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) CaCO3 decomposes at a lower temperature because Ca2+ is smaller and polarises the carbonate ion more strongly, weakening the C–O bonds. · (c)(i) calcium hydroxide, Ca(OH)2 · (c)(ii) The difference in lattice enthalpy between Ca(OH)2 and Ba(OH)2 is greater than the difference in hydration enthalpy, making ΔHsol for Ca(OH)2 less exothermic, so Ca(OH)2 is less soluble.

(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. (i) Complete the expression for $K_{\text{sp}}$ of $\text{Cr(OH)}_3$. Include the units. $$ K_{\text{sp}} = $$ units = .............................. (ii) Calculate the solubility, in $\text{g dm}^{-3}$, of $\text{Cr(OH)}_3$ in pure water at 298 K. Show your working. solubility =…

(a)(i) Ksp = [Cr3+][OH-]^3; units = mol^4 dm^-12 · (a)(ii) 2.54 × 10^-7 g dm^-3 · (a)(iii) Common ion effect · (b)(i) pH = 2.91 · (b)(ii) pH = 13.0 · (b)(iii) pH = 4.82 · (c) Kpc = 3.01

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; no exchangeable protons in E · (b)(iii) δ=1.3: 6H, CH3, triplet; δ=2.7: 4H, CH2, quartet; δ=7.1: 4H · (b)(iv) D: 5 peaks; E: 6 peaks · (c)(i) C10H14 + 12[O] -> C8H6O4 + 2CO2 + 4H2O · (c)(ii) See structures: G = phthaloyl chloride, J = cyclic diester, K = dimeric diester with two remaining COCl groups

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

(a)(i) Br2 + AlBr3 -> Br+ + AlBr4- · (a)(ii) 3-bromo-1-methylbenzene (meta-isomer); methyl group is ortho/para directing · (a)(iii) See mechanism diagram · (b)(i) Chlorobutane; white precipitate · (b)(ii) C4H9Cl + H2O -> C4H9OH + H+ + Cl-; Ag+ + Cl- -> AgCl · (b)(iii) Delocalization of Cl lone pair into benzene ring strengthens C-Cl bond

(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 · (a)(ii) propanenitrile; LiAlH$_4$ · (a)(iii) bromine(aq); phenylamine; see diagram for 2,4,6-tribromophenylamine · (b) phenylamine < ammonia < propylamine

9701/43 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

Group 2 carbonates (MCO₃) decompose on heating 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 — a single reactant breaks down into two simpler products under the influence of heat. The pattern holds for all Group 2 carbonates, including CaCO₃ and BaCO₃, as stated in the question stem.

Understanding the Question

This part asks for a balanced chemical equation for the thermal decomposition of calcium carbonate. The stem already tells us the products are "the metal oxide and a gas" — so we need calcium oxide and carbon dioxide. The single mark is for a correct, balanced equation with state symbols.

Approach

Identify the reactant (CaCO₃) and the two products (CaO and CO₂). Write the skeleton equation, then check that it balances. Add state symbols: the carbonate and oxide are solids, the gas is carbon dioxide.

Step-by-Step Reasoning

The reactant is solid calcium carbonate, CaCO₃(s). On heating it decomposes to solid calcium oxide, CaO(s), and carbon dioxide gas, CO₂(g):
$\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})$
Check the atom balance: left side has 1 Ca, 1 C, 3 O; right side has 1 Ca, 1 C, 3 O. The equation balances as written — no coefficients are needed. State symbols: (s) for both solids, (g) for the gas.

Key Takeaways

Thermal decomposition of Group 2 carbonates always gives the metal oxide and CO₂. A balanced equation must account for every atom and include state symbols.

Common Mistakes

Omitting state symbols — the mark scheme expects (s) and (g).
Writing CO instead of CO₂ (the gas is carbon dioxide, not carbon monoxide).
Writing an unbalanced equation such as CaCO₃ → CaO + CO (oxygen does not balance).

Things to Be Careful About

The equation must be balanced and include state symbols. The products are CaO(s) and CO₂(g).

Techniques used

write a balanced thermal decomposition equationinclude state symbols

(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

CaCO₃ decomposes at a lower temperature.

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

Final answer

CaCO3 decomposes at a lower temperature because Ca2+ is smaller and polarises the carbonate ion more strongly, weakening the C–O bonds.

Detailed explanation

Background Concept

The thermal stability of Group 2 carbonates increases down the group. This is explained by the polarising power of the cation. The carbonate ion, CO₃²⁻, is large and has a diffuse electron cloud that can be distorted (polarised) by a nearby cation. A small, highly charged cation (high charge density) polarises the carbonate ion strongly, pulling electron density away from the C–O bonds and weakening them. A weaker C–O bond means the carbonate breaks down (decomposes) more easily, i.e. at a lower temperature.

Going down Group 2, the cation gets larger (Ca²⁺ → Sr²⁺ → Ba²⁺), its charge density decreases, its polarising power decreases, the carbonate ion is less distorted, the C–O bonds are stronger, and the carbonate is more thermally stable.

Understanding the Question

We must state which of CaCO₃ and BaCO₃ decomposes at a lower temperature, then explain why. The explanation must link the size (or charge density) of the cation to the distortion of the carbonate ion and hence to thermal stability. The mark scheme awards one mark for identifying CaCO₃ AND stating that Ca²⁺ is smaller (or that ionic radius increases down the group), and a second mark for stating that the carbonate ion is more distorted/polarised by Ca²⁺.

Approach

Compare the ionic radii of Ca²⁺ and Ba²⁺. The smaller cation (Ca²⁺) has the higher charge density and the greater polarising power. It distorts the carbonate ion more, weakening the C–O bonds, so CaCO₃ is less thermally stable and decomposes at a lower temperature.

