to fall to half its initial value — is constant and independent of the starting concentration. Mathematically, $t_{1/2} = \frac{\ln 2}{k}$ . This constancy makes the half-life a powerful tool for predicting concentrations at future times.

Understanding the Question

The reaction X $\rightarrow$ 2Y is first order in X, with $t_{1/2} = 900$ s at 20°C. A solution is prepared with $[X] = 0.180$ mol dm $^{-3}$ . We must find the average rate over the first 1800 s. The average rate is the total change in concentration of X divided by the total time elapsed. Crucially, 1800 s is exactly two half-lives, so the concentration falls from 0.180 to 0.045 mol dm $^{-3}$ .

Approach

First recognise that 1800 s = 2 × 900 s, so two half-lives elapse. After one half-life the concentration halves; after two it quarters. Compute $[X]$ after 1800 s, subtract from the initial concentration to get the change, and divide by 1800 s. The result is the average rate of consumption of X.

Step-by-Step Reasoning

After one half-life (t = 900 s): $[X] = 0.180 / 2 = 0.090$ mol dm $^{-3}$ .

After two half-lives (t = 1800 s): $[X] = 0.090 / 2 = 0.045$ mol dm $^{-3}$ .

Change in $[X]$ = 0.180 − 0.045 = 0.135 mol dm $^{-3}$ .

Average rate = 0.135 / 1800 = $7.5 \times 10^{-5}$ mol dm $^{-3}$ s $^{-1}$ .

The rate is reported as a positive quantity even though $[X]$ is decreasing; the rate of reaction is conventionally positive.

Key Takeaways

For a first-order reaction, the half-life is constant regardless of concentration.
After $n$ half-lives, the remaining concentration is $(1/2)^n$ of the initial.
Average rate = total change in concentration ÷ total time.

Common Mistakes

Treating 1800 s as one half-life and halving $[X]$ only once — the correct answer requires two halvings.
Quoting the rate as negative — rates are always positive.
Forgetting units (mol dm $^{-3}$ s $^{-1}$ ).

Things to Be Careful About

The concentration after two half-lives is 0.045, not 0.090 mol dm $^{-3}$ .
The average rate over the whole interval is not the same as the instantaneous rate at any moment; it is the slope of the chord on a concentration–time graph.
The rate is measured with respect to X, the reactant consumed.

Techniques used

recognise that 1800 s equals two half-livescalculate the concentration remaining after two half-livesdivide the change in concentration by the time interval

(ii)

Complete the rate equation for reaction 1.

rate =

(iii)

Show that the rate constant, $k$ , is $7.70 \times 10^{-4}\text{ s}^{-1}$ at $20^\circ\text{C}$ .

DifficultyEasy

Worked solution

Working

For a first-order reaction, $t_{1/2} = \frac{\ln 2}{k}$ .

k = \frac{0.693}{900} = 7.70 \times 10^{-4} \text{ s}^{-1}

Answer

$k = 7.70 \times 10^{-4}$ s $^{-1}$ , as required.

Final answer

7.70 × 10^-4 s^-1

Detailed explanation

Background Concept

For a first-order reaction, the half-life is related to the rate constant by $t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$ . This relationship follows from integrating the first-order rate law. Rearranging gives $k = \frac{0.693}{t_{1/2}}$ . Since $t_{1/2}$ is constant for a first-order reaction, $k$ can be found from a single half-life measurement.

Understanding the Question

The half-life is given as 900 s at 20°C. We must show that $k = 7.70 \times 10^{-4}$ s $^{-1}$ . This is a direct substitution into the half-life–rate-constant relationship.

Approach

Use $k = \frac{0.693}{t_{1/2}}$ and substitute $t_{1/2} = 900$ s.

Step-by-Step Reasoning

k = \frac{0.693}{900} = 7.70 \times 10^{-4} \text{ s}^{-1}

This confirms the value given in the question. The units of $k$ for a first-order reaction are s $^{-1}$ .

Key Takeaways

$t_{1/2} = \frac{\ln 2}{k}$ for first-order reactions.
$k = \frac{0.693}{t_{1/2}}$ .
First-order rate constants have units s $^{-1}$ .

Common Mistakes

Using $k = \frac{t_{1/2}}{0.693}$ (inverted).
Using $\log 2 = 0.301$ instead of $\ln 2 = 0.693$ .

Things to Be Careful About

$\ln 2 \approx 0.693$ , not 0.301.
The units of $k$ must be s $^{-1}$ for a first-order reaction.

Techniques used

apply the half-life–rate constant relationship for a first-order reaction

(iv)

Calculate the initial rate of reaction 1 when the concentration of X is $0.150\text{ mol dm}^{-3}$ .

Include units.

rate = .............................. units ..............................

DifficultyMedium-Easy

Worked solution

Working

rate $= k[X] = 7.70 \times 10^{-4} \times 0.150 = 1.155 \times 10^{-4}$

Answer

rate $= 1.16 \times 10^{-4}$ mol dm $^{-3}$ s $^{-1}$

Final answer

1.16 × 10^-4 mol dm^-3 s^-1

Detailed explanation

Background Concept

The rate equation rate = $k[X]$ allows the instantaneous rate to be calculated at any concentration, provided $k$ is known. Here $k = 7.70 \times 10^{-4}$ s $^{-1}$ (from part iii) and $[X] = 0.150$ mol dm $^{-3}$ .

Understanding the Question

We must calculate the initial rate when $[X] = 0.150$ mol dm $^{-3}$ , and give units. This is a direct substitution into the rate equation from part (ii).

Approach

Substitute $k = 7.70 \times 10^{-4}$ s $^{-1}$ and $[X] = 0.150$ mol dm $^{-3}$ into rate = $k[X]$ .

Step-by-Step Reasoning

rate = $7.70 \times 10^{-4} \times 0.150 = 1.155 \times 10^{-4} \approx 1.16 \times 10^{-4}$ mol dm $^{-3}$ s $^{-1}$ .

The units: (s $^{-1}$ ) × (mol dm $^{-3}$ ) = mol dm $^{-3}$ s $^{-1}$ .

Key Takeaways

The rate equation can be used to find the rate at any concentration.
Units of rate are always concentration per time.

Common Mistakes

Forgetting to multiply by $k$ (just quoting $[X]$ ).
Wrong units (e.g. mol dm $^{-3}$ or s $^{-1}$ alone).
Rounding to the wrong number of significant figures (the mark scheme expects $1.16 \times 10^{-4}$ , i.e. 3 s.f.).

Things to Be Careful About

Use the value of $k$ from part (iii), $7.70 \times 10^{-4}$ s $^{-1}$ .
The answer should be given to 3 significant figures to match the data.

Techniques used

substitute values into the rate equationcalculate the initial rate with correct units

(b)

Catalysts may be homogeneous or heterogeneous.

5M

(i)

Platinum is a transition element. Explain why transition elements behave as catalysts.

DifficultyEasy

Worked solution

Answer

Transition elements behave as catalysts because they have variable oxidation states and possess vacant d orbitals, allowing them to accept electrons and form dative bonds with reactants, providing an alternative reaction pathway with a lower activation energy.

Final answer

Variable oxidation states and vacant d orbitals / ability to form dative bonds

Detailed explanation

Background Concept

Transition elements (d-block metals) are effective catalysts because they can adopt multiple oxidation states and have vacant d orbitals. This allows them to form intermediate complexes with reactants, providing an alternative reaction pathway with a lower activation energy. The catalyst is regenerated at the end of the reaction.

Understanding the Question

The question asks why transition elements behave as catalysts. The mark scheme credits any two of: variable oxidation state; vacant/empty/unfilled d orbitals; can form dative bonds / accept electrons.

Approach

State two distinct properties of transition elements that enable catalysis.

Step-by-Step Reasoning

Transition elements have variable oxidation states, so they can accept and donate electrons readily, shuttling between oxidation states during a reaction. They also have vacant d orbitals, which allow them to accept electron pairs from reactants, forming dative (coordinate) bonds and stabilising reaction intermediates. This lowers the activation energy and speeds up the reaction.

Key Takeaways

Variable oxidation state enables electron transfer during catalysis.
Vacant d orbitals allow dative bonding with reactants.

Common Mistakes

Saying "they have a large surface area" — that is a property of the finely divided metal, not of transition elements per se.
Giving only one reason when the mark scheme requires two.

Things to Be Careful About

The mark scheme requires TWO points for one mark. Give both.

Techniques used

explain catalytic activity of transition elements in terms of variable oxidation states and vacant d orbitals

(ii)

Name the metal catalyst in the Haber process and explain why it is a heterogeneous catalyst.

metal .........................................

(iii)

Platinum acts as a heterogeneous catalyst in the removal of nitrogen dioxide, $\text{NO}_2$ , from the exhaust gases of car engines.

Describe the mode of action of a platinum catalyst in this process.

DifficultyMedium-Easy

Worked solution

Answer

Reactant molecules (e.g. NO $_2$ and CO) are adsorbed onto the platinum surface
Bonds within the reactants weaken, lowering the activation energy
Product molecules are desorbed from the surface

Final answer

Adsorption of reactants, weakening of reactant bonds, desorption of products

Detailed explanation

Background Concept

Heterogeneous catalysis occurs on the surface of a solid catalyst. The mechanism involves: (1) adsorption of reactant molecules onto the catalyst surface; (2) weakening of bonds within the reactants (due to chemisorption), which lowers the activation energy; (3) reaction to form products; (4) desorption of products, freeing the surface for more reactant molecules.

Understanding the Question

Platinum acts as a heterogeneous catalyst in removing NO $_2$ from car exhaust gases. We must describe the mode of action: adsorption, bond weakening, desorption.

Approach

Describe the three key stages of heterogeneous catalysis on the platinum surface.

Step-by-Step Reasoning

Adsorption: NO $_2$ (and other exhaust gases such as CO) are adsorbed onto the platinum surface. The reactants are held on the surface by weak bonds.
Bond weakening: The bonds within the adsorbed reactant molecules weaken because of the interaction with the metal surface, lowering the activation energy for the reaction.
Desorption: The product molecules (e.g. N $_2$ , CO $_2$ ) are desorbed from the surface, leaving the catalyst available for further reaction.

Key Takeaways

Heterogeneous catalysis: adsorption $\rightarrow$ bond weakening $\rightarrow$ reaction $\rightarrow$ desorption.
The catalyst provides a surface and an alternative lower-energy pathway.

Common Mistakes

Omitting "desorption" — the mark scheme explicitly credits it.
Saying "the catalyst is used up" — it is regenerated.

Things to Be Careful About

The mark scheme gives 2 marks for 3 correct points: adsorption, bond weakening, desorption. Include all three.

Techniques used

describe adsorption of reactants onto the catalyst surfaceexplain bond weakening on the surfacedescribe desorption of products

(iv)

$\text{NO}_2$ acts as a homogeneous catalyst in the oxidation of atmospheric sulfur dioxide, $\text{SO}_2$ .

Write equations for the two reactions that occur.

equation 1 .........................................................................................................................

equation 2 .........................................................................................................................

DifficultyMedium-Easy

Worked solution

Answer

\text{NO}_2 + \text{SO}_2 \rightarrow \text{NO} + \text{SO}_3

2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2

Final answer

NO2 + SO2 -> NO + SO3; 2NO + O2 -> 2NO2

Detailed explanation

Background Concept

A homogeneous catalyst is in the same phase as the reactants. NO $_2$ catalyses the oxidation of SO $_2$ by O $_2$ in the atmosphere. NO $_2$ oxidises SO $_2$ to SO $_3$ , being reduced to NO. NO is then re-oxidised by O $_2$ back to NO $_2$ , completing the catalytic cycle. The NO $_2$ is regenerated, so it is not consumed overall.

Understanding the Question

Write two equations showing how NO $_2$ catalyses the oxidation of SO $_2$ . The two steps are: (1) NO $_2$ + SO $_2$ $\rightarrow$ NO + SO $_3$ ; (2) 2NO + O $_2$ $\rightarrow$ 2NO $_2$ .

Approach

Recognise the catalytic cycle: NO $_2$ is the oxidant in step 1, and NO is re-oxidised in step 2.

Step-by-Step Reasoning

Step 1: NO $_2$ oxidises SO $_2$ to SO $_3$ . NO $_2$ is reduced to NO.

\text{NO}_2 + \text{SO}_2 \rightarrow \text{NO} + \text{SO}_3

Step 2: NO is oxidised back to NO $_2$ by atmospheric oxygen.

2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2

Adding the two equations: 2SO $_2$ + O $_2$ $\rightarrow$ 2SO $_3$ , with NO $_2$ regenerated. This shows NO $_2$ is a catalyst — it appears in the first step and is regenerated in the second.

Key Takeaways

A catalyst is regenerated at the end of the reaction.
Homogeneous catalysis often involves a redox cycle.

Common Mistakes

Writing the overall equation instead of the two steps.
Not balancing the second equation (2NO + O $_2$ $\rightarrow$ 2NO $_2$ ).

Things to Be Careful About

Both equations must be balanced.
The mark scheme awards one mark for BOTH equations.

Techniques used

write the equation for NO2 oxidising SO2write the equation for regeneration of NO2recognise the catalytic cycle

(c)

$\text{SO}_2$ dissolves in water, forming $\text{H}_2\text{SO}_3$ .

$\text{H}_2\text{SO}_3$ can be oxidised under acidic conditions.

The relevant electrode reaction and its $E^\ominus$ value are shown.

\text{SO}_4^{2-} + 4\text{H}^+ + 2\text{e}^- \rightleftharpoons \text{H}_2\text{SO}_3 + \text{H}_2\text{O} \qquad E^\ominus = +0.17\text{ V}

Four more half-equations for reactions occurring under acidic conditions, and their $E^\ominus$ values, are shown.

\begin{aligned} \text{H}_3\text{BO}_3 + 3\text{H}^+ + 3\text{e}^- &\rightleftharpoons \text{B} + 3\text{H}_2\text{O} & E^\ominus &= -0.73\text{ V} \\ \text{BiO}^+ + 2\text{H}^+ + 3\text{e}^- &\rightleftharpoons \text{Bi} + \text{H}_2\text{O} & E^\ominus &= +0.28\text{ V} \\ \text{S} + 2\text{H}^+ + 2\text{e}^- &\rightleftharpoons \text{H}_2\text{S} & E^\ominus &= +0.14\text{ V} \\ \text{Sb} + 3\text{H}^+ + 3\text{e}^- &\rightleftharpoons \text{SbH}_3 & E^\ominus &= -0.51\text{ V} \end{aligned}

Select the oxidising agent that could oxidise $\text{H}_2\text{SO}_3$ to $\text{SO}_4^{2-}$ ions under acidic conditions.

Write an equation, and give the $E^\ominus_{\text{cell}}$ value, for the reaction that occurs.

oxidising agent .........................................

equation .........................................................................................................................

$E^\ominus_{\text{cell}} =$ .............................. V

DifficultyMedium

Worked solution

Working

To oxidise H $_2$ SO $_3$ to SO $_4^{2-}$ , the oxidising agent must have a more positive reduction potential than $E^\ominus = +0.17$ V. Only BiO $^+$ /Bi with $E^\ominus = +0.28$ V satisfies this.

Oxidation (reverse of the given half-reaction, ×3):

3\text{H}_2\text{SO}_3 + 3\text{H}_2\text{O} \rightarrow 3\text{SO}_4^{2-} + 12\text{H}^+ + 6\text{e}^-

Reduction (×2):

2\text{BiO}^+ + 4\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Bi} + 2\text{H}_2\text{O}

Overall:

3\text{H}_2\text{SO}_3 + 2\text{BiO}^+ + \text{H}_2\text{O} \rightarrow 3\text{SO}_4^{2-} + 8\text{H}^+ + 2\text{Bi}

E^\ominus_{\text{cell}} = 0.28 - 0.17 = 0.11 \text{ V}

Answer

oxidising agent: BiO $^+$

3\text{H}_2\text{SO}_3 + 2\text{BiO}^+ + \text{H}_2\text{O} \rightarrow 3\text{SO}_4^{2-} + 8\text{H}^+ + 2\text{Bi}

$E^\ominus_{\text{cell}} = 0.11$ V

Final answer

BiO+; 3H2SO3 + 2BiO+ + H2O -> 3SO4^2- + 8H+ + 2Bi; 0.11 V

Detailed explanation

Background Concept

Standard electrode potentials ( $E^\ominus$ ) measure the tendency of a half-reaction to occur as a reduction. A more positive $E^\ominus$ means a greater tendency to gain electrons (be reduced). An oxidising agent is itself reduced; it can oxidise another species if its $E^\ominus$ is more positive than the $E^\ominus$ of the reduction half-reaction of the species being oxidised. The cell potential is $E^\ominus_{\text{cell}} = E^\ominus(\text{reduction}) - E^\ominus(\text{oxidation})$ .

Understanding the Question

We must select, from four half-equations, the oxidising agent that can oxidise H $_2$ SO $_3$ to SO $_4^{2-}$ under acidic conditions. The relevant half-reaction is SO $_4^{2-}$ + 4H $^+$ + 2e $^-$ $\rightleftharpoons$ H $_2$ SO $_3$ + H $_2$ O, $E^\ominus = +0.17$ V. The oxidising agent must have $E^\ominus > +0.17$ V. Only BiO $^+$ /Bi (+0.28 V) qualifies. We must write the balanced overall equation and calculate $E^\ominus_{\text{cell}}$ .

Approach

Compare $E^\ominus$ values: the oxidising agent must have $E^\ominus > +0.17$ V.
Identify BiO $^+$ as the only candidate.
Write the oxidation half-reaction (reverse of the given reduction) and the reduction half-reaction.
Balance electrons and combine.
Calculate $E^\ominus_{\text{cell}} = 0.28 - 0.17 = 0.11$ V.

Step-by-Step Reasoning

The four candidate half-reactions have $E^\ominus$ values:

H $_3$ BO $_3$ /B: −0.73 V — cannot oxidise ( $E^\ominus < 0.17$ V)
BiO $^+$ /Bi: +0.28 V — can oxidise ( $E^\ominus > 0.17$ V)
S/H $_2$ S: +0.14 V — cannot ( $E^\ominus < 0.17$ V)
Sb/SbH $_3$ : −0.51 V — cannot ( $E^\ominus < 0.17$ V)

So BiO $^+$ is the oxidising agent.

