Both calcium carbonate, $\text{CaCO}_3$, and barium carbonate, $\text{BaCO}_3$, decompose when heated to form the metal oxide and a gas. (a) Write an equation for the thermal decomposition of $\text{CaCO}_3$. (b) State which of $\text{CaCO}_3$ and $\text{BaCO}_3$ decomposes at a lower temperature. Explain your answer. The compound that decomposes at a lower temperature is ................................................... explanation…

(a) CaCO3(s) -> CaO(s) + CO2(g) · (b) Calcium carbonate decomposes at a lower temperature; Ca2+ is smaller and polarises CO3^2- more strongly. · (c)(i) Ca(OH)2 (calcium hydroxide) · (c)(ii) A is Ca(OH)2; less soluble because its ΔH_sol is less exothermic than Ba(OH)2 due to greater difference in lattice enthalpy than hydration enthalpy.

A hydrocarbon is known to be either compound D or compound E . (a) Give the systematic name of E . (b) The proton ($^1\text{H}$) NMR spectra of D and E are compared. They are very similar. The proton ($^1\text{H}$) NMR spectrum of D is shown in Fig. 7.2. (i) Suggest a suitable solvent for obtaining the spectrum in Fig. 7.2. (ii) The proton ($^1\text{H}$) NMR spectrum of E is obtained twice, once before and once after shaking with $\text{D}_2\text{O}$. Describe any…

(a) 1,3-diethylbenzene · (b)(i) CDCl3 · (b)(ii) No difference; E has no H atoms that can be exchanged with D · (b)(iii) 1.3 ppm: 6H, CH3, triplet; 2.7 ppm: 4H, CH2, quartet; 7.1 ppm: 4H · (b)(iv) D: 5 peaks; E: 6 peaks · (c)(i) C10H14 + 12[O] → C8H6O4 + 2CO2 + 4H2O · (c)(ii) G: 1,2-C6H4(COCl)2; J: cyclic diester (benzene fused to 8-membered ring with two COO groups and CH2CH2); K: two C6H4(COCl) units linked by COOCH2CH2OOC bridge

(a) Phenylamine, $\text{C}_6\text{H}_5\text{NH}_2$, and propylamine, $\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$, can be produced by different reduction reactions. (i) Identify an organic compound that can be converted into $\text{C}_6\text{H}_5\text{NH}_2$ by a reduction reaction. State the reagents and conditions for this reaction. organic compound ............................................................................................................. reagents…

(a)(i) nitrobenzene; Sn and HCl; concentrated HCl and heat/reflux · (a)(ii) propanenitrile ($C_2H_5CN$); $LiAlH_4$ · (a)(iii) bromine (aq); phenylamine; 2,4,6-tribromophenylamine structure · (b) phenylamine < ammonia < propylamine

9701/41 May/June 2025 — Chemistry Solved Past Paper

Q1MediumGroup 2

Both calcium carbonate, $\text{CaCO}_3$ , and barium carbonate, $\text{BaCO}_3$ , decompose when heated to form the metal oxide and a gas.

(a)

Write an equation for the thermal decomposition of $\text{CaCO}_3$ .

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})$

Final answer

CaCO3(s) -> CaO(s) + CO2(g)

Detailed explanation

Background Concept

Calcium carbonate is a Group 2 carbonate. When heated, Group 2 carbonates decompose thermally to give the metal oxide and carbon dioxide gas. The general equation is $\text{MCO}_3(\text{s}) \rightarrow \text{MO}(\text{s}) + \text{CO}_2(\text{g})$ . This is a thermal decomposition reaction.

Understanding the Question

This part asks for the balanced symbol equation for the decomposition of calcium carbonate. The reactants and products are given in the stem: calcium carbonate decomposes to the metal oxide and a gas. The metal oxide is CaO and the gas is CO2.

Approach

Write the formula of calcium carbonate, then replace the carbonate group with oxide, and add carbon dioxide. Balance if necessary. Include state symbols.

Step-by-Step Reasoning

Calcium carbonate has the formula $\text{CaCO}_3$ .
On heating, it loses carbon dioxide to form calcium oxide.
The equation is $\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})$ .
The equation is already balanced: one Ca, one C, three O on each side.
State symbols: CaCO3 and CaO are solids; CO2 is a gas.

Key Takeaways

All Group 2 carbonates decompose on heating to the oxide and CO2.
The balanced equation must show correct formulae and state symbols.

Common Mistakes

Forgetting state symbols: the mark scheme often accepts without, but including them is good practice.
Writing CaO2 instead of CaO.
Unbalanced equation.

Things to Be Careful About

Use correct formulae: CaCO3, CaO, CO2.
Include state symbols if asked or for good practice.
The gas is CO2, not CO or O2.

Techniques used

write a balanced thermal decomposition equationinclude state symbols in the equation

(b)

State which of $\text{CaCO}_3$ and $\text{BaCO}_3$ decomposes at a lower temperature.

Explain your answer.

The compound that decomposes at a lower temperature is ...................................................

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

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Calcium carbonate decomposes at a lower temperature.

The $\text{Ca}^{2+}$ ion is smaller than $\text{Ba}^{2+}$ (ionic radius increases down Group 2), so $\text{Ca}^{2+}$ has a higher charge density and polarises the carbonate ion, $\text{CO}_3^{2-}$ , more strongly. This distorts the carbonate ion and weakens its C–O bonds, making $\text{CaCO}_3$ thermally less stable, so it decomposes at a lower temperature.

Final answer

Calcium carbonate decomposes at a lower temperature; Ca2+ is smaller and polarises CO3^2- more strongly.

Detailed explanation

Background Concept

The thermal stability of Group 2 carbonates depends on the polarising power of the metal cation. A cation with a high charge density (small radius, high charge) can distort the electron cloud of the large carbonate anion, $\text{CO}_3^{2-}$ . This distortion weakens the C–O bonds, making it easier for the carbonate to decompose on heating to the oxide and $\text{CO}_2$ .

Understanding the Question

This part asks you to state which of calcium carbonate and barium carbonate decomposes at a lower temperature, and to explain why. The explanation must compare the sizes of the cations and relate this to polarisation of the carbonate ion.

Approach

Recall that down Group 2, the ionic radius increases, so the charge density of the cation decreases. A lower charge density means less polarisation of the carbonate ion, so the carbonate is more thermally stable and decomposes at a higher temperature. Therefore, the smaller cation (Ca2+) gives a less stable carbonate that decomposes at a lower temperature.

Step-by-Step Reasoning

Ca is above Ba in Group 2, so Ca2+ is smaller than Ba2+.
Ionic radius increases down the group: $r(\text{Ca}^{2+}) < r(\text{Ba}^{2+})$ .
Since Ca2+ is smaller, it has a higher charge density and is a stronger polarising agent.
The carbonate ion is large and polarisable; the Ca2+ ion distorts the electron cloud of $\text{CO}_3^{2-}$ more than Ba2+ does.
This distortion weakens the C–O bonds, so less thermal energy is needed to break them and release $\text{CO}_2$ .
Hence CaCO3 decomposes at a lower temperature than BaCO3.

Key Takeaways

Down Group 2, thermal stability of carbonates increases because cation polarising power decreases.
The explanation must mention both the smaller cation and the greater polarisation/distortion of the carbonate ion.

Common Mistakes

Saying that CaCO3 is less stable because Ca is more reactive – this is not the reason.
Confusing ionic radius trend: it increases down the group, not decreases.
Omitting the effect on the carbonate ion (distortion/polarisation).

Things to Be Careful About

Use precise terms: 'polarise', 'distort', 'charge density'.
The mark scheme requires both the comparison of cation size and the effect on the carbonate ion.
Do not say 'Ca2+ is bigger' – it is smaller.

Techniques used

compare ionic radii down Group 2explain polarisation of carbonate ionrelate polarisation to thermal stability

(c)

Calcium oxide, $\text{CaO}$ , reacts with water to form compound A.

Barium oxide, $\text{BaO}$ , reacts with water to form compound B.

4M

(i)

Identify A.

1M Similar

(ii)

Explain why A is less soluble than B.

3M Similar

DifficultyMedium

Worked solution

Answer

A (calcium hydroxide) is less soluble than B (barium hydroxide) because $\text{Ca}^{2+}$ is smaller than $\text{Ba}^{2+}$ , so both the lattice enthalpy and the hydration enthalpy of $\text{Ca(OH)}_2$ are more exothermic than those of $\text{Ba(OH)}_2$ . However, the difference in lattice enthalpy is greater than the difference in hydration enthalpy, so the overall enthalpy change of solution, $\Delta H_{\text{sol}}$ , is less exothermic for $\text{Ca(OH)}_2$ than for $\text{Ba(OH)}_2$ . A less exothermic $\Delta H_{\text{sol}}$ makes dissolving less favourable, so $\text{Ca(OH)}_2$ is less soluble.

Final answer

A is Ca(OH)2; less soluble because its ΔH_sol is less exothermic than Ba(OH)2 due to greater difference in lattice enthalpy than hydration enthalpy.

Detailed explanation

Background Concept

The solubility of an ionic compound in water depends on the enthalpy change of solution, $\Delta H_{\text{sol}}$ . This is the enthalpy change when one mole of the compound dissolves in water. It can be considered as the sum of two contributions:

Lattice enthalpy, $\Delta H_{\text{latt}}$ : the enthalpy change when one mole of the solid lattice forms from its gaseous ions. This is exothermic (negative) for the formation of the lattice.
Hydration enthalpy, $\Delta H_{\text{hyd}}$ : the enthalpy change when gaseous ions are hydrated (surrounded by water molecules). This is also exothermic (negative).

Thus $\Delta H_{\text{sol}} = \Delta H_{\text{latt}} + \Delta H_{\text{hyd}}$ (using the convention that lattice enthalpy is exothermic). A more exothermic (more negative) $\Delta H_{\text{sol}}$ makes dissolving more energetically favourable and generally leads to greater solubility.

Understanding the Question

This part asks why calcium hydroxide (A) is less soluble than barium hydroxide (B). You need to compare the lattice and hydration enthalpies of the two compounds and explain how the difference affects the enthalpy of solution.

Approach

Use the relationship $\Delta H_{\text{sol}} = \Delta H_{\text{latt}} + \Delta H_{\text{hyd}}$ . Compare the ionic radii of Ca2+ and Ba2+. Since Ca2+ is smaller, both lattice and hydration enthalpies are more exothermic for Ca(OH)2. However, the key is that the lattice enthalpy difference is larger than the hydration enthalpy difference, so the net $\Delta H_{\text{sol}}$ is less exothermic for Ca(OH)2, making it less soluble.

Step-by-Step Reasoning

Ca2+ is smaller than Ba2+ (ionic radius increases down Group 2).
A smaller cation has a higher charge density, so it attracts the anion more strongly in the lattice and attracts water molecules more strongly during hydration.
Therefore, both $\Delta H_{\text{latt}}$ and $\Delta H_{\text{hyd}}$ are more exothermic (more negative) for Ca(OH)2 than for Ba(OH)2.
However, the magnitude of the difference in lattice enthalpy is greater than the magnitude of the difference in hydration enthalpy. This is because lattice enthalpy depends on the close approach of cation and anion in the solid, while hydration enthalpy involves the interaction of the cation with water molecules and is somewhat less sensitive to the cation size.
As a result, when you add $\Delta H_{\text{latt}}$ and $\Delta H_{\text{hyd}}$ to get $\Delta H_{\text{sol}}$ , the less exothermic lattice term dominates, so $\Delta H_{\text{sol}}$ for Ca(OH)2 is less exothermic (more positive) than for Ba(OH)2.
A less exothermic $\Delta H_{\text{sol}}$ means the dissolving process is less energetically favourable, so Ca(OH)2 is less soluble.

Key Takeaways

Down Group 2, the solubility of hydroxides increases (e.g. Mg(OH)2 is sparingly soluble, Ba(OH)2 is more soluble).
The trend is explained by the balance between lattice and hydration enthalpies.
Always compare the differences in lattice and hydration enthalpies, not just their individual signs.

Common Mistakes

Saying that Ca(OH)2 is less soluble because its lattice enthalpy is more exothermic – this would suggest it is more soluble; you must consider the difference.
Forgetting that hydration enthalpy is also more exothermic for the smaller ion.
Confusing the sign convention: in some contexts lattice enthalpy is endothermic (breaking the lattice). Here the mark scheme uses the exothermic convention (formation of lattice).
Not mentioning that the lattice enthalpy difference is greater than the hydration enthalpy difference.

Things to Be Careful About

Use the correct sign convention: more exothermic means more negative.
The mark scheme awards separate marks for (i) both enthalpies more exothermic, (ii) lattice difference greater, (iii) enthalpy of solution less exothermic.
Ensure you state that A is calcium hydroxide and B is barium hydroxide if you refer to them.
No need to calculate numerical values; a qualitative explanation is sufficient.

Techniques used

compare ionic radiicompare lattice and hydration enthalpiesdeduce enthalpy of solution trendrelate enthalpy of solution to solubility

Q2Medium-EasyReaction Kinetics

Three experiments are carried out to investigate the reaction of nitrogen oxide, $\text{NO}$ , with chlorine.

2\text{NO} + \text{Cl}_2 \rightarrow 2\text{NOCl}

The rate equation for this reaction is shown.

\text{rate} = k[\text{NO}]^2[\text{Cl}_2]

(a)

Under the conditions used in experiments 1 and 2, the value of $k$ is 26.4.

4M

(i)

The rate of the reaction is measured in $\text{mol dm}^{-3} \text{ s}^{-1}$ . State the units of $k$ .

1M Similar

DifficultyEasy

Worked solution

Answer

The overall order is 3, so the units of $k$ are $\text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}$ .

Final answer

mol^-2 dm^6 s^-1

Detailed explanation

Background Concept

A rate equation shows how the rate of a reaction depends on the concentrations of the reactants. The rate constant, $k$ , is the proportionality constant, and its units must make the two sides of the equation dimensionally consistent. Since rate is measured in $\text{mol dm}^{-3}\text{ s}^{-1}$ and each concentration term has units $\text{mol dm}^{-3}$ , the units of $k$ depend on the overall order of the reaction.

Understanding the Question

The rate equation is $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$ . The overall order is $2 + 1 = 3$ . The question asks for the units of $k$ when rate is expressed in $\text{mol dm}^{-3}\text{ s}^{-1}$ .

Approach

Use the general relationship: units of $k = \dfrac{\text{units of rate}}{(\text{units of concentration})^{\text{overall order}}}$ . Substitute the overall order 3.

Step-by-Step Reasoning

From the rate equation, $k = \dfrac{\text{rate}}{[\text{NO}]^2[\text{Cl}_2]}$ . Therefore the units are:

\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2(\text{mol dm}^{-3})} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{\text{mol}^3\text{ dm}^{-9}} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}.

Equivalently, the general formula $(\text{mol dm}^{-3})^{1-n}\text{ s}^{-1}$ with $n=3$ gives $(\text{mol dm}^{-3})^{-2}\text{ s}^{-1}$ .

Key Takeaways

The units of $k$ are determined by the overall order of the reaction. For overall order 3, the units are $\text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}$ . Always check dimensional consistency.

Common Mistakes

A common error is to give the units of rate, $\text{mol dm}^{-3}\text{ s}^{-1}$ , instead of the units of $k$ . Another is to use the stoichiometric coefficients rather than the orders from the rate equation.

Things to Be Careful About

Use the overall order from the rate equation, not from the balanced equation. Pay attention to negative exponents and the order of dm and mol in the final units.

Techniques used

deduce units of rate constant from rate equation

(ii)

In experiment 1, the initial concentrations of $\text{NO}$ and $\text{Cl}_2$ are equal.

The initial rate of the reaction in experiment 1 is $2.57 \times 10^{-6} \text{ mol dm}^{-3} \text{ s}^{-1}$ .

Calculate the initial concentration of $\text{NO}$ .

Show your working.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Let $c = [\text{NO}] = [\text{Cl}_2]$ .

2.57 \times 10^{-6} = 26.4 \times c^3

c^3 = \frac{2.57 \times 10^{-6}}{26.4} = 9.73 \times 10^{-8}

c = 4.60 \times 10^{-3}\ \text{mol dm}^{-3}

Answer

$[\text{NO}] = 4.60 \times 10^{-3}\ \text{mol dm}^{-3}$

Final answer

4.60 x 10^-3 mol dm^-3

Detailed explanation

Background Concept

The rate equation $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$ allows calculation of an unknown concentration when the rate and $k$ are known. If two concentrations are equal, they can be represented by a single variable.

Understanding the Question

In experiment 1, $[\text{NO}] = [\text{Cl}_2] = c$ . The initial rate is $2.57 \times 10^{-6}\ \text{mol dm}^{-3}\text{ s}^{-1}$ and $k = 26.4$ . We need to find $c$ .

Approach

Substitute $c$ for both concentrations in the rate equation, solve for $c^3$ , then take the cube root.

Step-by-Step Reasoning

Since $[\text{NO}] = [\text{Cl}_2] = c$ , the rate equation becomes:

2.57 \times 10^{-6} = 26.4 \times c^3.

Divide both sides by 26.4:

c^3 = \frac{2.57 \times 10^{-6}}{26.4} = 9.73 \times 10^{-8}.

Taking the cube root:

c = (9.73 \times 10^{-8})^{1/3} = 4.60 \times 10^{-3}\ \text{mol dm}^{-3}.

This is the initial concentration of $\text{NO}$ .

Key Takeaways

When concentrations are equal, combine them into one variable. The cube root is needed because the reaction is third order overall under these conditions.

Common Mistakes

Using $c^2$ instead of $c^3$ ; forgetting to take the cube root after dividing; giving the answer without units.

Things to Be Careful About

The final answer should be given to 3 significant figures to match the data. Include the unit $\text{mol dm}^{-3}$ .

Techniques used

substitute into rate equationtake cube root

(iii)

In experiment 2, the initial concentrations of $\text{NO}$ and $\text{Cl}_2$ are both ten times greater than the initial concentrations used in experiment 1.

Calculate the initial rate of the reaction in experiment 2.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Both concentrations are multiplied by 10, so the rate is multiplied by $10^2 \times 10 = 10^3$ .

\text{rate}_2 = 10^3 \times 2.57 \times 10^{-6} = 2.57 \times 10^{-3}\ \text{mol dm}^{-3}\text{ s}^{-1}

Answer

$2.57 \times 10^{-3}\ \text{mol dm}^{-3}\text{ s}^{-1}$

Final answer

2.57 x 10^-3 mol dm^-3 s^-1

Detailed explanation

Background Concept

In a rate equation, the order with respect to each reactant tells how the rate changes when that concentration changes. Here rate $\propto [\text{NO}]^2[\text{Cl}_2]$ .

Understanding the Question

Experiment 2 uses initial concentrations of both $\text{NO}$ and $\text{Cl}_2$ that are 10 times greater than in experiment 1. The rate constant is unchanged. We need the new initial rate.

Approach

Use the ratio method: multiply the original rate by the factor arising from each concentration change.

Step-by-Step Reasoning

Multiplying $[\text{NO}]$ by 10 increases the rate by $10^2 = 100$ . Multiplying $[\text{Cl}_2]$ by 10 increases the rate by $10^1 = 10$ . The combined factor is $100 \times 10 = 1000$ . Therefore:

\text{rate}_2 = 1000 \times 2.57 \times 10^{-6} = 2.57 \times 10^{-3}\ \text{mol dm}^{-3}\text{ s}^{-1}.

Key Takeaways

The order with respect to each reactant determines the factor by which the rate changes. A 10-fold increase in a second-order reactant gives a 100-fold rate increase.

Common Mistakes

Only applying the factor to one reactant; using $10^2$ for both reactants; forgetting that $\text{Cl}_2$ is first order.

Things to Be Careful About

Keep the units $\text{mol dm}^{-3}\text{ s}^{-1}$ . The rate constant does not need to be recalculated because the conditions are the same as experiment 1.

Techniques used

apply factor change to rate equation

(b)

Experiment 3 uses a large excess of $\text{NO}$ .

The initial concentration of $\text{Cl}_2$ is $2.00 \times 10^{-4} \text{ mol dm}^{-3}$ .

3M

(i)

The graph of $[\text{Cl}_2]$ against time shows that the reaction has a constant half-life, $t_{1/2}$ .

Explain this observation.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

With a large excess of $\text{NO}$ , $[\text{NO}]$ is effectively constant, so the rate equation becomes $\text{rate} = k'[\text{Cl}_2]$ (pseudo first order). A first-order reaction has a constant half-life.

Final answer

Pseudo first order; constant half-life because reaction is first order with respect to Cl2.

Detailed explanation

Background Concept

When one reactant is present in large excess, its concentration changes very little during the reaction and can be treated as constant. The rate equation then becomes first order in the other reactant; this is called pseudo first order. A first-order reaction has a half-life that is independent of concentration, so the half-life is constant.

Understanding the Question

Experiment 3 uses a large excess of $\text{NO}$ . The graph of $[\text{Cl}_2]$ against time shows a constant half-life. We need to explain why.

Approach

Show that the rate equation reduces to $\text{rate} = k'[\text{Cl}_2]$ and that constant half-life is the signature of first-order kinetics.

Step-by-Step Reasoning

With $\text{NO}$ in large excess, $[\text{NO}]$ is effectively constant throughout the reaction. The rate equation becomes:

\text{rate} = k[\text{NO}]^2[\text{Cl}_2] = k'[\text{Cl}_2],

where $k' = k[\text{NO}]^2$ is a pseudo first-order rate constant. The reaction is first order with respect to $\text{Cl}_2$ . For a first-order reaction, $t_{1/2} = \ln 2/k'$ , which does not depend on the initial concentration of $\text{Cl}_2$ . Hence each successive half-life is the same, giving a constant half-life on the graph.