Step-by-Step Reasoning

Ca²⁺ is smaller than Ba²⁺ — ionic radius increases down Group 2.
A smaller cation has a higher charge density (charge per unit volume).
Ca²⁺ therefore polarises (distorts) the large carbonate ion, CO₃²⁻, more strongly than Ba²⁺ does.
This distortion draws electron density away from the C–O bonds, weakening them.
Weaker C–O bonds mean the carbonate ion breaks apart more readily on heating.
Hence CaCO₃ decomposes at a lower temperature than BaCO₃.

Key Takeaways

Thermal stability of Group 2 carbonates increases down the group. The trend is controlled by the cation's polarising power: smaller cation → more polarisation → less stable carbonate → lower decomposition temperature.

Common Mistakes

Stating that BaCO₃ decomposes at the lower temperature (the direction is wrong).
Saying "the larger cation polarises more" — it is the smaller, more charge-dense cation that polarises more.
Giving only the observation without the explanation (e.g. "CaCO₃ is less stable" without mentioning polarisation of the carbonate ion).

Things to Be Careful About

The mark scheme awards two separate marks: (M1) identifying CaCO₃ AND stating Ca²⁺ is smaller / ionic radius increases down the group; (M2) stating the carbonate ion / CO₃²⁻ is more distorted/polarised by Ca²⁺. Both points must be present for full marks.

Techniques used

compare ionic radii down Group 2explain polarisation of the 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 ( $\text{Ca(OH)}_2$ ) has more exothermic lattice enthalpy and more exothermic hydration enthalpy than B ( $\text{Ba(OH)}_2$ ). However, the difference in lattice enthalpy between $\text{Ca(OH)}_2$ and $\text{Ba(OH)}_2$ is greater than the difference in hydration enthalpy. Therefore $\Delta H_{\text{sol}}$ for A is less exothermic than for B, so A is less soluble.

Final answer

The difference in lattice enthalpy between Ca(OH)2 and Ba(OH)2 is greater than the difference in hydration enthalpy, making ΔHsol for Ca(OH)2 less exothermic, so Ca(OH)2 is less soluble.

Detailed explanation

Background Concept

The solubility of an ionic compound in water is governed by the enthalpy change of solution, ΔHsol. Dissolving an ionic solid involves two energy terms:

Lattice enthalpy, ΔHlatt: the energy required to separate the ions in the solid lattice (endothermic, positive).
Hydration enthalpy, ΔHhyd: the energy released when the separated ions are surrounded by water molecules (exothermic, negative).

The enthalpy of solution is the sum:
$\Delta H_{\text{sol}} = \Delta H_{\text{latt}} + \Delta H_{\text{hyd}}$
A more exothermic (more negative) ΔHsol favours dissolution and correlates with greater solubility.

For Group 2 hydroxides, solubility increases down the group: Mg(OH)₂ is sparingly soluble, Ca(OH)₂ is slightly soluble, Sr(OH)₂ is more soluble, Ba(OH)₂ is noticeably soluble.

Why? As the cation gets larger down the group:

ΔHlatt becomes less exothermic (less negative) because the ions are farther apart and the lattice is held together less strongly.
ΔHhyd also becomes less exothermic (less negative) because larger ions are less strongly hydrated.

However, the decrease in ΔHlatt is greater than the decrease in ΔHhyd. So, going down the group, ΔHsol becomes more exothermic (more negative), favouring solubility.

Understanding the Question

We must explain why compound A (Ca(OH)₂) is less soluble than compound B (Ba(OH)₂). The explanation must compare the lattice and hydration enthalpies of the two hydroxides and identify which difference dominates. The mark scheme awards three separate marks: (M1) both ΔHlatt and ΔHhyd more exothermic for A; (M2) the lattice enthalpy difference is greater for A; (M3) ΔHsol is less exothermic for A.

Approach

Apply the relationship ΔHsol = ΔHlatt + ΔHhyd. Compare ΔHlatt and ΔHhyd for Ca(OH)₂ and Ba(OH)₂, then determine which difference is greater to decide the sign/magnitude of ΔHsol.

Step-by-Step Reasoning

Ca²⁺ is smaller than Ba²⁺, so the ions in Ca(OH)₂ are held together more tightly. The lattice enthalpy of Ca(OH)₂ is more exothermic (more negative) than that of Ba(OH)₂.
Ca²⁺ is also more strongly hydrated than Ba²⁺ (smaller ion, higher charge density, stronger ion–dipole attraction to water). The hydration enthalpy of Ca(OH)₂ is more exothermic (more negative) than that of Ba(OH)₂.
However, the difference in lattice enthalpy between Ca(OH)₂ and Ba(OH)₂ is greater than the difference in hydration enthalpy.
Substituting into ΔHsol = ΔHlatt + ΔHhyd: the more negative ΔHlatt of Ca(OH)₂ outweighs its more negative ΔHhyd, so ΔHsol for Ca(OH)₂ is less exothermic (less negative) than for Ba(OH)₂.
A less exothermic ΔHsol makes dissolution less thermodynamically favourable, so Ca(OH)₂ is less soluble than Ba(OH)₂.

Key Takeaways

The solubility of Group 2 hydroxides increases down the group. The key insight is that although both lattice and hydration enthalpies become less exothermic down the group, the lattice enthalpy change is the larger effect, making ΔHsol more exothermic (and solubility greater) for the larger cation.

Common Mistakes

Claiming the trend is due to hydration enthalpy alone.
Not stating that the lattice enthalpy difference is the greater one (this is the crucial point for the mark).
Confusing the direction: Ca(OH)₂ has more exothermic ΔHlatt AND ΔHhyd, but the lattice difference dominates, making its ΔHsol less exothermic.