Oxidation (H $_2$ SO $_3$ loses electrons — reverse of given reduction):

\text{H}_2\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4\text{H}^+ + 2\text{e}^-

Reduction (BiO $^+$ gains electrons):

\text{BiO}^+ + 2\text{H}^+ + 3\text{e}^- \rightarrow \text{Bi} + \text{H}_2\text{O}

Balance electrons (LCM of 2 and 3 = 6):

3(\text{H}_2\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4\text{H}^+ + 2\text{e}^-)

2(\text{BiO}^+ + 2\text{H}^+ + 3\text{e}^- \rightarrow \text{Bi} + \text{H}_2\text{O})

Sum:

3\text{H}_2\text{SO}_3 + 3\text{H}_2\text{O} + 2\text{BiO}^+ + 4\text{H}^+ \rightarrow 3\text{SO}_4^{2-} + 12\text{H}^+ + 2\text{Bi} + 2\text{H}_2\text{O}

Cancel 2H $_2$ O from both sides:

3\text{H}_2\text{SO}_3 + \text{H}_2\text{O} + 2\text{BiO}^+ \rightarrow 3\text{SO}_4^{2-} + 8\text{H}^+ + 2\text{Bi}

E^\ominus_{\text{cell}} = E^\ominus(\text{reduction}) - E^\ominus(\text{oxidation}) = 0.28 - 0.17 = +0.11 \text{ V}

Since $E^\ominus_{\text{cell}} > 0$ , the reaction is feasible.

Key Takeaways

A species with a more positive $E^\ominus$ can oxidise a species with a less positive $E^\ominus$ .
$E^\ominus_{\text{cell}} = E^\ominus(\text{reduction}) - E^\ominus(\text{oxidation})$ .
A positive $E^\ominus_{\text{cell}}$ indicates a spontaneous reaction.

Common Mistakes

Choosing S/H $_2$ S (+0.14 V) because it is close to +0.17 V — but it is still less positive, so it cannot oxidise H $_2$ SO $_3$ .
Forgetting to reverse the oxidation half-reaction.
Not balancing electrons before combining half-equations.
Using the wrong sign in the $E^\ominus_{\text{cell}}$ calculation.

Things to Be Careful About

The oxidising agent is the species that is reduced (BiO $^+$ $\rightarrow$ Bi), not the one that is oxidised.
The overall equation must be balanced for atoms AND charge.
$E^\ominus_{\text{cell}} = E^\ominus(\text{reduction}) - E^\ominus(\text{oxidation})$ , not the reverse.
The mark scheme gives 1 mark for each of: identifying BiO $^+$ , the balanced equation, and the $E^\ominus_{\text{cell}}$ value.

Techniques used

compare standard electrode potentials to select the oxidising agentreverse a half-equation for the oxidation stepbalance electrons to combine half-equationscalculate the standard cell potential

Q2MediumChemical Energetics

(a)

Predict and explain the variation in enthalpy change of hydration for the ions $\text{Na}^+$ , $\text{Mg}^{2+}$ and $\text{Al}^{3+}$ .

DifficultyMedium-Easy

Worked solution

Answer

$\Delta H_{\text{hyd}}$ becomes more negative from $\text{Na}^+$ to $\text{Mg}^{2+}$ to $\text{Al}^{3+}$ because:

The ionic charge increases from left to right (+1, +2, +3).
The ionic radius decreases from left to right.
This causes increased attractive force between the ion and the $\delta^-$ oxygen atoms of water molecules, releasing more energy on hydration.

Final answer

ΔH_hyd becomes more negative left to right due to increasing charge and decreasing radius, causing stronger attraction to water molecules.

Detailed explanation

Background Concept

Enthalpy change of hydration ( $\Delta H_{\text{hyd}}$ ) is the enthalpy change when one mole of gaseous ions is dissolved in sufficient water to form infinitely dilute solution with a concentration of 1 mol dm⁻³. It is always exothermic (negative) because energy is released when water molecules form ion–dipole bonds with the gaseous ion.

The magnitude of $\Delta H_{\text{hyd}}$ depends on the charge density of the ion — the ratio of charge to ionic radius. A higher charge density means stronger electrostatic attraction to the partial negative charge on the oxygen atom of water, resulting in more energy released (a more negative $\Delta H_{\text{hyd}}$ ).

Understanding the Question

The question asks for a prediction (direction of the trend) and an explanation (why the trend occurs) for the variation in $\Delta H_{\text{hyd}}$ across three period 3 cations: $\text{Na}^+$ , $\text{Mg}^{2+}$ , and $\text{Al}^{3+}$ . The command word "predict and explain" requires both the observation and the reasoning.

Approach

Identify the two structural factors that change across the period: ionic charge (increases) and ionic radius (decreases). Both contribute to increasing charge density, which strengthens ion–dipole interactions with water.

Step-by-Step Reasoning

Point 1 — Charge increases: Moving from Na to Mg to Al, the ionic charge goes from +1 to +2 to +3. A higher charge means a stronger electrostatic attraction to the $\delta^-$ end of water molecules.

Point 2 — Radius decreases: All three ions are isoelectronic with neon (10 electrons), but the nuclear charge increases from 11 to 12 to 13. The greater nuclear attraction pulls the same number of electrons closer, so ionic radius decreases: $\text{Na}^+$ > $\text{Mg}^{2+}$ > $\text{Al}^{3+}$ .

Point 3 — Combined effect: Higher charge and smaller radius both increase charge density, strengthening the ion–dipole attraction to water. More energy is released as water molecules orient around the ion, so $\Delta H_{\text{hyd}}$ becomes more negative.

Key Takeaways

Hydration enthalpy magnitude is governed by charge density (charge/radius ratio).
Across a period, both increasing charge and decreasing radius reinforce the trend toward more exothermic hydration.
Always give the direction of the trend AND the reason in "predict and explain" questions.

Common Mistakes

Stating only that "charge increases" without mentioning the radius effect — the mark scheme requires both factors.
Saying "the ion gets smaller" without linking it to increased attraction to water molecules.
Confusing $\Delta H_{\text{hyd}}$ with lattice enthalpy — hydration is about gaseous ions dissolving, not forming a lattice.

Things to Be Careful About

Use the phrase "more negative" or "increases in magnitude" rather than just "increases," which is ambiguous.
State that the attraction is specifically to the $\delta^-$ oxygen of water (or "water molecules"), not just "water."
All three marks must be present: charge, radius, and the resulting stronger attraction.

Techniques used

compare ionic charge across a periodrelate charge density to hydration enthalpy magnitudeexplain trend using Coulombic attraction

(b)

Fig. 2.1 shows an incomplete energy cycle.

Fig. 2.1

6M

(i)

Complete line C on Fig. 2.1. Include state symbols.

(ii)

Use both words and symbols to identify change 2 on Fig. 2.1.

Use changes 1 and 3 as examples of how this should be done.

DifficultyMedium-Easy

Worked solution

Answer

Words: Enthalpy change of hydration of magnesium ions and chloride ions

Symbols: $\Delta H_{\text{hyd}}\text{ Mg}^{2+} + 2\Delta H_{\text{hyd}}\text{ Cl}^-$

Final answer

ΔH_hyd Mg²⁺ + 2ΔH_hyd Cl⁻

Detailed explanation

Background Concept

Change 2 in the cycle goes from gaseous ions ( $\text{Mg}^{2+}\text{(g)} + 2\text{Cl}^-\text{(g)}$ ) directly to aqueous ions ( $\text{Mg}^{2+}\text{(aq)} + 2\text{Cl}^-\text{(aq)}$ ). This is the combined hydration of all the ions present in one formula unit of $\text{MgCl}_2$ .

The enthalpy change of hydration is defined for one mole of a single gaseous ion. Since $\text{MgCl}_2$ contains one $\text{Mg}^{2+}$ and two $\text{Cl}^-$ ions, the total hydration enthalpy is $\Delta H_{\text{hyd}}(\text{Mg}^{2+}) + 2\Delta H_{\text{hyd}}(\text{Cl}^-)$ .

Understanding the Question

The question asks to identify change 2 using both words and symbols, following the format of changes 1 and 3. Change 1 is identified as "lattice energy of magnesium chloride, $\Delta H_{\text{latt}}\text{ MgCl}_2\text{(s)}$ " and change 3 as "enthalpy change of solution of magnesium chloride, $\Delta H_{\text{sol}}\text{ MgCl}_2\text{(s)}$ ".

Approach

Name the process in words (what happens to the gaseous ions to become aqueous) and then write the symbolic expression showing the correct stoichiometry.

Step-by-Step Reasoning

Words mark: The process is the hydration of the gaseous ions — specifically the enthalpy change of hydration of magnesium ions AND chloride ions. Both must be mentioned.
Symbols mark: Since there are two chloride ions per formula unit of $\text{MgCl}_2$ , the expression is $\Delta H_{\text{hyd}}\text{ Mg}^{2+} + 2\Delta H_{\text{hyd}}\text{ Cl}^-$ . The coefficient of 2 before the chloride term is essential.

Key Takeaways

When expressing a combined enthalpy change from a cycle, the stoichiometric coefficients from the formula unit must appear.
"Enthalpy change of hydration" applies to individual ions, not to the compound as a whole.

Common Mistakes

Writing just " $\Delta H_{\text{hyd}}$ " without specifying which ions.
Forgetting the coefficient of 2 before $\Delta H_{\text{hyd}}\text{ Cl}^-$ .
Writing "lattice energy" or "solution energy" instead of "hydration."

Things to Be Careful About

The mark scheme requires BOTH words and symbols — omitting either loses a mark.
The word "and" between magnesium ions and chloride ions is important — both ions are hydrated.

Techniques used

name an enthalpy change in wordsexpress a combined enthalpy change using symbols

(iii)

Calculate a value for the lattice energy of magnesium chloride, $\Delta H_{\text{latt}}\text{ MgCl}_2(\text{s})$ , by selecting and using appropriate data from Table 2.1.

Table 2.1

energy change	value / $\text{kJ mol}^{-1}$
enthalpy change of solution of magnesium chloride	$-155$
enthalpy change of formation of magnesium chloride	$-642$
first ionisation energy of magnesium	$+736$
second ionisation energy of magnesium	$+1450$
electron affinity of chlorine	$-349$
enthalpy change of hydration of $\text{Mg}^{2+}$	$-1920$
enthalpy change of hydration of $\text{Cl}^-$	$-364$

\Delta H_{\text{latt}}\text{ MgCl}_2(\text{s}) = \text{.............................. kJ mol}^{-1}

DifficultyMedium

Worked solution

Working

From the cycle: Change 2 = Change 1 + Change 3

\Delta H_{\text{hyd}}\text{ Mg}^{2+} + 2\Delta H_{\text{hyd}}\text{ Cl}^- = \Delta H_{\text{latt}}\text{ MgCl}_2 + \Delta H_{\text{sol}}\text{ MgCl}_2

Rearranging:

\Delta H_{\text{latt}}\text{ MgCl}_2 = \Delta H_{\text{hyd}}\text{ Mg}^{2+} + 2\Delta H_{\text{hyd}}\text{ Cl}^- - \Delta H_{\text{sol}}\text{ MgCl}_2

Substituting:

\Delta H_{\text{latt}}\text{ MgCl}_2 = (-1920) + 2(-364) - (-155)

= -1920 - 728 + 155

= -2493 \text{ kJ mol}^{-1}

Answer

$\Delta H_{\text{latt}}\text{ MgCl}_2(\text{s}) = -2493 \text{ kJ mol}^{-1}$

Final answer

-2493 kJ mol⁻¹

Detailed explanation

Background Concept

A Born-Haber cycle (or in this case a solution cycle) relates different enthalpy changes through Hess's law. The key principle is that the total enthalpy change is the same regardless of the path taken between two states.

In this cycle, going from gaseous ions to aqueous ions (change 2) can be achieved either directly or via the solid (change 1 then change 3). Therefore:

\text{Change 2} = \text{Change 1} + \text{Change 3}

This allows the lattice energy to be calculated if the hydration and solution enthalpies are known.

Understanding the Question

The table provides seven values, but only three are needed: $\Delta H_{\text{sol}}$ , $\Delta H_{\text{hyd}}(\text{Mg}^{2+})$ , and $\Delta H_{\text{hyd}}(\text{Cl}^-)$ . The formation enthalpy, ionisation energies, and electron affinity are distractors (they would be needed for a full Born-Haber cycle but not for this solution cycle).

Approach

Write the Hess's law relationship from the cycle.
Rearrange to solve for $\Delta H_{\text{latt}}$ .
Select only the relevant data.
Substitute carefully, watching signs and the factor of 2 for chloride.

Step-by-Step Reasoning

Step 1 — Hess's law from the cycle:
Change 2 (gaseous ions → aqueous ions) = Change 1 (gaseous ions → solid) + Change 3 (solid → aqueous ions)

\Delta H_{\text{hyd}}(\text{Mg}^{2+}) + 2\Delta H_{\text{hyd}}(\text{Cl}^-) = \Delta H_{\text{latt}} + \Delta H_{\text{sol}}

Step 2 — Rearrange:

\Delta H_{\text{latt}} = \Delta H_{\text{hyd}}(\text{Mg}^{2+}) + 2\Delta H_{\text{hyd}}(\text{Cl}^-) - \Delta H_{\text{sol}}

Step 3 — Select data: Only $\Delta H_{\text{sol}} = -155$ , $\Delta H_{\text{hyd}}(\text{Mg}^{2+}) = -1920$ , and $\Delta H_{\text{hyd}}(\text{Cl}^-) = -364$ are needed. The mark scheme specifically awards a mark for selecting these three only.

Step 4 — Substitute:

\Delta H_{\text{latt}} = (-1920) + 2(-364) - (-155)

Step 5 — Calculate:

= -1920 - 728 + 155 = -2493 \text{ kJ mol}^{-1}

The factor of $2 \times 364 = 728$ is a separate method mark in the scheme.

Key Takeaways

In a solution cycle, the lattice energy can be found from hydration and solution enthalpies without needing formation, ionisation, or electron affinity data.
Always check which path the cycle takes and write the Hess's law equation accordingly.
The stoichiometric coefficient (×2 for chloride) is a common source of lost marks.

Common Mistakes

Including the formation enthalpy or ionisation energies in the calculation (they belong to a full Born-Haber cycle, not this solution cycle).
Forgetting the factor of 2 for $\text{Cl}^-$ .
Sign errors: subtracting $\Delta H_{\text{sol}}$ when it should be $-(-155) = +155$ .
Writing the answer as positive (lattice energy of formation from gaseous ions is always negative).

Things to Be Careful About

The mark scheme gives one mark specifically for selecting only the correct three values — do not use all seven.
The factor of 2 before $\Delta H_{\text{hyd}}\text{ Cl}^-$ earns a separate method mark.
Final answer must be negative and to the correct value (-2493, not -2129 which would result from forgetting the ×2).

Techniques used

apply Hess's law to an energy cycleselect appropriate data from a tablecalculate lattice energy from hydration and solution enthalpies

(c)

Define entropy.

(d)

At $25^\circ\text{C}$ the enthalpy change of solution of compound $\mathbf{Z}$ is $+26\text{ kJ mol}^{-1}$ . The entropy change of solution of $\mathbf{Z}$ at the same temperature is $+52\text{ J K}^{-1}\text{mol}^{-1}$ .

Calculate the value of the Gibbs free energy change, $\Delta G$ , for the solution of $\mathbf{Z}$ at $25^\circ\text{C}$ .

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

DifficultyMedium-Easy

Worked solution

Working

\Delta G = \Delta H - T\Delta S

\Delta G = 26 - 298 \times 0.052

\Delta G = 26 - 15.496

\Delta G = +10.5 \text{ kJ mol}^{-1}

Answer

$\Delta G = +10.5 \text{ kJ mol}^{-1}$

Final answer

+10.5 kJ mol⁻¹

Detailed explanation

Background Concept

The Gibbs free energy change ( $\Delta G$ ) determines whether a process is thermodynamically feasible (spontaneous) at a given temperature:

\Delta G = \Delta H - T\Delta S

If $\Delta G < 0$ , the process is feasible (spontaneous). If $\Delta G > 0$ , it is not feasible. If $\Delta G = 0$ , the system is at equilibrium.

Units must be consistent: $\Delta H$ is typically in kJ mol⁻¹, $\Delta S$ in J K⁻¹ mol⁻¹, so $\Delta S$ must be divided by 1000 to give kJ K⁻¹ mol⁻¹ before multiplying by $T$ (in K).

Understanding the Question

Given $\Delta H = +26$ kJ mol⁻¹ and $\Delta S = +52$ J K⁻¹ mol⁻¹ at 25°C, calculate $\Delta G$ .

Approach

Convert temperature: $25°\text{C} = 298 \text{ K}$ .
Convert $\Delta S$ to kJ: $52 \text{ J K}^{-1}\text{mol}^{-1} = 0.052 \text{ kJ K}^{-1}\text{mol}^{-1}$ .
Substitute into $\Delta G = \Delta H - T\Delta S$ .

Step-by-Step Reasoning

Step 1 — Equation:

\Delta G = \Delta H - T\Delta S

Step 2 — Convert units:

$T = 25 + 273 = 298 \text{ K}$
$\Delta S = 52 \text{ J K}^{-1}\text{mol}^{-1} = 0.052 \text{ kJ K}^{-1}\text{mol}^{-1}$

Step 3 — Substitute:

\Delta G = 26 - (298 \times 0.052) = 26 - 15.496 = +10.504 \approx +10.5 \text{ kJ mol}^{-1}

The answer is positive, meaning the dissolution of Z is not feasible at 25°C.

Key Takeaways

Always convert $\Delta S$ from J to kJ to match $\Delta H$ .
Always convert temperature to Kelvin.
A positive $\Delta G$ means the process is not spontaneous.

Common Mistakes

Forgetting to convert $\Delta S$ from J to kJ, giving $\Delta G = 26 - 298 \times 52 = -15474$ (wildly wrong).
Using 25 instead of 298 for temperature.
Sign error: writing $\Delta G = \Delta H + T\Delta S$ .

Things to Be Careful About

The mark scheme gives one mark for the equation and one for the answer. Both must be correct.
Significant figures: the answer should be given to 3 s.f. (+10.5).
Include the positive sign to show the process is non-feasible.

Techniques used

apply the Gibbs free energy equationconvert temperature to Kelvinconvert entropy units to match enthalpy

(e)

2M

(i)

Use your answer to (d) to predict whether or not $\mathbf{Z}$ is soluble in water at $25^\circ\text{C}$ . Explain your answer.