Key Takeaways

A large excess of one reactant creates pseudo first-order conditions. Constant half-life is diagnostic of first-order kinetics.

Common Mistakes

Saying the reaction is zero order in $\text{Cl}_2$ ; saying $[\text{NO}]$ is zero; not mentioning that the concentration of $\text{NO}$ is effectively constant.

Things to Be Careful About

Use the phrase “pseudo first order” and make clear that the order with respect to $\text{Cl}_2$ is first order.

Techniques used

recognise pseudo first-order conditionsrelate constant half-life to first order

(ii)

Under the conditions used in experiment 3, the value of the rate constant is 105.6.

Show that $t_{1/2}$ of $[\text{Cl}_2]$ is $6.56 \times 10^{-3} \text{ s}$ under these conditions.

1M Similar

(iii)

Calculate the time taken, in s, for $[\text{Cl}_2]$ to fall to $1.25 \times 10^{-5} \text{ mol dm}^{-3}$ in experiment 3.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

$[\text{Cl}_2]$ falls from $2.00 \times 10^{-4}$ to $1.25 \times 10^{-5}\ \text{mol dm}^{-3}$ , a factor of 16, i.e. 4 half-lives.

t = 4 \times 6.56 \times 10^{-3} = 2.62 \times 10^{-2}\ \text{s}

Answer

$2.62 \times 10^{-2}\ \text{s}$

Final answer

2.62 x 10^-2 s

Detailed explanation

Background Concept

In a first-order decay, after $n$ half-lives the concentration is reduced by a factor of $2^n$ . The time required is $n \times t_{1/2}$ .

Understanding the Question

The initial $[\text{Cl}_2]$ is $2.00 \times 10^{-4}\ \text{mol dm}^{-3}$ and we want the time for it to fall to $1.25 \times 10^{-5}\ \text{mol dm}^{-3}$ . The half-life is $6.56 \times 10^{-3}\ \text{s}$ .

Approach

Find the ratio of initial to final concentration, express it as a power of 2, and multiply the half-life by that number.

Step-by-Step Reasoning

\frac{2.00 \times 10^{-4}}{1.25 \times 10^{-5}} = 16 = 2^4.

So four half-lives are needed:

t = 4 \times 6.56 \times 10^{-3} = 2.62 \times 10^{-2}\ \text{s}.

Equivalently, using the integrated first-order law: $t = \dfrac{\ln(16)}{105.6} = 0.0262\ \text{s}$ .

Key Takeaways

Half-life reasoning is efficient when the concentration ratio is a power of 2. The integrated rate law is a general alternative.

Common Mistakes

Counting three half-lives instead of four; forgetting to multiply by the half-life; using the wrong initial concentration.

Things to Be Careful About

Give the final answer as $2.62 \times 10^{-2}\ \text{s}$ or $0.0262\ \text{s}$ with the correct unit.

Techniques used

apply half-life decay for first-order reaction

(c)

Sulfur dioxide, $\text{SO}_2$ , reacts very slowly with oxygen in the atmosphere, forming sulfur trioxide, $\text{SO}_3$ . This reaction is much faster in the presence of $\text{NO}$ .

Explain the role of $\text{NO}$ in this process.

Include chemical equations in your answer.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\text{NO}$ acts as a homogeneous catalyst. It provides an alternative route with a lower activation energy and is regenerated.

2\text{NO}(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{NO}_2(\text{g})

\text{NO}_2(\text{g}) + \text{SO}_2(\text{g}) \rightarrow \text{NO}(\text{g}) + \text{SO}_3(\text{g})

Final answer

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

Detailed explanation

Background Concept

A catalyst increases the rate of a reaction by providing an alternative pathway with a lower activation energy. It is chemically unchanged at the end of the reaction, so it is regenerated. A homogeneous catalyst is in the same physical state as the reactants.

Understanding the Question

The reaction $2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$ is slow in the atmosphere but faster in the presence of $\text{NO}$ . We need to explain the role of $\text{NO}$ and include equations.

Approach

Recognise that $\text{NO}$ is a homogeneous catalyst. Write a two-step cycle in which $\text{NO}$ is first consumed to form an intermediate, then regenerated in the second step.

Step-by-Step Reasoning

In the first step, $\text{NO}$ reacts with oxygen:

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

In the second step, the $\text{NO}_2$ formed oxidises $\text{SO}_2$ :

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

Adding the two steps (with the second doubled) gives the overall reaction $2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$ , and $\text{NO}$ is regenerated. Because $\text{NO}$ is not used up overall and provides a lower-energy pathway, it acts as a homogeneous catalyst.

Key Takeaways

A catalyst is regenerated in the reaction. Writing a catalytic cycle with balanced equations shows how the catalyst is consumed and reformed.

Common Mistakes

Saying $\text{NO}$ is a reactant and is consumed; writing unbalanced equations; omitting $\text{O}_2$ from the first step; calling $\text{NO}$ a heterogeneous catalyst.

Things to Be Careful About

Both equations must be balanced. The intermediate $\text{NO}_2$ is formed and then consumed, while $\text{NO}$ is regenerated.

Techniques used

identify homogeneous catalystwrite catalytic cycle equations

Q3MediumEquilibria

(a)

Chromium(III) hydroxide, $\text{Cr(OH)}_3$ , is only slightly soluble in water. The value of the solubility product, $K_{\text{sp}}$ , of $\text{Cr(OH)}_3$ is $1.0 \times 10^{-33}$ at 298 K.

6M

(i)

Complete the expression for $K_{\text{sp}}$ of $\text{Cr(OH)}_3$ . Include the units.

K_{\text{sp}} =

2M Similar

DifficultyEasy

Worked solution

Answer

$K_{\text{sp}} = [\text{Cr}^{3+}][\text{OH}^-]^3$

Units: $\text{mol}^4\text{ dm}^{-12}$

Final answer

Ksp = [Cr3+][OH-]^3; units: mol^4 dm^-12

Detailed explanation

Background Concept

The solubility product, $K_{\text{sp}}$ , is the equilibrium constant for the dissolution of a sparingly soluble ionic solid in water. For a general salt $A_mB_n$ that dissolves as
$A_mB_n(\text{s}) \rightleftharpoons m\text{A}^{n+}(\text{aq}) + n\text{B}^{m-}(\text{aq})$
the solubility product is
$K_{\text{sp}} = [\text{A}^{n+}]^m[\text{B}^{m-}]^n$
The solid itself is not included in the expression because its concentration is constant (it is a pure phase). The units of $K_{\text{sp}}$ depend on the total power of the concentration terms.

Understanding the Question

This part asks you to write the $K_{\text{sp}}$ expression for chromium(III) hydroxide and give its units. It is a recall task: no calculation is needed. The key is to translate the formula $\text{Cr(OH)}_3$ into the correct stoichiometry of dissolution and to combine the units correctly.

Approach

Write the dissolution equilibrium for $\text{Cr(OH)}_3$ .
Write the product of the ion concentrations, raising each to the power equal to its stoichiometric coefficient.
Multiply the units of each concentration term to get the units of $K_{\text{sp}}$ .

Step-by-Step Reasoning

Chromium(III) hydroxide dissolves as
$\text{Cr(OH)}_3(\text{s}) \rightleftharpoons \text{Cr}^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq})$
One formula unit gives one $\text{Cr}^{3+}$ and three $\text{OH}^-$ . Therefore
$K_{\text{sp}} = [\text{Cr}^{3+}][\text{OH}^-]^3$
The coefficient 3 on $\text{OH}^-$ becomes the power 3 in the expression.

For units: $[\text{Cr}^{3+}]$ has units mol dm $^{-3}$ , and $[\text{OH}^-]^3$ has units (mol dm $^{-3}$ ) $^3$ = mol $^3$ dm $^{-9}$ . Multiplying gives mol $^4$ dm $^{-12}$ .

Key Takeaways

Stoichiometric coefficients become powers in the $K_{\text{sp}}$ expression.
The solid is never included.
Units of $K_{\text{sp}}$ are mol $^{(m+n)}$ dm $^{-3(m+n)}$ for a salt giving $m+n$ ions.

Common Mistakes

Writing $K_{\text{sp}} = [\text{Cr}^{3+}][\text{OH}^-]$ (forgetting the power 3).
Including $\text{Cr(OH)}_3(\text{s})$ in the expression.
Giving units as mol dm $^{-3}$ instead of mol $^4$ dm $^{-12}$ .

Things to Be Careful About

The power applies to both the number and the units. Since there are four ions in total (1 Cr + 3 OH), the units are mol $^4$ dm $^{-12}$ .

Techniques used

write the Ksp expressionidentify ion stoichiometry from formuladerive units of Ksp

(ii)

Calculate the solubility, in $\text{g dm}^{-3}$ , of $\text{Cr(OH)}_3$ in pure water at 298 K.

Show your working.

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Let the solubility of $\text{Cr(OH)}_3$ be $s$ mol dm $^{-3}$ .

$\text{Cr(OH)}_3(\text{s}) \rightleftharpoons \text{Cr}^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq})$

$[\text{Cr}^{3+}] = s$ , $[\text{OH}^-] = 3s$

$K_{\text{sp}} = s(3s)^3 = 27s^4$

$s^4 = \frac{1.0 \times 10^{-33}}{27} = 3.70 \times 10^{-35}$

$s = 2.47 \times 10^{-9}\text{ mol dm}^{-3}$

$M_r(\text{Cr(OH)}_3) = 52.0 + 3(16.0 + 1.0) = 103.0$

Solubility $= 2.47 \times 10^{-9} \times 103.0 = 2.54 \times 10^{-7}\text{ g dm}^{-3}$

Answer

$2.54 \times 10^{-7}\text{ g dm}^{-3}$

Final answer

2.54 × 10^-7 g dm^-3

Detailed explanation

Background Concept

For a sparingly soluble salt, the molar solubility $s$ (in mol dm $^{-3}$ ) is the amount that dissolves per dm $^3$ of solution. The ion concentrations are related to $s$ by the stoichiometry of dissolution. For $\text{Cr(OH)}_3$ :
$[\text{Cr}^{3+}] = s, \quad [\text{OH}^-] = 3s$
Substituting into $K_{\text{sp}}$ gives a relationship that can be solved for $s$ . To express solubility in g dm $^{-3}$ , multiply the molar solubility by the molar mass.

Understanding the Question

We are given $K_{\text{sp}} = 1.0 \times 10^{-33}$ and asked for the solubility in g dm $^{-3}$ . The instruction "show your working" means the full chain — from $K_{\text{sp}}$ to molar solubility to mass concentration — must be shown.

Approach

Let $s$ = molar solubility of $\text{Cr(OH)}_3$ .
Express $[\text{Cr}^{3+}]$ and $[\text{OH}^-]$ in terms of $s$ .
Substitute into $K_{\text{sp}}$ and solve for $s$ .
Multiply by $M_r(\text{Cr(OH)}_3)$ to convert to g dm $^{-3}$ .

Step-by-Step Reasoning

From the dissolution equilibrium, $[\text{Cr}^{3+}] = s$ and $[\text{OH}^-] = 3s$ .
$K_{\text{sp}} = [\text{Cr}^{3+}][\text{OH}^-]^3 = s(3s)^3 = 27s^4$
$s^4 = \frac{K_{\text{sp}}}{27} = \frac{1.0 \times 10^{-33}}{27} = 3.70 \times 10^{-35}$
$s = (3.70 \times 10^{-35})^{1/4} = 2.47 \times 10^{-9}\text{ mol dm}^{-3}$

$M_r(\text{Cr(OH)}_3) = 52.0 + 3(16.0 + 1.0) = 52.0 + 51.0 = 103.0$ g mol $^{-1}$ .

Solubility in g dm $^{-3}$ = $2.47 \times 10^{-9} \times 103.0 = 2.54 \times 10^{-7}$ g dm $^{-3}$ .

Key Takeaways

The factor 27 arises from $(3)^3$ — it is easy to forget.
Always check what units the question asks for and convert accordingly.
The fourth root of $3.70 \times 10^{-35}$ is $2.47 \times 10^{-9}$ .

Common Mistakes

Writing $K_{\text{sp}} = s^4$ (forgetting the 27).
Taking the square root instead of the fourth root.
Using $M_r = 52$ (chromium only) instead of 103.
Reporting the answer in mol dm $^{-3}$ when g dm $^{-3}$ is asked.

Things to Be Careful About

Significant figures: the given $K_{\text{sp}}$ has 2 significant figures, so the answer is reasonably given as $2.5 \times 10^{-7}$ g dm $^{-3}$ ; the mark scheme accepts $2.54 \times 10^{-7}$ .

Techniques used

relate ion concentrations to molar solubilitysolve fourth-root equationconvert molar solubility to g dm^-3

(iii)

$\text{Cr(OH)}_3$ is less soluble in $0.100 \text{ mol dm}^{-3} \text{ NaOH}$ than it is in pure water.

Explain this observation.

1M Similar

DifficultyEasy

Worked solution

Answer

NaOH provides $\text{OH}^-$ ions, a common ion. This increases $[\text{OH}^-]$ , shifting the equilibrium $\text{Cr(OH)}_3(\text{s}) \rightleftharpoons \text{Cr}^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq})$ to the left, so less $\text{Cr(OH)}_3$ dissolves.

Final answer

Common ion effect (OH- from NaOH shifts equilibrium left)

Detailed explanation

Background Concept

The common ion effect is a direct application of Le Chatelier's principle to solubility equilibria. If a solution already contains one of the ions produced by a sparingly soluble salt, the presence of that ion (the "common ion") shifts the dissolution equilibrium to the left, reducing the amount of solid that can dissolve.

Understanding the Question

We are told that $\text{Cr(OH)}_3$ is less soluble in 0.100 mol dm $^{-3}$ NaOH than in pure water, and asked to explain why. NaOH is a strong base that fully dissociates into $\text{Na}^+$ and $\text{OH}^-$ . The $\text{OH}^-$ is the common ion.

Approach

Identify the ion that NaOH contributes ( $\text{OH}^-$ ), note it is already a product of $\text{Cr(OH)}_3$ dissolution, and apply Le Chatelier's principle: increased $[\text{OH}^-]$ shifts the equilibrium left.

Step-by-Step Reasoning

NaOH dissociates completely: $\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-$ . The $\text{OH}^-$ ion is common to the dissolution equilibrium $\text{Cr(OH)}_3(\text{s}) \rightleftharpoons \text{Cr}^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq})$ . Adding NaOH increases $[\text{OH}^-]$ , so the equilibrium shifts to the left, favouring the solid and reducing solubility.

Key Takeaways

A common ion reduces solubility (common ion effect).
This is Le Chatelier's principle applied to a solubility equilibrium.

Common Mistakes

Saying "NaOH reacts with $\text{Cr(OH)}_3$ " — there is no acid-base reaction here; the effect is purely the common ion.
Not mentioning the equilibrium shift.

Things to Be Careful About

The mark scheme awards one mark for "common ion effect". A clear statement that $\text{OH}^-$ is the common ion and shifts the equilibrium left is sufficient.

Techniques used

identify the common ionapply Le Chatelier's principle to solubility equilibrium

(b)

The value of the acid dissociation constant, $K_{\text{a}}$ , of butanoic acid, $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ , is $1.51 \times 10^{-5}$ at 298 K.

5M

(i)

Calculate the pH of $0.100 \text{ mol dm}^{-3} \text{ CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ at 298 K.

Show your working.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

$K_{\text{a}} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \approx \frac{[\text{H}^+]^2}{[\text{HA}]}$

$[\text{H}^+]^2 = K_{\text{a}}[\text{HA}] = 1.51 \times 10^{-5} \times 0.100 = 1.51 \times 10^{-6}$

$[\text{H}^+] = 1.23 \times 10^{-3}\text{ mol dm}^{-3}$

$\text{pH} = -\log_{10}(1.23 \times 10^{-3}) = 2.91$

Answer

$\text{pH} = 2.91$

Final answer

pH = 2.91

Detailed explanation

Background Concept

For a weak acid HA, the acid dissociation constant is
$K_{\text{a}} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$
For a weak acid, very little dissociates, so $[\text{HA}] \approx$ initial concentration and $[\text{H}^+] \approx [\text{A}^-]$ . This gives the approximation
$K_{\text{a}} \approx \frac{[\text{H}^+]^2}{[\text{HA}]} \quad \Rightarrow \quad [\text{H}^+] = \sqrt{K_{\text{a}}[\text{HA}]}$
Then $\text{pH} = -\log_{10}[\text{H}^+]$ .

Understanding the Question

Butanoic acid, $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ , is a weak acid with $K_{\text{a}} = 1.51 \times 10^{-5}$ . We are asked for the pH of a 0.100 mol dm $^{-3}$ solution. The weak acid approximation is appropriate because $K_{\text{a}}$ is small.

Approach

Apply $[\text{H}^+] = \sqrt{K_{\text{a}}[\text{HA}]}$ .
Take the negative logarithm to find pH.

Step-by-Step Reasoning

$[\text{H}^+]^2 = K_{\text{a}} \times [\text{HA}] = 1.51 \times 10^{-5} \times 0.100 = 1.51 \times 10^{-6}$
$[\text{H}^+] = \sqrt{1.51 \times 10^{-6}} = 1.23 \times 10^{-3}\text{ mol dm}^{-3}$
$\text{pH} = -\log_{10}(1.23 \times 10^{-3}) = 2.91$

Key Takeaways

Weak acid pH: $[\text{H}^+] = \sqrt{K_{\text{a}}[\text{HA}]}$ .
The approximation is valid when $K_{\text{a}}$ is small and the acid is not too dilute.

Common Mistakes

Using $[\text{H}^+] = [\text{HA}]$ (strong acid assumption).
Forgetting to take the square root.
Logarithm arithmetic errors.

Things to Be Careful About

pH is dimensionless. The answer 2.91 is acidic, which is consistent with a weak acid.

Techniques used

apply weak acid approximationcalculate [H+] from Kacompute pH from [H+]

(ii)

Calculate the pH of $0.100 \text{ mol dm}^{-3} \text{ NaOH}$ at 298 K.

1M Similar

(iii)

$5.00 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3} \text{ NaOH}$ is added to $10.00 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3} \text{ CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ .

Calculate the pH of the resulting solution.

Show your working.

2M Similar

DifficultyMedium

Worked solution

Working

$n(\text{NaOH}) = 5.00 \times 10^{-3} \times 0.100 = 5.00 \times 10^{-4}\text{ mol}$

$n(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}) = 10.00 \times 10^{-3} \times 0.100 = 1.00 \times 10^{-3}\text{ mol}$

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

$n(\text{acid remaining}) = 1.00 \times 10^{-3} - 5.00 \times 10^{-4} = 5.00 \times 10^{-4}\text{ mol}$

$n(\text{conjugate base}) = 5.00 \times 10^{-4}\text{ mol}$

Both are in the same total volume (15.00 cm $^3$ ), so $[\text{acid}] = [\text{conjugate base}]$ .

$\text{pH} = \text{p}K_{\text{a}} = -\log_{10}(1.51 \times 10^{-5}) = 4.82$

Answer

$\text{pH} = 4.82$

Final answer

pH = 4.82

Detailed explanation

Background Concept

When a strong base is added to a weak acid, the base neutralises some of the acid, producing its conjugate base. If both the weak acid and its conjugate base remain in solution, a buffer is formed. The pH of a buffer is given by the Henderson–Hasselbalch equation:
$\text{pH} = \text{p}K_{\text{a}} + \log_{10}\frac{[\text{A}^-]}{[\text{HA}]}$
When $[\text{A}^-] = [\text{HA}]$ , the log term is zero and pH = pKa.

Understanding the Question

5.00 cm $^3$ of 0.100 mol dm $^{-3}$ NaOH is added to 10.00 cm $^3$ of 0.100 mol dm $^{-3}$ butanoic acid. We need the pH of the resulting solution. This is a buffer calculation: neutralisation converts some acid to conjugate base.

Approach

Calculate moles of NaOH and acid.
Determine moles of acid remaining and conjugate base formed.
Since both are in the same total volume, compare their concentrations.
Apply the Henderson–Hasselbalch equation (or recognise pH = pKa when concentrations are equal).

Step-by-Step Reasoning

$n(\text{NaOH}) = 5.00 \times 10^{-3} \times 0.100 = 5.00 \times 10^{-4}\text{ mol}$
$n(\text{acid}) = 10.00 \times 10^{-3} \times 0.100 = 1.00 \times 10^{-3}\text{ mol}$

Neutralisation:
$\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{COO}^-\text{Na}^+ + \text{H}_2\text{O}$

Moles of acid consumed = moles of NaOH = $5.00 \times 10^{-4}$ mol.
Moles of acid remaining = $1.00 \times 10^{-3} - 5.00 \times 10^{-4} = 5.00 \times 10^{-4}$ mol.
Moles of conjugate base formed = $5.00 \times 10^{-4}$ mol.

Total volume = 15.00 cm $^3$ = 0.01500 dm $^3$ . Since both acid and conjugate base are in the same volume, $[\text{acid}] = [\text{conjugate base}]$ .

$\text{pH} = \text{p}K_{\text{a}} = -\log_{10}(1.51 \times 10^{-5}) = 4.82$

Key Takeaways

When $[\text{A}^-] = [\text{HA}]$ , pH = pKa.
The volume cancels when both species are in the same solution; only the mole ratio matters.
This is the half-neutralisation point of the acid.

Common Mistakes

Using initial concentrations instead of moles.
Forgetting that NaOH consumes acid.
Not recognising that acid and conjugate base concentrations are equal.