Things to Be Careful About

The mark scheme language: "more exothermic ΔHlatt" refers to lattice formation being more exothermic (more negative). Be consistent with this convention. Note the three separate marking points and ensure all three appear in the answer.

Techniques used

compare lattice and hydration enthalpiesapply the relationship ΔHsol = ΔHlatt + ΔHhyddeduce solubility from enthalpy of solution

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 $k$ .

units of $k$ = ..............................

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{mol}^{-2}\text{ dm}^{6}\text{ s}^{-1}$

Final answer

mol^-2 dm^6 s^-1

Detailed explanation

Background Concept

The rate constant $k$ is the proportionality constant linking the concentrations of reactants (each raised to its order) to the rate of reaction. Its units depend entirely on the overall order of the reaction, which is the sum of the individual orders in the rate equation. For a general rate equation $\text{rate} = k[\text{A}]^m[\text{B}]^n$ , the units of $k$ are found by rearranging: $k = \text{rate} / ([\text{A}]^m[\text{B}]^n)$ , then substituting the units of rate ( $\text{mol dm}^{-3}\text{ s}^{-1}$ ) and concentration ( $\text{mol dm}^{-3}$ ).

Understanding the Question

The question gives the rate equation $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$ and states that rate is measured in $\text{mol dm}^{-3}\text{ s}^{-1}$ . It asks for the units of $k$ . The overall order is $2 + 1 = 3$ , so $k$ must carry units that make the right-hand side come out in $\text{mol dm}^{-3}\text{ s}^{-1}$ .

Approach

Rearrange the rate equation to make $k$ the subject, then substitute the units of rate and of the concentration product. The key is that $[\text{NO}]^2[\text{Cl}_2]$ has units $(\text{mol dm}^{-3})^3 = \text{mol}^3\text{ dm}^{-9}$ .

Step-by-Step Reasoning

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

Units of rate $= \text{mol dm}^{-3}\text{ s}^{-1}$ .

Units of $[\text{NO}]^2[\text{Cl}_2] = (\text{mol dm}^{-3})^3 = \text{mol}^3\text{ dm}^{-9}$ .

So units of $k = \dfrac{\text{mol dm}^{-3}\text{ s}^{-1}}{\text{mol}^3\text{ dm}^{-9}} = \text{mol}^{-2}\text{ dm}^{6}\text{ s}^{-1}$ .

The negative exponent on mol appears because the concentration term is cubed — a higher power than the single $\text{mol dm}^{-3}$ in the rate. The $\text{s}^{-1}$ comes straight from the rate.

Key Takeaways

For a third-order reaction, $k$ has units $\text{mol}^{-2}\text{ dm}^{6}\text{ s}^{-1}$ . In general, for a reaction of overall order $n$ , the units of $k$ are $\text{mol}^{(1-n)}\text{ dm}^{3(n-1)}\text{ s}^{-1}$ . Each extra order beyond first adds a factor of $\text{mol}^{-1}\text{ dm}^{3}$ .

Common Mistakes

Forgetting that the concentration term is cubed and writing units of $\text{mol dm}^{-3}\text{ s}^{-1} / \text{mol dm}^{-3} = \text{s}^{-1}$ (this would be the units for a first-order reaction).
Getting the sign of the mol exponent wrong, or writing positive exponents on dm.
Confusing the stoichiometric equation with the rate equation — the stoichiometry has nothing to do with the orders.

Things to Be Careful About

Always derive units from the rate equation given, not from the stoichiometric equation. The stoichiometry ( $2\text{NO} + \text{Cl}_2 \rightarrow 2\text{NOCl}$ ) does not determine the orders. Cube the whole concentration term including its units.

Techniques used

derive units of rate constant from rate equationsubstitute units of rate and concentration

(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.

initial concentration of $\text{NO}$ = .............................. $\text{mol dm}^{-3}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Let $c$ = initial concentration of NO = initial concentration of Cl $_2$ . Since the concentrations are equal:

$\text{rate} = k[\text{NO}]^2[\text{Cl}_2] = kc^3$

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

$c^3 = \frac{2.57 \times 10^{-6}}{26.4} = 9.735 \times 10^{-8}$

$c = (9.735 \times 10^{-8})^{1/3} = 4.60 \times 10^{-3}$

Answer

initial concentration of NO = $4.60 \times 10^{-3}\text{ mol dm}^{-3}$

Final answer

4.60 × 10^-3 mol dm^-3

Detailed explanation

Background Concept

The rate equation $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$ tells us how the rate depends on the concentrations of the two reactants. The orders (2 for NO, 1 for Cl $_2$ ) must be determined experimentally — they are not related to the stoichiometric coefficients. When the initial concentrations of NO and Cl $_2$ are equal, we can replace both $[\text{NO}]$ and $[\text{Cl}_2]$ with a single variable $c$ , so the rate equation becomes $\text{rate} = kc^3$ .

Understanding the Question

In experiment 1, $[\text{NO}] = [\text{Cl}_2]$ initially, $k = 26.4$ , and the initial rate is $2.57 \times 10^{-6}\text{ mol dm}^{-3}\text{ s}^{-1}$ . We must find the initial concentration of NO. The 2 marks are awarded for: M1 writing the correct equation with $c^3$ , and M2 correctly evaluating the cube root.

Approach

Substitute the known values into $\text{rate} = kc^3$ , solve for $c^3$ , then take the cube root. Keep the working clear so that both method and accuracy marks are earned.