(ii)

Predict whether $\mathbf{Z}$ becomes more or less soluble as the water is heated from $25^\circ\text{C}$ to $95^\circ\text{C}$ . Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

$\mathbf{Z}$ becomes more soluble as the water is heated because $\Delta G$ becomes more negative (less positive / closer to zero) as temperature increases, since $\Delta S$ is positive.

Final answer

More soluble because ΔG becomes more negative (less positive) as T increases since ΔS is positive.

Detailed explanation

Background Concept

From $\Delta G = \Delta H - T\Delta S$ , when $\Delta S$ is positive, increasing $T$ makes the $-T\Delta S$ term more negative. This reduces $\Delta G$ (makes it less positive or more negative), moving the system toward feasibility.

This is why many endothermic dissolutions become more soluble at higher temperatures — the entropy gain of dissolving becomes more important at higher $T$ .

Understanding the Question

Given that $\Delta H = +26$ kJ mol⁻¹ and $\Delta S = +52$ J K⁻¹ mol¹ (both positive), predict the effect of heating from 25°C to 95°C on solubility.

Approach

Consider what happens to $\Delta G$ as $T$ increases when $\Delta S > 0$ :

$-T\Delta S$ becomes more negative
So $\Delta G = \Delta H - T\Delta S$ decreases (becomes less positive)
Eventually $\Delta G$ could become negative, making dissolution feasible

Step-by-Step Reasoning

At 25°C (298 K): $\Delta G = 26 - 298(0.052) = +10.5$ kJ mol⁻¹ (not feasible)
At 95°C (368 K): $\Delta G = 26 - 368(0.052) = 26 - 19.1 = +6.9$ kJ mol⁻¹ (still positive but smaller)
The trend is that $\Delta G$ is decreasing toward zero as $T$ increases.
Since $\Delta G$ is becoming less positive (closer to zero / more negative), the dissolution becomes more feasible.
Therefore Z becomes more soluble.

The mark scheme accepts: "more soluble" + reason "because $\Delta G$ becomes more negative / less positive / smaller / closer to zero."

Key Takeaways

When $\Delta S > 0$ , increasing temperature always makes $\Delta G$ more negative (or less positive).
This is the thermodynamic basis for the general rule that endothermic dissolutions with positive entropy changes become more soluble when heated.
The direction of change in $\Delta G$ determines the direction of change in feasibility.

Common Mistakes

Saying "more soluble because the reaction is endothermic" — while Le Chatelier's principle gives the same answer, the question asks you to use $\Delta G$ .
Saying "less soluble" — this would be correct only if $\Delta S$ were negative.
Not providing the reason — the mark requires both the prediction AND the explanation.

Things to Be Careful About

The explanation must reference $\Delta G$ becoming more negative/less positive/closer to zero, not just "temperature increases."
The word "because" is essential — the mark is for the causal link between changing $\Delta G$ and changing solubility.

Techniques used

predict effect of temperature on ΔG when ΔS is positivelink changing ΔG to changing solubility

Q3MediumEquilibriaGroup 2Chemical Energetics

(a)

The pH of a saturated solution of calcium hydroxide is 12.35 at $298\text{ K}$ .

11M

(i)

Show that the concentration of hydroxide ions in a saturated solution of calcium hydroxide is $0.0224\text{ mol dm}^{-3}$ at $298\text{ K}$ .

DifficultyMedium-Easy

Worked solution

Working

pOH = 14 − 12.35 = 1.65

$[\text{OH}^-] = 10^{-\text{pOH}} = 10^{-1.65} = 0.0224\text{ mol dm}^{-3}$

Answer

$[\text{OH}^-] = 0.0224\text{ mol dm}^{-3}$

Final answer

[OH⁻] = 0.0224 mol dm⁻³

Detailed explanation

Background Concept

pH is defined as $\text{pH} = -\log_{10}[\text{H}^+]$ . In aqueous solution at 298 K, the ionic product of water is $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\text{ mol}^2\text{ dm}^{-6}$ , which means $\text{pH} + \text{pOH} = 14$ . Given either $[\text{H}^+]$ or $[\text{OH}^-]$ , the other can be found. Calcium hydroxide is a strong base that fully dissociates, so the hydroxide concentration in a saturated solution directly reflects the solubility of $\text{Ca(OH)}_2$ .

Understanding the Question

The question gives the pH of a saturated $\text{Ca(OH)}_2$ solution as 12.35 and asks you to show that $[\text{OH}^-] = 0.0224\text{ mol dm}^{-3}$ . The command word "show that" means you must demonstrate the calculation, not merely restate the value.

Approach

There are two equivalent routes: (1) find $[\text{H}^+]$ from pH, then use $K_w$ to find $[\text{OH}^-]$ ; or (2) find pOH = 14 − pH, then $[\text{OH}^-] = 10^{-\text{pOH}}$ . Route 2 is quicker and cleaner here.

Step-by-Step Reasoning

Calculate pOH: $\text{pOH} = 14 - 12.35 = 1.65$ .
Convert to concentration: $[\text{OH}^-] = 10^{-1.65} = 0.0224\text{ mol dm}^{-3}$ .

This exactly matches the value given in the question, so the statement is verified.

(Alternative route: $[\text{H}^+] = 10^{-12.35} = 4.47 \times 10^{-13}$ , then $[\text{OH}^-] = K_w/[\text{H}^+] = 1.0 \times 10^{-14}/4.47 \times 10^{-13} = 0.0224\text{ mol dm}^{-3}$ .)

Key Takeaways

At 298 K, $\text{pH} + \text{pOH} = 14$ . Converting from pH to a concentration requires $10^{-\text{pH}}$ . $K_w$ links $[\text{H}^+]$ and $[\text{OH}^-]$ in any aqueous solution.

Common Mistakes

Writing $[\text{OH}^-] = 10^{-12.35}$ instead of $10^{-1.65}$ — this gives the hydrogen ion concentration, not hydroxide.
Forgetting that pOH = 14 − pH (not pOH = pH).
Rounding the pH to 12.4 before taking the antilog, which loses accuracy.

Things to Be Careful About

Use the exact pH value in the exponent.
The answer must be quoted to 4 significant figures (0.0224) to match the given value — the mark scheme expects this exact figure.

Techniques used

convert pH to pOHcalculate hydroxide ion concentration from pOHapply the ionic product of water

(ii)

Use data given in (i) to calculate the solubility product, $K_{\text{sp}}$ , of calcium hydroxide at $298\text{ K}$ .

Include the units of $K_{\text{sp}}$ in your answer.

$K_{\text{sp}} =$ .............................. units ..............................

DifficultyMedium-Easy

Worked solution

Working

$\text{Ca(OH)}_2(\text{s}) \rightleftharpoons \text{Ca}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})$

$K_{\text{sp}} = [\text{Ca}^{2+}][\text{OH}^-]^2$

$[\text{Ca}^{2+}] = \tfrac{1}{2}[\text{OH}^-] = \tfrac{1}{2} \times 0.0224 = 0.0112\text{ mol dm}^{-3}$

$K_{\text{sp}} = (0.0112)(0.0224)^2 = 5.62 \times 10^{-6}$

Answer

$K_{\text{sp}} = 5.62 \times 10^{-6}\text{ mol}^3\text{ dm}^{-9}$

Final answer

Ksp = 5.62 × 10⁻⁶ mol³ dm⁻⁹

Detailed explanation

Background Concept

The solubility product $K_{\text{sp}}$ is the equilibrium constant for the dissolution of a sparingly soluble salt. For $\text{Ca(OH)}_2(\text{s}) \rightleftharpoons \text{Ca}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})$ , the expression is $K_{\text{sp}} = [\text{Ca}^{2+}][\text{OH}^-]^2$ . The stoichiometry tells us that for every mole of $\text{Ca(OH)}_2$ that dissolves, one $\text{Ca}^{2+}$ and two $\text{OH}^-$ are produced, so $[\text{Ca}^{2+}] = \tfrac{1}{2}[\text{OH}^-]$ in a solution made only by dissolving the solid.

Understanding the Question

We now take the $[\text{OH}^-]$ found in part (i) and use it to calculate $K_{\text{sp}}$ . The question explicitly asks for units — these arise from the powers in the expression: $\text{mol dm}^{-3} \times (\text{mol dm}^{-3})^2 = \text{mol}^3\text{ dm}^{-9}$ .

Approach

Write the dissociation equilibrium and its $K_{\text{sp}}$ expression, substitute $[\text{OH}^-] = 0.0224$ and $[\text{Ca}^{2+}] = 0.0112$ , then calculate the numerical value and attach the correct units.

Step-by-Step Reasoning

Dissociation: $\text{Ca(OH)}_2(\text{s}) \rightleftharpoons \text{Ca}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})$ .
$K_{\text{sp}} = [\text{Ca}^{2+}][\text{OH}^-]^2$ .
From stoichiometry, $[\text{Ca}^{2+}] = \tfrac{1}{2}[\text{OH}^-] = 0.0112\text{ mol dm}^{-3}$ .
Substitute: $K_{\text{sp}} = 0.0112 \times (0.0224)^2 = 0.0112 \times 5.018 \times 10^{-4} = 5.62 \times 10^{-6}$ .
Units: $\text{mol dm}^{-3} \times (\text{mol dm}^{-3})^2 = \text{mol}^3\text{ dm}^{-9}$ .

Key Takeaways

The coefficient in the formula becomes the exponent in the $K_{\text{sp}}$ expression. The units of $K_{\text{sp}}$ depend on the total number of ions produced — here three ions, so units are $\text{mol}^3\text{ dm}^{-9}$ .

Common Mistakes

Using $[\text{Ca}^{2+}] = 0.0224$ (forgetting to halve) — this doubles the answer.
Forgetting to square $[\text{OH}^-]$ .
Giving units as $\text{mol dm}^{-3}$ only — the mark scheme requires $\text{mol}^3\text{ dm}^{-9}$ .

Things to Be Careful About

The coefficient 2 in the formula becomes the exponent 2 in the $K_{\text{sp}}$ expression.
Units must be stated explicitly — the question prints a separate line for them.
Quote the answer to 3 significant figures (5.62 × 10⁻⁶) to match the mark scheme.

Techniques used

write the Ksp expressionuse stoichiometric ratio between ionscalculate Ksp with units

(iii)

A spatula measure of solid calcium chloride is stirred into a sample of saturated calcium hydroxide solution. All of the calcium chloride dissolves.

Describe one other observation that would be made and give an estimated value of the pH of the solution obtained.

Explain both your answers.

observation ........................................................................................................................

pH of solution .................................

explanation ........................................................................................................................

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

DifficultyMedium

Worked solution

Answer

Observation: a white precipitate of $\text{Ca(OH)}_2$ forms.
pH of solution: between 7 and 12.35 (lower than the original 12.35).
Explanation: $\text{CaCl}_2$ provides $\text{Ca}^{2+}$ , a common ion. By Le Chatelier's principle, the equilibrium $\text{Ca(OH)}_2(\text{s}) \rightleftharpoons \text{Ca}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})$ shifts to the left, precipitating $\text{Ca(OH)}_2$ . This removes $\text{OH}^-$ from solution, so $[\text{OH}^-]$ falls and pH decreases below 12.35.

Final answer

White precipitate of Ca(OH)₂ forms; pH between 7 and 12.35 (common ion effect)

Detailed explanation

Background Concept

The common ion effect is a direct consequence of Le Chatelier's principle applied to solubility equilibria. When a salt is added to a saturated solution of another salt that shares a common ion, the concentration of that ion increases, so the equilibrium shifts to the left, precipitating more of the sparingly soluble salt. Here, $\text{CaCl}_2$ dissociates to give $\text{Ca}^{2+}$ , which is the same cation as in $\text{Ca(OH)}_2$ .

Understanding the Question

A spatula measure of solid $\text{CaCl}_2$ is stirred into a saturated $\text{Ca(OH)}_2$ solution and all dissolves. You must give (1) one observation, (2) an estimated pH of the resulting solution, and (3) an explanation of both. The key insight is that adding $\text{Ca}^{2+}$ forces $\text{Ca(OH)}_2$ out of solution, removing $\text{OH}^-$ and lowering the pH.

Approach

Identify the common ion ( $\text{Ca}^{2+}$ ), apply Le Chatelier's principle to predict the shift of the $\text{Ca(OH)}_2$ equilibrium, deduce what is observed (precipitation), and reason about how $[\text{OH}^-]$ — and hence pH — changes.

Step-by-Step Reasoning

$\text{CaCl}_2$ dissolves: $\text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^-$ . This raises $[\text{Ca}^{2+}]$ in the solution.
The solubility equilibrium is $\text{Ca(OH)}_2(\text{s}) \rightleftharpoons \text{Ca}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})$ .
Increasing $[\text{Ca}^{2+}]$ shifts the equilibrium to the left (Le Chatelier's principle), so solid $\text{Ca(OH)}_2$ precipitates — this is the white precipitate observed.
As $\text{Ca(OH)}_2$ precipitates, $\text{OH}^-$ is removed from solution, so $[\text{OH}^-]$ decreases.
Lower $[\text{OH}^-]$ means higher pOH and therefore lower pH. The solution is still basic (it contains $\text{Ca(OH)}_2$ ), so pH stays above 7. The mark scheme accepts any pH strictly between 7.01 and 12.34.

Key Takeaways

The common ion effect reduces the solubility of a sparingly soluble salt. Adding a soluble salt sharing the cation of a sparingly soluble base lowers the pH of the saturated solution because hydroxide is precipitated out.

Common Mistakes

Saying "the pH stays the same" — wrong, because $\text{OH}^-$ is removed by precipitation.
Saying the solution becomes acidic — it remains basic, just less so.
Not identifying the precipitate as $\text{Ca(OH)}_2$ — the mark scheme requires this.
Giving a pH of exactly 12.35 or exactly 7 — the mark scheme requires a value strictly between them.

Things to Be Careful About

The observation must be a white precipitate — $\text{Ca(OH)}_2$ is white.
The pH estimate must be a single value within the accepted range (7.01–12.34).
The explanation must mention the common ion effect explicitly to score the second mark.

Techniques used

apply the common ion effectapply Le Chatelier's principle to a solubility equilibriumpredict the effect on pH

(iv)

Calcium hydroxide reacts with dilute sulfuric acid to form calcium sulfate. Barium hydroxide behaves in a similar way, forming barium sulfate.

Explain why calcium sulfate is more soluble in water than barium sulfate.

DifficultyMedium

Worked solution

Answer

Both lattice energy and hydration energy decrease down Group 2 as ionic radius increases.
The hydration energy decreases more than the lattice energy.
Hence the enthalpy of solution (lattice + hydration) is more endothermic for $\text{BaSO}_4$ than for $\text{CaSO}_4$ , so $\text{BaSO}_4$ is less soluble.

Final answer

Hydration energy decreases more than lattice energy down Group 2, so the enthalpy of solution is more endothermic for BaSO₄ than CaSO₄, making BaSO₄ less soluble.

Detailed explanation

Background Concept

The solubility of a Group 2 sulfate is governed by the enthalpy of solution, $\Delta H_{\text{sol}} = \text{lattice energy} + \text{hydration energy}$ . The lattice energy is the endothermic energy required to break the ionic lattice into gaseous ions. The hydration energy is the exothermic energy released when those gaseous ions are surrounded by water molecules. Dissolution is favoured when the exothermic hydration exceeds the endothermic lattice term, giving a negative (exothermic) $\Delta H_{\text{sol}}$ .

Understanding the Question

We must explain why $\text{CaSO}_4$ is more soluble than $\text{BaSO}_4$ . The answer lies in how the two energy terms change down Group 2 as the cation gets larger.

Approach

State how both lattice and hydration energies change down the group (they both decrease in magnitude as ionic radius increases), identify which change is larger (hydration), and connect the result to the enthalpy of solution and hence solubility.

Step-by-Step Reasoning

Down Group 2, the ionic radius increases ( $\text{Ca}^{2+} < \text{Ba}^{2+}$ ).
Lattice energy decreases in magnitude down the group: $\text{CaSO}_4$ has a more exothermic (more negative) lattice energy than $\text{BaSO}_4$ .
Hydration energy also decreases in magnitude down the group: the smaller $\text{Ca}^{2+}$ ion has a higher charge density and attracts water molecules more strongly, so its hydration energy is more exothermic.
The key point: hydration energy decreases more than lattice energy down the group.
Therefore the net enthalpy of solution becomes more endothermic for $\text{BaSO}_4$ . A more endothermic $\Delta H_{\text{sol}}$ makes dissolution less energetically favourable, so $\text{BaSO}_4$ is less soluble than $\text{CaSO}_4$ .

Key Takeaways

Solubility of Group 2 sulfates decreases down the group. The deciding factor is the balance between lattice and hydration energies, with hydration energy being the dominant term because it changes more rapidly with ionic radius.

Common Mistakes

Saying lattice energy is the dominant factor — it is hydration energy that decreases more.
Confusing the direction: saying lattice energy increases down the group (it decreases).
Only mentioning one of the two energies — both must be discussed.
Not linking the energy argument to solubility (the conclusion must connect $\Delta H_{\text{sol}}$ to solubility).

Things to Be Careful About

Both lattice energy and hydration energy must be mentioned to score the first mark.
The phrase "hydration energy decreases more" (or "is the dominant factor") is required for the second mark.
The final mark requires the explicit conclusion that $\Delta H_{\text{sol}}$ is more endothermic for $\text{BaSO}_4$ .

Techniques used

compare lattice and hydration energies down Group 2relate enthalpy of solution to solubilityidentify the dominant energetic factor

(b)

Some solid calcium is added to an excess of aqueous ethanoic acid, $\text{CH}_3\text{COOH}$ , and left until all the calcium has reacted. The resulting mixture, mixture D, contains no undissolved solids.

8M

(i)

Write an equation for the reaction of calcium with $\text{CH}_3\text{COOH}$ .

DifficultyEasy

Worked solution

Answer

$\text{Ca} + 2\text{CH}_3\text{COOH} \rightarrow \text{Ca}(\text{CH}_3\text{COO})_2 + \text{H}_2$

Final answer

Ca + 2CH3COOH -> Ca(CH3COO)2 + H2

Detailed explanation

Background Concept

Calcium is a reactive Group 2 metal that reacts with acids to form a salt and hydrogen gas. Ethanoic acid is a weak carboxylic acid, but it still reacts with reactive metals. Calcium forms a +2 ion, so two ethanoate ions ( $\text{CH}_3\text{COO}^-$ ) are needed to balance the charge in the salt $\text{Ca}(\text{CH}_3\text{COO})_2$ .