Things to Be Careful About

The mark scheme explicitly notes M1: $[\text{acid}] = [\text{conjugate base}]$ and M2: pH = 4.82. Both must be stated.

Techniques used

calculate moles of acid and baseapply neutralisation stoichiometryuse Henderson-Hasselbalch equation

(c)

$80.0 \text{ cm}^3$ of an aqueous solution containing $0.704 \text{ g}$ of $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ is shaken with $100 \text{ cm}^3$ of benzene, $\text{C}_6\text{H}_6$ .

There is $0.556 \text{ g}$ of $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ in the $100 \text{ cm}^3$ of $\text{C}_6\text{H}_6$ at equilibrium.

Calculate the partition coefficient, $K_{\text{pc}}$ , of $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ between $\text{C}_6\text{H}_6$ and water.

Show your working.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Mass of acid in water $= 0.704 - 0.556 = 0.148\text{ g}$ (in 80.0 cm $^3$ )

$K_{\text{pc}} = \frac{[\text{acid in }\text{C}_6\text{H}_6]}{[\text{acid in water}]} = \frac{0.556/100}{0.148/80} = \frac{0.00556}{0.00185} = 3.01$

Answer

$K_{\text{pc}} = 3.01$

Final answer

Kpc = 3.01

Detailed explanation

Background Concept

The partition (or distribution) coefficient, $K_{\text{pc}}$ , describes how a solute distributes between two immiscible solvents at equilibrium:
$K_{\text{pc}} = \frac{\text{concentration of solute in solvent A}}{\text{concentration of solute in solvent B}}$
Here the two solvents are benzene ( $\text{C}_6\text{H}_6$ , organic) and water (aqueous). The question specifies "between $\text{C}_6\text{H}_6$ and water", so the ratio is concentration in benzene divided by concentration in water.

Understanding the Question

0.704 g of butanoic acid is distributed between 80.0 cm $^3$ water and 100 cm $^3$ benzene. At equilibrium, 0.556 g is in the benzene layer. We need $K_{\text{pc}}$ .

Approach

Find the mass of acid remaining in the water layer by subtracting the benzene mass from the total.
Convert both masses to concentrations (g cm $^{-3}$ works fine since units cancel).
Divide the benzene concentration by the water concentration.

Step-by-Step Reasoning

Mass of acid in water = 0.704 - 0.556 = 0.148 g (in 80.0 cm $^3$ ).

Concentration in benzene = 0.556/100 = 0.00556 g cm $^{-3}$ .
Concentration in water = 0.148/80 = 0.00185 g cm $^{-3}$ .

$K_{\text{pc}} = \frac{0.00556}{0.00185} = 3.01$

Key Takeaways

Mass conservation: mass in one phase = total - mass in the other phase.
Concentrations, not masses, go into the ratio.
The units cancel, so $K_{\text{pc}}$ is dimensionless.

Common Mistakes

Using masses (0.556/0.148) instead of concentrations — wrong because the volumes differ.
Reversing the ratio.
Forgetting to subtract to find the mass in water.

Things to Be Careful About

The mark scheme accepts 3.01 OR 0.333 (the inverse ratio). The question's phrasing "between $\text{C}_6\text{H}_6$ and water" makes benzene over water the primary answer, 3.01.

Techniques used

calculate mass remaining in aqueous phasecompute concentration ratio for partition coefficient

Q4Medium-EasyChemical Energetics

Table 4.1 gives the enthalpy changes of hydration, $\Delta H_{\text{hyd}}$ , of three ions, $\text{F}^-$ , $\text{K}^+$ and $\text{Ca}^{2+}$ .

Table 4.1

ion	$\Delta H_{\text{hyd}} / \text{kJ mol}^{-1}$
$\text{F}^-$	$-506$
$\text{K}^+$	$-322$
$\text{Ca}^{2+}$	$-1650$

(a)

6M

(i)

Define enthalpy change of hydration.

1M Similar

DifficultyEasy

Worked solution

Answer

The enthalpy change of hydration is the energy change when one mole of gaseous ions dissolves in water.

Final answer

The energy change when one mole of gaseous ions dissolves in water.

Detailed explanation

Background Concept

The enthalpy change of hydration, $\Delta H_{\text{hyd}}$ , is a thermodynamic quantity that measures the energy change when gaseous ions are surrounded by water molecules. When an ionic compound dissolves, the process can be thought of in two conceptual stages: the lattice breaks apart (endothermic) and the freed ions become hydrated (exothermic). The hydration stage involves ion–dipole attractions: a cation attracts the δ− oxygen atoms of water molecules, while an anion attracts the δ+ hydrogen atoms. Because these attractions form spontaneously, energy is released, and $\Delta H_{\text{hyd}}$ is always negative (exothermic).

Understanding the Question

This is a straightforward recall question. The command word "Define" requires a precise, exam-ready statement. The definition must capture three things: the amount of substance (one mole), the physical state of the starting material (gaseous ions), and the process (dissolving in water). The mark scheme rewards the exact phrasing: "energy change when one mole of gaseous ions dissolves in water".

Approach

There is no calculation or reasoning needed — just state the definition exactly as the examiners expect. The three key phrases are "one mole", "gaseous ions", and "dissolves in water".

Step-by-Step Reasoning

The definition has three essential components:

"One mole of gaseous ions" — the starting state is one mole of the ion in question, in the gas phase. The energy change depends on the amount of substance.
"Dissolves in water" — the ions are surrounded by water molecules, forming an infinitely dilute solution so that ion–ion interactions are negligible.
"Energy change" — the enthalpy change at constant pressure, which is negative because forming ion–dipole attractions releases energy.

For example, for Ca²⁺: $\text{Ca}^{2+}(\text{g}) \rightarrow \text{Ca}^{2+}(\text{aq})$ , with $\Delta H = -1650 \text{ kJ mol}^{-1}$ .

Key Takeaways

Thermodynamic definitions must be precise about physical states (gaseous vs aqueous), quantities (one mole), and the direction of the process. This definition is frequently examined and pairs with lattice energy and enthalpy of solution in energy cycles.

Common Mistakes

Omitting "gaseous" — the ions must start in the gas phase.
Saying "one mole of ions" without specifying "gaseous".
Confusing hydration enthalpy with enthalpy of solution (solution includes lattice breaking).
Writing "energy released" without framing it as the enthalpy change.

Things to Be Careful About

The definition applies to ONE mole of a specific ion, not one mole of the compound.
Hydration enthalpy is always negative.
This definition is used in Born–Haber cycles and enthalpy-of-solution calculations.

Techniques used

state the definition of enthalpy change of hydration

(ii)

Explain the relative magnitudes of the enthalpy changes of hydration of $\text{K}^+$ and $\text{Ca}^{2+}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\text{K}^+$ has a larger ionic radius than $\text{Ca}^{2+}$ AND $\text{K}^+$ has a smaller charge (+1) than $\text{Ca}^{2+}$ (+2).
$\text{Ca}^{2+}$ therefore has a much larger attraction for water molecules (higher charge density), so its hydration enthalpy is more exothermic (more negative).

Final answer

Ca2+ has a smaller radius and higher charge than K+, giving it a higher charge density and stronger attraction for water molecules, so its hydration enthalpy is more exothermic.

Detailed explanation

Background Concept

The enthalpy change of hydration depends on the charge density of the ion — the ratio of its charge to its ionic radius. A higher charge density produces a stronger electrostatic attraction between the ion and the surrounding water molecules. This attraction is an ion–dipole interaction: a cation attracts the δ− oxygen end of a water molecule, and an anion attracts the δ+ hydrogen ends. The stronger the attraction, the more energy is released on hydration and the more negative $\Delta H_{\text{hyd}}$ becomes.

Understanding the Question

The question asks you to explain why Ca²⁺ has a much more exothermic hydration enthalpy (−1650 kJ mol⁻¹) than K⁺ (−322 kJ mol⁻¹). Both ions are cations, so the difference must come from their charge and/or size. The command word "Explain" requires a reason, not just a restatement of the data.

Approach

Compare the two ions on both charge and ionic radius, then connect these to the strength of attraction for water molecules. The unifying concept is charge density (charge ÷ radius).

Step-by-Step Reasoning

Charge: Ca²⁺ carries a +2 charge; K⁺ carries only +1.
Radius: Ca²⁺ is smaller than K⁺. Although both are in Period 4, Ca²⁺ has a higher nuclear charge (20 protons vs 19) pulling the remaining electrons in more tightly.
Charge density: Because Ca²⁺ has a higher charge AND a smaller radius, its charge density (charge/radius ratio) is much larger than that of K⁺.
Attraction to water: The higher charge density of Ca²⁺ means it attracts the δ− oxygen atoms of water molecules much more strongly than K⁺ does.
Energy released: The stronger ion–dipole attraction releases more energy when Ca²⁺ is hydrated, so $\Delta H_{\text{hyd}}(\text{Ca}^{2+})$ is much more negative.

The mark scheme awards M1 for stating BOTH that K⁺ has a larger radius AND a smaller charge, and M2 for stating that Ca²⁺ has a larger attraction for H₂O.

Key Takeaways

Charge density (charge/radius ratio) is the unifying concept that explains trends in hydration enthalpy, lattice energy, and many other ionic properties. Whenever you compare two ions, always consider both charge and size.

Common Mistakes

Mentioning only charge or only radius — the mark scheme requires BOTH.
Saying "Ca²⁺ attracts more water molecules" — the number of water molecules is not the reason; the strength of attraction is.
Using vague language like "Ca²⁺ is more reactive" — the answer must link charge density to attraction for water.

Things to Be Careful About

The mark scheme requires both charge and radius for the first mark.
Use precise language: "attraction for water molecules" rather than "reactivity".
More exothermic = more negative = stronger ion–dipole attraction.

Techniques used

compare ionic charge and radiusrelate charge density to strength of ion-dipole attraction

(iii)

Define lattice energy.

1M Similar

DifficultyEasy

Worked solution

Answer

The lattice energy is the energy change when one mole of an ionic solid forms from its gaseous ions.

Final answer

The energy change when one mole of an ionic solid forms from its gaseous ions.

Detailed explanation

Background Concept

Lattice energy, $\Delta H_{\text{latt}}$ , is the energy change when one mole of an ionic solid forms from its gaseous ions. For example, for calcium fluoride: $\text{Ca}^{2+}(\text{g}) + 2\text{F}^-(\text{g}) \rightarrow \text{CaF}_2(\text{s})$ . This process is always exothermic (negative) because forming ionic bonds releases energy. Lattice energy is a direct measure of the strength of ionic bonding in a crystal — the more negative the value, the stronger the ionic lattice.

Understanding the Question

This is a recall question. "Define lattice energy" requires the precise definition. The direction of the process matters: gaseous ions → solid lattice.

Approach

State the definition exactly. The key phrases are "one mole", "ionic solid", "forms from", and "gaseous ions".

Step-by-Step Reasoning

The definition has three components:

"One mole of ionic solid" — the product is one mole of the solid compound.
"Forms from gaseous ions" — the reactants are the constituent ions in the gas phase.
"Energy change" — the enthalpy change at constant pressure.

The value is negative because forming ionic bonds releases energy. For CaF₂, the lattice energy is −2602 kJ mol⁻¹, reflecting the strong attraction between Ca²⁺ and two F⁻ ions.

Key Takeaways

Lattice energy is a formation enthalpy — the direction is from gaseous ions to solid. This is the opposite of lattice dissociation energy (solid → gaseous ions). The two are equal in magnitude but opposite in sign.

Common Mistakes

Saying "energy to break the lattice" — that is lattice dissociation energy, not lattice energy.
Omitting "gaseous ions".
Omitting "one mole".
Confusing lattice energy with hydration or solution enthalpy.

Things to Be Careful About

The direction of the definition: gaseous ions → solid.
Lattice energy is always negative (exothermic).
This definition is used in Born–Haber cycles and enthalpy-of-solution calculations.

Techniques used

state the definition of lattice energy

(iv)

The lattice energy, $\Delta H_{\text{latt}}$ , of calcium fluoride, $\text{CaF}_2$ , is $-2602 \text{ kJ mol}^{-1}$ .

Calculate the enthalpy change of solution, $\Delta H_{\text{sol}}$ , in $\text{kJ mol}^{-1}$ , of $\text{CaF}_2$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

\Delta H_{\text{sol}} = \Delta H_{\text{hyd}}(\text{Ca}^{2+}) + 2 \times \Delta H_{\text{hyd}}(\text{F}^-) + \Delta H_{\text{latt}}

= -1650 + 2(-506) + 2602 = -1650 - 1012 + 2602 = -60 \text{ kJ mol}^{-1}

Answer

$\Delta H_{\text{sol}} = -60 \text{ kJ mol}^{-1}$

Final answer

-60 kJ mol^-1

Detailed explanation

Background Concept

The enthalpy change of solution, $\Delta H_{\text{sol}}$ , is the energy change when one mole of an ionic compound dissolves in water. It can be calculated from a thermodynamic cycle with two conceptual steps:

Breaking the lattice: $\text{CaF}_2(\text{s}) \rightarrow \text{Ca}^{2+}(\text{g}) + 2\text{F}^-(\text{g})$ . This is the reverse of lattice formation, so it requires +2602 kJ mol⁻¹.
Hydrating the ions: $\text{Ca}^{2+}(\text{g}) \rightarrow \text{Ca}^{2+}(\text{aq})$ and $2\text{F}^-(\text{g}) \rightarrow 2\text{F}^-(\text{aq})$ , releasing −1650 kJ mol⁻¹ and 2 × (−506) kJ mol⁻¹ respectively.

Summing these gives $\Delta H_{\text{sol}}$ .

Understanding the Question

The question provides the hydration enthalpies of Ca²⁺ (−1650 kJ mol⁻¹) and F⁻ (−506 kJ mol⁻¹) from Table 4.1, and the lattice energy of CaF₂ (−2602 kJ mol⁻¹). You must calculate the enthalpy change of solution of CaF₂. Note that CaF₂ contains TWO fluoride ions, so the hydration enthalpy of F⁻ must be doubled.

Approach

Use the energy cycle relationship:

\Delta H_{\text{sol}} = \Delta H_{\text{hyd}}(\text{Ca}^{2+}) + 2 \times \Delta H_{\text{hyd}}(\text{F}^-) + \Delta H_{\text{latt}}

where $\Delta H_{\text{latt}}$ here is the lattice dissociation energy (positive, +2602 kJ mol⁻¹), which is the reverse of the lattice formation energy given (−2602 kJ mol⁻¹).

Step-by-Step Reasoning

Identify the lattice dissociation term: +2602 kJ mol⁻¹ (reverse of lattice formation).
Hydration of Ca²⁺: −1650 kJ mol⁻¹.
Hydration of 2F⁻: 2 × (−506) = −1012 kJ mol⁻¹.
Sum: +2602 + (−1650) + (−1012) = −60 kJ mol⁻¹.

The negative value means dissolving CaF₂ is slightly exothermic.

Key Takeaways

The enthalpy of solution is the sum of the lattice dissociation energy and the hydration enthalpies of all ions present. Always multiply the hydration enthalpy of an ion by its stoichiometric coefficient in the formula.

Common Mistakes

Forgetting to double the hydration enthalpy of F⁻ (using −506 instead of −1012).
Using the lattice energy with the wrong sign (using −2602 instead of +2602).
Adding the values without considering signs.

Things to Be Careful About

The lattice energy given is a formation enthalpy; breaking the lattice requires the positive value.
Watch the stoichiometry: CaF₂ has two F⁻ ions.
The final answer must include the unit (kJ mol⁻¹) and the negative sign.

Techniques used

apply the enthalpy of solution energy cyclesum hydration and lattice enthalpy terms with correct signs

(b)

The formation of $\text{CaF}_2$ at 298 K is shown.

\text{Ca(s)} + \text{F}_2\text{(g)} \rightarrow \text{CaF}_2\text{(s)} \quad \Delta H^\ominus = -1214 \text{ kJ mol}^{-1}, \ \Delta G^\ominus = -1162 \text{ kJ mol}^{-1}

Calculate the entropy change, $\Delta S^\ominus$ , in $\text{J K}^{-1} \text{ mol}^{-1}$ , for this reaction.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

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

-1162 = -1214 - 298 \times \Delta S

298 \times \Delta S = -1214 + 1162 = -52

\Delta S = \frac{-52}{298} = -0.1745 \text{ kJ K}^{-1} \text{ mol}^{-1}

Answer

$\Delta S = -174.5 \text{ J K}^{-1} \text{ mol}^{-1}$

Final answer

-174.5 J K^-1 mol^-1

Detailed explanation

Background Concept

The Gibbs free energy change, $\Delta G$ , relates enthalpy change, $\Delta H$ , and entropy change, $\Delta S$ , through the equation:

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

At a given temperature, if any two of the three quantities are known, the third can be calculated. Entropy is a measure of disorder; a negative $\Delta S$ means the system becomes more ordered.

Understanding the Question

The reaction $\text{Ca}(\text{s}) + \text{F}_2(\text{g}) \rightarrow \text{CaF}_2(\text{s})$ has $\Delta H^\ominus = -1214$ kJ mol⁻¹ and $\Delta G^\ominus = -1162$ kJ mol⁻¹ at 298 K. You must calculate $\Delta S^\ominus$ in J K⁻¹ mol⁻¹. Note the required units — the answer must be converted from kJ to J.

Approach

Rearrange $\Delta G = \Delta H - T\Delta S$ to solve for $\Delta S$ , substitute the given values, and convert the result to J K⁻¹ mol⁻¹.

Step-by-Step Reasoning

Write the equation: $\Delta G = \Delta H - T\Delta S$ .
Substitute: −1162 = −1214 − 298 × $\Delta S$ .
Rearrange: 298 × $\Delta S$ = −1214 + 1162 = −52.
Solve: $\Delta S$ = −52 / 298 = −0.1745 kJ K⁻¹ mol⁻¹.
Convert to J: −0.1745 × 1000 = −174.5 J K⁻¹ mol⁻¹.

The negative entropy change makes chemical sense: the reaction converts 1 mole of solid Ca and 1 mole of gaseous F₂ into 1 mole of solid CaF₂. The gas is consumed, so the system becomes more ordered (less entropy).

Key Takeaways

The Gibbs equation links $\Delta G$ , $\Delta H$ , and $\Delta S$ ; any one can be found if the other two are known.
Always check units: $\Delta H$ and $\Delta G$ are usually in kJ mol⁻¹, while $\Delta S$ is often required in J K⁻¹ mol⁻¹.
A negative $\Delta S$ is consistent with a reaction that consumes gas.

Common Mistakes

Forgetting to convert kJ to J (giving −0.1745 instead of −174.5).
Sign errors in rearranging the equation.
Using the wrong temperature (e.g., 273 K instead of 298 K).

Things to Be Careful About

The answer must be in J K⁻¹ mol⁻¹ — convert from kJ.
Keep track of signs: both $\Delta H$ and $\Delta G$ are negative here.
The calculation is straightforward but the unit conversion is the trap.

Techniques used

rearrange the Gibbs free energy equationconvert units from kJ to J

Q5MediumTransition ElementsElectrochemistry

Copper is a transition element.

(a)

2M

(i)

Complete the electronic configurations of a $\text{Cu}^+$ ion and a $\text{Cu}^{2+}$ ion.

$\text{Cu}^+$ ion: [Ar] ..............................

$\text{Cu}^{2+}$ ion: [Ar] ..............................

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{Cu}^+$ ion: $[\text{Ar}]\, 3\text{d}^{10}$

$\text{Cu}^{2+}$ ion: $[\text{Ar}]\, 3\text{d}^{9}$

Final answer

[Ar] 3d10 and [Ar] 3d9

Detailed explanation

Background Concept

Copper is a transition element in Period 4. Its neutral atom has the configuration $[\text{Ar}]\, 3\text{d}^{10}\, 4\text{s}^1$ , an anomaly from the expected $[\text{Ar}]\, 3\text{d}^9\, 4\text{s}^2$ because a filled 3d subshell is particularly stable.

When transition metals form positive ions, electrons are removed from the 4s subshell before the 3d subshell. Although the 4s fills first in the neutral atom (it is lower in energy when empty), once the 3d subshell is occupied the 4s electrons sit at slightly higher energy and are lost first on ionisation.

Understanding the Question

The question asks you to complete the electron configurations of the $\text{Cu}^+$ and $\text{Cu}^{2+}$ ions, given the $[\text{Ar}]$ core. The $[\text{Ar}]$ core represents $1\text{s}^2\, 2\text{s}^2\, 2\text{p}^6\, 3\text{s}^2\, 3\text{p}^6$ .

Approach

Write the neutral copper configuration, then remove electrons in the correct order: remove 4s first, then 3d.

Step-by-Step Reasoning

Neutral Cu: $[\text{Ar}]\, 3\text{d}^{10}\, 4\text{s}^1$
$\text{Cu}^+$ (loses one electron): remove the 4s electron $\rightarrow [\text{Ar}]\, 3\text{d}^{10}$
$\text{Cu}^{2+}$ (loses two electrons): remove the 4s electron and one 3d electron $\rightarrow [\text{Ar}]\, 3\text{d}^{9}$

The 4s electron is always removed before any 3d electron, so $\text{Cu}^+$ retains a full 3d subshell.

Key Takeaways

The 4s subshell fills before 3d but empties first on ionisation.
$\text{Cu}^+$ has a full $3\text{d}^{10}$ ; $\text{Cu}^{2+}$ has $3\text{d}^9$ .

Common Mistakes

Writing $[\text{Ar}]\, 3\text{d}^9\, 4\text{s}^1$ for $\text{Cu}^+$ — the 4s electron is removed first.
Writing $[\text{Ar}]\, 3\text{d}^{10}\, 4\text{s}^1$ for $\text{Cu}^{2+}$ — two electrons must be removed.