Step-by-Step Reasoning

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

$\text{rate} = kc^3$
$2.57 \times 10^{-6} = 26.4 \times c^3$
$c^3 = \frac{2.57 \times 10^{-6}}{26.4} = 9.735 \times 10^{-8}$
$c = (9.735 \times 10^{-8})^{1/3} = 4.60 \times 10^{-3}\text{ mol dm}^{-3}$

Check: $4.60 \times 10^{-3}$ cubed $= 9.73 \times 10^{-8}$ ; $\times 26.4 = 2.57 \times 10^{-6}$ . ✓

Key Takeaways

When concentrations are equal, collapse the rate equation to a single power of one variable. Always check your answer by substituting back into the rate equation.

Common Mistakes

Using the stoichiometric coefficients (2 and 1) as the orders instead of the orders in the given rate equation.
Forgetting to cube the concentration — writing $\text{rate} = kc$ instead of $\text{rate} = kc^3$ .
Taking the square root instead of the cube root.
Dropping units in intermediate steps.

Things to Be Careful About

The value of $k$ is 26.4 with units $\text{mol}^{-2}\text{ dm}^{6}\text{ s}^{-1}$ , so the final concentration comes out in $\text{mol dm}^{-3}$ . Keep 3 significant figures in the final answer to match the data given.

Techniques used

substitute values into rate equationsolve cubic 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.

initial rate of reaction in experiment 2 = .............................. $\text{mol dm}^{-3}\text{ s}^{-1}$

1M Similar

DifficultyEasy

Worked solution

Working

Rate $\propto [\text{NO}]^2[\text{Cl}_2]$ . Both concentrations are multiplied by 10, so the rate is multiplied by $10^2 \times 10 = 1000$ .

$\text{rate}_2 = 2.57 \times 10^{-6} \times 1000 = 2.57 \times 10^{-3}$

Answer

initial rate of reaction in experiment 2 = $2.57 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}$

Final answer

2.57 × 10^-3 mol dm^-3 s^-1

Detailed explanation

Background Concept

The rate equation shows how rate scales with concentration. If a concentration is multiplied by a factor $f$ , the rate is multiplied by $f$ raised to the power of its order. Since $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$ , increasing $[\text{NO}]$ by 10 multiplies rate by $10^2 = 100$ , and increasing $[\text{Cl}_2]$ by 10 multiplies rate by $10^1 = 10$ . The combined effect is $\times 1000$ .

Understanding the Question

Experiment 2 uses concentrations of both NO and Cl $_2$ that are ten times those of experiment 1. We know the rate of experiment 1 ( $2.57 \times 10^{-6}\text{ mol dm}^{-3}\text{ s}^{-1}$ ) and must find the rate of experiment 2. No new calculation of concentrations is needed — just scale the rate.

Approach

Multiply the experiment 1 rate by the factor $10^2 \times 10 = 1000$ .

Step-by-Step Reasoning

Rate is proportional to $[\text{NO}]^2[\text{Cl}_2]$ .

$[\text{NO}] \times 10$ → rate $\times 10^2 = \times 100$
$[\text{Cl}_2] \times 10$ → rate $\times 10^1 = \times 10$

Combined: rate $\times 100 \times 10 = \times 1000$ .

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

Key Takeaways

Scaling arguments from the rate equation save time: the rate factor is the product of each concentration factor raised to its order.

Common Mistakes

Multiplying by 10 only once (treating the reaction as first order overall).
Multiplying by 20 (adding the factors instead of multiplying).
Confusing the order of NO (2) with the stoichiometric coefficient.

Things to Be Careful About

The units of rate stay the same ( $\text{mol dm}^{-3}\text{ s}^{-1}$ ). The answer has 3 significant figures, matching the given rate.

Techniques used

apply proportionality of rate to concentration changesmultiply rate by scaling factors

(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

[NO] is in large excess and so its concentration remains effectively constant. The rate therefore depends only on [Cl $_2$ ], giving pseudo first-order kinetics with respect to Cl $_2$ . A first-order reaction has a constant half-life.

Final answer

Pseudo first order: [NO] constant, first order in Cl2, hence constant half-life.

Detailed explanation

Background Concept

When one reactant is present in large excess, its concentration changes negligibly during the reaction and can be treated as constant. The rate equation then simplifies: $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$ becomes $\text{rate} = k'[\text{Cl}_2]$ , where $k' = k[\text{NO}]^2$ is a new (pseudo) rate constant. The reaction is said to be pseudo first order with respect to Cl $_2$ . A key property of first-order reactions is that the half-life is constant — it does not depend on the initial concentration.

Understanding the Question

Experiment 3 uses a large excess of NO. The graph of $[\text{Cl}_2]$ against time shows a constant half-life. We must explain why. The mark is awarded for recognising that the reaction is overall first order (pseudo first order) under these conditions, with Cl $_2$ being the only reactant whose concentration changes appreciably.

Approach

State that [NO] is effectively constant, so the rate depends only on [Cl $_2$ ] (first order in Cl $_2$ ), and first-order reactions have constant half-life.

Step-by-Step Reasoning

[NO] is in large excess, so its concentration stays essentially constant throughout the reaction.
The rate equation $\text{rate} = k[\text{NO}]^2[\text{Cl}_2]$ therefore simplifies to $\text{rate} = k'[\text{Cl}_2]$ , where $k' = k[\text{NO}]^2$ is constant.
This is first order with respect to Cl $_2$ (pseudo first order overall).
For a first-order reaction, the half-life is independent of concentration, hence constant.

Key Takeaways

"Large excess" is a standard technique to isolate the order with respect to one reactant. A constant half-life is the signature of first-order kinetics.

Common Mistakes

Saying "the reaction is first order" without mentioning the large excess of NO / that [NO] is constant.
Confusing pseudo first order with true first order — the reaction is second order in NO but that order is masked by the excess.
Saying the half-life is constant because the reaction is "slow" or "fast" — that is not the reason.