Understanding the Question

Write the balanced equation for the reaction of solid calcium with aqueous ethanoic acid. This is a straightforward equation-writing task worth 1 mark.

Approach

Identify the products — calcium ethanoate and hydrogen gas — then balance the equation by matching the number of ethanoic acid molecules to the stoichiometry.

Step-by-Step Reasoning

Calcium is oxidised: $\text{Ca} \rightarrow \text{Ca}^{2+} + 2\text{e}^-$ .
Each $\text{H}^+$ from the acid is reduced: $2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2$ .
Two ethanoic acid molecules supply two $\text{H}^+$ and leave two $\text{CH}_3\text{COO}^-$ ions, which combine with $\text{Ca}^{2+}$ to give $\text{Ca}(\text{CH}_3\text{COO})_2$ .
Balanced equation: $\text{Ca} + 2\text{CH}_3\text{COOH} \rightarrow \text{Ca}(\text{CH}_3\text{COO})_2 + \text{H}_2$ .

Key Takeaways

Balancing an acid-metal reaction requires matching the metal's charge with the number of anions. Calcium is divalent, so two ethanoate ions are needed.

Common Mistakes

Writing $\text{CaCH}_3\text{COO}$ instead of $\text{Ca}(\text{CH}_3\text{COO})_2$ — the charge on Ca²⁺ requires two ethanoate ions.
Wrong stoichiometry, e.g. 1:1 instead of 1:2.
Forgetting hydrogen gas as the other product.

Things to Be Careful About

The equation must be balanced — atoms of each element on both sides.
The mark scheme accepts the equation exactly as written above; no state symbols are required.

Techniques used

balance a metal-acid equationidentify the products of an acid-metal reaction

(ii)

Use formulae of molecules and ions to identify two conjugate acid–base pairs present in mixture D.

Pair 1 should consist of organic species.

pair 1: ............................................ ............................................
conjugate acid conjugate base

pair 2: ............................................ ............................................
conjugate acid conjugate base

DifficultyMedium-Easy

Worked solution

Answer

Pair 1: $\text{CH}_3\text{COOH}$ (conjugate acid) / $\text{CH}_3\text{COO}^-$ (conjugate base)
Pair 2: $\text{H}_3\text{O}^+$ (conjugate acid) / $\text{H}_2\text{O}$ (conjugate base)

Final answer

Pair 1: CH3COOH / CH3COO⁻; Pair 2: H3O⁺ / H2O

Detailed explanation

Background Concept

A conjugate acid-base pair consists of two species that differ by exactly one proton ( $\text{H}^+$ ). The acid is the species with the extra proton; the base is the species after the proton is removed. For example, $\text{CH}_3\text{COOH}$ and $\text{CH}_3\text{COO}^-$ form a pair because $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$ .

Understanding the Question

In mixture D, after calcium has reacted with excess ethanoic acid, the solution contains $\text{CH}_3\text{COOH}$ (excess acid), $\text{CH}_3\text{COO}^-$ (from $\text{Ca}(\text{CH}_3\text{COO})_2$ ), $\text{H}_2\text{O}$ , and small amounts of $\text{H}_3\text{O}^+$ and $\text{OH}^-$ . You must identify two conjugate pairs — pair 1 must be organic (both species contain carbon).

Approach

Recall that a conjugate pair differs by one proton. Look at the species present and find pairs that differ by H⁺.

Step-by-Step Reasoning

$\text{CH}_3\text{COOH}$ and $\text{CH}_3\text{COO}^-$ differ by one proton: $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$ . This is the organic pair.
$\text{H}_3\text{O}^+$ and $\text{H}_2\text{O}$ differ by one proton: $\text{H}_3\text{O}^+ \rightleftharpoons \text{H}_2\text{O} + \text{H}^+$ . This is the second pair.
(The mark scheme also accepts $\text{H}_2\text{O}$ / $\text{OH}^-$ as pair 2, since $\text{H}_2\text{O} \rightleftharpoons \text{OH}^- + \text{H}^+$ .)

Key Takeaways

Conjugate acid-base pairs always differ by exactly one proton. The acid is the species with the extra H⁺.

Common Mistakes

Writing $\text{CH}_3\text{COOH}$ and $\text{H}_2\text{O}$ as a pair — they do not differ by one proton.
Writing $\text{Ca}^{2+}$ and $\text{Ca}$ as a pair — this is a redox pair, not an acid-base pair.
Forgetting that $\text{H}_3\text{O}^+$ / $\text{H}_2\text{O}$ is a valid conjugate pair.

Things to Be Careful About

Pair 1 must be organic — both species must contain carbon.
The acid is the species with the extra proton; the base is the deprotonated form.

Techniques used

identify conjugate acid-base pairsrecognise proton transfer relationships

(iii)

Write the expression for the $K_{\text{a}}$ of $\text{CH}_3\text{COOH}$ .

$K_{\text{a}} =$

DifficultyEasy

Worked solution

Answer

$K_{\text{a}} = \dfrac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$

Final answer

Ka = [H⁺][CH3COO⁻] / [CH3COOH]

Detailed explanation

Background Concept

The acid dissociation constant $K_{\text{a}}$ is the equilibrium constant for the dissociation of a weak acid. For a general weak acid $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$ , the expression is $K_{\text{a}} = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$ . For ethanoic acid, $\text{HA} = \text{CH}_3\text{COOH}$ and $\text{A}^- = \text{CH}_3\text{COO}^-$ .

Understanding the Question

Write the $K_{\text{a}}$ expression for ethanoic acid. This is a pure recall task worth 1 mark.

Approach

Apply the general formula for $K_{\text{a}}$ to ethanoic acid, placing products (H⁺ and the ethanoate ion) over the reactant (undissociated ethanoic acid).

Step-by-Step Reasoning

Dissociation: $\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^-$ .
$K_{\text{a}} = \dfrac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$ .

Key Takeaways

The $K_{\text{a}}$ expression always has the form: concentration of products over concentration of reactant, with each coefficient becoming an exponent (here all coefficients are 1).

Common Mistakes

Omitting $[\text{H}^+]$ from the numerator.
Writing the reciprocal (reactant over products).
Including $[\text{H}_2\text{O}]$ in the expression — water is not included because it is in large excess.

Things to Be Careful About

The question only asks for the expression, not units — do not add units.
Square brackets denote concentration in mol dm⁻³.

Techniques used

write the Ka expression for a weak acid

(iv)

The concentration of calcium ethanoate, $(\text{CH}_3\text{COO})_2\text{Ca}$ , in mixture D is $0.394\text{ mol dm}^{-3}$ .
The concentration of $\text{CH}_3\text{COOH}$ in mixture D is $0.270\text{ mol dm}^{-3}$ .

The $K_{\text{a}}$ of $\text{CH}_3\text{COOH}$ is $1.74 \times 10^{-5}\text{ mol dm}^{-3}$ at $298\text{ K}$ .

Calculate the pH of mixture D.

pH =

DifficultyMedium

Worked solution

Working

Each $\text{Ca}(\text{CH}_3\text{COO})_2$ provides two ethanoate ions:

$[\text{CH}_3\text{COO}^-] = 2 \times 0.394 = 0.788\text{ mol dm}^{-3}$

$K_{\text{a}} = \dfrac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$

$[\text{H}^+] = K_{\text{a}} \times \dfrac{[\text{CH}_3\text{COOH}]}{[\text{CH}_3\text{COO}^-]} = 1.74 \times 10^{-5} \times \dfrac{0.270}{0.788} = 5.96 \times 10^{-6}\text{ mol dm}^{-3}$

$\text{pH} = -\log(5.96 \times 10^{-6}) = 5.22$

Answer

pH = 5.22

Final answer

pH = 5.22

Detailed explanation

Background Concept

A buffer solution contains a weak acid and its conjugate base in comparable concentrations. The pH is governed by the acid dissociation equilibrium. Rearranging $K_{\text{a}} = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$ gives $[\text{H}^+] = K_{\text{a}} \times \dfrac{[\text{HA}]}{[\text{A}^-]}$ , which is the basis of the Henderson–Hasselbalch equation. The conjugate base here comes from calcium ethanoate, $\text{Ca}(\text{CH}_3\text{COO})_2$ , which dissociates to give two ethanoate ions per formula unit.

Understanding the Question

Mixture D is a buffer containing 0.394 mol dm⁻³ $\text{Ca}(\text{CH}_3\text{COO})_2$ and 0.270 mol dm⁻³ $\text{CH}_3\text{COOH}$ . The critical trap is that the concentration of the ethanoate ion is double the concentration of the calcium salt, because each formula unit provides two $\text{CH}_3\text{COO}^-$ ions.

Approach

Calculate $[\text{CH}_3\text{COO}^-]$ from the salt concentration.
Substitute into the rearranged $K_{\text{a}}$ expression.
Take the negative logarithm to find pH.

Step-by-Step Reasoning

$[\text{CH}_3\text{COO}^-] = 2 \times 0.394 = 0.788\text{ mol dm}^{-3}$ .
Rearrange $K_{\text{a}}$ : $[\text{H}^+] = K_{\text{a}} \times \dfrac{[\text{CH}_3\text{COOH}]}{[\text{CH}_3\text{COO}^-]}$ .
Substitute: $[\text{H}^+] = 1.74 \times 10^{-5} \times \dfrac{0.270}{0.788} = 5.96 \times 10^{-6}\text{ mol dm}^{-3}$ .
$\text{pH} = -\log(5.96 \times 10^{-6}) = 5.22$ .

Key Takeaways

In buffer calculations, always check how many conjugate-base ions each formula unit of the salt provides. Divalent metal salts of monobasic acids give a factor of 2.

Common Mistakes

Using $[\text{CH}_3\text{COO}^-] = 0.394$ instead of 0.788 — this is the most common error and gives pH = 5.49 instead of 5.22.
Forgetting to take the negative logarithm at the end.
Mixing up which concentration goes in the numerator (it is the acid, not the base).

Things to Be Careful About

The factor of 2 from $\text{Ca}(\text{CH}_3\text{COO})_2$ is the main trap in this question.
Quote the pH to 2 decimal places (5.22) to match the mark scheme.
The $K_{\text{a}}$ value is given to 3 significant figures, so the answer should reflect that precision.

Techniques used

calculate conjugate base concentration from salt concentrationrearrange the Ka expression for a buffercalculate pH from [H⁺]

(v)

Write two equations to show how mixture D can act as a buffer solution.

equation 1 .........................................................................................................................

equation 2 .........................................................................................................................

DifficultyMedium-Easy

Worked solution

Answer

$\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$
$\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$

Final answer

CH3COO⁻ + H⁺ -> CH3COOH; CH3COOH + OH⁻ -> CH3COO⁻ + H2O

Detailed explanation

Background Concept

A buffer solution resists changes in pH when small amounts of acid or base are added. It contains a weak acid ( $\text{CH}_3\text{COOH}$ ) and its conjugate base ( $\text{CH}_3\text{COO}^-$ ). The conjugate base neutralises added acid; the weak acid neutralises added base. The pH stays roughly constant because the ratio $[\text{acid}]/[\text{base}]$ changes only slightly.

Understanding the Question

Write two equations showing how mixture D acts as a buffer. One equation must show how the buffer removes added $\text{H}^+$ (acid), and the other how it removes added $\text{OH}^-$ (base).

Approach

Write the neutralisation of $\text{H}^+$ by the conjugate base, and the neutralisation of $\text{OH}^-$ by the weak acid.

Step-by-Step Reasoning

Removing added acid: the ethanoate ion reacts with $\text{H}^+$ : $\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$ . The added $\text{H}^+$ is consumed, so pH barely changes.
Removing added base: ethanoic acid reacts with $\text{OH}^-$ : $\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$ . The added $\text{OH}^-$ is consumed, so pH barely changes.

Key Takeaways

Buffer action is described by two equations: the conjugate base mops up added $\text{H}^+$ , and the weak acid mops up added $\text{OH}^-$ . The buffer works because these reactions consume the added species without significantly changing the acid/base ratio.

Common Mistakes

Writing only one equation — the question asks for two.
Getting the direction wrong (e.g. $\text{CH}_3\text{COOH} + \text{H}^+$ makes no sense).
Writing $\text{H}_2\text{O} + \text{H}^+ \rightarrow \text{H}_3\text{O}^+$ — this does not show buffer action by the ethanoate system.

Things to Be Careful About

The mark scheme accepts $\text{Ca}(\text{CH}_3\text{COO})_2 + 2\text{H}^+ \rightarrow 2\text{CH}_3\text{COOH} + \text{Ca}^{2+}$ as an alternative for equation 1.
Both equations must be balanced.

Techniques used

write buffer action equations for added acid and added base

Q4MediumTransition ElementsEquilibria

Transition metal atoms and transition metal ions form complexes by combining with species called ligands.

(a)

When $\text{NaOH}(\text{aq})$ is added to an aqueous solution containing $[\text{Co}(\text{H}_2\text{O})_6]^{2+}$ a precipitation reaction occurs accompanied by a colour change.

In this reaction two of the water ligands each lose one $\text{H}^+$ ion. The $\text{H}^+$ ions are gained by $\text{OH}^-$ ions from the $\text{NaOH}(\text{aq})$ .

3M

(i)

State the colour change seen in this precipitation reaction.

from ............................................................. to .............................................................

DifficultyEasy

Worked solution

Answer

from pink to blue

Final answer

pink to blue

Detailed explanation

Background Concept

Transition metal complexes are coloured because electrons in partially filled d-orbitals can absorb visible light and be promoted between d-orbital energy levels. The colour observed depends on the size of the d-orbital splitting, which in turn depends on the ligand attached to the metal ion. Changing the ligand environment changes the splitting and therefore the colour.

Understanding the Question

This part asks for the observed colour change when $\text{NaOH}(\text{aq})$ is added to an aqueous solution containing $[\text{Co(H}_2\text{O)}_6]^{2+}$ . The reaction is a precipitation: two water ligands lose $\text{H}^+$ to $\text{OH}^-$ , forming a neutral cobalt(II) hydroxide complex. You only need to state the initial and final colours.

Approach

Recall the standard colours of aqueous cobalt(II) complexes: the hexaaqua ion is pink, and the hydroxide-containing complex formed on adding alkali is blue. No calculation is needed.

Step-by-Step Reasoning

In aqueous solution, $\text{Co}^{2+}$ exists as the pink hexaaqua complex $[\text{Co(H}_2\text{O)}_6]^{2+}$ . When $\text{OH}^-$ is added, two coordinated water molecules are deprotonated to give the neutral complex $\text{Co(H}_2\text{O)}_4(\text{OH})_2$ , which is insoluble and blue. The mark scheme accepts red/pink for the initial colour and blue for the final colour.

Key Takeaways

A change in ligand around a transition metal ion changes the d-orbital splitting and hence the colour. Standard colours such as pink aqueous $\text{Co}^{2+}$ and blue cobalt(II) hydroxide-type precipitates should be remembered.

Common Mistakes

Reversing the order: the change is from pink to blue, not blue to pink.
Saying 'colourless' or 'white'; the precipitate is blue.
Giving only one colour; both initial and final colours are required.

Things to Be Careful About

The mark scheme allows 'red/pink' for the initial colour. Make sure the two colours are linked by 'to' in the correct order.

Techniques used

recall characteristic colours of transition metal complexesrelate colour change to a change in ligand environment

(ii)

Complete the ionic equation for this precipitation reaction.

[\text{Co}(\text{H}_2\text{O})_6]^{2+} + \text{.................................} \rightarrow \text{.................................} + \text{.................................}

DifficultyMedium-Easy

Worked solution

Answer

[\text{Co(H}_2\text{O)}_6]^{2+} + 2\text{OH}^- \rightarrow \text{Co(H}_2\text{O)}_4(\text{OH})_2 + 2\text{H}_2\text{O}

Final answer

[Co(H2O)6]2+ + 2OH- -> Co(H2O)4(OH)2 + 2H2O

Detailed explanation

Background Concept

Coordinated water molecules in aqua complexes can behave as Brønsted acids, releasing $\text{H}^+$ ions. When a base such as $\text{OH}^-$ is added, it accepts these protons. Here, each $\text{OH}^-$ removes one $\text{H}^+$ from a coordinated water molecule, so the water ligand becomes a hydroxide ligand. The overall process is a deprotonation of the complex.

Understanding the Question

The equation is already started with $[\text{Co(H}_2\text{O)}_6]^{2+}$ on the left. You need to fill in the missing reactant and the two products. The stem tells you that two water ligands each lose one $\text{H}^+$ , and that the $\text{H}^+$ ions are gained by $\text{OH}^-$ ions.

Approach

Identify that the missing reactant is $\text{OH}^-$ . Since two $\text{H}^+$ ions are released, two $\text{OH}^-$ ions are needed. Each $\text{H}^+ + \text{OH}^-$ forms $\text{H}_2\text{O}$ , so two water molecules are produced. The complex loses two protons from two coordinated water molecules, leaving four water ligands and two hydroxide ligands.

Step-by-Step Reasoning

Two coordinated water molecules lose one $\text{H}^+$ each: $[\text{Co(H}_2\text{O)}_6]^{2+} \rightarrow \text{Co(H}_2\text{O)}_4(\text{OH})_2 + 2\text{H}^+$ .
Each $\text{H}^+$ reacts with $\text{OH}^-$ : $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$ .
Overall: $[\text{Co(H}_2\text{O)}_6]^{2+} + 2\text{OH}^- \rightarrow \text{Co(H}_2\text{O)}_4(\text{OH})_2 + 2\text{H}_2\text{O}$ .
Check atoms and charge: left has Co 1, O 8, H 14, charge $+2 - 2 = 0$ ; right has Co 1, O 6 + 2 = 8, H 8 + 2 + 4 = 14, charge 0. Balanced.

Key Takeaways

Aqua complexes can be deprotonated by bases; the stoichiometry is controlled by the number of $\text{H}^+$ ions released. Balancing ionic equations requires checking both atoms and charge.

Common Mistakes

Writing only one $\text{OH}^-$ because only one precipitate forms; two are required because two $\text{H}^+$ are released.
Writing $\text{Co(OH)}_2$ as the product; the product still contains four water ligands.
Forgetting the two water molecules formed from $\text{H}^+ + \text{OH}^-$ .