Things to Be Careful About

The $[\text{Ar}]$ core must not be expanded.
The superscripts must be correct: 10 for $\text{Cu}^+$ , 9 for $\text{Cu}^{2+}$ .

Techniques used

write the electron configuration of a transition metal ionremove electrons from the correct subshells in ion formation

(ii)

Explain why transition elements have variable oxidation states.

1M Similar

DifficultyEasy

Worked solution

Answer

The 3d and 4s subshells have very similar energies, so different numbers of electrons can be removed from them, giving rise to different oxidation states.

Final answer

Similar energy of 3d and 4s subshells

Detailed explanation

Background Concept

Transition elements are defined as elements with a partially filled d subshell in at least one of their oxidation states. A defining feature is that they commonly show variable oxidation states.

The root cause is the closeness in energy of the 3d and 4s subshells. Because the energy gap is small, it costs little extra energy to remove additional electrons from the 3d subshell once the 4s electrons are gone. This allows a range of oxidation states to be stable in different compounds.

Understanding the Question

The question asks you to explain why transition elements have variable oxidation states. This is a one-mark recall-style question — the examiner is looking for the single key reason.

Approach

State the fundamental reason: the 3d and 4s subshells have similar energies.

Step-by-Step Reasoning

The 3d and 4s subshells are very close in energy. As a result, different numbers of electrons (from both subshells) can be removed or involved in bonding without a large energy penalty. This gives a range of stable oxidation states — for example, iron shows +2 and +3, copper shows +1 and +2, manganese shows +2, +4, +6 and +7.

Key Takeaways

The similar energy of the 3d and 4s subshells is the fundamental reason for variable oxidation states.

Common Mistakes

Saying "different numbers of electrons in the d subshell" without mentioning the energy similarity — the mark requires the energy point.
Confusing this with the reason for colour or catalytic behaviour.

Things to Be Careful About

The answer must mention both 3d and 4s subshells and their similar energies.
Keep it concise — one mark, one point.

Techniques used

explain variable oxidation states via subshell energy similarity

(b)

Aqueous copper(II) sulfate, $\text{CuSO}_4$ , contains the $[\text{Cu(H}_2\text{O)}_6]^{2+}$ complex ion.

2M

(i)

A few drops of $\text{NH}_3\text{(aq)}$ are added to $\text{CuSO}_4\text{(aq)}$ .

Describe any observations made.

1M Similar

DifficultyEasy

Worked solution

Answer

A pale blue precipitate forms.

Final answer

Pale blue precipitate

Detailed explanation

Background Concept

The hexaaquacopper(II) ion, $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ , is a pale blue complex in aqueous solution. Ammonia is a weak base: $\text{NH}_3(\text{aq}) + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$ . The hydroxide ions produced deprotonate the hexaaqua complex, forming an insoluble copper(II) hydroxide precipitate.

Understanding the Question

A few drops of $\text{NH}_3(\text{aq})$ are added to $\text{CuSO}_4(\text{aq})$ . The question asks for the observation. With only a few drops, the $\text{OH}^-$ concentration is sufficient to precipitate the hydroxide but not to dissolve it.

Approach

Recognise that $\text{NH}_3$ supplies $\text{OH}^-$ ; $\text{OH}^-$ deprotonates the hexaaqua complex to form the hydroxide precipitate.

Step-by-Step Reasoning

A few drops of $\text{NH}_3(\text{aq})$ produce $\text{OH}^-$ ions.
$\text{OH}^-$ deprotonates $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ to form insoluble $\text{Cu}(\text{OH})_2(\text{H}_2\text{O})_4$ .
The observation is a pale blue precipitate.

Key Takeaways

$\text{NH}_3$ acts as a base, providing $\text{OH}^-$ that precipitates copper(II) hydroxide.
With few drops, a precipitate forms; with excess, the precipitate dissolves (see part (c)).

Common Mistakes

Saying "a blue solution" — with few drops, a precipitate forms, not a solution.
Saying "deep blue" — deep blue only appears with excess $\text{NH}_3$ .

Things to Be Careful About

The precipitate is pale blue, not deep blue.
The observation must be stated as a precipitate, not a colour change of the solution.

Techniques used

predict the observation on adding OH- to the hexaaqua complex

(ii)

Write an equation for the reaction taking place.

1M Similar

DifficultyEasy

Worked solution

Answer

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

Final answer

[Cu(H2O)6]2+ + 2OH- → Cu(OH)2(H2O)4 + 2H2O

Detailed explanation

Background Concept

The reaction of the hexaaqua complex with hydroxide is a Brønsted–Lowry acid–base reaction. The water ligands act as acids, donating protons to the hydroxide ions. Two water ligands lose a proton each, becoming hydroxide ligands, and the released protons combine with $\text{OH}^-$ to form water molecules.

Understanding the Question

Write the balanced equation for the reaction in (b)(i), where a few drops of $\text{NH}_3(\text{aq})$ are added to $\text{CuSO}_4(\text{aq})$ .

Approach

The hexaaqua complex loses two protons (from two water ligands), forming the hydroxide complex.

Step-by-Step Reasoning

The $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ complex has six water ligands.
Two of these water ligands are deprotonated: each loses an $\text{H}^+$ to an $\text{OH}^-$ .
The product is $\text{Cu}(\text{OH})_2(\text{H}_2\text{O})_4$ — four water ligands remain, two hydroxide ligands replace the deprotonated waters.
Two water molecules are formed from the $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$ .

Key Takeaways

The equation must be balanced in both atoms and charge.
The coordination number of copper remains 6 throughout.

Common Mistakes

Writing $\text{Cu}(\text{OH})_2 + 4\text{H}_2\text{O}$ instead of $\text{Cu}(\text{OH})_2(\text{H}_2\text{O})_4$ — the mark scheme accepts the complex form.
Forgetting the coefficient 2 on $\text{OH}^-$ .
Omitting charges on the complex ions.

Things to Be Careful About

State symbols are not strictly required here but must be correct if included.
The charge on the left is $2+ + 2(-1) = 0$ ; on the right, the neutral $\text{Cu}(\text{OH})_2(\text{H}_2\text{O})_4$ plus neutral water — balanced.

Techniques used

write the equation for hydroxide precipitation

(c)

An excess of $\text{NH}_3\text{(aq)}$ is added to $\text{CuSO}_4\text{(aq)}$ .

3M

(i)

Describe any further observations made.

1M Similar

(ii)

Write an equation for the reaction taking place.

1M Similar

DifficultyEasy

Worked solution

Answer

[\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}

Final answer

[Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 4H2O

Detailed explanation

Background Concept

Ligand exchange (ligand substitution) is a reaction in which one ligand is replaced by another. Here, ammonia replaces four of the six water ligands in the hexaaquacopper(II) complex. The coordination number of copper remains 6 — four ammonia ligands and two water ligands.

Understanding the Question

Write the equation for the reaction in (c)(i), where excess $\text{NH}_3$ is added.

Approach

Replace four water ligands with four ammonia ligands, keeping the coordination number at 6.

Step-by-Step Reasoning

The starting complex is $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ .
Four $\text{NH}_3$ molecules replace four $\text{H}_2\text{O}$ ligands.
The product is $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ — four ammonia and two water ligands.
Four water molecules are released.
Charge is conserved: $2+$ on both sides.

Key Takeaways

The coordination number stays 6 during ligand exchange.
$\text{NH}_3$ is a stronger ligand than $\text{H}_2\text{O}$ and displaces it.

Common Mistakes

Writing $[\text{Cu}(\text{NH}_3)_6]^{2+}$ — only four $\text{NH}_3$ replace water; two water ligands remain.
Forgetting the coefficient 4 on $\text{NH}_3$ and $\text{H}_2\text{O}$ .
Omitting the charge $2+$ on the complex.

Things to Be Careful About

The equation must be balanced: 4 $\text{NH}_3$ on the left, 4 $\text{H}_2\text{O}$ on the right.
The coordination number (6) is unchanged.

Techniques used

write the equation for ligand exchange with NH3

(iii)

State the name for the type of reaction taking place.

1M Similar

(d)

Copper metal can be oxidised by acidified $\text{KMnO}_4$ . The relevant half-equations and their standard electrode potentials, $E^\ominus$ , are shown.

\begin{aligned} \text{Cu}^{2+} + 2\text{e}^- &\rightleftharpoons \text{Cu} & E^\ominus &= +0.34 \text{ V} \\ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- &\rightleftharpoons \text{Mn}^{2+} + 4\text{H}_2\text{O} & E^\ominus &= +1.52 \text{ V} \end{aligned}

5M

(i)

A $\text{MnO}_4^-/\text{Mn}^{2+}$ electrode is constructed using $0.0020 \text{ mol dm}^{-3} \text{ MnO}_4^-$ , $1.0 \text{ mol dm}^{-3} \text{ Mn}^{2+}$ and $1.0 \text{ mol dm}^{-3} \text{ H}^+$ . The temperature used is 298 K.

Use the Nernst equation to show that the $E$ value for this $\text{MnO}_4^-/\text{Mn}^{2+}$ electrode is $+1.49 \text{ V}$ .

2M Similar

DifficultyMedium

Worked solution

Working

The Nernst equation at 298 K is:

E = E^\ominus + \frac{0.059}{z}\log\frac{[\text{oxidised}]}{[\text{reduced}]}

For the $\text{MnO}_4^-/\text{Mn}^{2+}$ electrode, $z = 5$ :

E = 1.52 + \frac{0.059}{5}\log\left(\frac{0.0020 \times 1^8}{1.0}\right)

E = 1.52 + 0.0118 \times \log(0.0020) = 1.52 + 0.0118 \times (-2.699)

E = 1.52 - 0.0318 = 1.49 \text{ V}

Answer

$E = +1.49 \text{ V}$ , as required.

Final answer

+1.49 V

Detailed explanation

Background Concept

The Nernst equation relates the electrode potential of a half-cell to the concentrations of the species involved when conditions are not standard. At 298 K it is written:

E = E^\ominus + \frac{0.059}{z}\log\frac{[\text{oxidised}]}{[\text{reduced}]}

where $z$ is the number of electrons transferred in the half-equation.

For the $\text{MnO}_4^-/\text{Mn}^{2+}$ half-cell:

\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightleftharpoons \text{Mn}^{2+} + 4\text{H}_2\text{O}

$z = 5$ . The oxidised form is $\text{MnO}_4^-$ and the reduced form is $\text{Mn}^{2+}$ . The full Nernst expression includes the $[\text{H}^+]^8$ term, but since $[\text{H}^+] = 1.0 \text{ mol dm}^{-3}$ , this term equals 1 and drops out of the logarithm.

Understanding the Question

The electrode is constructed with $[\text{MnO}_4^-] = 0.0020 \text{ mol dm}^{-3}$ , $[\text{Mn}^{2+}] = 1.0 \text{ mol dm}^{-3}$ and $[\text{H}^+] = 1.0 \text{ mol dm}^{-3}$ at 298 K. You must use the Nernst equation to show that $E = +1.49 \text{ V}$ .

Approach

Substitute the given concentrations into the Nernst equation. Since $[\text{H}^+] = 1.0$ , the $[\text{H}^+]^8$ term is 1 and can be omitted. The ratio is $[\text{MnO}_4^-]/[\text{Mn}^{2+}] = 0.0020/1.0 = 0.0020$ .

Step-by-Step Reasoning

Write the Nernst equation: $E = E^\ominus + \frac{0.059}{z}\log\frac{[\text{oxidised}]}{[\text{reduced}]}$ .
Identify $z = 5$ from the half-equation.
Substitute: $E = 1.52 + \frac{0.059}{5}\log(0.0020)$ .
Calculate $\frac{0.059}{5} = 0.0118$ .
Evaluate $\log(0.0020) = -2.699$ (using base-10 log).
$E = 1.52 + 0.0118 \times (-2.699) = 1.52 - 0.0318 = 1.49 \text{ V}$ .

Key Takeaways

The Nernst equation accounts for the effect of concentration on electrode potential.
When $[\text{H}^+] = 1.0 \text{ mol dm}^{-3}$ , the $[\text{H}^+]^8$ term vanishes from the logarithm.
$z$ is the number of electrons in the half-equation — here 5.

Common Mistakes

Forgetting to divide 0.059 by $z$ (5) — using 0.059 instead of 0.0118.
Inverting the ratio (using $[\text{Mn}^{2+}]/[\text{MnO}_4^-]$ instead of $[\text{MnO}_4^-]/[\text{Mn}^{2+}]$ ).
Using natural log instead of base-10 log.
Omitting the $[\text{H}^+]^8$ term even though it is 1 here — it is safe to omit it only because $[\text{H}^+] = 1.0$ .

Things to Be Careful About

The sign of the log term: since $[\text{MnO}_4^-] < 1.0$ , the log is negative, so $E < E^\ominus$ .
The final answer must be quoted to two decimal places to match the given value of +1.49 V.

Techniques used

apply the Nernst equationcalculate log of concentration ratiosubstitute into the Nernst equation

(ii)

An electrochemical cell is constructed using a standard $\text{Cu}^{2+}/\text{Cu}$ electrode and the $\text{MnO}_4^-/\text{Mn}^{2+}$ electrode described in (d)(i).

Calculate the value of $E_{\text{cell}}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

The $\text{MnO}_4^-/\text{Mn}^{2+}$ electrode has the higher potential (+1.49 V), so it is the cathode. The $\text{Cu}^{2+}/\text{Cu}$ electrode is the anode.

E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 1.49 - 0.34 = 1.15 \text{ V}

Answer

$E_{\text{cell}} = +1.15 \text{ V}$

Final answer

+1.15 V

Detailed explanation

Background Concept

The electromotive force (emf) of an electrochemical cell is the difference between the electrode potentials of the two half-cells:

E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}

The electrode with the higher (more positive) potential is the cathode (where reduction occurs); the electrode with the lower potential is the anode (where oxidation occurs).

Understanding the Question

The cell uses a standard $\text{Cu}^{2+}/\text{Cu}$ electrode ( $E^\ominus = +0.34 \text{ V}$ ) and the $\text{MnO}_4^-/\text{Mn}^{2+}$ electrode from (d)(i) with $E = +1.49 \text{ V}$ . Calculate $E_{\text{cell}}$ .

Approach

Identify the cathode (higher potential) and anode (lower potential), then subtract.

Step-by-Step Reasoning

$E(\text{MnO}_4^-/\text{Mn}^{2+}) = +1.49 \text{ V}$ (from part (d)(i)).
$E(\text{Cu}^{2+}/\text{Cu}) = +0.34 \text{ V}$ .
The $\text{MnO}_4^-/\text{Mn}^{2+}$ electrode is the cathode (higher potential).
$E_{\text{cell}} = 1.49 - 0.34 = 1.15 \text{ V}$ .

Key Takeaways

The cathode is the electrode with the higher (more positive) potential.
$E_{\text{cell}}$ is always positive for a spontaneous reaction.

Common Mistakes

Subtracting the wrong way: $0.34 - 1.49 = -1.15 \text{ V}$ — wrong sign.
Forgetting to use the non-standard value (+1.49 V) for the $\text{MnO}_4^-/\text{Mn}^{2+}$ electrode instead of the standard +1.52 V.

Things to Be Careful About

Use the calculated value from (d)(i), not the standard value $E^\ominus = +1.52 \text{ V}$ .
The answer should be positive: $+1.15 \text{ V}$ .

Techniques used

calculate Ecell from electrode potentials

(iii)

Write an equation for the reaction taking place in the electrochemical cell described in (d)(ii).

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

5\text{Cu} + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 5\text{Cu}^{2+} + 2\text{Mn}^{2+} + 8\text{H}_2\text{O}

Final answer

5Cu + 2MnO4- + 16H+ → 5Cu2+ + 2Mn2+ + 8H2O

Detailed explanation

Background Concept

In an electrochemical cell, oxidation occurs at the anode and reduction at the cathode. Here, copper is oxidised (Cu → Cu2+ + 2e-) and permanganate is reduced (MnO4- + 8H+ + 5e- → Mn2+ + 4H2O). To write the overall cell reaction, the two half-equations must be combined so that the electrons cancel.

Understanding the Question

Write the equation for the reaction taking place in the cell from (d)(ii). The cell uses a $\text{Cu}^{2+}/\text{Cu}$ electrode and a $\text{MnO}_4^-/\text{Mn}^{2+}$ electrode.

Approach

Write the oxidation half-equation (Cu → Cu2+ + 2e-) and the reduction half-equation (MnO4- + 8H+ + 5e- → Mn2+ + 4H2O), then balance electrons by multiplying.

Step-by-Step Reasoning

Oxidation at anode: $\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^-$ (×5)
Reduction at cathode: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ (×2)
Multiply the copper half-equation by 5 (10 e-) and the permanganate half-equation by 2 (10 e-).
Add: $5\text{Cu} + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 5\text{Cu}^{2+} + 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$ .

Check: atoms — 5 Cu, 2 Mn, 8 O, 16 H on each side. Charge — left: $2(-1) + 16(+1) = +14$ ; right: $5(+2) + 2(+2) = +14$ . Balanced.

Key Takeaways

To combine half-equations, multiply each so the electrons cancel.
The overall equation must be balanced in both atoms and charge.

Common Mistakes

Not multiplying the copper half-equation by 5 — leaving unbalanced electrons.
Forgetting the coefficient 16 on $\text{H}^+$ .
Writing $\text{Cu}^{2+}$ without the charge.

Things to Be Careful About

The equation must be fully balanced — check atoms and charge.
State symbols are not required here but must be correct if included.

Techniques used

combine the two half-equationsbalance the redox equation

(iv)

Complete the sentences for the electrochemical cell described in (d)(ii).

The .............................. electrode is the negative electrode. Electrons flow from the .............................. electrode to the .............................. electrode when the cell is in use.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The $\text{Cu}^{2+}/\text{Cu}$ electrode is the negative electrode. Electrons flow from the $\text{Cu}^{2+}/\text{Cu}$ electrode to the $\text{MnO}_4^-/\text{Mn}^{2+}$ /platinum electrode when the cell is in use.

Final answer

Cu2+/Cu; Cu2+/Cu; MnO4-/Mn2+/platinum

Detailed explanation

Background Concept

In an electrochemical cell, the negative electrode is the anode, where oxidation occurs. Electrons are produced at the anode and flow through the external circuit to the cathode, where reduction occurs. The electrode with the lower (less positive) potential is the anode (negative electrode).

Understanding the Question

Complete the sentences identifying the negative electrode and the direction of electron flow.

Approach

Identify which electrode has the lower potential (Cu2+/Cu at +0.34 V is lower than MnO4-/Mn2+ at +1.49 V), so it is the negative electrode. Electrons flow from the negative electrode to the positive electrode.

Step-by-Step Reasoning

The $\text{Cu}^{2+}/\text{Cu}$ electrode has $E = +0.34 \text{ V}$ ; the $\text{MnO}_4^-/\text{Mn}^{2+}$ electrode has $E = +1.49 \text{ V}$ .
The lower-potential electrode is the negative electrode: $\text{Cu}^{2+}/\text{Cu}$ .
Oxidation occurs there: $\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^-$ .
Electrons flow from the negative electrode ( $\text{Cu}^{2+}/\text{Cu}$ ) to the positive electrode ( $\text{MnO}_4^-/\text{Mn}^{2+}$ /platinum).

Note: the $\text{MnO}_4^-/\text{Mn}^{2+}$ electrode requires an inert platinum electrode because neither species is a metal.

Key Takeaways

The negative electrode is the anode (oxidation).
Electrons always flow from the negative electrode to the positive electrode in the external circuit.

Common Mistakes

Saying the $\text{MnO}_4^-/\text{Mn}^{2+}$ electrode is negative — it has the higher potential, so it is positive.
Saying electrons flow from positive to negative — the opposite is true.

Things to Be Careful About

The first blank is the negative electrode: $\text{Cu}^{2+}/\text{Cu}$ .
The second and third blanks describe electron flow: from $\text{Cu}^{2+}/\text{Cu}$ to $\text{MnO}_4^-/\text{Mn}^{2+}$ /platinum.

Techniques used

identify the negative electrodededuce the direction of electron flow

(e)

A solution containing $[\text{Cu(H}_2\text{O)}_6]^{2+}$ is electrolysed for 5.00 hours using a constant electric current. $0.764 \text{ g}$ of copper metal is formed at the cathode. No other reduction reaction takes place.

Calculate the electric current, in A, used. Give your answer to three significant figures.

3M Similar

DifficultyMedium

Worked solution

Working

\text{Moles of Cu} = \frac{0.764}{63.5} = 0.0120 \text{ mol}

The cathode reaction is $\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$ , so 2 mol of electrons are needed per mol of Cu.

\text{Charge} = 0.0120 \times 2 \times 96500 = 2322 \text{ C}

\text{Time} = 5.00 \times 3600 = 18000 \text{ s}

\text{Current} = \frac{2322}{18000} = 0.129 \text{ A}

Answer

$0.129 \text{ A}$ (3 s.f.)

Final answer

0.129 A

Detailed explanation

Background Concept

Faraday's laws of electrolysis relate the amount of substance deposited at an electrode to the charge passed. The charge $Q$ (in coulombs) is given by:

Q = n \times F

where $n$ is the number of moles of electrons and $F = 96500 \text{ C mol}^{-1}$ is the Faraday constant. The current is then $I = Q/t$ , where $t$ is the time in seconds.

For copper deposition at the cathode:

\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}

so 2 moles of electrons are required per mole of copper deposited.

Understanding the Question

A solution containing $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ is electrolysed for 5.00 hours at constant current. 0.764 g of copper is deposited. Calculate the current in amperes to three significant figures.