Things to Be Careful About

The mark scheme accepts either "overall 1st order / pseudo 1st order" OR "1st order with respect to Cl $_2$ AND concentration of NO doesn't change". Make sure your answer includes the excess condition.

Techniques used

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

(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

DifficultyEasy

Worked solution

Working

For a first-order reaction:

$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{105.6} = 6.56 \times 10^{-3}\text{ s}$

Answer

$t_{1/2} = 6.56 \times 10^{-3}\text{ s}$ , as required.

Final answer

6.56 × 10^-3 s

Detailed explanation

Background Concept

For a first-order reaction, the half-life is given by $t_{1/2} = \dfrac{\ln 2}{k}$ , where $k$ is the first-order rate constant. Here, under the pseudo first-order conditions of experiment 3, the effective rate constant is $k' = 105.6$ (with units $\text{s}^{-1}$ , since the reaction is now first order overall). The question asks us to show that $t_{1/2} = 6.56 \times 10^{-3}\text{ s}$ .

Understanding the Question

We are told that under experiment 3 conditions, the value of the rate constant is 105.6. We must show that $t_{1/2} = 6.56 \times 10^{-3}\text{ s}$ . This is a "show that" question — the working must be presented clearly.

Approach

Use the first-order half-life formula $t_{1/2} = \ln 2 / k = 0.693 / 105.6$ and evaluate.

Step-by-Step Reasoning

$t_{1/2} = \frac{\ln 2}{105.6} = \frac{0.693}{105.6} = 6.56 \times 10^{-3}\text{ s}$

This matches the given value, so the result is shown.

Key Takeaways

$\ln 2 \approx 0.693$ is a constant worth remembering for first-order half-life calculations.

Common Mistakes

Using $\ln 2 = 0.301$ (that is $\log_{10} 2$ ).
Using the original $k = 26.4$ instead of the pseudo rate constant 105.6.
Forgetting units (s).

Things to Be Careful About

This is a "show that" question — the final answer is given, so the working must be shown. The value 105.6 is the pseudo first-order rate constant $k' = k[\text{NO}]^2$ , not the original $k$ .

Techniques used

apply first-order half-life formulasubstitute rate constant value

(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.

time = .............................. s

1M Similar

DifficultyMedium-Easy

Worked solution

Working

$\frac{[\text{Cl}_2]_t}{[\text{Cl}_2]_0} = \frac{1.25 \times 10^{-5}}{2.00 \times 10^{-4}} = 0.0625 = \frac{1}{16} = \left(\frac{1}{2}\right)^4$

So 4 half-lives have elapsed.

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

Answer

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

Final answer

2.62 × 10^-2 s

Detailed explanation

Background Concept

For a first-order reaction, after $n$ half-lives the concentration falls to $(1/2)^n$ of its initial value. We can use this to find how many half-lives have elapsed, then multiply by the half-life to get the time.

Understanding the Question

Initial $[\text{Cl}_2] = 2.00 \times 10^{-4}\text{ mol dm}^{-3}$ . We need the time for $[\text{Cl}_2]$ to fall to $1.25 \times 10^{-5}\text{ mol dm}^{-3}$ . The half-life is $6.56 \times 10^{-3}\text{ s}$ (from part (b)(ii)).

Approach

Find the fraction remaining: $1.25 \times 10^{-5} / 2.00 \times 10^{-4} = 0.0625 = 1/16 = (1/2)^4$ . So 4 half-lives have elapsed. Time $= 4 \times 6.56 \times 10^{-3} = 2.62 \times 10^{-2}\text{ s}$ .

Step-by-Step Reasoning

Fraction remaining:
$\frac{1.25 \times 10^{-5}}{2.00 \times 10^{-4}} = 0.0625 = \frac{1}{16} = \left(\frac{1}{2}\right)^4$

So 4 half-lives have elapsed.

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

Key Takeaways

The half-life approach is the quickest way to find time for a first-order decay when the fraction remaining is a power of 1/2. Alternatively, use the integrated rate law $[\text{A}] = [\text{A}]_0 e^{-kt}$ .

Common Mistakes

Using the wrong initial concentration (e.g. $2.00 \times 10^{-4}$ vs $1.25 \times 10^{-5}$ ).
Calculating the fraction as $1.25/2.00 = 0.625$ instead of $0.0625$ (off by a factor of 10).
Multiplying by the wrong number of half-lives.

Things to Be Careful About

The answer $2.62 \times 10^{-2}\text{ s}$ has 3 significant figures. Note that $0.0625 = 1/16$ is exactly $(1/2)^4$ , so the half-life method gives an exact answer.

Techniques used

calculate fraction remainingdetermine number of half-livesmultiply by half-life

(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

NO acts as a homogeneous catalyst: it provides an alternative pathway of lower activation energy and is regenerated at the end of the reaction.

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

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

Final answer

NO is a homogeneous catalyst; 2NO + O2 → 2NO2; NO2 + SO2 → NO + SO3

Detailed explanation

Background Concept

A catalyst provides an alternative reaction pathway with a lower activation energy, increasing the rate without being consumed overall. A homogeneous catalyst is in the same phase as the reactants. NO is a homogeneous catalyst in the oxidation of SO $_2$ by O $_2$ : it participates in the reaction but is regenerated, so it is not used up.

Understanding the Question

SO $_2$ reacts very slowly with O $_2$ to form SO $_3$ , but the reaction is much faster in the presence of NO. We must explain NO's role and include chemical equations. 3 marks: M1 for identifying NO as a homogeneous catalyst / regenerated, M2 for $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$ , M3 for $\text{NO}_2 + \text{SO}_2 \rightarrow \text{NO} + \text{SO}_3$ .

Approach

State that NO is a catalyst (specifically homogeneous), then write the two-step catalytic cycle showing NO being consumed in one step and regenerated in the next.