Things to Be Careful About

State symbols are not required by the mark scheme for this equation. Ensure the charges on the complex and hydroxide are correct, and that the equation is balanced in both atoms and charge.

Techniques used

balance an ionic equationidentify deprotonation of aqua ligandsapply acid-base neutralisation of H+ by OH-

(iii)

This precipitation reaction can also be described as a different type of reaction.

Name this type of reaction.

(b)

$\mathbf{L}$ is an uncharged tridentate ligand. $\mathbf{L}$ donates three lone pairs to a metal atom or ion.

Cobalt forms an octahedral complex ion, $\mathbf{E}$ , with $\mathbf{L}$ . Complex ion $\mathbf{E}$ has a $2+$ charge.

4M

(i)

Give the formula of $\mathbf{E}$ .

(ii)

Identify the oxidation state of cobalt in $\mathbf{E}$ .

(iii)

The d-orbitals of the cobalt atom or ion present in $\mathbf{E}$ are split in energy.

State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level.


number of d-orbitals at a higher energy level
number of d-orbitals at a lower energy level

(iv)

Define the term non-degenerate d-orbitals.

(c)

The mineral chromite contains a compound which has the formula $\text{FeCr}_n\text{O}_4$ . The oxidation state of iron in $\text{FeCr}_n\text{O}_4$ is $+2$ .

A sample of $4.18\text{ g}$ of $\text{FeCr}_n\text{O}_4$ is dissolved in an excess of sulfuric acid. The resulting solution is made up to $250\text{ cm}^3$ . This is solution F.

All the $\text{Fe}^{2+}$ ions in $25.0\text{ cm}^3$ of solution F are oxidised to $\text{Fe}^{3+}$ ions by exactly $18.7\text{ cm}^3$ of $0.0200\text{ mol dm}^{-3}\text{ KMnO}_4$ .

One $\text{MnO}_4^-$ ion reacts with five $\text{Fe}^{2+}$ ions. Assume no other oxidation reaction occurs.

5M

(i)

Write an equation for the reaction of $\text{Fe}^{2+}$ ions with $\text{MnO}_4^-$ ions in acid solution.

DifficultyMedium-Easy

Worked solution

Answer

5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}

Final answer

5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O

Detailed explanation

Background Concept

In acid solution, manganate(VII), $\text{MnO}_4^-$ , is a strong oxidising agent. It is reduced to $\text{Mn}^{2+}$ . Iron(II) is oxidised to iron(III). Redox equations can be built from half-equations: an oxidation half-equation and a reduction half-equation, then combined so that the number of electrons lost equals the number gained.

Understanding the Question

You need to write the balanced ionic equation for the reaction of $\text{Fe}^{2+}$ with $\text{MnO}_4^-$ in acid. The stem also tells you that one $\text{MnO}_4^-$ reacts with five $\text{Fe}^{2+}$ , which is a useful check.

Approach

Write the two half-equations, balance electrons, and add them.

Step-by-Step Reasoning

Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-$ .
Reduction: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ .
Multiply the oxidation half-equation by 5 so that 5 electrons are released: $5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5\text{e}^-$ .
Add the two half-equations: $5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}$ .
Check: atoms balance (Fe 5, Mn 1, O 4, H 8) and charge balances (left $+10 - 1 + 8 = +17$ ; right $+15 + 2 = +17$ ).

Key Takeaways

The half-equation method is reliable for redox equations. $\text{MnO}_4^-$ in acid always gives $\text{Mn}^{2+}$ and water, with 5 electrons accepted.

Common Mistakes

Forgetting the $8\text{H}^+$ and $4\text{H}_2\text{O}$ .
Writing $\text{MnO}_2$ or $\text{Mn}^{4+}$ as the reduction product.
Not multiplying the Fe half-equation by 5.
Leaving the equation unbalanced in charge.

Things to Be Careful About

Use $\text{H}^+$ rather than $\text{H}_2\text{O}$ as the acid source. The coefficient 5 on $\text{Fe}^{2+}$ and $\text{Fe}^{3+}$ is essential because $\text{MnO}_4^-$ accepts 5 electrons.

Techniques used

write oxidation and reduction half-equationsbalance electrons to combine half-equationsbalance atoms and charge

(ii)

Calculate the number of moles of $\text{Fe}^{2+}$ ions in $25.0\text{ cm}^3$ of solution F.

number of moles of $\text{Fe}^{2+}$ ions =

DifficultyMedium-Easy

Worked solution

Working

n(\text{MnO}_4^-) = 0.0200 \times \frac{18.7}{1000} = 3.74 \times 10^{-4} \text{ mol}

n(\text{Fe}^{2+}) = 5 \times 3.74 \times 10^{-4} = 1.87 \times 10^{-3} \text{ mol}

Answer

$1.87 \times 10^{-3}$ mol

Final answer

1.87 × 10^-3 mol

Detailed explanation

Background Concept

Titration calculations use concentration × volume (in dm³) to find moles of the known reagent, then the stoichiometric ratio from the balanced equation to find moles of the unknown. Here the known reagent is $\text{KMnO}_4$ , which provides $\text{MnO}_4^-$ ions. The ratio from part (c)(i) is 5 $\text{Fe}^{2+}$ : 1 $\text{MnO}_4^-$ .

Understanding the Question

25.0 cm³ of solution F required 18.7 cm³ of 0.0200 mol dm⁻³ $\text{KMnO}_4$ . You need the moles of $\text{Fe}^{2+}$ in that 25.0 cm³ aliquot.

Approach

Convert the $\text{KMnO}_4$ volume to dm³, calculate moles of $\text{MnO}_4^-$ , then multiply by 5.

Step-by-Step Reasoning

Volume of $\text{KMnO}_4$ = 18.7 cm³ = 18.7/1000 = 0.0187 dm³.
Moles $\text{MnO}_4^-$ = 0.0200 × 0.0187 = $3.74 \times 10^{-4}$ mol.
From the equation, 1 $\text{MnO}_4^-$ reacts with 5 $\text{Fe}^{2+}$ , so moles $\text{Fe}^{2+}$ = 5 × $3.74 \times 10^{-4}$ = $1.87 \times 10^{-3}$ mol.
This is the amount in the 25.0 cm³ portion titrated.

Key Takeaways

Always convert cm³ to dm³ by dividing by 1000. Use the mole ratio from the balanced equation, not the ratio of volumes.

Common Mistakes

Forgetting to divide 18.7 by 1000.
Using the ratio as 1 $\text{Fe}^{2+}$ : 5 $\text{MnO}_4^-$ instead of 5 $\text{Fe}^{2+}$ : 1 $\text{MnO}_4^-$ .
Using 250 cm³ instead of 25.0 cm³; the question asks for moles in the 25.0 cm³ aliquot.

Things to Be Careful About

$\text{KMnO}_4$ dissociates to give one $\text{MnO}_4^-$ per formula unit, so moles of $\text{KMnO}_4$ equals moles of $\text{MnO}_4^-$ . Keep three significant figures, matching the data.

Techniques used

calculate moles from concentration and volumeapply stoichiometric ratio from balanced equation

(iii)

Calculate the $M_{\text{r}}$ of $\text{FeCr}_n\text{O}_4$ and use your answer to deduce the value of $n$ .

$M_{\text{r}}$ of $\text{FeCr}_n\text{O}_4$ = ..............................

value of $n$ = ..............................

DifficultyMedium

Worked solution

Working

Moles of $\text{Fe}^{2+}$ in $250\text{ cm}^3$ = $1.87 \times 10^{-3} \times 10 = 1.87 \times 10^{-2}$ mol.

Moles of $\text{FeCr}_n\text{O}_4$ = $1.87 \times 10^{-2}$ mol (1 mol compound gives 1 mol $\text{Fe}^{2+}$ ).

M_{\text{r}} = \frac{4.18}{1.87 \times 10^{-2}} = 224 \text{ (or } 223.5)

M_{\text{r}} = 55.8 + 52.0n + 4(16.0) = 119.8 + 52.0n

119.8 + 52.0n = 223.5 \Rightarrow n = \frac{103.7}{52.0} = 1.99 \approx 2

Answer

$M_{\text{r}} = 224$ (or $223.5$ ); $n = 2$

Final answer

Mr = 224 (or 223.5); n = 2

Detailed explanation

Background Concept

To find a relative molecular mass from a titration, you first find the amount of substance in the portion analysed, scale it to the whole sample, and then use $M_{\text{r}} = \text{mass} / \text{moles}$ . The formula $\text{FeCr}_n\text{O}_4$ contains one Fe per formula unit, so the moles of $\text{Fe}^{2+}$ in the whole solution equal the moles of $\text{FeCr}_n\text{O}_4$ dissolved. The subscript $n$ is then found by comparing the experimental $M_{\text{r}}$ with the sum of relative atomic masses.

Understanding the Question

4.18 g of $\text{FeCr}_n\text{O}_4$ was dissolved and made up to 250 cm³. A 25.0 cm³ portion contained $1.87 \times 10^{-3}$ mol $\text{Fe}^{2+}$ from part (c)(ii). You need the $M_{\text{r}}$ of the compound and the value of $n$ .

Approach

Scale the moles from the aliquot to the whole 250 cm³, use mass/moles to get $M_{\text{r}}$ , then set up an expression for $M_{\text{r}}$ in terms of $n$ and solve.

Step-by-Step Reasoning

Moles $\text{Fe}^{2+}$ in 25.0 cm³ = $1.87 \times 10^{-3}$ mol.
The whole solution is 250 cm³, which is 10 times larger, so moles $\text{Fe}^{2+}$ in 250 cm³ = $1.87 \times 10^{-3} \times 10 = 1.87 \times 10^{-2}$ mol.
Each formula unit of $\text{FeCr}_n\text{O}_4$ provides one $\text{Fe}^{2+}$ , so moles of $\text{FeCr}_n\text{O}_4$ = $1.87 \times 10^{-2}$ mol.
$M_{\text{r}}$ = mass / moles = 4.18 / ( $1.87 \times 10^{-2}$ ) = 223.5, which rounds to 224.
Using relative atomic masses Fe = 55.8, Cr = 52.0, O = 16.0: $M_{\text{r}}$ = 55.8 + 52.0n + 4(16.0) = 119.8 + 52.0n.
Equate to 223.5: 119.8 + 52.0n = 223.5, so 52.0n = 103.7, n = 1.99 ≈ 2.

Key Takeaways

A dilution factor is needed when only an aliquot is titrated. $M_{\text{r}}$ = mass/moles. The formula subscript is found by solving a simple algebraic equation.

Common Mistakes

Forgetting to multiply by 10 to scale from 25 cm³ to 250 cm³.
Using the mass of the aliquot rather than the whole 4.18 g sample.
Using incorrect relative atomic masses.
Not rounding n to a whole number; n must be an integer.

Things to Be Careful About

The mark scheme allows $M_{\text{r}}$ = 224 or 223.5. Show the scaling step clearly because it carries a method mark. Use consistent significant figures.

Techniques used

scale moles by dilution factorcalculate relative molecular mass from mass and molessolve for subscript n using atomic masses

Q5Medium-HardTransition Elements

$\text{Ni}^{2+}$ ions form a number of different complex ions, including $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ , $[\text{Ni}(\text{NH}_3)_6]^{2+}$ and $[\text{Ni}(\text{en})_3]^{2+}$ .

The abbreviation en represents 1,2-diaminoethane. The numerical values of two stability constants, $K_{\text{stab}}$ , are given in Table 5.1.

Table 5.1

complex	$K_{\text{stab}}$
$[\text{Ni}(\text{NH}_3)_6]^{2+}$	$4.8 \times 10^7$
$[\text{Ni}(\text{en})_3]^{2+}$	$2.0 \times 10^{18}$

(a)

Complete the expression for the $K_{\text{stab}}$ of $[\text{Ni}(\text{en})_3]^{2+}$ .

$K_{\text{stab}} =$

DifficultyEasy

Worked solution

Answer

K_{\text{stab}} = \frac{[[\text{Ni}(\text{en})_3]^{2+}]}{[\text{Ni}^{2+}][\text{en}]^3}

Alternatively:

K_{\text{stab}} = \frac{[[\text{Ni}(\text{en})_3]^{2+}]}{[[\text{Ni}(\text{H}_2\text{O})_6]^{2+}][\text{en}]^3}

Final answer

K_stab = [[Ni(en)3]2+] / ([Ni2+][en]3)

Detailed explanation

Background Concept

The stability constant, $K_{\text{stab}}$ , quantifies the thermodynamic stability of a complex ion relative to its constituent metal ion and free ligands. For the overall formation reaction:

\text{M}^{n+} + x\text{L} \rightleftharpoons [\text{ML}_x]^{n+}

the expression is:

K_{\text{stab}} = \frac{[[\text{ML}_x]^{n+}]}{[\text{M}^{n+}][\text{L}]^x}

Water molecules coordinated to the metal in the hexaaqua complex, $[\text{M}(\text{H}_2\text{O})_6]^{n+}$ , are not included in the denominator when the reaction is written from $\text{M}^{n+}(\text{aq})$ , because the free water concentration is essentially constant (~55.5 mol dm $^{-3}$ ) and is absorbed into the constant. If the reaction is written starting from the hexaaqua complex:

[\text{M}(\text{H}_2\text{O})_6]^{n+} + x\text{L} \rightleftharpoons [\text{ML}_x]^{n+} + 6\text{H}_2\text{O}

then $[[\text{M}(\text{H}_2\text{O})_6]^{n+}]$ appears in the denominator instead of $[\text{M}^{n+}]$ , and $[\text{H}_2\text{O}]$ is omitted. Both forms are acceptable and give the same numerical $K_{\text{stab}}$ value.

Understanding the Question

Part (a) asks for the $K_{\text{stab}}$ expression for $[\text{Ni}(\text{en})_3]^{2+}$ . The complex forms from $\text{Ni}^{2+}$ and three bidentate ethylenediamine (en) ligands. We need to write the equilibrium constant expression correctly, with the complex concentration in the numerator and the free metal ion and free ligand concentrations in the denominator, raised to their stoichiometric powers.

Approach

Identify the formation reaction: $\text{Ni}^{2+} + 3\text{en} \rightleftharpoons [\text{Ni}(\text{en})_3]^{2+}$ . Write $K_{\text{stab}}$ as products over reactants, omitting water. The ligand 'en' appears with a power of 3 because three en molecules are required.

Step-by-Step Reasoning

The overall ligand exchange reaction is:

\text{Ni}^{2+} + 3\text{en} \rightleftharpoons [\text{Ni}(\text{en})_3]^{2+}

Applying the definition of the equilibrium constant:

K_{\text{stab}} = \frac{[[\text{Ni}(\text{en})_3]^{2+}]}{[\text{Ni}^{2+}][\text{en}]^3}

If instead we start from the hexaaqua ion:

[\text{Ni}(\text{H}_2\text{O})_6]^{2+} + 3\text{en} \rightleftharpoons [\text{Ni}(\text{en})_3]^{2+} + 6\text{H}_2\text{O}

then:

K_{\text{stab}} = \frac{[[\text{Ni}(\text{en})_3]^{2+}]}{[[\text{Ni}(\text{H}_2\text{O})_6]^{2+}][\text{en}]^3}

Both expressions are correct. Water is never included in $K_{\text{stab}}$ expressions.

Key Takeaways

$K_{\text{stab}}$ expressions always have the complex in the numerator and free metal ion + free ligands in the denominator.
Water of coordination is omitted from $K_{\text{stab}}$ expressions.
The ligand concentration is raised to the power equal to the number of ligand molecules (or the number of coordination sites it occupies, for monodentate ligands).

Common Mistakes

Including $[\text{H}_2\text{O}]$ in the expression — water is the solvent and its concentration is effectively constant.
Writing $[\text{en}]$ instead of $[\text{en}]^3$ — forgetting the stoichiometric coefficient.
Including the charge in the concentration brackets — write $[[\text{Ni}(\text{en})_3]^{2+}]$ , not $[[\text{Ni}(\text{en})_3]^{2+}]^{2+}$ or similar.

Things to Be Careful About

The expression must be balanced: numerator has one complex, denominator has one metal ion and three ligand molecules.
Both forms (from $\text{Ni}^{2+}$ or from $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ ) are acceptable; do not lose marks by writing only one if the question does not specify.
Use square brackets for concentrations; do not write $K_{\text{stab}} = \frac{\text{Ni}(\text{en})_3^{2+}}{\text{Ni}^{2+} \cdot \text{en}^3}$ without brackets.

Techniques used

write the stability constant expressionidentify ligands and metal ion in complex formation

(b)

A solution of $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ is added to a solution that contains $0.10\text{ mol dm}^{-3}\text{ NH}_3$ and $0.10\text{ mol dm}^{-3}\text{ en}$ .

2M

(i)

Predict which complex ion, $[\text{Ni}(\text{NH}_3)_6]^{2+}$ or $[\text{Ni}(\text{en})_3]^{2+}$ , is present in the resulting mixture in the highest concentration. Explain your answer.

complex ion present in largest concentration = ............................................

explanation ........................................................................................................................

DifficultyMedium-Easy

Worked solution

Answer

complex ion present in largest concentration = $[\text{Ni}(\text{en})_3]^{2+}$

explanation: $[\text{Ni}(\text{en})_3]^{2+}$ has a larger $K_{\text{stab}}$ ( $2.0 \times 10^{18}$ compared with $4.8 \times 10^7$ ), so it is more stable and the equilibrium lies further to the right.

Final answer

[Ni(en)3]2+; larger K_stab means more stable

Detailed explanation

Background Concept

The stability constant $K_{\text{stab}}$ is an equilibrium constant for the formation of a complex ion from the free metal ion and ligands. A larger $K_{\text{stab}}$ means the equilibrium lies further to the right, i.e., more complex is formed relative to the free components. This reflects greater thermodynamic stability of the complex.

When two different ligands compete for the same metal ion, the ligand that produces the complex with the larger $K_{\text{stab}}$ will be preferentially bound, provided concentrations are comparable.

Understanding the Question

A solution containing $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ is mixed with equal concentrations (0.10 mol dm $^{-3}$ ) of both NH $_3$ and en. We must predict which complex — $[\text{Ni}(\text{NH}_3)_6]^{2+}$ or $[\text{Ni}(\text{en})_3]^{2+}$ — will be present at the highest concentration.

Approach

Compare the two $K_{\text{stab}}$ values. The complex with the larger $K_{\text{stab}}$ is more stable and will be favoured at equilibrium. Since both ligands are present at the same concentration, the $K_{\text{stab}}$ value alone determines which complex dominates.