Approach

Follow the chain: mass of Cu → moles of Cu → moles of electrons → charge (C) → current (A).

Step-by-Step Reasoning

Moles of Cu: $\frac{0.764}{63.5} = 0.01203 \approx 0.0120 \text{ mol}$ .
Moles of electrons: $0.01203 \times 2 = 0.02406 \text{ mol}$ (2 e- per Cu).
Charge: $Q = 0.02406 \times 96500 = 2322 \text{ C}$ .
Time in seconds: $5.00 \times 3600 = 18000 \text{ s}$ .
Current: $I = \frac{2322}{18000} = 0.129 \text{ A}$ .

Key Takeaways

$Q = n_{\text{e}^-} \times F$ and $I = Q/t$ .
The stoichiometry of the cathode reaction determines the moles of electrons.
Time must be converted from hours to seconds.

Common Mistakes

Using 1 electron per Cu instead of 2.
Forgetting to convert 5.00 hours to seconds.
Using $M_r$ of copper as 63.5 but misplacing the decimal in the division.
Quoting more or fewer than three significant figures.

Things to Be Careful About

The answer must be to three significant figures: 0.129 A (not 0.1290 or 0.13).
The Faraday constant is 96500 C mol-1.
The unit is amperes (A).

Techniques used

calculate moles of copper from massapply Faraday's law with 2 electrons per Cucalculate current from charge and time

Q6Medium-EasyTransition Elements

Iron forms complex ions with the monodentate ligand, $\text{CN}^-$ .

(a)

4M

(i)

Complex ion A contains one $\text{Fe}^{3+}$ ion and six $\text{CN}^-$ ligands.

Complex ion B contains one $\text{Fe}^{2+}$ ion and six $\text{CN}^-$ ligands.

State the formulae of these two complex ions. Include the overall charge of each complex ion.

complex ion A .................................................

complex ion B .................................................

1M Similar

DifficultyEasy

Worked solution

Answer

complex ion A: $[\text{Fe}(\text{CN})_6]^{3-}$
complex ion B: $[\text{Fe}(\text{CN})_6]^{4-}$

Final answer

[Fe(CN)6]3– and [Fe(CN)6]4–

Detailed explanation

Background Concept

A complex ion consists of a central metal ion surrounded by ligands. Ligands are ions or molecules that donate a lone pair of electrons to the metal ion to form coordinate (dative covalent) bonds. The overall charge of the complex ion is the sum of the charge on the central metal ion and the charges on all the ligands.

Understanding the Question

The question asks for the formulae and overall charges of two complex ions, A and B. Both have six cyanide ( $\text{CN}^-$ ) ligands. Complex ion A has an $\text{Fe}^{3+}$ centre, and complex ion B has an $\text{Fe}^{2+}$ centre.

Approach

To find the overall charge, add the charge of the central metal ion to the total charge contributed by the ligands. The formula is written with the metal first, followed by the ligands in brackets, and the overall charge as a superscript outside.

Step-by-Step Reasoning

Complex ion A:

Central ion: $\text{Fe}^{3+}$ (charge $+3$ )
Ligands: 6 $\text{CN}^-$ (each has charge $-1$ , total $-6$ )
Overall charge: $(+3) + 6(-1) = +3 - 6 = -3$
Formula: $[\text{Fe}(\text{CN})_6]^{3-}$

Complex ion B:

Central ion: $\text{Fe}^{2+}$ (charge $+2$ )
Ligands: 6 $\text{CN}^-$ (total $-6$ )
Overall charge: $(+2) + 6(-1) = +2 - 6 = -4$
Formula: $[\text{Fe}(\text{CN})_6]^{4-}$

Key Takeaways

The overall charge of a complex ion is the algebraic sum of the metal ion's charge and the ligands' charges. The coordination number (number of ligands) is indicated by the subscript in the formula.

Common Mistakes

Forgetting to include the overall charge or getting the sign wrong (e.g., writing $3-$ instead of $-3$ , though both are often accepted, the standard is $3-$ ).
Miscounting the total ligand charge (e.g., thinking 6 $\text{CN}^-$ gives $-6$ but adding it incorrectly).

Things to Be Careful About

Ensure the brackets enclose the entire complex $[\text{Fe}(\text{CN})_6]$ and the charge is outside.
The charge of cyanide is $-1$ , not $+1$ .

Techniques used

calculate overall charge of a complex ion

(ii)

Explain why a solution containing complex ion A and a solution containing complex ion B are different colours.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The $\text{Fe}^{3+}$ and $\text{Fe}^{2+}$ ions have different charges (different oxidation states).
This results in different d-orbital splitting energies ( $\Delta E$ ).
Therefore, they absorb different frequencies (or wavelengths) of visible light.

Final answer

Different d-orbital splitting energies ( $\Delta E$ ) lead to absorption of different frequencies of visible light.

Detailed explanation

Background Concept

Transition metal complexes are coloured because of the absorption of visible light. In an octahedral complex, the five d-orbitals of the metal ion split into two energy levels: a lower energy set ( $t_{2g}$ ) and a higher energy set ( $e_g$ ). The energy difference between these sets is called the crystal field splitting energy, $\Delta E$ .

Electrons in the lower d-orbitals can absorb photons of light to jump to the higher d-orbitals. The energy of the photon must match the splitting energy: $\Delta E = hf = \frac{hc}{\lambda}$ , where $h$ is Planck's constant, $f$ is frequency, and $\lambda$ is wavelength.

The colour we see is the complementary colour to the light absorbed. If the splitting energy changes, the frequency/wavelength of light absorbed changes, and thus the observed colour changes.

Understanding the Question

We need to explain why solutions of $[\text{Fe}(\text{CN})_6]^{3-}$ (containing $\text{Fe}^{3+}$ ) and $[\text{Fe}(\text{CN})_6]^{4-}$ (containing $\text{Fe}^{2+}$ ) are different colours. Both have the same ligand ( $\text{CN}^-$ ) and the same geometry (octahedral).

Approach

The key difference between the two complexes is the oxidation state of the iron ion ( $+3$ vs $+2$ ). The charge of the central metal ion affects the d-orbital splitting energy. A higher charge on the metal ion pulls the ligands closer, increasing the splitting energy $\Delta E$ . Different $\Delta E$ means different energy photons are absorbed, leading to different colours.

Step-by-Step Reasoning

Identify the difference: Complex ion A contains $\text{Fe}^{3+}$ , while complex ion B contains $\text{Fe}^{2+}$ . They have different oxidation states/charges.
Link to splitting energy: The different charge on the central ion results in different d-orbital splitting energies ( $\Delta E$ ). (Higher charge $\text{Fe}^{3+}$ generally causes a larger splitting than $\text{Fe}^{2+}$ with the same ligand).
Link to light absorption: The energy gap $\Delta E$ corresponds to the energy of the absorbed photon ( $E = hf$ ). Since $\Delta E$ is different, the frequency (or wavelength) of visible light absorbed is different.
Conclusion: Different absorption leads to different transmitted/reflected colours.

Key Takeaways

The colour of a complex ion depends on the d-orbital splitting energy ( $\Delta E$ ). This splitting is influenced by the oxidation state of the metal ion, the nature of the ligands, and the geometry of the complex. Changing the metal's oxidation state (even with the same ligands) changes the colour.

Common Mistakes

Saying "different number of d-electrons" without linking it to the splitting energy. While the number of d-electrons differs ( $d^5$ vs $d^6$ ), the primary reason for the difference in absorption energy here is the different nuclear charge/pulling power of $\text{Fe}^{3+}$ vs $\text{Fe}^{2+}$ affecting the splitting.
Forgetting to mention "frequency" or "wavelength" of light. Just saying "absorb different light" is often not enough; specify visible light frequency/wavelength.
Saying "different energy levels" without specifying d-orbitals or splitting energy.

Things to Be Careful About

Use the term "d-orbital splitting energy" or $\Delta E$ .
Ensure you link the energy difference to the frequency or wavelength of light absorbed ( $E=hf$ ).

Techniques used

explain colour using d-orbital splittingrelate energy gap to light frequency

(iii)

In complex ion A, the carbon atom of each $\text{CN}^-$ ligand bonds to the $\text{Fe}^{3+}$ ion.

State the type of bonding involved.

1M Similar

(b)

Complex ions have different geometries.

Complex ion A is octahedral.

$\text{Ag}^+$ ions form a linear complex with ammonia.

Ni atoms form a tetrahedral complex with carbon monoxide molecules. The carbon atom in the monodentate carbon monoxide ligand bonds to the nickel atom.

$\text{Pd}^{2+}$ ions form a square planar complex with chloride ions.

Complete Fig. 6.1 to show the geometry of each of these four ions, using three-dimensional bonds where necessary. Label one bond angle on each complex ion.

Fig. 6.1

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

See diagram below for the completed Fig. 6.1.

Complex ion A ( $[\text{Fe}(\text{CN})_6]^{3-}$ ): Octahedral geometry. Bond angle $90^\circ$ .
Ag $^+$ with ammonia ( $[\text{Ag}(\text{NH}_3)_2]^+$ ): Linear geometry. Bond angle $180^\circ$ .
Ni with CO ( $\text{Ni}(\text{CO})_4$ ): Tetrahedral geometry. Bond angle $109.5^\circ$ .
Pd $^{2+}$ with chloride ( $[\text{Pd}\text{Cl}_4]^{2-}$ ): Square planar geometry. Bond angle $90^\circ$ .

Final answer

Octahedral (90°), Linear (180°), Tetrahedral (109.5°), Square planar (90°). See diagram.

Detailed explanation

Background Concept

Complex ions can have different geometries depending on the coordination number (number of ligands) and the electronic configuration of the metal ion (specifically $d^8$ configurations often form square planar complexes).

Coordination number 6: Octahedral geometry. Bond angles are $90^\circ$ (and $180^\circ$ for opposite ligands). Represented with two plain lines (in plane), two wedges (coming out), and two dashes (going in), or similar 3D representation.
Coordination number 4: Can be tetrahedral or square planar.
- Tetrahedral: Bond angles $109.5^\circ$ . Represented with a 3D cross or pyramid shape.
- Square planar: Bond angles $90^\circ$ . All ligands in one plane (usually represented as a cross with plain lines).
Coordination number 2: Linear geometry. Bond angle $180^\circ$ .

Understanding the Question

We need to draw the 3D structures and label bond angles for four specific complexes in the boxes provided in Fig. 6.1.

Complex ion A: $[\text{Fe}(\text{CN})_6]^{3-}$ . Coordination number 6. Geometry: Octahedral.
Ag $^+$ with ammonia: Ag $^+$ typically forms linear complexes with coordination number 2 ( $d^{10}$ configuration). Ligands are NH $_3$ . Formula: $[\text{Ag}(\text{NH}_3)_2]^+$ .
Ni with CO: Ni is a transition metal, CO is a neutral ligand. Ni(0) is $d^{10}$ . Coordination number 4. Geometry: Tetrahedral. Formula: $\text{Ni}(\text{CO})_4$ .
Pd $^{2+}$ with chloride: Pd $^{2+}$ is a $4d^8$ ion. $d^8$ ions (especially 4d and 5d) strongly prefer square planar geometry with coordination number 4. Ligands are Cl $^-$ . Formula: $[\text{Pd}\text{Cl}_4]^{2-}$ .

Approach

For each box, identify the central atom/ion, the ligands, the coordination number, and the geometry. Draw the structure using appropriate bond types (plain, wedge, dash) to show 3D shape. Label one bond angle clearly.

Step-by-Step Reasoning

Box 1: Complex ion A ( $[\text{Fe}(\text{CN})_6]^{3-}$ )

Central ion: Fe.
Ligands: 6 CN $^-$ . The question states C bonds to Fe. So draw Fe-C bonds. The ligand is written as CN or NC. To show C bonding to Fe, write NC on the outside or use an arrow, but in simple structural formulas, often just CN with the understanding C is the donor, or NC-Fe. The mark scheme shows 'CN' and 'NC' to indicate orientation. Let's follow the mark scheme style: Central Fe, with 6 bonds. Top/Bottom/Right-Back/Left-Back/Front-Left/Front-Right.
Geometry: Octahedral.
Drawing: Draw Fe in the centre. Draw 6 bonds. Two vertical (up/down) as plain lines. Two horizontal (left/right) as dashed lines (going back). Two diagonal (front-left/front-right) as wedges (coming forward). Attach CN/NC groups. Note: C must bond to Fe. So if drawing NC, the bond is from C (right side of NC) to Fe. If drawing CN, bond is from C (left side of CN) to Fe. Mark scheme uses 'NC' with dashed bond (N is outer, C is inner) and 'CN' with wedge (C is inner, N is outer? Wait. C bonds to Fe. So bond is Fe-C. If label is 'CN', C is next to Fe. If label is 'NC', N is next to outside, C is next to Fe. So bond is Fe-C.
Angle: Label $90^\circ$ between adjacent ligands (e.g., top and right-back).

Box 2: Ag $^+$ with ammonia

Central ion: Ag.
Ligands: 2 NH $_3$ (ammonia). N donates lone pair.
Geometry: Linear (coordination number 2 for Ag $^+$ ).
Drawing: H $_3$ N — Ag — NH $_3$ . All in a straight line.
Angle: Label $180^\circ$ between the two N-Ag-N bonds.

Box 3: Ni with carbon monoxide

Central atom: Ni.
Ligands: 4 CO. C bonds to Ni.
Geometry: Tetrahedral.
Drawing: Ni in centre. One bond up (plain), one bond down-left (plain), one bond down-right (wedge), one bond back-right (dashed). Attach CO groups. Ensure C is bonded to Ni. Mark scheme shows 'CO' (C next to Ni) and 'OC' (O next to outside, C next to Ni).
Angle: Label $109.5^\circ$ between two adjacent bonds (e.g., down-left and down-right).

Box 4: Pd $^{2+}$ with chloride

Central ion: Pd.
Ligands: 4 Cl $^-$ .
Geometry: Square planar (characteristic of Pd $^{2+}$ $d^8$ ).
Drawing: Pd in centre. Four bonds in a cross shape (up, down, left, right), all plain lines (all in plane).
Angle: Label $90^\circ$ between adjacent bonds (e.g., up and right).

Key Takeaways

Coordination number 6 is almost always octahedral ( $90^\circ$ ).
Coordination number 2 is linear ( $180^\circ$ ), common for Ag $^+$ , Au $^+$ , Cu $^+$ .
Coordination number 4 can be tetrahedral ( $109.5^\circ$ ) or square planar ( $90^\circ$ ). $d^8$ ions like Pd $^{2+}$ , Pt $^{2+}$ , Ni $^{2+}$ (with strong field ligands like CN $^-$ ) are square planar. Ni(CO) $_4$ is tetrahedral.
3D representation uses plain lines (in plane), wedges (forward), dashes (back).

Common Mistakes

Drawing square planar for Ni(CO) $_4$ (it is tetrahedral).
Drawing tetrahedral for [PdCl $_4$ ] $^{2-}$ (it is square planar).
Forgetting to label the bond angle.
Drawing the wrong number of ligands (e.g., 4 for Fe complex).
Not showing 3D bonds (wedges/dashes) for octahedral and tetrahedral shapes. For square planar and linear, plain lines are sufficient.
Bonding the wrong atom: In CN $^-$ , C bonds to metal. In CO, C bonds to metal. In NH $_3$ , N bonds to metal. In Cl $^-$ , Cl bonds to metal.

Things to Be Careful About

Bond Angles: Octahedral ( $90^\circ$ ), Linear ( $180^\circ$ ), Tetrahedral ( $109.5^\circ$ ), Square Planar ( $90^\circ$ ). Note that octahedral and square planar both have $90^\circ$ angles, so the geometry must be clear from the drawing.
3D Bonds: For octahedral and tetrahedral, you must use wedges and dashes to show the 3D shape. For square planar and linear, the structure is flat, so plain lines are fine (though a slight perspective can be used, plain cross is standard for square planar).
Ligand Orientation: In CN, C is the donor. So the bond should be to C. Writing 'NC-' with bond to C is correct. Writing '-CN' with bond to C is correct. The mark scheme is flexible but shows C connecting to the metal.

Techniques used

draw complex ion geometriesidentify bond angles for different coordination numbers

Q7Medium-HardIntroduction to A Level Organic ChemistryAnalytical TechniquesHydrocarbonsCarboxylic Acids and Derivatives

A hydrocarbon is known to be either compound D or compound E.

Fig. 7.1

(a)

Give the systematic name of E.

1M Similar

DifficultyEasy

Worked solution

Answer

1,3-diethylbenzene

Final answer

1,3-diethylbenzene

Detailed explanation

Background Concept

Disubstituted benzene rings are named by assigning the lowest possible locants to the substituents. When two identical substituents are present, the positions are numbered to give the smallest set of numbers. The prefixes 1,2- (ortho), 1,3- (meta), and 1,4- (para) describe the relative positions on the ring.

Understanding the Question

Compound E has two ethyl groups ( $-\text{CH}_2\text{CH}_3$ ) attached to a benzene ring. From the structure shown, the ethyl groups are separated by one carbon between them, giving a 1,3-relationship.

Approach

Identify the parent structure (benzene), note the substituents (two ethyl groups), and assign locants to give the lowest numbers. Starting from one ethyl group at position 1, the other is at position 3.

Step-by-Step Reasoning

The parent is a benzene ring with two identical substituents.
The substituents are ethyl groups ( $-\text{CH}_2\text{CH}_3$ ).
Numbering from one ethyl group (position 1), the next carbon is position 2, and the second ethyl group is at position 3.
This gives the lowest locant set: 1,3 (rather than 1,5 which would result from numbering the other way).
The name is 1,3-diethylbenzene.

Key Takeaways

For disubstituted benzenes, always number to give the lowest locant set.
1,2- = ortho, 1,3- = meta, 1,4- = para (though systematic names use numbers).

Common Mistakes

Writing "m-diethylbenzene" — while acceptable in some contexts, the question asks for the systematic name.
Numbering incorrectly to give 1,5 instead of 1,3.

Things to Be Careful About

Ensure the name uses "diethyl" (not "ethyl") since there are two identical substituents.
The comma between numbers and hyphen between numbers and letters must be correct.

Techniques used

identify positions of substituents on benzene ringapply IUPAC naming rules for disubstituted benzenes

(b)

The proton ( $^1\text{H}$ ) NMR spectra of D and E are compared. They are very similar. The proton ( $^1\text{H}$ ) NMR spectrum of D is shown in Fig. 7.2.

Fig. 7.2

6M

(i)

Suggest a suitable solvent for obtaining the spectrum in Fig. 7.2.

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{CDCl}_3$ (deuterated chloroform)

Final answer

CDCl3

Detailed explanation

Background Concept

Proton NMR spectroscopy requires a solvent that does not contain hydrogen atoms (or whose hydrogen signals are well outside the region of interest), because the solvent would otherwise overwhelm the sample signals. Deuterated solvents are used because deuterium ( $^2\text{H}$ ) does not produce a signal in a $^1\text{H}$ NMR experiment.

Understanding the Question

The question asks for a suitable solvent for obtaining a $^1\text{H}$ NMR spectrum. The sample is a hydrocarbon (diethylbenzene), which is non-polar, so the solvent must dissolve it and must not interfere with the proton signals.

Approach

Recall the most commonly used deuterated solvent for organic NMR — $\text{CDCl}_3$ — which is suitable for most non-polar and moderately polar organic compounds.

Step-by-Step Reasoning

The solvent must be deuterated so it does not produce $^1\text{H}$ signals in the spectrum.
$\text{CDCl}_3$ is the standard solvent for most organic compounds, including hydrocarbons.
It dissolves non-polar compounds well and its residual $\text{CHCl}_3$ peak at $\delta$ 7.26 ppm is easily identifiable and does not interfere.

Key Takeaways

$\text{CDCl}_3$ is the default solvent for $^1\text{H}$ NMR of organic compounds.
Other acceptable answers include $\text{CD}_3\text{COCD}_3$ (deuterated acetone) or $\text{C}_6\text{D}_6$ (deuterated benzene), but $\text{CDCl}_3$ is most common.

Common Mistakes

Writing "chloroform" instead of "deuterated chloroform" or " $\text{CDCl}_3$ " — ordinary $\text{CHCl}_3$ would give a large proton signal.
Writing a non-deuterated solvent like $\text{CCl}_4$ (which has no H at all but is not deuterated; while technically it wouldn't interfere, the mark scheme expects $\text{CDCl}_3$ ).

Things to Be Careful About

The answer must specify the deuterated form, not the protonated form.

Techniques used

identify suitable NMR solvent

(ii)

The proton ( $^1\text{H}$ ) NMR spectrum of E is obtained twice, once before and once after shaking with $\text{D}_2\text{O}$ .

Describe any differences between these two spectra. Explain your answer.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

There is no difference between the two spectra because compound E (1,3-diethylbenzene) has no O $-$ H, N $-$ H, or S $-$ H protons that can be exchanged with deuterium from $\text{D}_2\text{O}$ .

Final answer

No difference; E has no H atoms that can be exchanged with D

Detailed explanation

Background Concept

The $\text{D}_2\text{O}$ shake test is used in $^1\text{H}$ NMR to identify exchangeable protons — those attached to electronegative atoms such as oxygen (in alcohols, phenols, carboxylic acids), nitrogen (in amines, amides), or sulfur. When $\text{D}_2\text{O}$ is added, these protons are replaced by deuterium ( $^2\text{H}$ ), which does not appear in a $^1\text{H}$ NMR spectrum, so the corresponding peaks disappear. C $-$ H protons are not exchanged under these mild conditions.