Step-by-Step Reasoning

NO is a homogeneous catalyst: it is in the gas phase like the reactants, and it is regenerated at the end of the cycle.
Step 1: NO reacts with O $_2$ to form NO $_2$ : $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$ .
Step 2: NO $_2$ oxidises SO $_2$ to SO $_3$ , regenerating NO: $\text{NO}_2 + \text{SO}_2 \rightarrow \text{NO} + \text{SO}_3$ .
Overall: adding the two steps gives $2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$ (NO cancels), confirming NO is not consumed.

Key Takeaways

A catalyst is identified by its regeneration. The catalytic cycle shows consumption and regeneration explicitly. Homogeneous catalysts are in the same phase as the reactants.

Common Mistakes

Writing $\text{NO} + \text{O}_2 \rightarrow \text{NO}_2$ (unbalanced — must be $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$ ).
Writing the second equation with $\text{SO}_2 + \text{NO}_2 \rightarrow \text{SO}_3 + \text{NO}$ but forgetting to balance.
Saying NO is a heterogeneous catalyst (it is in the same gas phase).
Saying NO "increases the rate" without mentioning the lower activation energy pathway or regeneration.

Things to Be Careful About

Both equations must be balanced. The mark scheme gives M1 for "homogeneous catalyst OR catalyst because it is regenerated" — so make sure you state regeneration. The overall reaction is $2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$ , and NO is regenerated, so it does not appear in the overall equation.

Techniques used

identify homogeneous catalystwrite catalytic cycle equationsverify catalyst regeneration

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}} =

units = ..............................

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 compound. For a general salt $\text{A}_x\text{B}_y$ dissociating as $\text{A}_x\text{B}_y(\text{s}) \rightleftharpoons x\text{A}^{y+}(\text{aq}) + y\text{B}^{x-}(\text{aq})$ , the expression is $K_{\text{sp}} = [\text{A}^{y+}]^x[\text{B}^{x-}]^y$ . Only aqueous ion concentrations (in mol dm $^{-3}$ ) appear; the solid is excluded because its concentration is constant. The units of $K_{\text{sp}}$ depend on the total power of the concentration terms.

Understanding the Question

The question provides the formula $\text{Cr(OH)}_3$ and its $K_{\text{sp}}$ value, and asks for the equilibrium expression in terms of ion concentrations, with units. The solid $\text{Cr(OH)}_3$ is excluded from the expression.

Approach

Write the dissociation equation, then express $K_{\text{sp}}$ as the product of ion concentrations raised to their stoichiometric coefficients. For units, count the total power of concentration terms.

Step-by-Step Reasoning

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

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

Each concentration has units mol dm $^{-3}$ . The expression has one power of $[\text{Cr}^{3+}]$ and three powers of $[\text{OH}^-]$ , so the total power is $1 + 3 = 4$ . Units = $(\text{mol dm}^{-3})^4 = \text{mol}^4\text{ dm}^{-12}$ .

Key Takeaways

The $K_{\text{sp}}$ expression excludes the solid; the exponent on each ion equals its stoichiometric coefficient; units are obtained by summing the powers of concentration.

Common Mistakes

Including $\text{Cr(OH)}_3(\text{s})$ in the expression (it is a solid, excluded).
Writing $[\text{OH}^-]$ without the power 3.
Giving units as mol dm $^{-3}$ instead of mol $^4$ dm $^{-12}$ .

Things to Be Careful About

Count the total exponent carefully: 1 from $\text{Cr}^{3+}$ and 3 from $\text{OH}^-$ , giving 4. The mark scheme awards one mark for the correct expression and one for the correct units.

Techniques used

write the Ksp expression from the dissociation equationdetermine the units of Ksp from the powers of concentration

(ii)

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

Show your working.

solubility = .............................. $\text{g dm}^{-3}$

3M Similar

DifficultyMedium

Worked solution

Working

Let the solubility be $s$ mol dm $^{-3}$ .
$[\text{Cr}^{3+}] = s, \quad [\text{OH}^-] = 3s$
$K_{\text{sp}} = s(3s)^3 = 27s^4$
$s^4 = \frac{1.0 \times 10^{-33}}{27} = 3.70 \times 10^{-35}$
$s = 2.47 \times 10^{-9}\text{ mol dm}^{-3}$
$M_r(\text{Cr(OH)}_3) = 52 + 3(16 + 1) = 103$
$\text{solubility} = 2.47 \times 10^{-9} \times 103 = 2.54 \times 10^{-7}\text{ g dm}^{-3}$

Answer

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

Final answer

2.54 × 10^-7 g dm^-3

Detailed explanation

Background Concept

For a sparingly soluble salt, if $s$ is the solubility in mol dm $^{-3}$ , each formula unit of $\text{Cr(OH)}_3$ that dissolves releases one $\text{Cr}^{3+}$ and three $\text{OH}^-$ ions. So $[\text{Cr}^{3+}] = s$ and $[\text{OH}^-] = 3s$ . Substituting into $K_{\text{sp}}$ gives a relationship that can be solved for $s$ . The solubility in g dm $^{-3}$ is obtained by multiplying $s$ by the molar mass.

Understanding the Question

We must find the solubility of $\text{Cr(OH)}_3$ in pure water, i.e. the number of moles (and grams) that dissolve per dm $^3$ . The $K_{\text{sp}}$ value is given as $1.0 \times 10^{-33}$ .

Approach

Set up the stoichiometric relationship between ion concentrations and solubility $s$ , substitute into $K_{\text{sp}}$ , solve for $s$ , then convert to g dm $^{-3}$ using $M_r$ .

Step-by-Step Reasoning

Let $s$ = solubility in mol dm $^{-3}$ .