Step-by-Step Reasoning

From Table 5.1:

$K_{\text{stab}}$ for $[\text{Ni}(\text{NH}_3)_6]^{2+} = 4.8 \times 10^7$
$K_{\text{stab}}$ for $[\text{Ni}(\text{en})_3]^{2+} = 2.0 \times 10^{18}$

$K_{\text{stab}}$ for the en complex is about $10^{11}$ times larger. This enormous difference means that at equilibrium, virtually all the Ni $^{2+}$ will be bound as $[\text{Ni}(\text{en})_3]^{2+}$ . The en ligand displaces both water and ammonia because the resulting complex is thermodynamically far more stable.

This is partly due to the chelate effect: en is a bidentate ligand (it forms two bonds to the metal per molecule), so replacing six water molecules with three en molecules increases the number of particles in solution (from 7 to 9), giving a favourable entropy change that drives the equilibrium to the right.

Key Takeaways

A larger $K_{\text{stab}}$ means a more stable complex and equilibrium lies further to the right.
When competing ligands are present at equal concentrations, the complex with the larger $K_{\text{stab}}$ dominates.
Chelate ligands (bidentate or polydentate) form more stable complexes than monodentate ligands at the same concentration — this is the chelate effect.

Common Mistakes

Saying 'en is a stronger ligand' without reference to $K_{\text{stab}}$ or stability — must mention the larger $K_{\text{stab}}$ or greater stability.
Confusing $K_{\text{stab}}$ with $K_d$ (dissociation constant) — larger $K_{\text{stab}}$ means more stable, not less.
Not stating which complex is present in largest concentration before giving the explanation.

Things to Be Careful About

The mark scheme requires BOTH the name of the complex AND the explanation (larger $K_{\text{stab}}$ / more stable). Giving only one may lose a mark.
Do not say 'higher concentration of en' — both ligands are at 0.10 mol dm $^{-3}$ .
The explanation must link $K_{\text{stab}}$ to stability or equilibrium position.

Techniques used

compare stability constants to predict equilibrium positionrelate K_stab to thermodynamic stability

(ii)

Complete the equation for the ligand exchange reaction occurring in (i).

[\text{Ni}(\text{H}_2\text{O})_6]^{2+} + \text{.................................} \rightarrow \text{.................................} + \text{.................................}

DifficultyMedium-Easy

Worked solution

Answer

[\text{Ni}(\text{H}_2\text{O})_6]^{2+} + 3\text{en} \rightarrow [\text{Ni}(\text{en})_3]^{2+} + 6\text{H}_2\text{O}

Final answer

[Ni(H2O)6]2+ + 3en -> [Ni(en)3]2+ + 6H2O

Detailed explanation

Background Concept

Ligand exchange (substitution) reactions in coordination chemistry involve replacement of one set of ligands by another. When a hexaaqua metal ion reacts with a ligand L, water molecules are displaced:

[\text{M}(\text{H}_2\text{O})_6]^{n+} + x\text{L} \rightarrow [\text{ML}_x]^{n+} + 6\text{H}_2\text{O}

For bidentate ligands like en (1,2-diaminoethane, H $_2$ NCH $_2$ CH $_2$ NH $_2$ ), each en molecule occupies two coordination sites. To fill six coordination sites around Ni $^{2+}$ , three en molecules are needed:

3 \times 2 = 6 \text{ coordination sites}

Understanding the Question

Part (b)(ii) asks for the balanced equation for the ligand exchange reaction that occurs when $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ is added to a solution containing en. From part (b)(i), we know that $[\text{Ni}(\text{en})_3]^{2+}$ is the product. We must write the full balanced equation showing reactants and products.

Approach

Start with the hexaaqua complex on the left. Add three en molecules (since each en is bidentate and 3 × 2 = 6 sites). The products are the new complex $[\text{Ni}(\text{en})_3]^{2+}$ and six water molecules released.

Step-by-Step Reasoning

Reactants:

$[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ : the starting complex with six water ligands
$3\text{en}$ : three ethylenediamine molecules, each bidentate

Products:

$[\text{Ni}(\text{en})_3]^{2+}$ : the new complex with three en ligands
$6\text{H}_2\text{O}$ : six water molecules displaced

Balancing check:

Ni: 1 on each side ✓
Charge: +2 on each side ✓
N atoms: 3 en × 2 N = 6 N on left; 3 en × 2 N = 6 N on right ✓
C atoms: 3 en × 2 C = 6 C on left; 3 en × 2 C = 6 C on right ✓
H $_2$ O: 6 on left (as ligands); 6 on right (released) ✓

The equation is:

[\text{Ni}(\text{H}_2\text{O})_6]^{2+} + 3\text{en} \rightarrow [\text{Ni}(\text{en})_3]^{2+} + 6\text{H}_2\text{O}

Key Takeaways

Bidentate ligands occupy two coordination sites each; count sites carefully when balancing.
Water molecules coordinated to the metal are released as free H $_2$ O in the products.
The overall charge is conserved: +2 on both sides.

Common Mistakes

Writing '6en' instead of '3en' — en is bidentate, so only 3 are needed for 6 coordination sites.
Forgetting the 6H $_2$ O on the product side.
Writing the equation with NH $_3$ instead of en — the question asks about the reaction with en specifically.
Not balancing: the number of water molecules displaced must equal the number originally coordinated.

Things to Be Careful About

Use '→' (or '⟶') for the arrow; an equilibrium arrow is not required here since the reaction goes essentially to completion (very large $K_{\text{stab}}$ ).
Write 'en' not the full formula — the question uses the abbreviation en.
State symbols are not required for this equation in the mark scheme, but if included, H $_2$ O should be (l) and the complexes (aq).

Techniques used

write balanced ligand exchange equationbalance atoms and charges in complex reactions

(c)

Complete Fig. 5.1 to show the three-dimensional structures of the two isomers of $[\text{Ni}(\text{en})_3]^{2+}$ .

Fig. 5.1

Name the type of isomerism shown.

type of isomerism shown ..........................................................................................................

DifficultyMedium-Hard

Worked solution

Answer

type of isomerism shown: optical isomerism

Final answer

optical isomerism

Detailed explanation

Background Concept

Optical isomerism occurs when a molecule (or complex ion) has a non-superimposable mirror image. These two forms are called enantiomers. For a complex to show optical isomerism, it must lack a plane of symmetry and a centre of inversion.

In octahedral complexes, optical isomerism is common when there are three bidentate ligands, giving a structure like $[\text{M}(\text{bidentate})_3]$ . The three bidentate ligands can wrap around the metal centre in a right-handed (Δ, delta) or left-handed (Λ, lambda) propeller arrangement. These two arrangements are non-superimposable mirror images — exactly like your left and right hands.

The en ligand (1,2-diaminoethane, $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$ ) is bidentate: each molecule uses its two nitrogen lone pairs to form two coordinate bonds to the metal, creating a five-membered chelate ring (Ni–N–C–C–N). In $[\text{Ni}(\text{en})_3]^{2+}$ , three en ligands occupy all six coordination positions, with each en spanning two adjacent (cis) positions.

Understanding the Question

Part (c) asks us to draw the two 3D structures of $[\text{Ni}(\text{en})_3]^{2+}$ that are optical isomers, using the given representation where 'N⌒N' denotes an en ligand (two N atoms connected by a curved line representing the –CH $_2$ CH $_2$ – bridge). We must also name the type of isomerism.

Approach

Draw an octahedral Ni centre with six coordination positions.
Place three en ligands, each occupying two adjacent (cis) positions.
Draw the first isomer with the three chelate rings arranged in a right-handed (or left-handed) fashion.
Draw the second isomer as the non-superimposable mirror image.
Use wedge (solid triangle) and dash (dotted lines) bonds to show 3D perspective.
Name the isomerism as 'optical isomerism'.

Step-by-Step Reasoning

Drawing the first isomer:

Place Ni at the centre of an octahedron.
Use the convention: plain vertical lines for bonds in the plane of the paper (top and bottom positions), solid wedges for bonds coming out of the plane towards the viewer, and dashed lines for bonds going behind the plane.
Assign the three en ligands:
- Ligand 1: occupies the top position and one right-hand position (e.g., top-right, coming out of the plane). Connect the two N atoms with a curved line.
- Ligand 2: occupies the bottom position and one left-hand position (e.g., bottom-left, going behind the plane). Connect with a curved line.
- Ligand 3: occupies the remaining two positions (front-right and back-left, or similar cis arrangement). Connect with a curved line.
Ensure all three en ligands are arranged so the complex has no plane of symmetry.

Drawing the second isomer:

Draw the mirror image of the first structure. If the first has the chelate rings winding clockwise (Δ), the second winds anticlockwise (Λ).
The second structure must be non-superimposable on the first — you cannot rotate it to match.

Naming the isomerism:

These are optical isomers (also called enantiomers). They are non-superimposable mirror images of each other.
The type of isomerism is optical isomerism.

Mark scheme image description:
The marking scheme shows:

Left structure: Ni at centre, with N atoms at six octahedral positions. Three curved lines (en ligands) connect pairs of N atoms. Bonds to N atoms use: plain vertical lines (top/bottom), solid wedges (front), dashed lines (back). The three chelate rings form a right-handed propeller.
Right structure: mirror image — same arrangement but with wedges and dashes swapped on the left/right sides, forming a left-handed propeller.

Key Takeaways

$[\text{M}(\text{bidentate})_3]$ octahedral complexes show optical isomerism when all three ligands are identical.
The two isomers are non-superimposable mirror images (enantiomers).
Use wedge-dash notation to convey 3D structure: solid wedge = coming towards you, dashed = going away, plain line = in the plane.
Optical isomers have identical physical properties except for the direction they rotate plane-polarised light.

Common Mistakes

Drawing geometric isomers (cis/trans) instead of optical isomers — with three identical bidentate ligands in an octahedral complex, there are no cis/trans isomers, only optical isomers.
Drawing superimposable structures — the two drawings must be non-superimposable mirror images. Check by mentally trying to rotate one to match the other.
Forgetting the curved lines connecting the two N atoms of each en ligand — the en is bidentate, so both N atoms must be shown bonded to Ni and connected to each other.
Drawing en as monodentate (only one N bonded to Ni) — en always bonds through both nitrogens.
Not using 3D notation (wedges/dashes) — a flat 2D drawing does not show the chirality.
Naming the isomerism as 'geometric' or 'structural' — it is specifically optical isomerism.

Things to Be Careful About

The mark scheme awards 1 mark for a correct 3D octahedral structure, 1 mark for the second optical isomer (mirror image), and 1 mark for naming 'optical isomerism'. All three must be present.
The en ligand must be drawn as N⌒N (two N atoms connected by a curved/bridged line), not as a straight line or as separate monodentate ligands.
Both structures must be clearly distinguishable as mirror images. If they look identical, you have drawn the same structure twice.
Do not write 'enantiomerism' — the mark scheme accepts 'optical isomerism'. 'Enantiomerism' may not be credited at CIE.

Techniques used

draw 3D octahedral complex structuresidentify optical isomersuse wedge-dash notation for 3D representation

Q6MediumCarboxylic Acids and DerivativesHydrocarbonsPolymerisation

Fig. 6.1 shows two reactions of ethanedioic acid, $\text{HOOCCOOH}$ .

Fig. 6.1

(a)

2M

(i)

Draw the organic product G in the box in Fig. 6.1.

DifficultyEasy

Worked solution

Answer

The product G is ethanedioyl dichloride (oxalyl chloride):

\text{ClOCCOCl}

Final answer

ClOCCOCl (ethanedioyl dichloride / oxalyl chloride)

Detailed explanation

Background Concept

Thionyl chloride (SOCl₂) is a common reagent used to convert carboxylic acids (–COOH) into acyl chlorides (–COCl). The –OH group of the carboxylic acid is replaced by a chlorine atom, with SO₂ and HCl produced as by-products. When a dicarboxylic acid such as ethanedioic acid (HOOCCOOH) reacts with an excess of SOCl₂, both carboxylic acid groups are converted to acyl chloride groups.

Understanding the Question

The question asks you to draw the organic product G formed when ethanedioic acid reacts with an excess of SOCl₂. The key word is 'excess', meaning both –COOH groups will react.

Approach

Identify the functional group transformation: –COOH → –COCl. Apply it to both ends of the molecule since SOCl₂ is in excess.

Step-by-Step Reasoning

Ethanedioic acid has the structure HOOC–COOH (two carboxylic acid groups joined directly).
SOCl₂ replaces the –OH of each –COOH with –Cl, forming –COCl.
With excess SOCl₂, both groups react.
The product is ClOC–COCl (ethanedioyl dichloride, commonly called oxalyl chloride).

Key Takeaways

SOCl₂ converts –COOH to –COCl.
Excess reagent ensures all reactive groups are converted.
Oxalyl chloride (ClOCCOCl) is a useful reagent in organic synthesis.

Common Mistakes

Drawing only one –COCl group (forgetting the 'excess' means both react).
Writing the formula as ClCO–COCl without showing the structure clearly.
Confusing SOCl₂ with PCl₅ (both give acyl chlorides but the by-products differ).

Things to Be Careful About

Ensure both carbonyl groups are shown with the =O and the –Cl attached to the carbon.
The two carbonyl carbons are directly bonded (no CH₂ between them in ethanedioic acid derivatives).

Techniques used

convert carboxylic acid to acyl chloride using SOCl2draw displayed formula of dicarboxylic acid derivative

(ii)

In Fig. 6.1, $\text{SOCl}_2$ is given as the reagent that reacts with $\text{HOOCCOOH}$ to produce G.

Identify a different reagent that also reacts with $\text{HOOCCOOH}$ to produce G.

(b)

Identify two different reagents that oxidise $\text{HOOCCOOH}$ to form carbon dioxide and water.

DifficultyMedium-Easy

Worked solution

Answer

Oxygen ( $\text{O}_2$ )
Acidified potassium manganate(VII) ( $\text{KMnO}_4$ )

Final answer

Oxygen and acidified KMnO4

Detailed explanation

Background Concept

Ethanedioic acid (oxalic acid, HOOCCOOH) can be oxidised to carbon dioxide and water. This is because the carbon atoms in ethanedioic acid are in the +3 oxidation state, and can be further oxidised to +4 (as in CO₂). Two common oxidising agents achieve this: combustion in oxygen (complete oxidation) and acidified potassium manganate(VII) (a strong oxidising agent in acidic solution that is reduced from MnO₄⁻ to Mn²⁺).

Understanding the Question

The question asks for two different reagents that oxidise HOOCCOOH to form CO₂ and H₂O. The products given (2CO₂ + H₂O) indicate complete oxidation of both carbon atoms.

Approach

Identify oxidising agents capable of fully oxidising a carboxylic acid. Combustion (oxygen) is the most obvious. For a chemical oxidising agent, acidified KMnO₄ is the standard choice for oxidising ethanedioic acid (this reaction is also used in kinetics experiments due to the colour change from purple to colourless).

Step-by-Step Reasoning

Oxygen (O₂) — combustion of ethanedioic acid gives 2CO₂ + H₂O. This is a complete oxidation.
Acidified KMnO₄ — the MnO₄⁻ ion in acidic solution is a powerful oxidising agent. Ethanedioic acid reduces MnO₄⁻ to Mn²⁺ while being oxidised to CO₂. The ionic equation is: 5HOOCCOOH + 2MnO₄⁻ + 6H⁺ → 10CO₂ + 2Mn²⁺ + 8H₂O.
Both give the products stated in the question (CO₂ and H₂O).

Key Takeaways

Ethanedioic acid is unusual among carboxylic acids in that it can be further oxidised (the carbon is already at +3, below the +4 maximum).
Acidified KMnO₄ is the classic reagent for this oxidation and is used in rate-of-reaction experiments.
Combustion in oxygen also achieves complete oxidation.

Common Mistakes

Writing alkaline KMnO₄ instead of acidified (the question specifies acidified conditions for this oxidation).
Writing an oxidising agent that would not fully oxidise to CO₂ (e.g. mild oxidants).
Confusing this with the oxidation of alcohols (where K₂Cr₂O₇ is more commonly cited).

Things to Be Careful About

The question asks for reagents, not conditions — 'oxygen' and 'acidified KMnO₄' are the expected answers.
K₂Cr₂O₇ could also be accepted as an oxidising agent, but the mark scheme specifically lists oxygen and acidified KMnO₄.

Techniques used

recall oxidising agents for ethanedioic acididentify redox reagents that fully oxidise a dicarboxylic acid

(c)

$\text{HOOCCOOH}$ ionises as shown.

\text{HOOCCOOH} \rightleftharpoons \text{HOOCCOO}^- + \text{H}^+

$\text{HOOCCOOH}$ is a much stronger acid than methanoic acid, $\text{HCOOH}$ .

Suggest an explanation for this difference in acidity.

DifficultyMedium

Worked solution

Answer

The second C=O group (carbonyl) is electron-withdrawing (strongly electronegative), exerting a negative inductive effect ( $-I$ ).
This weakens the O–H bond (making H⁺ easier to lose) and/or stabilises the resulting anion (HOOCCOO⁻) by dispersing the negative charge.

Final answer

The electron-withdrawing C=O group weakens the O-H bond and stabilises the anion, making HOOCCOOH a stronger acid than HCOOH.

Detailed explanation

Background Concept

The strength of a carboxylic acid depends on how readily it donates a proton (H⁺) and how stable the resulting carboxylate anion is. Electron-withdrawing groups (EWGs) attached to or near the carboxyl group increase acidity by two mechanisms: (1) they pull electron density away from the O–H bond through the sigma framework (negative inductive effect, −I), weakening the bond and making proton loss easier; and (2) they stabilise the negative charge on the conjugate base (carboxylate anion) by dispersing it over a larger region of electron density.

Ethanedioic acid (HOOCCOOH) has a second –COOH group directly attached to the first. Methanoic acid (HCOOH) has only a hydrogen on the other side. The adjacent C=O group in ethanedioic acid is strongly electron-withdrawing due to the electronegativity of oxygen, whereas the H in methanoic acid has negligible inductive effect.

Understanding the Question

The question asks you to explain why HOOCCOOH is a much stronger acid than HCOOH. You need to identify the structural difference (the second C=O group) and explain how it affects acidity. Two marks are available, so you need two linked points.