Understanding the Question

Compound E is 1,3-diethylbenzene. The question asks what happens to its $^1\text{H}$ NMR spectrum when it is shaken with $\text{D}_2\text{O}$ , and requires an explanation.

Approach

Examine the structure of E for any O $-$ H, N $-$ H, or S $-$ H groups. Since E is a pure hydrocarbon (only C and H atoms), it has no such groups, so no exchange can occur.

Step-by-Step Reasoning

Compound E is 1,3-diethylbenzene: $\text{C}_6\text{H}_4(\text{CH}_2\text{CH}_3)_2$ .
All hydrogen atoms are bonded to carbon (C $-$ H bonds).
C $-$ H protons are not acidic enough to be exchanged with $\text{D}_2\text{O}$ under the mild conditions of the shake test.
Therefore, the spectrum before and after $\text{D}_2\text{O}$ addition is identical — no peaks disappear.

Key Takeaways

$\text{D}_2\text{O}$ exchange only removes peaks from O $-$ H, N $-$ H, and S $-$ H protons.
A hydrocarbon will show no change in its $^1\text{H}$ NMR spectrum upon $\text{D}_2\text{O}$ addition.

Common Mistakes

Saying "the peak disappears" without checking whether the compound actually has exchangeable protons.
Confusing this with the reaction of $\text{D}_2\text{O}$ with acyl chlorides or other reactive species.

Things to Be Careful About

The answer must include BOTH parts: the observation (no difference) AND the explanation (no H that can be exchanged with D). Missing either half loses the mark.

Techniques used

apply D2O exchange test to identify exchangeable protonsanalyse molecular structure for O-H/N-H/S-H groups

(iii)

Complete Table 7.1 for the proton ( $^1\text{H}$ ) NMR spectrum of D.

Table 7.1

chemical shift $\delta$ / ppm	group responsible for the peak	splitting pattern
1.3
2.7
7.1	[shaded]	[shaded]

Table 7.2

environment of proton	example	chemical shift range $\delta$ / ppm
alkane	$-\text{CH}_3$ , $-\text{CH}_2-$ , $>\text{CH}-$	0.9–1.7
alkyl next to C=O	$\text{CH}_3-\text{C=O}$ , $-\text{CH}_2-\text{C=O}$ , $>\text{CH}-\text{C=O}$	2.2–3.0
alkyl next to aromatic ring	$\text{CH}_3-\text{Ar}$ , $-\text{CH}_2-\text{Ar}$ , $>\text{CH}-\text{Ar}$	2.3–3.0
alkyl next to electronegative atom	$\text{CH}_3-\text{O}$ , $-\text{CH}_2-\text{O}$ , $-\text{CH}_2-\text{Cl}$	3.2–4.0
attached to alkene	$=\text{CHR}$	4.5–6.0
attached to aromatic ring	$\text{H}-\text{Ar}$	6.0–9.0
aldehyde	$\text{HCOR}$	9.3–10.5
alcohol	$\text{ROH}$	0.5–6.0
phenol	$\text{Ar}-\text{OH}$	4.5–7.0
carboxylic acid	$\text{RCOOH}$	9.0–13.0
alkyl amine	$\text{R}-\text{NH}-$	1.0–5.0
aryl amine	$\text{Ar}-\text{NH}_2$	3.0–6.0
amide	$\text{RCONHR}$	5.0–12.0

3M Similar

DifficultyMedium

Worked solution

Answer

chemical shift $\delta$ / ppm	number of $^1\text{H}$ atoms	group responsible	splitting pattern
1.3	6	$\text{CH}_3$	triplet
2.7	4	$\text{CH}_2$	quartet
7.1	4	—	—

Explanation of splitting:

$\text{CH}_3$ (6H): adjacent to $\text{CH}_2$ (2H), so $n+1 = 3$ → triplet
$\text{CH}_2$ (4H): adjacent to $\text{CH}_3$ (3H), so $n+1 = 4$ → quartet

Final answer

1.3 ppm: 6H, CH3, triplet; 2.7 ppm: 4H, CH2, quartet; 7.1 ppm: 4H

Detailed explanation

Background Concept

In $^1\text{H}$ NMR, equivalent protons give one signal whose integration reflects the number of protons in that environment. The chemical shift identifies the type of environment (from data tables), and the splitting pattern follows the $n+1$ rule: a set of equivalent protons is split by $n$ equivalent protons on the adjacent carbon into $n+1$ peaks.

Compound D is 1,2-diethylbenzene. The molecule has a plane of symmetry passing through the midpoint of the C1 $-$ C2 bond and the midpoint of the C4 $-$ C5 bond, making the two ethyl groups equivalent and the aromatic protons fall into two equivalent pairs.

Understanding the Question

Complete Table 7.1 for compound D by assigning: the number of protons responsible for each peak, the group (CH3 or CH2), and the splitting pattern. The 7.1 ppm row only requires the number of protons (group and splitting are shaded/not required).

Approach

Identify all proton environments in D using symmetry.
Count protons in each environment.
Use the data table to assign chemical shifts.
Apply the $n+1$ rule to determine splitting.

Step-by-Step Reasoning

Environment 1: CH3 protons

Two equivalent $\text{CH}_3$ groups (one on each ethyl) → 6H total.
Chemical shift ~1.3 ppm (alkane region, 0.9 $-$ 1.7 ppm from data table).
Adjacent carbon has 2H (the $\text{CH}_2$ ), so $n+1 = 3$ → triplet.

Environment 2: CH2 protons

Two equivalent $\text{CH}_2$ groups → 4H total.
Chemical shift ~2.7 ppm (alkyl next to aromatic ring, 2.3 $-$ 3.0 ppm from data table).
Adjacent carbon has 3H (the $\text{CH}_3$ ), so $n+1 = 4$ → quartet.

Environment 3: Aromatic protons

Four aromatic H atoms (two pairs of equivalent protons at positions 3,6 and 4,5) → 4H.
Chemical shift ~7.1 ppm (aromatic region, 6.0 $-$ 9.0 ppm from data table).
The group and splitting columns are shaded (not required for this row).

Key Takeaways

Symmetry reduces the number of distinct proton environments.
The $n+1$ rule applies only to protons on adjacent carbons (usually $^3J$ coupling).
Integration ratios must sum to the total number of H atoms in the molecule (6 + 4 + 4 = 14 = $\text{C}_{10}\text{H}_{14}$ ✓).

Common Mistakes

Writing 3H for the CH3 peak instead of 6H (forgetting there are two equivalent CH3 groups).
Writing 2H for the CH2 peak instead of 4H.
Confusing which group causes which splitting (CH3 is split by CH2, not vice versa).
Writing "multiplet" for the aromatic region when the question only asks for the number of protons in that row.

Things to Be Careful About

The mark scheme awards one mark per correct column (number of H, group, splitting). All three must be correct for a row to score.
"Quadruplet" is also accepted for "quartet".

Techniques used

assign chemical shifts to proton environmentscount equivalent protons from molecular symmetryapply n+1 rule for splitting patterns

(iv)

Compounds D and E can be distinguished by carbon-13 NMR spectroscopy.

State the number of peaks in each spectrum.

The carbon-13 NMR spectrum of D has .............................. peaks.

The carbon-13 NMR spectrum of E has .............................. peaks.

1M Similar

DifficultyMedium

Worked solution

Answer

The carbon-13 NMR spectrum of D has 5 peaks.

The carbon-13 NMR spectrum of E has 6 peaks.

Final answer

D: 5 peaks; E: 6 peaks

Detailed explanation

Background Concept

In $^{13}\text{C}$ NMR, each chemically distinct carbon environment gives one peak. The number of peaks is determined by the symmetry of the molecule. For disubstituted benzenes, the symmetry plane (or planes) determines which aromatic carbons are equivalent.

Understanding the Question

Compounds D (1,2-diethylbenzene) and E (1,3-diethylbenzene) must be distinguished by counting the number of unique carbon environments in each.

Approach

Identify all symmetry elements in each molecule, then count the number of unique carbon environments.

Step-by-Step Reasoning

Compound D (1,2-diethylbenzene):

Plane of symmetry bisects the C1 $-$ C2 bond and the C4 $-$ C5 bond.
Unique carbons:
1. $\text{CH}_3$ (both equivalent by symmetry) → 1 signal
2. $\text{CH}_2$ (both equivalent) → 1 signal
3. Aromatic C1 and C2 (equivalent, bearing ethyl groups) → 1 signal
4. Aromatic C3 and C6 (equivalent) → 1 signal
5. Aromatic C4 and C5 (equivalent) → 1 signal
Total: 5 peaks

Compound E (1,3-diethylbenzene):

Plane of symmetry passes through C2 and C5.
Unique carbons:
1. $\text{CH}_3$ (both equivalent) → 1 signal
2. $\text{CH}_2$ (both equivalent) → 1 signal
3. Aromatic C1 and C3 (equivalent, bearing ethyl groups) → 1 signal
4. Aromatic C2 (unique, between the two substituted carbons) → 1 signal
5. Aromatic C4 and C6 (equivalent) → 1 signal
6. Aromatic C5 (unique, opposite C2) → 1 signal
Total: 6 peaks

Key Takeaways

The number of $^{13}\text{C}$ peaks equals the number of chemically distinct carbon environments.
Symmetry planes in disubstituted benzenes make certain ring carbons equivalent.
1,2-disubstituted benzenes with identical substituents have fewer peaks than 1,3-disubstituted ones because of different symmetry.

Common Mistakes

Forgetting that the two ethyl groups are equivalent in both isomers (both have a symmetry plane making them equivalent).
Miscounting aromatic carbons by not identifying which are equivalent under the symmetry plane.
Confusing the number of peaks with the total number of carbon atoms (both have 10 carbons, but symmetry reduces the number of signals).

Things to Be Careful About

The answer requires BOTH numbers (5 for D and 6 for E). Getting one wrong loses the mark.
Note that D has fewer peaks than E despite having the same molecular formula — this is the basis of the distinction.

Techniques used

identify symmetry elements in aromatic moleculescount unique carbon environments

(c)

Compound D can be oxidised to compound F by alkaline $\text{KMnO}_4$ followed by dilute acid.

Fig. 7.3

4M

(i)

Write an equation for this reaction using molecular formulae for D and F. The products of this reaction are F, water and carbon dioxide.

Use [O] to represent one atom of oxygen from the oxidising agent.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

D is 1,2-diethylbenzene: $\text{C}_{10}\text{H}_{14}$
F is benzene-1,2-dicarboxylic acid (phthalic acid): $\text{C}_8\text{H}_6\text{O}_4$

Each ethyl group ( $-\text{CH}_2\text{CH}_3$ ) is oxidised to $-\text{COOH}$ on the ring, with the terminal carbon lost as $\text{CO}_2$ .

\text{C}_{10}\text{H}_{14} + 12[\text{O}] \rightarrow \text{C}_8\text{H}_6\text{O}_4 + 2\text{CO}_2 + 4\text{H}_2\text{O}

Check: Left: C = 10, H = 14, O = 12. Right: C = 8 + 2 = 10, H = 6 + 8 = 14, O = 4 + 4 + 4 = 12. ✓

Answer

$\text{C}_{10}\text{H}_{14} + 12[\text{O}] \rightarrow \text{C}_8\text{H}_6\text{O}_4 + 2\text{CO}_2 + 4\text{H}_2\text{O}$

Final answer

C10H14 + 12[O] → C8H6O4 + 2CO2 + 4H2O

Detailed explanation

Background Concept

Alkaline $\text{KMnO}_4$ followed by dilute acid is a vigorous oxidising condition that cleaves alkyl side chains on aromatic rings. Any alkyl group with at least one benzylic hydrogen is oxidised to a carboxylic acid group ( $-\text{COOH}$ ) on the ring. The carbon atoms beyond the benzylic position are lost as $\text{CO}_2$ .

For an ethyl group ( $-\text{CH}_2\text{CH}_3$ ): the $\text{CH}_2$ (benzylic) carbon becomes the carboxyl carbon, and the $\text{CH}_3$ carbon is oxidised to $\text{CO}_2$ .

Understanding the Question

Compound D (1,2-diethylbenzene, $\text{C}_{10}\text{H}_{14}$ ) is oxidised to compound F (phthalic acid, $\text{C}_8\text{H}_6\text{O}_4$ ). The products are F, $\text{CO}_2$ , and $\text{H}_2\text{O}$ . The equation must use [O] to represent oxygen from the oxidising agent.

Approach

Write the molecular formulae of D and F.
Determine how many [O] atoms are needed by balancing C, H, and O on both sides.
The two terminal methyl carbons become 2 $\text{CO}_2$ .
Balance hydrogen: D has 14H, F has 6H, so 8H go to water → 4 $\text{H}_2\text{O}$ .
Count oxygen on the right: 4 (from F) + 4 (from 2CO2) + 4 (from 4H2O) = 12, so 12[O] needed.

Step-by-Step Reasoning

D: $\text{C}_{10}\text{H}_{14}$ (benzene ring C6H4 + two ethyl groups C2H5 each = C6H4 + C4H10 = C10H14)
F: $\text{C}_8\text{H}_6\text{O}_4$ (benzene ring C6H4 + two COOH groups = C6H4 + C2H2O4 = C8H6O4)
Carbon balance: 10 = 8 + 2(1) → 2 CO2 ✓
Hydrogen balance: 14 = 6 + 4(2) → 4 H2O ✓
Oxygen balance: RHS has 4 + 4 + 4 = 12 O atoms → 12[O] on LHS ✓

Key Takeaways

Each ethyl side chain requires 6[O] to become one COOH + one CO2 + two H2O.
The [O] notation represents one oxygen atom from the oxidising agent.
Always verify the balance of all three elements.

Common Mistakes

Forgetting that the terminal carbon of each ethyl group becomes CO2 (writing only F as the organic product).
Incorrectly writing 6[O] instead of 12[O] (forgetting there are two ethyl groups).
Writing O2 instead of [O] (the question specifies [O]).

Things to Be Careful About

Use molecular formulae (not structural formulae) as instructed.
The coefficient of [O] must be correct — it is the key mark.

Techniques used

balance oxidation equation using [O] notationdeduce molecular formulae from structures

(ii)

F reacts with an excess of $\text{SOCl}_2$ to form compound G. The molecular formula of G is $\text{C}_8\text{H}_4\text{Cl}_2\text{O}_2$ .

G reacts with ethane-1,2-diol, $\text{HOCH}_2\text{CH}_2\text{OH}$ , to form a mixture of products that includes compounds J, molecular formula $\text{C}_{10}\text{H}_8\text{O}_4$ , and K, molecular formula $\text{C}_{18}\text{H}_{12}\text{Cl}_2\text{O}_6$ .

Draw the structures of compounds G, J and K in Fig. 7.4.

Fig. 7.4

3M Similar

DifficultyMedium-Hard

Worked solution

Answer

Compound G ( $\text{C}_8\text{H}_4\text{Cl}_2\text{O}_2$ ): benzene-1,2-dicarbonyl dichloride (phthaloyl chloride) — benzene ring with two adjacent $-\text{COCl}$ groups.

Compound J ( $\text{C}_{10}\text{H}_8\text{O}_4$ ): cyclic diester — benzene ring fused to an 8-membered ring containing two ester linkages ( $-\text{COO}-\text{CH}_2-\text{CH}_2-\text{OOC}-$ ).

Compound K ( $\text{C}_{18}\text{H}_{12}\text{Cl}_2\text{O}_6$ ): two benzene rings, each bearing one remaining $-\text{COCl}$ group, linked by a $-\text{COOCH}_2\text{CH}_2\text{OOC}-$ bridge (one ester from each ring's other $-\text{COCl}$ reacting with the two OH groups of the diol).

Final answer

G: 1,2-C6H4(COCl)2; J: cyclic diester (benzene fused to 8-membered ring with two COO groups and CH2CH2); K: two C6H4(COCl) units linked by COOCH2CH2OOC bridge

Detailed explanation

Background Concept

Carboxylic acids react with excess $\text{SOCl}_2$ to form acyl chlorides ( $-\text{COCl}$ ), releasing $\text{SO}_2$ and $\text{HCl}$ . A dicarboxylic acid forms a diacyl chloride. Acyl chlorides react with alcohols (including diols) via nucleophilic addition-elimination to form esters. When a diol reacts with a diacyl chloride, both ends can react intramolecularly (forming a cyclic diester) or intermolecularly (forming a larger product).

Understanding the Question

F (phthalic acid, $\text{C}_8\text{H}_6\text{O}_4$ ) reacts with excess $\text{SOCl}_2$ to give G ( $\text{C}_8\text{H}_4\text{Cl}_2\text{O}_2$ ). G then reacts with ethane-1,2-diol to give J ( $\text{C}_{10}\text{H}_8\text{O}_4$ ) and K ( $\text{C}_{18}\text{H}_{12}\text{Cl}_2\text{O}_6$ ). Three structures must be drawn.

Approach

G: F has two $-\text{COOH}$ groups; replacing both $-\text{OH}$ with $-\text{Cl}$ gives two $-\text{COCl}$ groups. Formula check: $\text{C}_8\text{H}_4\text{Cl}_2\text{O}_2$ ✓ (loss of 2H and 2O, gain of 2Cl relative to F).
J: G + $\text{HOCH}_2\text{CH}_2\text{OH}$ → cyclic diester + 2HCl. Both $-\text{COCl}$ groups react with the two $-\text{OH}$ groups of the diol, forming an 8-membered ring fused to the benzene ring. Formula: $\text{C}_8\text{H}_4\text{Cl}_2\text{O}_2$ + $\text{C}_2\text{H}_6\text{O}_2$ - 2HCl = $\text{C}_{10}\text{H}_8\text{O}_4$ ✓.
K: Two molecules of G react with one molecule of diol, but only one $-\text{COCl}$ from each G reacts (the other remains). Formula: 2( $\text{C}_8\text{H}_4\text{Cl}_2\text{O}_2$ ) + $\text{C}_2\text{H}_6\text{O}_2$ - 2HCl = $\text{C}_{18}\text{H}_{12}\text{Cl}_2\text{O}_6$ ✓.

Step-by-Step Reasoning

Structure G:

Start with phthalic acid: benzene-1,2-dicarboxylic acid.
Replace both $-\text{OH}$ of the $-\text{COOH}$ groups with $-\text{Cl}$ .
Result: 1,2-benzenedicarbonyl dichloride (phthaloyl chloride).
Benzene ring with two adjacent $-\text{COCl}$ groups.

Structure J:

The diol's two $-\text{OH}$ groups each attack a $-\text{COCl}$ on the same G molecule.
This forms two ester bonds, creating a ring: $-\text{COO}-\text{CH}_2-\text{CH}_2-\text{OOC}-$ .
The ring is 8-membered (2 from each C=O carbon, 2 from each O, 2 from each CH2) fused to the benzene ring.
No Cl remains (both COCl groups reacted).

Structure K:

One $-\text{COCl}$ from each of two G molecules reacts with the two $-\text{OH}$ groups of one diol molecule.
Each G retains one unreacted $-\text{COCl}$ group.
The bridge between the two benzene rings is $-\text{COO}-\text{CH}_2-\text{CH}_2-\text{OOC}-$ .
Each benzene ring also has one remaining $-\text{COCl}$ adjacent to the ester linkage.

Key Takeaways

Excess $\text{SOCl}_2$ converts all $-\text{COOH}$ to $-\text{COCl}$ .
A diol + diacyl chloride can form a cyclic diester (intramolecular) or a bridged product (intermolecular).
Molecular formulae are the key to deducing structures: count atoms and verify against the proposed structure.
K is a mixture product where only partial reaction occurs (one COCl per G reacts).

Common Mistakes

Drawing G as a mono-acyl chloride (forgetting excess $\text{SOCl}_2$ converts both $-\text{COOH}$ groups).
Drawing J as a linear diester rather than a cyclic one (the formula $\text{C}_{10}\text{H}_8\text{O}_4$ with no Cl indicates both COCl groups reacted).
Forgetting that K retains two COCl groups (one on each ring), as shown by the two Cl atoms in the formula.
Drawing incorrect ring size for J (it is an 8-membered ring, not 6 or 7).

Things to Be Careful About

All structures must show the correct connectivity and the benzene ring with appropriate substitution pattern (1,2-relationship preserved throughout).
For J, the ester linkages must be drawn correctly: $-\text{COO}-$ (not $-\text{OOC}-$ in the wrong orientation relative to the ring).
Verify each structure's molecular formula matches what is given.

Techniques used

deduce structure from molecular formula after SOCl2 reactionidentify cyclic diester from diol reacting with diacyl chloriderecognise partial reaction product retaining one COCl group

Q8MediumHydrocarbonsHalogen Compounds

Bromine reacts with methylbenzene in the dark in the presence of a suitable catalyst to form HBr and compound L, $\text{C}_7\text{H}_7\text{Br}$ . L is one of three isomers that can form in this reaction.

(a)

6M

(i)

The mechanism for the reaction involves methylbenzene reacting with a $\text{Br}^+$ ion. This ion is produced when bromine reacts with the catalyst.

Complete the equation for the reaction of $\text{Br}_2$ with the catalyst.

\text{Br}_2 + \text{..............................} \rightarrow \text{Br}^+ + \text{..............................}

1M Similar

DifficultyEasy

Worked solution

Answer

\text{Br}_2 + \text{AlBr}_3 \rightarrow \text{Br}^+ + \text{AlBr}_4^-

Final answer

Br2 + AlBr3 -> Br+ + AlBr4–

Detailed explanation

Background Concept

In electrophilic aromatic substitution (EAS) reactions involving benzene or arenes, the aromatic ring is not nucleophilic enough to react with a non-polar halogen molecule like $\text{Br}_2$ alone. A Lewis acid catalyst (such as $\text{AlBr}_3$ , $\text{FeBr}_3$ , or $\text{AlCl}_3$ ) is required to polarise the halogen molecule and generate a stronger electrophile, typically the halogen cation ( $\text{Br}^+$ or $\text{Cl}^+$ ).