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

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

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

$s = (3.70 \times 10^{-35})^{1/4} = 2.47 \times 10^{-9}$ mol dm $^{-3}$ .

$M_r(\text{Cr(OH)}_3) = 52 + 3(16 + 1) = 52 + 51 = 103$ g mol $^{-1}$ .

Solubility = $2.47 \times 10^{-9} \times 103 = 2.54 \times 10^{-7}$ g dm $^{-3}$ .

Key Takeaways

The stoichiometric coefficient of $\text{OH}^-$ (3) appears both as the exponent in $K_{\text{sp}}$ and as the factor relating $[\text{OH}^-]$ to $s$ . Converting mol dm $^{-3}$ to g dm $^{-3}$ requires the molar mass.

Common Mistakes

Forgetting the factor 3: writing $K_{\text{sp}} = s^4$ instead of $27s^4$ .
Using $[\text{OH}^-] = s$ instead of $3s$ .
Not converting to g dm $^{-3}$ , or using the wrong $M_r$ .

Things to Be Careful About

Raise $3s$ to the power 3 correctly: $(3s)^3 = 27s^3$ . Take the fourth root, not the square root. Use $M_r = 103$ g mol $^{-1}$ for $\text{Cr(OH)}_3$ . The mark scheme gives $2.47 \times 10^{-9}$ mol dm $^{-3}$ (M2) and $2.54 \times 10^{-7}$ g dm $^{-3}$ (M3).

Techniques used

relate ion concentrations to solubility ssubstitute into the Ksp expression and solve for sconvert mol dm^-3 to g dm^-3 using molar mass

(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

(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.

pH = ..............................

2M Similar

DifficultyMedium-Easy

Worked solution

Working

$K_{\text{a}} = \frac{[\text{H}^+][\text{CH}_3\text{CH}_2\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}]} \approx \frac{[\text{H}^+]^2}{[\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}]}$
$[\text{H}^+]^2 = 1.51 \times 10^{-5} \times 0.100 = 1.51 \times 10^{-6}$
$[\text{H}^+] = 1.23 \times 10^{-3}\text{ mol dm}^{-3}$
$\text{pH} = -\log_{10}(1.23 \times 10^{-3}) = 2.91$

Answer

pH = 2.91

Final answer

pH = 2.91

Detailed explanation

Background Concept

A weak acid partially dissociates: $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$ . $K_{\text{a}} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$ . For a weak acid where dissociation is small, $[\text{HA}] \approx$ initial concentration, and $[\text{H}^+] = [\text{A}^-]$ , so $K_{\text{a}} \approx \frac{[\text{H}^+]^2}{[\text{HA}]}$ .

Understanding the Question

Calculate the pH of $0.100$ mol dm $^{-3}$ butanoic acid, a weak acid with $K_{\text{a}} = 1.51 \times 10^{-5}$ .

Approach

Use the weak acid approximation to find $[\text{H}^+]$ , then $\text{pH} = -\log_{10}[\text{H}^+]$ .

Step-by-Step Reasoning

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

$[\text{H}^+] = \sqrt{1.51 \times 10^{-6}} = 1.23 \times 10^{-3}$ mol dm $^{-3}$ .

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

Key Takeaways

For a weak acid, $[\text{H}^+] = \sqrt{K_{\text{a}}[\text{HA}]}$ . The approximation $[\text{HA}] \approx$ initial concentration holds because $K_{\text{a}}$ is small.

Common Mistakes

Forgetting to take the square root.
Using $\text{pH} = -\log_{10}(K_{\text{a}})$ directly.
Confusing $K_{\text{a}}$ with $[\text{H}^+]$ .

Things to Be Careful About

Report pH to 2 decimal places. The approximation is valid here since $K_{\text{a}} \ll [\text{HA}]$ .

Techniques used

apply the weak acid approximationcalculate [H+] from Kaconvert [H+] to pH

(ii)

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

pH = ..............................

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.

pH = ..............................

2M Similar

DifficultyMedium

Worked solution

Working

Moles of NaOH added $= \frac{5.00}{1000} \times 0.100 = 5.00 \times 10^{-4}$ mol

Moles of acid initially $= \frac{10.00}{1000} \times 0.100 = 1.00 \times 10^{-3}$ mol

Moles of acid remaining $= 1.00 \times 10^{-3} - 5.00 \times 10^{-4} = 5.00 \times 10^{-4}$ mol

Moles of butanoate formed $= 5.00 \times 10^{-4}$ mol

Equal concentrations of acid and its conjugate base $\rightarrow$ buffer at half-neutralisation:
$\text{pH} = \text{p}K_{\text{a}} = -\log_{10}(1.51 \times 10^{-5}) = 4.82$

Answer

pH = 4.82

Final answer

pH = 4.82

Detailed explanation

Background Concept

When a strong base is added to a weak acid, it neutralises some of the acid, forming the conjugate base (salt). If both the weak acid and its conjugate base are present in comparable amounts, the solution is a buffer. For a buffer, $\text{pH} = \text{p}K_{\text{a}} + \log\frac{[\text{A}^-]}{[\text{HA}]}$ (Henderson-Hasselbalch). When $[\text{A}^-] = [\text{HA}]$ , $\text{pH} = \text{p}K_{\text{a}}$ .

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. Calculate the pH of the resulting solution.

Approach

Calculate moles of NaOH and acid. The NaOH neutralises an equal number of moles of acid, forming the same number of moles of butanoate. Compare remaining acid and formed salt.

Step-by-Step Reasoning

Moles NaOH $= \frac{5.00}{1000} \times 0.100 = 5.00 \times 10^{-4}$ mol.

Moles acid $= \frac{10.00}{1000} \times 0.100 = 1.00 \times 10^{-3}$ mol.