Approach

Identify the structural difference: HOOCCOOH has an additional C=O group compared to HCOOH.
Explain the effect of this group: it is electron-withdrawing (−I effect).
State the consequence: this either weakens the O–H bond (making proton loss easier) or stabilises the anion (making the equilibrium lie further to the right), or both.

Step-by-Step Reasoning

In HOOCCOOH, the carbon bearing the –COOH group is also bonded to another C=O group. In HCOOH, that carbon is bonded to H.
The C=O group contains a highly electronegative oxygen atom, making it strongly electron-withdrawing through the sigma bond (−I inductive effect).
This electron withdrawal pulls electron density away from the O–H bond, weakening it and making it easier for H⁺ to dissociate.
Additionally, the negative charge on the conjugate base (⁻OOCCOOH) is stabilised by the electron-withdrawing effect of the adjacent C=O, which disperses the charge.
Both effects shift the equilibrium further to the right, giving a lower pKa (stronger acid).

The mark scheme accepts either 'weakens O–H bond' OR 'stabilises anion' for the second mark, so either reasoning is sufficient.

Key Takeaways

Electron-withdrawing groups increase carboxylic acid acidity.
The −I effect operates through sigma bonds and diminishes with distance.
Two valid explanations exist: bond weakening and anion stabilisation (both are correct; either scores).

Common Mistakes

Saying 'the second COOH is electron-withdrawing' without specifying the C=O or the inductive effect — the mark scheme wants the C=O identified as the electron-withdrawing feature.
Confusing this with resonance stabilisation (the −I effect is through sigma bonds, not pi delocalisation, in this context).
Writing 'HOOCCOOH has more oxygen atoms' without explaining the mechanism.

Things to Be Careful About

You must mention the electron-withdrawing nature of the C=O (or the −I effect) for the first mark.
For the second mark, you must connect this to either O–H bond weakening OR anion stabilisation — just saying 'it's more acidic' is circular and scores zero.

Techniques used

explain electron-withdrawing effect on acidityrelate inductive effect to O-H bond weakening and anion stabilisation

(d)

Benzene-1,4-dicarboxylic acid, $\text{HOOCC}_6\text{H}_4\text{COOH}$ , can be made from benzene, $\text{C}_6\text{H}_6$ , in two steps as shown in Fig. 6.2.

Fig. 6.2

6M

(i)

Suggest the identity of J by drawing its structure in the box in Fig. 6.2.

DifficultyMedium-Easy

Worked solution

Answer

J is 1,4-dimethylbenzene (para-xylene):

Final answer

1,4-dimethylbenzene (p-xylene)

Detailed explanation

Background Concept

To synthesise benzene-1,4-dicarboxylic acid from benzene in two steps, we need to work backwards. The final product has two –COOH groups at para positions on the benzene ring. The second step (oxidation) converts alkyl side chains to carboxylic acid groups. Therefore, the intermediate J must have two methyl groups at para positions — 1,4-dimethylbenzene (p-xylene). The first step is a Friedel-Crafts alkylation introducing methyl groups.

Understanding the Question

The question asks you to identify the intermediate J in a two-step synthesis from benzene to benzene-1,4-dicarboxylic acid. You need to recognise what functional group transformation occurs in step 2 (oxidation of side chains to –COOH) and deduce what J must be before that oxidation.

Approach

Step 2 must be oxidation (side-chain oxidation converts –CH₃ to –COOH).
Therefore J must already have the carbon skeleton with two alkyl groups at para positions.
Step 1 must be Friedel-Crafts alkylation to introduce two methyl groups.
J = 1,4-dimethylbenzene.

Step-by-Step Reasoning

Benzene-1,4-dicarboxylic acid has –COOH groups at positions 1 and 4 (para).
Alkyl side chains on benzene rings can be oxidised to –COOH using hot alkaline KMnO₄ followed by acidification.
Therefore, before oxidation, the ring must have –CH₃ groups at positions 1 and 4.
J is 1,4-dimethylbenzene (p-xylene): a benzene ring with CH₃ groups at opposite (para) positions.

Key Takeaways

Side-chain oxidation (KMnO₄) converts any alkyl group on benzene to –COOH.
Working backwards from the product is essential in multi-step synthesis.
Friedel-Crafts alkylation introduces alkyl groups onto the benzene ring.

Common Mistakes

Drawing 1,2-dimethylbenzene (ortho) or 1,3-dimethylbenzene (meta) instead of 1,4 (para).
Drawing J as the dicarboxylic acid itself (not recognising that oxidation is step 2).
Drawing an intermediate with only one methyl group.

Things to Be Careful About

The methyl groups must be at para positions (1,4) to give the 1,4-dicarboxylic acid after oxidation.
Draw the benzene ring correctly (hexagon with circle or alternating double bonds).

Techniques used

deduce intermediate from multi-step synthesisidentify Friedel-Crafts alkylation product

(ii)

Identify the reagents and conditions for step 1 and step 2.

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

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

DifficultyMedium-Easy

Worked solution

Answer

Step 1: $\text{CH}_3\text{Cl}$ and $\text{AlCl}_3$ (Friedel-Crafts alkylation)

Step 2: Hot alkaline $\text{KMnO}_4$ followed by acid (side-chain oxidation)

Final answer

Step 1: CH3Cl + AlCl3; Step 2: hot alkaline KMnO4

Detailed explanation

Background Concept

Friedel-Crafts alkylation uses a halogenoalkane (e.g. CH₃Cl) in the presence of a Lewis acid catalyst (AlCl₃) to introduce an alkyl group onto a benzene ring via electrophilic substitution. The electrophile is generated by the catalyst polarising the C–Cl bond.

Side-chain oxidation uses hot alkaline potassium manganate(VII) (KMnO₄) to oxidise any alkyl group attached to a benzene ring to a carboxylic acid group (–COOH). The reaction requires heating under reflux. After oxidation, the solution is acidified to protonate the carboxylate salt to the free carboxylic acid.

Understanding the Question

The question asks for reagents AND conditions for each step. For step 1, you need the alkylating agent and catalyst. For step 2, you need the oxidising agent and the condition (hot/heat under reflux, alkaline).

Approach

Step 1 introduces methyl groups → Friedel-Crafts alkylation → CH₃Cl + AlCl₃.
Step 2 oxidises methyl groups to carboxylic acid groups → hot alkaline KMnO₄.

Step-by-Step Reasoning

Step 1:

Benzene → 1,4-dimethylbenzene requires two methyl groups to be added.
Friedel-Crafts alkylation: CH₃Cl + AlCl₃ generates CH₃⁺ electrophile.
Excess CH₃Cl (or two equivalents) needed for disubstitution.
The methyl group is an ortho/para director, so the second substitution goes to the para position (giving 1,4).

Step 2:

1,4-dimethylbenzene → benzene-1,4-dicarboxylic acid requires oxidation of both –CH₃ groups.
Reagent: KMnO₄ (potassium manganate(VII)).
Conditions: hot (heat under reflux) and alkaline (NaOH or KOH present).
Followed by acidification (H⁺/H₃O⁺) to give the free dicarboxylic acid.

Key Takeaways

Friedel-Crafts alkylation: CH₃Cl + AlCl₃ (anhydrous conditions).
Side-chain oxidation: hot alkaline KMnO₄, then acidify.
Methyl groups are activating and ortho/para-directing in electrophilic substitution.

Common Mistakes

Writing 'AlCl₃' alone for step 1 without CH₃Cl (both are needed).
Writing 'cold dilute KMnO₄' for step 2 (must be hot and concentrated/alkaline for side-chain oxidation).
Forgetting to mention 'alkaline' or 'hot' for step 2.
Writing PCl₅ for step 2 (wrong reaction entirely).

Things to Be Careful About

Both reagent AND conditions are needed for each step to earn the mark.
For step 2, 'hot' or 'heat under reflux' and 'alkaline' are the key conditions.
The mark scheme accepts 'hot alkaline KMnO₄' as a single phrase covering both reagent and conditions.

Techniques used

specify reagents and conditions for Friedel-Crafts alkylationspecify reagents and conditions for side-chain oxidation

(iii)

Draw the structure of exactly one repeat unit of the polymer formed when benzene-1,4-dicarboxylic acid reacts with ethane-1,2-diol, $\text{HOCH}_2\text{CH}_2\text{OH}$ . The linkage formed between the monomers should be shown fully displayed.

DifficultyMedium

Worked solution

Answer

Final answer

Repeat unit: -[C(=O)-C6H4-C(=O)-O-CH2-CH2-O]- with fully displayed ester linkage

Detailed explanation

Background Concept

Condensation polymerisation occurs between monomers with two complementary functional groups, eliminating a small molecule (usually water) at each linkage. A dicarboxylic acid reacts with a diol to form a polyester, with ester linkages (–COO–) connecting the monomer units. The repeat unit shows one unit of each monomer joined by the ester linkage, with open bonds at each end indicating continuation of the chain.

Understanding the Question

The question asks for exactly one repeat unit of the polyester formed from benzene-1,4-dicarboxylic acid and ethane-1,2-diol. The ester linkage must be fully displayed (showing all atoms and bonds in the –COO– group explicitly). Two marks: one for the correctly displayed ester linkage, one for the rest of the structure.

Approach

Identify the monomers: benzene-1,4-dicarboxylic acid (HOOC–C₆H₄–COOH) and ethane-1,2-diol (HO–CH₂–CH₂–OH).
The ester forms between –COOH and –OH, eliminating H₂O.
Draw the repeat unit showing: –C(=O)–C₆H₄–C(=O)–O–CH₂–CH₂–O– with open bonds at both ends.
Fully display the ester linkage: show C=O and C–O– explicitly.

Step-by-Step Reasoning

Benzene-1,4-dicarboxylic acid provides: –C(=O)–C₆H₄–C(=O)– (the diacid portion in the chain).
Ethane-1,2-diol provides: –O–CH₂–CH₂–O– (the diol portion in the chain).
The ester linkage is –C(=O)–O– formed by loss of H₂O from –COOH + HO–.
The repeat unit is: –[C(=O)–C₆H₄–C(=O)–O–CH₂–CH₂–O]–
The ester linkage must be fully displayed: show the C=O double bond and the C–O single bond explicitly.
Open bonds at each end show the chain continues.

Key Takeaways

Polyester from diacid + diol: the repeat unit contains both monomer fragments joined by ester linkages.
'Fully displayed' means showing all atoms and bonds in the functional group (not condensed).
Open bonds at the ends of the repeat unit indicate the polymer continues.

Common Mistakes

Drawing two repeat units instead of one.
Not fully displaying the ester linkage (e.g. writing –COO– instead of showing C=O and C–O separately).
Forgetting the benzene ring in the diacid portion.
Drawing the wrong connectivity (e.g. putting the O from the diol on the wrong side).
Not showing open bonds at the ends of the repeat unit.

Things to Be Careful About

The ester linkage must be fully displayed with the =O shown explicitly.
The benzene ring should be drawn correctly (hexagon with circle or alternating bonds).
The –CH₂–CH₂– from the diol must be shown clearly.
Only ONE repeat unit is required (the question says 'exactly one').

Techniques used

draw repeat unit of a polyester from a dicarboxylic acid and a diolfully display the ester linkage

(iv)

State the type of polymerisation that occurs when benzene-1,4-dicarboxylic acid reacts with ethane-1,2-diol and name the linkage formed between the monomers.

type of polymerisation .......................................................................................................

linkage ...............................................................................................................................

DifficultyEasy

Worked solution

Answer

Type of polymerisation: Condensation

Linkage: Ester

Final answer

Condensation polymerisation; ester linkage

Detailed explanation

Background Concept

Polymerisation reactions are classified as either addition (monomers join without loss of any atoms, typically involving C=C double bonds) or condensation (monomers join with the elimination of a small molecule such as H₂O or HCl). When a dicarboxylic acid reacts with a diol, an ester linkage is formed with the loss of water — this is condensation polymerisation producing a polyester.

Understanding the Question

The question asks for the type of polymerisation and the name of the linkage. Both must be correct to earn the single mark (the mark scheme says 'condensation AND ester').

Approach

A diacid + diol → ester linkages + water eliminated → condensation polymerisation.
The linkage between monomer units is an ester group (–COO–).

Step-by-Step Reasoning

The reaction involves –COOH groups reacting with –OH groups.
Each reaction eliminates H₂O (one H from –OH of the acid, one OH from the alcohol, or vice versa).
Since a small molecule is lost, this is condensation polymerisation (not addition).
The functional group linking the monomers is –C(=O)–O–, which is an ester linkage.

Key Takeaways

Diacid + diol → polyester (condensation, ester linkage).
Diacid + diamine → polyamide (condensation, amide linkage).
The type of polymerisation is determined by whether a small molecule is eliminated.

Common Mistakes

Writing 'addition' instead of 'condensation'.
Writing 'amide' instead of 'ester' (confusing with polyamide formation).
Writing 'carboxylate' instead of 'ester' for the linkage.
Giving only one of the two required answers (both 'condensation' AND 'ester' are needed for the mark).

Things to Be Careful About

The mark scheme requires BOTH answers to be correct for the single mark — 'condensation AND ester'.
'Ester' is the name of the linkage; 'esterification' is the reaction type (do not confuse them).

Techniques used

identify type of polymerisationname the linkage formed

Q7MediumHydrocarbonsHalogen Compounds

Benzene reacts with chlorine gas to form chlorobenzene. This reaction can be described as the reaction between benzene molecules and $\text{Cl}^+$ ions. The $\text{Cl}^+$ ions are formed by adding a suitable catalyst to the chlorine gas.

(a)

Give the name or formula of a catalyst that can be used for this reaction.

DifficultyEasy

Worked solution

Answer

aluminium chloride (or $\text{AlCl}_3$ )

Final answer

aluminium chloride (or AlCl3)

Detailed explanation

Background Concept

Benzene is unusually stable due to its delocalised pi electron system. Because of this stability, benzene does not undergo addition reactions with halogens like alkenes do. Instead, it undergoes electrophilic substitution, but only when a strong electrophile is generated. A Lewis acid catalyst is required to polarise the halogen molecule and generate the electrophile (e.g., $\text{Cl}^+$ from $\text{Cl}_2$ ).

Understanding the Question

The question asks for the name or formula of the catalyst used when benzene reacts with chlorine gas to form chlorobenzene.

Approach

Recall the standard conditions for the halogenation of benzene. The reaction requires a Lewis acid catalyst to generate the electrophile.

Step-by-Step Reasoning

For the chlorination of benzene, chlorine gas ( $\text{Cl}_2$ ) is not a strong enough electrophile on its own. A Lewis acid catalyst such as aluminium chloride ( $\text{AlCl}_3$ ) or iron(III) chloride ( $\text{FeCl}_3$ ) is added. The catalyst reacts with $\text{Cl}_2$ to form a complex that generates the $\text{Cl}^+$ electrophile. The mark scheme accepts aluminium chloride or $\text{AlCl}_3$ .

Key Takeaways

Benzene requires a Lewis acid catalyst (like $\text{AlCl}_3$ or $\text{FeCl}_3$ ) for electrophilic substitution with halogens to generate the necessary electrophile.

Common Mistakes

Writing "iron" instead of "iron(III) chloride" or "aluminium" instead of "aluminium chloride". The catalyst must be the Lewis acid, not the bare metal.

Things to Be Careful About

Ensure the catalyst is a Lewis acid (e.g., $\text{AlCl}_3$ , $\text{FeCl}_3$ ), not a transition metal catalyst used for addition reactions (like Ni or Pt for hydrogenation).

Techniques used

identify the catalyst for electrophilic substitution of benzene

(b)

The mechanism for this reaction is shown.

Fig. 7.1

4M

(i)

The movement of a pair of electrons is represented by $x$ in diagram 1.

State where this pair of electrons is before step 1 takes place.
State where this pair of electrons is after step 1 has taken place.

DifficultyMedium-Easy

Worked solution

Answer

Before step 1: delocalised pi system (or delocalised ring)
After step 1: C–Cl bond

Final answer

Before: delocalised pi system; After: C-Cl bond

Detailed explanation

Background Concept

In the first step of electrophilic aromatic substitution, the electron-rich delocalised pi system of the benzene ring acts as a nucleophile. It attacks the electrophile ( $\text{Cl}^+$ ), forming a new sigma bond. This disrupts the delocalised system, leaving a positive charge on the ring (an arenium ion or sigma complex).

Understanding the Question

The question asks to identify the location of the pair of electrons represented by curly arrow $x$ before and after step 1 in the mechanism diagram.

Approach

Arrow $x$ starts from the benzene ring and points to $\text{Cl}^+$ . Before the step, these electrons are in the delocalised pi system. After the step, they have formed a new bond between carbon and chlorine.

Step-by-Step Reasoning

Before step 1: The curly arrow $x$ originates from the delocalised pi system (the circle inside the benzene ring). These are the delocalised pi electrons.
After step 1: The arrow points to the $\text{Cl}^+$ ion. The pair of electrons has been used to form a new covalent bond between a carbon atom of the ring and the chlorine atom. Thus, the electrons are now in the C–Cl bond.

Key Takeaways

Curly arrows in mechanism diagrams show the movement of electron pairs. An arrow starting from a pi system and ending at an electrophile indicates the formation of a new sigma bond at the expense of the pi system.

Common Mistakes

Saying "the electrons come from a double bond". Benzene does not have localized double bonds; the electrons are in a delocalised pi system.

Things to Be Careful About

Use precise terminology: "delocalised pi system" or "delocalised ring" rather than "double bond" or "pi bond" (singular).

Techniques used

trace curly arrow x from pi system to electrophileidentify the new bond formed in the arenium ion

(ii)

The movement of another pair of electrons is represented by $y$ in diagram 2.

State where this pair of electrons is before step 2 takes place.
State where this pair of electrons is after step 2 has taken place.

(c)

There are six carbon atoms in diagram 2.

State how many of these carbon atoms are $\text{sp}$ hybridised, $\text{sp}^2$ hybridised, and $\text{sp}^3$ hybridised.

$\text{sp}$ hybridised .................................

$\text{sp}^2$ hybridised .................................

$\text{sp}^3$ hybridised .................................

DifficultyMedium-Easy

Worked solution

Answer

$\text{sp}$ hybridised: 0
$\text{sp}^2$ hybridised: 5
$\text{sp}^3$ hybridised: 1

Final answer

sp: 0, sp2: 5, sp3: 1

Detailed explanation

Background Concept

In benzene, all six carbon atoms are $sp^2$ hybridised, forming a planar ring with bond angles of 120°. Each carbon has a p-orbital perpendicular to the ring that contributes to the delocalised pi system.