Understanding the Question

The question asks for the balanced equation showing how the bromine molecule reacts with the catalyst (aluminium bromide) to produce the bromonium ion ( $\text{Br}^+$ ) electrophile. The catalyst acts as a Lewis acid by accepting a lone pair from one of the bromine atoms.

Approach

Recall the standard mechanism for generating the electrophile in the bromination of arenes. The $\text{AlBr}_3$ accepts a lone pair from $\text{Br}_2$ , forming a complex that dissociates into $\text{Br}^+$ and the tetrabromoaluminate anion $\text{AlBr}_4^-$ .

Step-by-Step Reasoning

Reactants: $\text{Br}_2$ and $\text{AlBr}_3$ . The aluminium in $\text{AlBr}_3$ has an incomplete octet and acts as a Lewis acid.
Interaction: A lone pair from one bromine atom in $\text{Br}_2$ forms a dative bond with the aluminium atom.
Dissociation: The Br-Br bond breaks heterolytically. One bromine takes the bonding electrons and becomes part of the $\text{AlBr}_4^-$ ion, while the other bromine leaves as the $\text{Br}^+$ electrophile.
Equation: $\text{Br}_2 + \text{AlBr}_3 \rightarrow \text{Br}^+ + \text{AlBr}_4^-$ .

Key Takeaways

Halogenation of arenes requires a Lewis acid catalyst ( $\text{AlX}_3$ or $\text{FeX}_3$ ) to generate the electrophile $\text{X}^+$ .
The byproduct is always the tetrahalometallate(III) anion, e.g., $\text{AlCl}_4^-$ or $\text{AlBr}_4^-$ .

Common Mistakes

Writing $\text{Fe}$ or $\text{FeBr}_3$ instead of $\text{AlBr}_3$ (the question implies $\text{AlBr}_3$ by the product $\text{AlBr}_4^-$ usually, though $\text{FeBr}_3$ is also acceptable in some contexts, but $\text{AlBr}_3$ is the standard match for $\text{AlBr}_4^-$ ). Actually, the mark scheme specifies $\text{AlBr}_3$ .
Forgetting the negative charge on the $\text{AlBr}_4^-$ ion.
Unbalanced equation.

Things to Be Careful About

Ensure the equation is balanced in terms of atoms and charge.
The catalyst is often written as $\text{AlBr}_3$ or $\text{FeBr}_3$ ; here $\text{AlBr}_3$ is required to form $\text{AlBr}_4^-$ .

Techniques used

identify the catalyst for electrophilic aromatic substitutionwrite the generation of the electrophile

(ii)

One of the isomers of L forms in much smaller amounts than the other two isomers.

Draw the structure of this isomer and explain why it forms in the smallest amount.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Structure:
1-bromo-3-methylbenzene (meta-bromotoluene).

Explanation:
The methyl group is an activating group and directs incoming electrophiles to the 2, 4, and 6 positions (ortho and para positions). Therefore, substitution at the 3 position (meta) is the least favourable, resulting in the smallest amount of product.

(Note: The structure is a benzene ring with a $\text{CH}_3$ group at position 1 and a $\text{Br}$ atom at position 3.)

Final answer

See structure: 1-bromo-3-methylbenzene. Explanation: methyl group is 2,4,6-directing.

Detailed explanation

Background Concept

In electrophilic aromatic substitution, substituents already on the benzene ring influence the position of the new substituent. Alkyl groups like methyl ( $-\text{CH}_3$ ) are electron-donating by induction. They activate the ring and direct incoming electrophiles to the ortho (2, 6) and para (4) positions. This is because the intermediate carbocation formed by attack at these positions is more stable (has more resonance structures stabilising the positive charge) than the intermediate formed by attack at the meta position.

Understanding the Question

The question asks for the isomer of $\text{C}_7\text{H}_7\text{Br}$ that forms in the smallest amount. Since methyl is an ortho/para director, the major products are 2-bromotoluene (ortho) and 4-bromotoluene (para). The minor product is 3-bromotoluene (meta). We need to draw this structure and explain why it is minor.

Approach

Identify the directing effect of the methyl group (ortho/para or 2,4,6 directing).
Determine that the meta position (3) is the least favoured.
Draw 1-bromo-3-methylbenzene.
Explain that the methyl group directs to 2, 4, 6, making 3 the minor product.

Step-by-Step Reasoning

Isomer identification: The three isomers are 2-bromo (ortho), 3-bromo (meta), and 4-bromo (para). Methyl is an ortho/para director. Thus, 2- and 4-bromotoluene are major, and 3-bromotoluene is minor.
Structure: Draw a benzene ring. Put $\text{CH}_3$ at the top (position 1). Put $\text{Br}$ at position 3 (meta to methyl).
Explanation: The methyl group donates electron density into the ring (inductive effect), activating it. It stabilises the arenium ion intermediate for ortho and para attack more than for meta attack. Therefore, the methyl group is described as a 2,4,6-directing group. Attack at the 3-position is least likely, so this isomer forms in the smallest amount.

Key Takeaways

Alkyl groups are ortho/para directors (2,4,6-directing).
Meta products are minor in electrophilic substitution of alkylbenzenes.

Common Mistakes

Drawing the ortho or para isomer as the minor product.
Saying "methyl is a meta director" (incorrect).
Not drawing the structure correctly (e.g., wrong positions of substituents).

Things to Be Careful About

The question asks for the isomer in the smallest amount (meta), not the major one.
Ensure the explanation links the directing effect to the specific positions (2, 4, 6 vs 3).

Techniques used

identify the minor isomer in electrophilic substitutionexplain directing effects of alkyl groups

(iii)

Complete the mechanism in Fig. 8.1 for the reaction between methylbenzene and the $\text{Br}^+$ ion.

Include all relevant curly arrows and charges.

Fig. 8.1

3M Similar

DifficultyMedium

Worked solution

Answer

Step 1: A curly arrow starts from the centre of the benzene ring (the delocalised $\pi$ system) and points to the $\text{Br}^+$ ion.

Step 2 (Intermediate): Draw a cyclohexadienyl cation (arenium ion). The ring has a $\text{CH}_3$ group at position 1. At position 2 (ortho), there is both a $\text{Br}$ atom and an $\text{H}$ atom attached to the same carbon (which is now $sp^3$ hybridised). The remaining five carbons form a delocalised system with a positive charge (drawn as a $+$ inside the ring or with a horseshoe arrow indicating delocalisation).

Step 3: A curly arrow starts from the $\text{C-H}$ bond at position 2 and points into the ring (towards the delocalised system).

Final Products: 2-bromotoluene (1-bromo-2-methylbenzene) and $\text{H}^+$ .

(Diagram reference: See Fig. 8.2 in marking scheme for exact arrow placement and intermediate structure.)

Final answer

See mechanism: arrow from ring to Br+, arenium ion intermediate, arrow from C-H bond to ring, products 2-bromotoluene + H+.

Detailed explanation

Background Concept

The mechanism of electrophilic aromatic substitution involves two main steps:

Electrophilic attack: The electrophile ( $\text{Br}^+$ ) attacks the electron-rich delocalised $\pi$ system of the benzene ring, forming a sigma complex (arenium ion). This breaks the aromaticity. The carbon attacked becomes $sp^3$ hybridised and bears both the new substituent and the original hydrogen.
Deprotonation (restoration of aromaticity): A base (often the $\text{AlBr}_4^-$ ion, though here simplified to loss of $\text{H}^+$ ) removes the proton from the $sp^3$ carbon. The electrons from the C-H bond return to the ring to restore the delocalised $\pi$ system.

Understanding the Question

The question provides a template (Fig 8.1) with methylbenzene + $\text{Br}^+$ going to an intermediate box, then to products. We must fill in the curly arrows, the intermediate structure, and the final products. The marking scheme diagram (Fig 8.2) shows attack at the ortho position.

Approach

Draw the first arrow: from the $\pi$ electrons to the electrophile.
Draw the intermediate: $sp^3$ carbon with Br and H, positive charge delocalised over the ring.
Draw the second arrow: from the C-H bond to the ring to restore aromaticity.
Draw the final organic product and the $\text{H}^+$ ion.

Step-by-Step Reasoning

M1 (First arrow): A curly arrow must originate from the delocalised $\pi$ system (inside the hexagon) and point to the $\text{Br}^+$ . Do not start from a specific double bond if a circle is used, or from the center of the ring.
Intermediate: The carbon attacked (let's say C2, ortho to methyl) now has 4 single bonds: to C1, C3, H, and Br. The ring is no longer fully aromatic. Draw a positive charge delocalised over the remaining 5 carbons (often drawn as a $+$ in the ring with a curved arrow showing delocalisation, or a circle with a $+$ ).
M3 (Second arrow and products): A curly arrow starts from the bond between C2 and H and points into the ring (towards the delocalised system). This restores the double bond/delocalisation. The products are the substituted arene (2-bromotoluene) and a proton ( $\text{H}^+$ ).

Key Takeaways

Electrophilic attack breaks aromaticity, forming a carbocation intermediate.
Loss of a proton restores aromaticity.
Curly arrows must show the movement of electron pairs.

Common Mistakes

Starting the first arrow from a specific $\text{C=C}$ bond instead of the delocalised system (though often allowed if consistent, the centre is safer).
Forgetting the positive charge on the intermediate.
Drawing the intermediate with the charge on the carbon that has Br and H attached (it should be delocalised in the ring).
Forgetting the $\text{H}^+$ in the final products.

Things to Be Careful About

The intermediate is an $sp^3$ carbon at the site of attack; ensure Br and H are both shown on that carbon.
The final organic product must show the restored aromatic ring (circle or alternating double bonds).

Techniques used

draw curly arrows for electrophilic aromatic substitutiondraw the arenium ion intermediate

(b)

Chlorobutane and chlorobenzene are added separately to samples of warm aqueous $\text{AgNO}_3$ .

One of the chloro-compounds reacts slowly and the other does not react.

5M

(i)

Identify the chloro-compound that reacts and describe any observations.

1M Similar

DifficultyEasy

Worked solution

Answer

Chloro-compound that reacts: chlorobutane ( $\text{C}_4\text{H}_9\text{Cl}$ ).
Observation: a white precipitate forms.

Final answer

chlorobutane; white precipitate

Detailed explanation

Background Concept

Halogenoalkanes (alkyl halides) undergo nucleophilic substitution reactions. When warmed with aqueous silver nitrate ( $\text{AgNO}_3$ ) in the presence of ethanol (usually, though the question says warm aqueous), the halogen is hydrolysed to form a halide ion ( $\text{X}^-$ ). The halide ion then reacts with silver ions to form an insoluble silver halide precipitate. The colour of the precipitate depends on the halide: $\text{AgCl}$ is white, $\text{AgBr}$ is cream, $\text{AgI}$ is yellow.

Halogenoarenes (aryl halides) like chlorobenzene do not undergo nucleophilic substitution under these conditions because the lone pair on the chlorine is delocalised into the benzene ring, strengthening the C-Cl bond.

Understanding the Question

We have chlorobutane (a halogenoalkane) and chlorobenzene (a halogenoarene). We add them separately to warm aqueous $\text{AgNO}_3$ . We need to identify which reacts and describe the observation.

Approach

Recall that halogenoalkanes react to release halide ions, while halogenoarenes do not.
Chlorobutane will react, releasing $\text{Cl}^-$ .
$\text{Cl}^-$ reacts with $\text{Ag}^+$ to form $\text{AgCl}$ , a white precipitate.

Step-by-Step Reasoning

Reactivity: Chlorobutane is a halogenoalkane. The C-Cl bond is polar and the carbon is electrophilic. Water (or $\text{OH}^-$ ) can act as a nucleophile. Chlorobenzene has a C-Cl bond that is strengthened by resonance with the ring, so it does not react.
Observation: The reaction of chlorobutane produces chloride ions. $\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})$ . Silver chloride is a white precipitate.

Key Takeaways

Halogenoalkanes react with aqueous $\text{AgNO}_3$ to give a precipitate.
Halogenoarenes do not react under these conditions.
$\text{AgCl}$ is a white precipitate.

Common Mistakes

Saying chlorobenzene reacts.
Describing the precipitate as cream or yellow (that's for Br or I).
Forgetting to mention the precipitate is white.

Things to Be Careful About

The question specifies "warm aqueous $\text{AgNO}_3$ ". Usually ethanol is added to dissolve the organic compound, but the key observation is the white precipitate.

Techniques used

predict reactivity of halogeno compounds with AgNO3identify precipitate from halide ions

(ii)

Write two equations to explain any observations in (b)(i).

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Equation 1 (Hydrolysis):

\text{C}_4\text{H}_9\text{Cl} + \text{H}_2\text{O} \rightarrow \text{C}_4\text{H}_9\text{OH} + \text{H}^+ + \text{Cl}^-

(Alternatively: $\text{C}_4\text{H}_9\text{Cl} + \text{OH}^- \rightarrow \text{C}_4\text{H}_9\text{OH} + \text{Cl}^-$ )

Equation 2 (Precipitation):

\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}

Final answer

C4H9Cl + H2O -> C4H9OH + H+ + Cl-; Ag+ + Cl- -> AgCl

Detailed explanation

Background Concept

The reaction of a halogenoalkane with water (hydrolysis) is a nucleophilic substitution ( $S_N1$ or $S_N2$ ). Water attacks the electron-deficient carbon, displacing the chloride ion. The chloride ion is then free to react with silver ions.

Understanding the Question

We need two equations to explain the observation in (b)(i): the formation of the white precipitate from chlorobutane.

Approach

Write the hydrolysis of chlorobutane to show the release of $\text{Cl}^-$ .
Write the ionic equation for the formation of silver chloride from silver ions and chloride ions.

Step-by-Step Reasoning

Hydrolysis: Chlorobutane reacts with water. The C-Cl bond breaks. Products are butanol, hydrogen ions, and chloride ions. $\text{C}_4\text{H}_9\text{Cl} + \text{H}_2\text{O} \rightarrow \text{C}_4\text{H}_9\text{OH} + \text{H}^+ + \text{Cl}^-$ . (Note: using $\text{OH}^-$ is also acceptable if the solution is basic, but aqueous $\text{AgNO}_3$ is slightly acidic/neutral, so water hydrolysis is standard).
Precipitation: The released $\text{Cl}^-$ reacts with $\text{Ag}^+$ from the silver nitrate. $\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})$ .

Key Takeaways

Hydrolysis releases halide ions.
Silver ions precipitate halide ions as insoluble silver halides.

Common Mistakes

Writing the hydrolysis equation with $\text{NaOH}$ instead of water (the question says aqueous $\text{AgNO}_3$ , so water is the nucleophile primarily, though $\text{OH}^-$ can be used).
Forgetting to balance the hydrolysis equation (producing $\text{H}^+$ ).
Not writing the ionic equation for precipitation (writing molecular equation like $\text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3$ is often not accepted; ionic is preferred).

Things to Be Careful About

The mark scheme accepts $\text{C}_4\text{H}_9\text{Cl} + \text{H}_2\text{O} \rightarrow \text{C}_4\text{H}_9\text{OH} + \text{H}^+ + \text{Cl}^-$ .
The precipitation equation is $\text{Cl}^- + \text{Ag}^+ \rightarrow \text{AgCl}$ .

Techniques used

write hydrolysis equation for halogenoalkanewrite precipitation equation for silver halide

(iii)

Explain the difference in reactivity of chlorobutane and chlorobenzene with warm aqueous $\text{AgNO}_3$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

In chlorobenzene, the lone pair on the chlorine atom is delocalised into the $\pi$ system of the benzene ring. This delocalisation gives the C–Cl bond partial double-bond character, making it stronger and shorter than the C–Cl bond in chlorobutane. Therefore, chlorobenzene does not undergo nucleophilic substitution under these conditions.

In chlorobutane, the C–Cl bond is a pure single bond and is more easily broken (hydrolysed) by nucleophiles like water.

Final answer

Lone pair on Cl delocalised into ring in chlorobenzene; strengthens C-Cl bond (partial double bond character).

Detailed explanation

Background Concept

In halogenoarenes (aryl halides), the halogen atom is attached directly to an $sp^2$ hybridised carbon of the benzene ring. The lone pair on the halogen can overlap with the delocalised $\pi$ system of the ring. This is a resonance effect (specifically, $p\pi-\pi$ conjugation).

Understanding the Question

Explain why chlorobutane reacts (slowly) but chlorobenzene does not react with warm aqueous $\text{AgNO}_3$ .

Approach

Compare the nature of the C-Cl bond in both compounds.
In chlorobutane, it's a polar single bond.
In chlorobenzene, resonance delocalisation strengthens the bond.

Step-by-Step Reasoning

Chlorobenzene: The lone pair on the chlorine atom is delocalised into the benzene ring's $\pi$ system. This creates partial double-bond character between the carbon and chlorine.
Bond strength: The partial double bond is stronger and shorter than a pure single bond. It requires more energy to break.
Chlorobutane: The C-Cl bond is a standard polar single bond. It is easier for a nucleophile (water) to attack the carbon and displace the chloride ion.
Conclusion: The delocalisation in chlorobenzene prevents the reaction (hydrolysis) that occurs in chlorobutane.

Key Takeaways

Lone pair delocalisation into the ring strengthens the C-X bond in aryl halides.
This makes aryl halides unreactive towards nucleophilic substitution under mild conditions.

Common Mistakes

Saying "the ring is stable" (vague; must mention delocalisation of the lone pair).
Saying "the bond is ionic" (it's covalent).
Not mentioning partial double bond character or bond strength.

Things to Be Careful About

Use precise terminology: "delocalisation of the lone pair", " $\pi$ system", "partial double bond character".

Techniques used

explain resonance effect in halogenoarenescompare bond strengths

Q9MediumNitrogen Compounds

(a)

Phenylamine, $\text{C}_6\text{H}_5\text{NH}_2$ , and propylamine, $\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$ , can be produced by different reduction reactions.

6M

(i)

Identify an organic compound that can be converted into $\text{C}_6\text{H}_5\text{NH}_2$ by a reduction reaction. State the reagents and conditions for this reaction.

organic compound .............................................................................................................

reagents ............................................................................................................................

conditions ..........................................................................................................................

2M Similar

DifficultyEasy

Worked solution

Answer

organic compound: nitrobenzene / $C_6H_5NO_2$
reagents: Sn and HCl
conditions: concentrated HCl and heat / boil / reflux

Final answer

nitrobenzene; Sn and HCl; concentrated HCl and heat/reflux

Detailed explanation

Background Concept

Aromatic amines like phenylamine (aniline) are typically prepared by the reduction of nitro compounds. The nitro group ( $-NO_2$ ) attached to an aromatic ring is reduced to an amino group ( $-NH_2$ ). This is a fundamental transformation in organic synthesis for introducing the amine functionality onto an aromatic ring.

Understanding the Question

The question asks for the organic precursor to phenylamine ( $C_6H_5NH_2$ ) via a reduction reaction, along with the specific reagents and conditions required for this transformation.

Approach

Recall the standard laboratory method for reducing a nitroarene to an arylamine. This involves a metal-acid reduction using tin and concentrated hydrochloric acid under heating.

Step-by-Step Reasoning

The functional group in phenylamine is $-NH_2$ attached to a benzene ring. The direct precursor with the same carbon skeleton and a reducible group is nitrobenzene ( $C_6H_5NO_2$ ).
The reduction of nitrobenzene to phenylamine is classically carried out using tin (Sn) and concentrated hydrochloric acid (HCl).
The reaction requires heating (boiling or refluxing) to proceed at a reasonable rate. The initial product is the phenylammonium ion ( $C_6H_5NH_3^+$ ), and a subsequent addition of base (like NaOH) is needed to liberate the free amine, though the mark scheme only asks for the reduction reagents and conditions.

Key Takeaways

Nitroarenes are reduced to arylamines using Sn/HCl and heat. This is a fundamental preparation method in organic chemistry.

Common Mistakes

Forgetting the need for concentrated acid and heat; dilute acid or room temperature will not drive the reaction to completion.
Writing 'H2 and Ni/Pt catalyst' which is used for aliphatic nitriles/amides but is not the standard method taught for nitroarenes in this syllabus.

Things to Be Careful About

Ensure the organic compound is correctly named or formulated as nitrobenzene / $C_6H_5NO_2$ .
State 'Sn and HCl' for reagents and 'concentrated HCl and heat' for conditions; partial credit is given for two correct points out of three.

Techniques used

identify reduction precursor for aromatic aminestate reagents and conditions for nitro reduction

(ii)

Identify an organic compound that can be converted into $\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$ by a reduction reaction. State the reagent for this reaction.

organic compound .............................................................................................................

reagent ..............................................................................................................................

2M Similar

DifficultyEasy

Worked solution

Answer

organic compound: propanenitrile / $C_2H_5CN$ (or propylamide / $C_2H_5CONH_2$ )
reagent: $LiAlH_4$

Final answer

propanenitrile ( $C_2H_5CN$ ); $LiAlH_4$

Detailed explanation

Background Concept

Aliphatic amines like propylamine can be prepared by the reduction of nitriles or amides. Nitriles ( $R-C\equiv N$ ) are reduced to primary amines ( $R-CH_2NH_2$ ) using a strong reducing agent. Amides ( $R-CONH_2$ ) can also be reduced to amines ( $R-CH_2NH_2$ ) using the same reagent.

Understanding the Question

The question asks for an organic precursor to propylamine ( $CH_3CH_2CH_2NH_2$ ) via reduction, and the specific reagent required.

Approach

Identify a compound with the same carbon skeleton that contains a reducible nitrogen functional group (nitrile or amide), then state the appropriate reducing agent.

Step-by-Step Reasoning

Propylamine has a three-carbon chain. The corresponding nitrile is propanenitrile ( $CH_3CH_2CN$ or $C_2H_5CN$ ). The corresponding amide is propanamide ( $CH_3CH_2CONH_2$ or $C_2H_5CONH_2$ ).
The reduction of nitriles or amides to amines requires a strong reducing agent. Lithium aluminium hydride ( $LiAlH_4$ ) is the standard reagent for this transformation.
Note that $LiAlH_4$ is typically used in dry ether, followed by acid workup, but the mark scheme only requires the reagent name.

Key Takeaways

Nitriles and amides are reduced to primary amines using $LiAlH_4$ . This is a key method for extending the carbon chain by one carbon when starting from a haloalkane (via nucleophilic substitution with $CN^-$ ) followed by reduction.

Common Mistakes

Suggesting $H_2$ with a Ni or Pt catalyst. While this can reduce nitriles in some contexts, $LiAlH_4$ is the specific reagent expected in this syllabus for nitrile/amide reduction to amines.
Writing the wrong carbon chain length (e.g., butanenitrile).

Things to Be Careful About

Ensure the organic compound has exactly three carbons: propanenitrile ( $C_2H_5CN$ ) or propanamide ( $C_2H_5CONH_2$ ).
State $LiAlH_4$ as the reagent; do not include solvent or workup unless asked.

Techniques used

identify reduction precursor for aliphatic aminestate reducing agent for nitrile or amide

(iii)

Identify a single test that will distinguish between $\text{C}_6\text{H}_5\text{NH}_2$ and $\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$ by producing a white precipitate with only one of these amines.

Draw the structure of the compound that is precipitated.

testing reagent ..................................................................................................................

amine that gives a precipitate ...........................................................................................

structure of the compound that is precipitated:

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

testing reagent: bromine (aq)
amine that gives a precipitate: phenylamine / $C_6H_5NH_2$
structure of the compound that is precipitated:

Final answer

bromine (aq); phenylamine; 2,4,6-tribromophenylamine structure

Detailed explanation

Background Concept

The amino group ( $-NH_2$ ) in phenylamine is a strongly activating group that donates electron density into the benzene ring via resonance (delocalisation of the nitrogen lone pair into the $\pi$ -system). This makes the ring highly susceptible to electrophilic substitution, much more so than benzene itself. When phenylamine reacts with bromine water at room temperature, it undergoes immediate tri-substitution at the 2, 4, and 6 positions, forming a white precipitate of 2,4,6-tribromophenylamine.

Aliphatic amines like propylamine do not have an aromatic ring and do not undergo this electrophilic aromatic substitution reaction with bromine water.

Understanding the Question

The question asks for a single test to distinguish between phenylamine (an aromatic amine) and propylamine (an aliphatic amine) that produces a white precipitate with only one of them. It also requires drawing the structure of the precipitated compound.

Approach

Use the difference in reactivity between the aromatic ring in phenylamine and the aliphatic chain in propylamine. Bromine water is a standard test for activated aromatic rings.

Step-by-Step Reasoning

Bromine water ( $Br_2(aq)$ ) is added to both amines.
Phenylamine reacts rapidly with bromine water at room temperature without a catalyst. The $-NH_2$ group activates the ring, causing substitution at all ortho and para positions (2, 4, and 6), forming 2,4,6-tribromophenylamine as a white precipitate.
Propylamine does not react with bromine water to form a precipitate (it may undergo slow oxidation or acid-base reaction, but no white precipitate forms).
The structure of 2,4,6-tribromophenylamine is a benzene ring with an $-NH_2$ group at position 1 and bromine atoms at positions 2, 4, and 6.

Key Takeaways

Phenylamine is so reactive towards electrophilic substitution that it decolourises bromine water and forms a white precipitate of 2,4,6-tribromophenylamine without a catalyst. This is a characteristic test for activated aromatic rings.

Common Mistakes

Suggesting $Br_2$ in $CH_2Cl_2$ or $Br_2$ vapour: these conditions give mono-substitution or no reaction without a catalyst, not the precipitate.
Forgetting to draw all three bromine atoms at the 2, 4, and 6 positions; mono- or di-bromination is incorrect for this reaction.
Drawing the precipitate from propylamine (which doesn't form one).

Things to Be Careful About

Specify 'bromine (aq)' or 'bromine water'; elemental bromine alone may not give the same immediate precipitate.
The structure must clearly show the benzene ring (with circle or alternating double bonds) and the correct positions of the substituents: $-NH_2$ at top, $-Br$ at 2, 4, 6.

Techniques used

identify electrophilic substitution test for activated arenedraw substitution product on activated ring

(b)

Describe the relative basicities of $\text{C}_6\text{H}_5\text{NH}_2$ , $\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$ and $\text{NH}_3$ .

Explain your answer.

\text{..............................................} < \text{..............................................} < \text{..............................................}

\text{least basic} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{most basic}

4M Similar

DifficultyMedium

Worked solution

Answer

phenylamine < ammonia < propylamine
least basic ........................................................... most basic

Explanation:

A base has a lone pair on the nitrogen atom that can accept a proton / $H^+$ .
In phenylamine, the lone pair on the nitrogen atom is delocalised into the benzene ring / $\pi$ -system, making it less available to accept a proton.
In propylamine, the electron-donating alkyl group (propyl group) increases the electron density on the nitrogen atom, making the lone pair more available to accept a proton.

Final answer

phenylamine < ammonia < propylamine

Detailed explanation

Background Concept

The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton ( $H^+$ ). Factors that affect this availability include:

Delocalisation: If the lone pair is delocalised into a $\pi$ -system (like a benzene ring), it is less available to accept a proton, decreasing basicity.
Inductive effect: Electron-donating groups (like alkyl groups) push electron density towards the nitrogen atom, increasing the electron density on the lone pair and making it more available to accept a proton, increasing basicity.
Solvation effects: In aqueous solution, the stability of the conjugate acid (ammonium ion) also plays a role, but for this comparison, electronic effects dominate.

Understanding the Question

The question asks to order phenylamine, ammonia, and propylamine in order of increasing basicity and explain the reasoning based on their structures.

Approach

Compare the availability of the nitrogen lone pair in each compound:

Phenylamine: lone pair delocalised into the ring.
Ammonia: no delocalisation, no alkyl groups (baseline).
Propylamine: lone pair enhanced by electron-donating alkyl group.

Step-by-Step Reasoning

Order: phenylamine < ammonia < propylamine.
Ammonia vs. Propylamine: In propylamine ( $CH_3CH_2CH_2NH_2$ ), the propyl group is an electron-donating alkyl group. Through the +I (positive inductive) effect, it pushes electron density towards the nitrogen atom. This increases the electron density on the lone pair, making it more available to accept a proton compared to ammonia ( $NH_3$ ). Thus, propylamine is more basic than ammonia.
Ammonia vs. Phenylamine: In phenylamine ( $C_6H_5NH_2$ ), the lone pair on the nitrogen atom is delocalised into the $\pi$ -system of the benzene ring. This resonance delocalisation means the lone pair is shared with the ring and is less available to accept a proton. Therefore, phenylamine is less basic than ammonia.
Combined order: Since phenylamine is less basic than ammonia and propylamine is more basic than ammonia, the order is phenylamine < ammonia < propylamine.

Key Takeaways

Alkyl groups increase basicity via the +I inductive effect.
Delocalisation of the lone pair (e.g., into an aromatic ring) decreases basicity.
Ammonia serves as a baseline: aliphatic amines are more basic, aromatic amines are less basic.

Common Mistakes

Stating that phenylamine is more basic because the ring is 'electron-donating': the ring actually accepts electron density from the lone pair via resonance, making the lone pair less available.
Forgetting to mention the lone pair's role in accepting a proton; basicity is defined by proton acceptance.
Confusing the inductive effect of alkyl groups with the delocalisation effect in phenylamine.

Things to Be Careful About

Use precise terminology: 'delocalised into the benzene ring / $\pi$ -system' and 'electron-donating alkyl group'.
Ensure the order is correctly placed in the inequality: least basic on the left, most basic on the right.
Explain why each amine is more or less basic than ammonia; simply stating the order without explanation scores no marks.

Techniques used

compare basicity using electronic effectsexplain lone pair availability in amines

Q10MediumCarboxylic Acids and Derivatives

Propanoic acid, methanoic acid and ethanedioic acid are all weak acids.

(a)

Draw the displayed formula of ethanedioic acid.

1M Similar

DifficultyEasy

Worked solution

Answer

Final answer

Displayed formula of ethanedioic acid: HOOC–COOH with all atoms and bonds shown explicitly (see diagram).

Detailed explanation

Background Concept

A displayed (or full structural) formula shows every atom and every bond in a molecule. For carboxylic acids, the functional group is –COOH, which contains a carbonyl group (C=O) and a hydroxyl group (–OH) attached to the same carbon. Ethanedioic acid (also called oxalic acid) is the simplest dicarboxylic acid, with two carboxyl groups bonded directly to each other: HOOC–COOH.

Understanding the Question

The question asks for the displayed formula of ethanedioic acid. This means every atom must be drawn and every bond (single, double) must be shown as a line. No condensed or skeletal formula is acceptable.

Approach

Draw two carbon atoms joined by a single bond. Attach a C=O double bond and a C–OH single bond to each carbon. Ensure all hydrogen atoms in the hydroxyl groups are shown.

Step-by-Step Reasoning

Ethanedioic acid has the molecular formula C₂H₂O₄ and the structural formula HOOC–COOH.
Place two carbon atoms side by side, connected by a single bond (C–C).
On each carbon, draw a double bond to an oxygen atom (C=O).
On each carbon, draw a single bond to an oxygen atom of a hydroxyl group (C–O–H).
All atoms (C, O, H) and all bonds (one C–C, two C=O, two C–O, two O–H) must be visible.

Key Takeaways

Displayed formulas must show every atom and every bond. For dicarboxylic acids like ethanedioic acid, the two –COOH groups are directly bonded to each other with no carbon chain between them.

Common Mistakes

Drawing a skeletal formula or condensed formula instead of a displayed formula. Mark schemes reject these.
Forgetting the hydrogen atoms on the hydroxyl groups.
Drawing the two carboxyl groups separated by a carbon chain (that would be a different acid).

Things to Be Careful About

The question says "displayed formula" — this is a specific term meaning all atoms and all bonds must be shown. A structural formula (HOOC–COOH) or a skeletal formula would not score. Check that there are exactly 2 C, 2 O (double-bonded), 2 O (single-bonded in –OH), and 2 H atoms, with the correct bond types.

Techniques used

draw displayed formulashow all atoms and all bonds

(b)

The three acids, propanoic acid, methanoic acid and ethanedioic acid, can be distinguished using a combination of two chemical tests. Neither testing reagent is an acid–base indicator.

Identify two suitable testing reagents and complete Table 10.1 to show the observations from each test.

reagent 1 ........................................................ reagent 2 ........................................................

Table 10.1

	observation when treated with reagent 1	observation when treated with reagent 2
propanoic acid
methanoic acid
ethanedioic acid

4M Similar

DifficultyMedium

Worked solution

Answer

reagent 1: Tollens' reagent (or Fehling's solution)

reagent 2: acidified potassium manganate(VII) (KMnO₄ / MnO₄⁻)

Table 10.1

	observation when treated with reagent 1 (Tollens')	observation when treated with reagent 2 (acidified MnO₄⁻)
propanoic acid	no change	no change
methanoic acid	silver mirror (or red precipitate)	colourless (decolourises)
ethanedioic acid	no change	colourless (decolourises)

Note: If Fehling's solution is used as reagent 1, the observation for methanoic acid is a red precipitate.

Final answer

reagent 1: Tollens' reagent (or Fehling's solution); reagent 2: acidified KMnO₄. Observations: propanoic acid — no change / no change; methanoic acid — silver mirror (or red ppt) / colourless; ethanedioic acid — no change / colourless.

Detailed explanation

Background Concept

Most carboxylic acids are not easily oxidised. However, two exceptions are well known:

Methanoic acid (HCOOH) has a hydrogen atom directly attached to the carbonyl carbon, giving it an aldehyde-like structure (H–C(=O)–OH). This allows it to act as a reducing agent, reducing Tollens' reagent (ammoniacal silver nitrate) to give a silver mirror, and Fehling's solution (Cu²⁺) to give a red precipitate of Cu₂O. It also reduces acidified MnO₄⁻ (purple) to Mn²⁺ (colourless).
Ethanedioic acid (HOOC–COOH) can be oxidised to carbon dioxide and water. It reduces acidified MnO₄⁻ (purple to colourless) but does NOT reduce Tollens' reagent or Fehling's solution.

Propanoic acid (CH₃CH₂COOH) has neither of these features and is not oxidised by any of these reagents under normal conditions.

Understanding the Question

We need two chemical tests (neither an acid-base indicator) that together can distinguish all three acids. The test must give a different observation for at least some of the acids.

Approach

Use Tollens' reagent (or Fehling's) to identify methanoic acid (the only one that gives a positive test).
Use acidified KMnO₄ to distinguish ethanedioic acid from propanoic acid (both methanoic and ethanedioic decolourise it, propanoic does not).
Together, these two tests give a unique observation pattern for each acid.

Step-by-Step Reasoning

Reagent 1 — Tollens' reagent (or Fehling's solution):

Propanoic acid: no reaction → no change.
Methanoic acid: oxidised to CO₂; Ag⁺ reduced to Ag → silver mirror (Tollens') or Cu₂O red precipitate (Fehling's).
Ethanedioic acid: no reaction → no change.

Reagent 2 — Acidified KMnO₄ (MnO₄⁻ / H⁺):

Propanoic acid: no reaction → purple solution remains (no change).
Methanoic acid: oxidised to CO₂; MnO₄⁻ reduced to Mn²⁺ → purple solution becomes colourless.
Ethanedioic acid: oxidised to CO₂; MnO₄⁻ reduced to Mn²⁺ → purple solution becomes colourless.

Combined observations:

Propanoic acid: no change / no change → uniquely identified.
Methanoic acid: silver mirror (or red ppt) / colourless → uniquely identified.
Ethanedioic acid: no change / colourless → uniquely identified.

All three acids give a distinct combination of observations.

Key Takeaways

Methanoic acid is the only carboxylic acid that acts as a reducing agent towards Tollens' and Fehling's reagents (due to its aldehyde-like H–COOH structure). Ethanedioic acid reduces acidified MnO₄⁻ but not Tollens'/Fehling's. Propanoic acid does neither. Combining these tests gives a unique fingerprint for each acid.

Common Mistakes

Suggesting an acid-base indicator (e.g., universal indicator, pH paper) — the question explicitly excludes these.
Using bromine water or 2,4-DNPH — these do not distinguish these acids appropriately.
Claiming ethanedioic acid gives a positive Tollens' test — it does not.
Forgetting that methanoic acid also decolourises acidified MnO₄⁻ (both methanoic and ethanedioic do).
Writing "decolourises" without specifying from what (purple MnO₄⁻ to colourless Mn²⁺).

Things to Be Careful About

The mark scheme accepts either Tollens' or Fehling's as reagent 1, but not both (they give redundant information for distinguishing all three).
"Colourless" for the MnO₄⁻ test means the purple colour disappears; write "decolourises" or "purple to colourless".
For Tollens', write "silver mirror"; for Fehling's, write "red precipitate". Do not mix the two.
Both columns must be completely correct to earn full marks (M3/M4 in the mark scheme). One correct column gives 1 mark; both give 2 marks.

Techniques used

identify reducing properties of methanoic acidrecognise ethanedioic acid as a reducing agent for MnO4-distinguish carboxylic acids by chemical tests

(c)

When propanoic acid is treated with chlorine gas in the presence of ultraviolet light, a mixture of products is formed. One of these products is 2,2-dichloropropanoic acid.

Explain why 2,2-dichloropropanoic acid is stronger than propanoic acid. Refer to the structure of each compound in your answer.

2M Similar

DifficultyMedium

Worked solution

Answer

The two chlorine atoms in 2,2-dichloropropanoic acid are electron-withdrawing (electronegative) and exert a positive inductive effect through the carbon chain.

This inductive effect:

weakens the O–H bond in the carboxyl group, making it easier to release H⁺; and/or
stabilises the conjugate base (the carboxylate ion, –COO⁻) by dispersing/delocalising the negative charge.

In propanoic acid (CH₃CH₂COOH), the alkyl group is electron-donating, which destabilises the conjugate base and makes the acid weaker.

Working

No calculation required — this is an explanation.

Answer (summary)

2,2-dichloropropanoic acid is stronger because the electronegative Cl atoms withdraw electron density via the inductive effect, weakening the O–H bond and/or stabilising the carboxylate conjugate base.

Final answer

The electronegative chlorine atoms withdraw electron density through the carbon chain (inductive effect), weakening the O–H bond and/or stabilising the conjugate base (carboxylate ion), making proton release easier.

Detailed explanation

Background Concept

The strength of a carboxylic acid depends on how readily it donates a proton (H⁺) from the –COOH group. This is governed by the stability of the resulting conjugate base (the carboxylate ion, R–COO⁻). If the conjugate base is stabilised, the equilibrium lies further to the right and the acid is stronger.

Substituents on the carbon chain can affect this through the inductive effect:

Electron-withdrawing groups (e.g., halogens like –Cl, –F) pull electron density away from the carboxylate group through the σ-bonds. This stabilises the negative charge on the carboxylate ion and weakens the O–H bond, making the acid stronger.
Electron-donating groups (e.g., alkyl groups like –CH₃, –CH₂CH₃) push electron density toward the carboxylate group, destabilising the negative charge and making the acid weaker.

The inductive effect operates through bonds and diminishes with distance. More electronegative atoms and more of them have a greater effect.

Understanding the Question

We are comparing propanoic acid (CH₃CH₂COOH) with 2,2-dichloropropanoic acid (CH₃CCl₂COOH). The latter has two chlorine atoms on carbon 2 (the carbon adjacent to the carboxyl group). We must explain why the dichloro derivative is the stronger acid, referring to the structure of each compound.

Approach

Identify the structural difference: two Cl atoms on C2 in 2,2-dichloropropanoic acid vs. two H atoms on C2 in propanoic acid.
State that Cl is electronegative / electron-withdrawing.
Explain the consequence: weakening of the O–H bond and/or stabilisation of the conjugate base.

Step-by-Step Reasoning

Structure comparison:

Propanoic acid: CH₃–CH₂–COOH. The ethyl group (CH₃CH₂–) is electron-donating by the inductive effect.
2,2-Dichloropropanoic acid: CH₃–CCl₂–COOH. The two Cl atoms on C2 are highly electronegative (Cl electronegativity ≈ 3.16 vs C ≈ 2.55).

Inductive effect:

The two Cl atoms pull electron density away from the carboxyl group through the σ-bond framework. This is a positive inductive effect (+I is electron-donating; here we describe the Cl effect as electron-withdrawing, sometimes called –I effect).
Note: In CIE terminology, halogens are described as "electron-withdrawing" or "electronegative". The mark scheme accepts "chlorine atoms are electron withdrawing / electronegative".

Effect on acid strength (two acceptable explanations, either or both):

Weakening of the O–H bond: The electron withdrawal reduces electron density on the oxygen of the O–H bond, making it more polar and easier to break, releasing H⁺ more readily.
Stabilisation of the conjugate base: The carboxylate ion (R–COO⁻) has a negative charge delocalised over two oxygen atoms. The electron-withdrawing Cl atoms help to分散 (disperse) this negative charge through the inductive effect, stabilising the ion. A more stable conjugate base means the acid more readily gives up H⁺.

Both explanations are acceptable and often go together. The mark scheme awards one mark for identifying the electron-withdrawing nature of Cl, and one mark for either weakening the O–H bond or stabilising the conjugate base.

Reference to structure:

The question requires referring to the structure of each compound. Mention that propanoic acid has an electron-donating alkyl group (or simply no electron-withdrawing groups), while 2,2-dichloropropanoic acid has two electronegative Cl atoms on the α-carbon.

Key Takeaways

Electron-withdrawing substituents (especially halogens on the α-carbon) increase carboxylic acid strength via the inductive effect. This can be explained either by weakening the O–H bond or by stabilising the conjugate base. Both explanations are chemically valid and often work together.

Common Mistakes

Saying "chlorine is more electronegative" without linking it to the inductive effect or its consequence on acid strength.
Not referring to the structure of both compounds (the question explicitly asks for this).
Saying the Cl atoms "attract electrons" without specifying through the σ-bonds (inductive effect through bonds, not through space).
Confusing the inductive effect with resonance (Cl does not participate in resonance with the carboxylate here; it's purely inductive through σ-bonds).
Writing "the Cl atoms make the acid stronger" without explaining why.
Saying "the conjugate base is more stable because of resonance" — the resonance is the same in both acids; the difference is the inductive stabilisation.

Things to Be Careful About

The mark scheme gives two marks: M1 for "chlorine atoms are electron withdrawing / electronegative" and M2 for "weakening O–H bond OR stabilising conjugate base". Either explanation in M2 is acceptable.
You must refer to the structure of each compound. Simply saying "Cl is electron-withdrawing" without comparing to propanoic acid may not earn full marks.
Use precise terminology: "inductive effect", "electron-withdrawing", "conjugate base", "O–H bond".
Do not say "Cl pulls electrons through the double bond" — the effect is through σ-bonds (the carbon chain).

Techniques used

explain inductive electron-withdrawing effectrelate to O-H bond weakening and conjugate base stabilisation

Chemistry 9701/41 — May/June 2025

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question