In the arenium ion intermediate (diagram 2), one carbon atom has formed a new sigma bond with chlorine and still retains its bond to hydrogen. This carbon is now bonded to four groups (two ring carbons, one H, one Cl) and is $sp^3$ hybridised with tetrahedral geometry (bond angles ~109.5°). The other five carbon atoms remain $sp^2$ hybridised and continue to contribute to the delocalised pi system (though it is now broken over only five carbons).

Understanding the Question

The question asks for the number of $sp$ , $sp^2$ , and $sp^3$ hybridised carbon atoms in diagram 2 (the arenium ion intermediate).

Approach

Identify the hybridisation of each carbon in the intermediate. Five carbons are part of the remaining conjugated system and are $sp^2$ . One carbon is saturated (bonded to H, Cl, and two ring carbons) and is $sp^3$ . No carbons are $sp$ hybridised.

Step-by-Step Reasoning

Diagram 2 shows the arenium ion. Five carbon atoms are still part of the delocalised pi system (indicated by the broken horseshoe and positive charge). These five carbons are $sp^2$ hybridised.
One carbon atom is bonded to both H and Cl, and has no p-orbital contribution to the pi system. This carbon is $sp^3$ hybridised.
There are no triple bonds or linear geometries, so 0 carbons are $sp$ hybridised.
Therefore: $sp$ = 0, $sp^2$ = 5, $sp^3$ = 1.

Key Takeaways

In electrophilic substitution intermediates, the carbon that bonds to the electrophile becomes $sp^3$ hybridised, while the rest of the ring remains $sp^2$ hybridised.

Common Mistakes

Assuming all carbons are $sp^2$ because the product is aromatic. Remember to look at the intermediate (diagram 2), not the final product (diagram 3).

Things to Be Careful About

Count carefully: there are 6 carbons total. 5 + 1 = 6.

Techniques used

determine hybridisation of carbons in the arenium ion intermediate

(d)

Complete the equation for this reaction between benzene and chlorine.

\text{C}_6\text{H}_6 + \text{.................................} \rightarrow \text{.................................} + \text{.................................}

(e)

The mechanism for this reaction is electrophilic substitution.

Complete the following sentence. Write formulae in the gaps provided.

During this reaction, the electrophile is ................................. and a ................................. atom in benzene is substituted by a ................................. atom.

(f)

Chloroethane reacts with $\text{NaOH}(\text{aq})$ . Chlorobenzene does not.

3M

(i)

Name the mechanism of the reaction that chloroethane undergoes with $\text{NaOH}(\text{aq})$ , and identify the major organic product that is formed.

mechanism ........................................................................................................................

major organic product .......................................................................................................

DifficultyMedium-Easy

Worked solution

Answer

mechanism: nucleophilic substitution
major organic product: ethanol

Final answer

nucleophilic substitution; ethanol

Detailed explanation

Background Concept

Chloroethane ( $\text{CH}_3\text{CH}_2\text{Cl}$ ) is a primary halogenoalkane. When heated with aqueous sodium hydroxide ( $\text{NaOH}$ ), it undergoes nucleophilic substitution. The hydroxide ion ( $\text{OH}^-$ ) acts as a nucleophile, attacking the electron-deficient carbon bonded to chlorine. The C–Cl bond breaks, and a C–OH bond forms, producing ethanol ( $\text{CH}_3\text{CH}_2\text{OH}$ ) and a chloride ion.

Understanding the Question

Name the mechanism and identify the major organic product when chloroethane reacts with $\text{NaOH}(\text{aq})$ .

Approach

Recall the reaction of primary halogenoalkanes with aqueous hydroxide ions. The mechanism is nucleophilic substitution, and the product is an alcohol.

Step-by-Step Reasoning

Chloroethane is $\text{CH}_3\text{CH}_2\text{Cl}$ .
Aqueous $\text{NaOH}$ provides $\text{OH}^-$ ions, which are nucleophiles.
The $\text{OH}^-$ attacks the $\text{C}$ atom bonded to $\text{Cl}$ , displacing $\text{Cl}^-$ .
Mechanism: nucleophilic substitution.
Product: $\text{CH}_3\text{CH}_2\text{OH}$ , which is ethanol.

Key Takeaways

Primary halogenoalkanes undergo nucleophilic substitution with aqueous hydroxide to form primary alcohols.

Common Mistakes

Confusing the conditions: aqueous $\text{NaOH}$ gives substitution (alcohol), while ethanolic $\text{NaOH}$ gives elimination (alkene). The question specifies $\text{NaOH}(\text{aq})$ .

Things to Be Careful About

Ensure the mechanism name is exactly "nucleophilic substitution". Do not just write "substitution".

Techniques used

identify the mechanism and product for SN2/SN1 reaction of a primary halogenoalkane with aqueous hydroxide

(ii)

Explain the difference in reactivity of chloroethane and chlorobenzene when treated with $\text{NaOH}(\text{aq})$ .

DifficultyMedium

Worked solution

Answer

In chlorobenzene, the lone pair on the chlorine atom is delocalised into the delocalised pi system of the benzene ring. This gives the C–Cl bond partial double-bond character, making it stronger and harder to break than the C–Cl bond in chloroethane.

Final answer

See explanation

Detailed explanation

Background Concept

The reactivity of halogenoalkanes vs halogenoarenes towards nucleophilic substitution depends on the strength of the C–X bond and the availability of the carbon atom for nucleophilic attack.

In halogenoalkanes (like chloroethane), the C–Cl bond is a pure sigma bond. The chlorine is more electronegative than carbon, making the carbon partially positive and susceptible to nucleophilic attack.

In halogenoarenes (like chlorobenzene), the lone pair on the chlorine atom is in a p-orbital that overlaps with the delocalised pi system of the benzene ring. This delocalisation gives the C–Cl bond partial double-bond character, making it shorter and stronger than a typical C–Cl sigma bond. Additionally, the electron density of the ring repels nucleophiles.

Understanding the Question

Explain why chloroethane reacts with $\text{NaOH}(\text{aq})$ but chlorobenzene does not.

Approach

Compare the C–Cl bonds in both molecules. Highlight the delocalisation of the chlorine lone pair in chlorobenzene and its effect on bond strength.

Step-by-Step Reasoning

Chloroethane: The C–Cl bond is a standard sigma bond. The carbon is partially positive ( $\delta^+$ ) and is readily attacked by the nucleophile ( $\text{OH}^-$ ). The bond is relatively easy to break.
Chlorobenzene: The lone pair on the chlorine atom is delocalised into the delocalised pi system of the benzene ring (resonance). This delocalisation gives the C–Cl bond partial double-bond character.
Because of the partial double-bond character, the C–Cl bond in chlorobenzene is stronger (has a higher bond enthalpy) and is harder to break than the C–Cl bond in chloroethane.
Therefore, nucleophilic substitution does not occur under normal conditions with chlorobenzene.

Key Takeaways

Lone pair delocalisation into an aromatic ring strengthens the C–X bond in halogenoarenes, making them unreactive towards nucleophilic substitution compared to halogenoalkanes.

Common Mistakes

Saying "the ring is electron-rich and repels nucleophiles". While true, the primary reason for the lack of reactivity in standard nucleophilic substitution conditions is the strength of the C–Cl bond due to delocalisation. The mark scheme specifically looks for delocalisation and bond strength.

Things to Be Careful About

Be precise: say "delocalisation of the lone pair of Cl with the pi system" and "C–Cl bond is stronger / partly double". Do not just say "the bond is stronger" without explaining why.

Techniques used

compare C-Cl bond strength and electron delocalisation in halogenoalkanes vs halogenoarenes

Q8Medium-HardIntroduction to A Level Organic ChemistryAnalytical TechniquesNitrogen Compounds

The amino acid serine, $\text{HOCH}_2\text{CH}(\text{NH}_2)\text{COOH}$ , exists in two optically active forms. These optical isomers, isomer P and isomer Q, are shown in Fig. 8.1.

Fig. 8.1

(a)

Isomer $\mathbf{P}$ and isomer $\mathbf{Q}$ have identical physical and chemical properties, with the exception of two specific properties. One of these two properties is their differing effect on plane polarised light.

State the other property by which they differ.

DifficultyEasy

Worked solution

Answer

Biological activity.

Final answer

Biological activity

Detailed explanation

Background Concept

Optical isomers (enantiomers) are non-superimposable mirror images of each other. Because they have the same connectivity and the same physical properties (boiling point, melting point, density, etc.), they are indistinguishable in an achiral environment. However, they interact differently with other chiral entities. In biological systems, receptors, enzymes, and other biomolecules are themselves chiral. Therefore, one enantiomer of a drug or amino acid may fit perfectly into a biological receptor (like a key in a lock), while its mirror image may not fit at all, or may even cause an adverse effect.

Understanding the Question

The question asks for the second property that differs between optical isomers, given that their differing effect on plane polarised light is the first. This is a direct recall question about the fundamental characteristics of enantiomers.

Approach

Recall the standard list of differences between enantiomers: rotation of plane polarised light (opposite directions) and interaction with other chiral molecules (biological activity, taste, smell).

Step-by-Step Reasoning

The mark scheme specifically looks for "biological activity". While they also differ in taste and smell (e.g., limonene), biological activity is the most significant and universally cited difference in the context of pharmaceuticals and amino acids.

Key Takeaways

Enantiomers have identical physical and chemical properties in achiral environments but differ in two key ways: the direction they rotate plane polarised light and their biological activity (interaction with chiral biological molecules).

Common Mistakes

Students often write "chemical properties" as a whole, which is incorrect because their chemical properties are identical towards achiral reagents. The difference only arises in chiral environments.

Things to Be Careful About

Be precise with terminology. "Biological activity" is the expected answer. Avoid vague terms like "how they react" without specifying the chiral nature of the interaction.

Techniques used

recall properties of optical isomers

(b)

A solution of pure isomer $\mathbf{P}$ of a particular concentration rotates plane polarised light by $5.0^\circ$ in a clockwise direction.

Describe how a solution of pure isomer $\mathbf{Q}$ of the same concentration affects plane polarised light.

(c)

State another term, in addition to stereoisomers, optical isomers and non-superimposable mirror images, which can be used to describe this pair of chiral compounds, isomer $\mathbf{P}$ and isomer $\mathbf{Q}$ .

(d)

Give the term used to describe a mixture containing equal amounts of isomer $\mathbf{P}$ and isomer $\mathbf{Q}$ .

(e)

Describe one way in which a single pure optical isomer of serine can be produced, instead of making a mixture of isomer $\mathbf{P}$ and isomer $\mathbf{Q}$ .

(f)

Complete Table 8.1 to describe the peaks seen in the proton ( $^1\text{H}$ ) NMR spectrum of $\text{HOCH}_2\text{CH}(\text{NH}_2)\text{COOH}$ dissolved in $\text{D}_2\text{O}$ .

Use as many rows in Table 8.1 as you need to, leaving the other rows blank.

Table 8.1

DifficultyMedium

Worked solution

Answer

group responsible for peak	name of splitting pattern shown by peak	explanation for splitting pattern
$\text{CH}_2$	doublet	due to 1 proton on neighbouring carbon ( $\text{CH}$ )
$\text{CH}$	triplet	due to 2 protons on neighbouring carbon ( $\text{CH}_2$ )

(Note: $\text{OH}$ and $\text{NH}_2$ protons exchange with $\text{D}_2\text{O}$ and do not appear in the spectrum or cause splitting.)

Final answer

See table above

Detailed explanation

Background Concept

In $^1$ H NMR spectroscopy, the splitting pattern (multiplicity) of a peak is determined by the number of protons on adjacent carbons using the $n+1$ rule, where $n$ is the number of neighbouring protons. However, protons attached to heteroatoms like oxygen ( $\text{OH}$ ) and nitrogen ( $\text{NH}_2$ , $\text{NH}$ ) are labile. When the solvent is $\text{D}_2\text{O}$ (deuterium oxide), these labile protons undergo rapid exchange with deuterium ( $\text{D}$ ). Deuterium does not produce a signal in the standard $^1$ H NMR range (its resonance is at a very different frequency), and it does not cause splitting of adjacent protons in typical routine spectra (or the coupling is not resolved). Therefore, $\text{OH}$ and $\text{NH}$ peaks disappear or broaden out, and they do not contribute to the splitting of neighbouring protons.

Understanding the Question

We need to complete a table for the $^1$ H NMR spectrum of serine ( $\text{HOCH}_2\text{CH}(\text{NH}_2)\text{COOH}$ ) dissolved in $\text{D}_2\text{O}$ . We must identify the groups responsible for peaks, the splitting pattern, and the explanation.

Approach

Identify all proton environments in serine: $\text{CH}_2\text{OH}$ (2H), $\text{CH}$ (1H), $\text{NH}_2$ (2H), $\text{COOH}$ (1H).
Apply $\text{D}_2\text{O}$ exchange: $\text{OH}$ and $\text{NH}_2$ protons exchange with D. The $\text{COOH}$ proton also exchanges. Only the $\text{C-H}$ protons remain.
Analyze the remaining protons: $\text{CH}_2$ group and $\text{CH}$ group.
Apply the $n+1$ rule to each.

Step-by-Step Reasoning

Protons in $\text{D}_2\text{O}$ : The protons on $\text{OH}$ (alcohol and carboxylic acid) and $\text{NH}_2$ exchange with deuterium. They do not appear as sharp peaks and do not cause splitting. Thus, only the $\text{CH}_2$ and $\text{CH}$ protons are relevant.
$\text{CH}_2$ group: This group is attached to the chiral $\text{CH}$ carbon. The $\text{CH}$ carbon has 1 proton. Using the $n+1$ rule: $1 + 1 = 2$ . So, the $\text{CH}_2$ peak is a doublet. Explanation: splitting due to 1 proton on the neighbouring carbon.
$\text{CH}$ group: This carbon is attached to the $\text{CH}_2$ group. The $\text{CH}_2$ carbon has 2 protons. Using the $n+1$ rule: $2 + 1 = 3$ . So, the $\text{CH}$ peak is a triplet. Explanation: splitting due to 2 protons on the neighbouring carbon.

The mark scheme specifically requires: " $\text{CH}_2$ and $\text{CH}$ only in column one", " $\text{CH}_2$ gives a doublet, $\text{CH}$ gives a triplet", and the explanations for splitting.

Key Takeaways

Always consider the solvent. In $\text{D}_2\text{O}$ , labile protons ( $\text{OH}$ , $\text{NH}$ , $\text{COOH}$ ) exchange and disappear. Only $\text{C-H}$ protons contribute to splitting and appear in the spectrum. Apply the $n+1$ rule only to neighbouring $\text{C-H}$ protons.

Common Mistakes

Including $\text{OH}$ , $\text{NH}_2$ , or $\text{COOH}$ in the table. These exchange with $\text{D}_2\text{O}$ .
Forgetting that $\text{D}$ does not split $\text{H}$ signals in this context.
Miscounting neighbouring protons. Remember to look at the adjacent carbon's protons, not the total number of protons in the molecule.

Things to Be Careful About

The question specifies $\text{D}_2\text{O}$ as the solvent. If it were $\text{CDCl}_3$ , the $\text{OH}$ and $\text{NH}_2$ peaks would appear and would cause splitting (though $\text{OH}$ splitting is often not resolved due to exchange, in ideal theory they would split). Here, $\text{D}_2\text{O}$ is key.

Techniques used

account for D2O exchange in NMRapply n+1 splitting rule

(g)

Proline is a naturally occurring amino acid. The skeletal formula of proline is shown.

Fig. 8.2

State the number of peaks in the carbon-13 ( $^{13}\text{C}$ ) NMR spectrum of proline.

DifficultyMedium-Easy

Worked solution

Answer

5

Final answer

5

Detailed explanation

Background Concept

In $^{13}$ C NMR spectroscopy, each chemically distinct carbon environment gives a separate peak. Symmetry in a molecule can make different carbons equivalent, reducing the number of peaks. We must count the number of unique carbon atoms.

Understanding the Question

We are given the skeletal formula of proline and asked for the number of peaks in its $^{13}$ C NMR spectrum.

Approach

Draw out the full structure of proline, label each carbon, and determine if any are equivalent by symmetry.

Step-by-Step Reasoning

Proline is a cyclic amino acid. The structure is a pyrrolidine ring (5-membered ring containing N) with a carboxylic acid group attached to the carbon adjacent to the nitrogen (the alpha carbon).

Let's number the carbons:

Carboxyl carbon ( $\text{C=O}$ ): Unique environment.
Alpha carbon ( $\text{CH}$ ): Attached to the carboxyl group, the ring nitrogen, and a ring $\text{CH}_2$ . Unique.
Ring $\text{CH}_2$ (beta carbon, adjacent to alpha): Attached to alpha carbon and another ring $\text{CH}_2$ . Unique.
Ring $\text{CH}_2$ (gamma carbon, middle of the chain): Attached to beta $\text{CH}_2$ and delta $\text{CH}_2$ . Unique.
Ring $\text{CH}_2$ (delta carbon, adjacent to nitrogen): Attached to gamma $\text{CH}_2$ and the nitrogen. Unique.

There is no plane of symmetry in proline that makes any carbons equivalent. All 5 carbons are in different chemical environments.

Therefore, there are 5 peaks in the $^{13}$ C NMR spectrum.

Key Takeaways

Always check for symmetry. In cyclic molecules without a plane of symmetry, all carbons are usually unique. Proline has 5 carbons and no symmetry, so 5 peaks.

Common Mistakes

Assuming the three $\text{CH}_2$ groups in the ring are equivalent. They are not, because one is next to $\text{C=O}$ , one is next to $\text{N}$ , and one is in the middle.
Forgetting the carboxyl carbon.

Things to Be Careful About

The skeletal formula might make it look like there are fewer carbons if you don't count the carboxyl carbon or misinterpret the ring vertices. Count every vertex and end of line that represents a carbon.

Techniques used

identify unique carbon environments in a cyclic structure

(h)

Glutamic acid is a naturally occurring amino acid.

The skeletal formula of glutamic acid is shown.

Fig. 8.3

The isoelectric point of glutamic acid is pH 3.

A sample of glutamic acid is dissolved in a solution of pH 1. A strong alkali is then added until the pH of the mixture reaches pH 14. During this process all possible ionised forms of glutamic acid are present at different times, depending on the pH of the solution.

Complete the boxes below to show four different ionised forms of glutamic acid that are present at the stated pH values.

Fig. 8.4