After neutralisation: acid remaining $= 1.00 \times 10^{-3} - 5.00 \times 10^{-4} = 5.00 \times 10^{-4}$ mol.

Butanoate formed $= 5.00 \times 10^{-4}$ mol.

Since $[\text{acid}] = [\text{butanoate}]$ , $\text{pH} = \text{p}K_{\text{a}} = -\log_{10}(1.51 \times 10^{-5}) = 4.82$ .

Key Takeaways

Exactly half-neutralisation gives $[\text{HA}] = [\text{A}^-]$ , so $\text{pH} = \text{p}K_{\text{a}}$ . This is the buffer region of maximum capacity.

Common Mistakes

Forgetting to convert cm $^3$ to dm $^3$ .
Not recognising the buffer situation and trying to use $[\text{H}^+] = \sqrt{K_{\text{a}}[\text{HA}]}$ directly.
Ignoring the salt concentration.

Things to Be Careful About

Since both species are in the same total volume, the volume cancels — equal moles mean equal concentrations. The answer $\text{pH} = \text{p}K_{\text{a}} = 4.82$ .

Techniques used

calculate moles of acid and basedetermine the buffer composition after neutralisationapply pH = pKa at half-neutralisation

(c)

$80.0\text{ cm}^3$ of an aqueous solution containing 0.704 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 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.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Mass of acid remaining in water $= 0.704 - 0.556 = 0.148$ g

$[\text{acid}]_{\text{benzene}} = \frac{0.556}{0.100} = 5.56\text{ g dm}^{-3}$
$[\text{acid}]_{\text{water}} = \frac{0.148}{0.080} = 1.85\text{ g dm}^{-3}$
$K_{\text{pc}} = \frac{[\text{acid}]_{\text{benzene}}}{[\text{acid}]_{\text{water}}} = \frac{5.56}{1.85} = 3.01$

Answer

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

Final answer

Kpc = 3.01

Detailed explanation

Background Concept

The partition coefficient $K_{\text{pc}}$ describes the distribution of a solute between two immiscible solvents at equilibrium: $K_{\text{pc}} = \frac{\text{concentration in solvent A}}{\text{concentration in solvent B}}$ . Here, between benzene and water.

Understanding the Question

$0.704$ g of butanoic acid in $80.0$ cm $^3$ aqueous solution is shaken with $100$ cm $^3$ benzene. At equilibrium, $0.556$ g is in the benzene layer. Calculate $K_{\text{pc}}$ between benzene and water.

Approach

Find the mass remaining in water by subtraction. Convert both masses to concentrations (g dm $^{-3}$ ) using the respective volumes. Divide benzene concentration by water concentration.

Step-by-Step Reasoning

Mass in water $= 0.704 - 0.556 = 0.148$ g.

$[\text{acid}]_{\text{benzene}} = \frac{0.556}{0.100} = 5.56$ g dm $^{-3}$ .

$[\text{acid}]_{\text{water}} = \frac{0.148}{0.080} = 1.85$ g dm $^{-3}$ .

$K_{\text{pc}} = \frac{5.56}{1.85} = 3.01$ .

Key Takeaways

$K_{\text{pc}} = \frac{\text{concentration in organic layer}}{\text{concentration in aqueous layer}}$ . The reciprocal ( $0.333$ ) would be $K_{\text{pc}}$ for water/benzene.

Common Mistakes

Using mass instead of concentration.
Forgetting to convert volumes to dm $^3$ .
Dividing in the wrong order (giving 0.333 instead of 3.01).

Things to Be Careful About

The question specifies "between $\text{C}_6\text{H}_6$ and water", so $K_{\text{pc}} = \frac{[\text{benzene}]}{[\text{water}]} = 3.01$ . The reciprocal 0.333 is also accepted per the mark scheme, but 3.01 is the direct answer.

Techniques used

find the mass left in water by subtractionconvert masses to concentrations in g dm^-3divide concentrations to find the 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 is dissolved in water to form an infinitely dilute solution.

Final answer

Energy change when one mole of gaseous ions dissolves in water to form an infinitely dilute solution.

Detailed explanation

Background Concept

When a gaseous ion comes into contact with water, polar water molecules arrange around the ion and are attracted to it by ion–dipole forces. Energy is released in this process, so enthalpy changes of hydration are exothermic and have negative values. A formal definition must be precise about the amount of substance (one mole), the initial state of the ions (gaseous), and the final state (aqueous, ideally infinitely dilute so that the ions no longer interact with each other).

Understanding the Question

The stem gives a table of hydration enthalpies, but this part simply asks for the definition. No calculation is needed; quoting the wording accurately earns the mark.

Approach

Recall the definition word by word. The essential ideas are: energy change, one mole, gaseous ions, water, infinitely dilute solution.

Step-by-Step Reasoning

The definition should state:

an energy or enthalpy change;
when one mole of gaseous ions;
is dissolved in water;
giving an infinitely dilute solution.

All four components are needed because each one makes the quantity unambiguous. “One mole” fixes the amount, “gaseous ions” fixes the starting state, and “water” fixes the solvent. “Infinitely dilute” ensures that the final ions are separated enough that ion–ion interactions can be neglected.

Key Takeaways

Hydration is an exothermic ion–dipole process. Its magnitude depends on the charge density of the ion, which is why values such as those given in Table 4.1 differ from ion to ion.

Common Mistakes

Missing “one mole” or saying “an ion” instead of “gaseous ions.”
Saying “energy released when an ion is dissolved” without specifying the gaseous state.
Stating that hydration enthalpy is always positive; for gaseous ions dissolving in water it is exothermic and negative.

Things to Be Careful About

Use “gaseous ions,” not “atoms” or “molecules.” A precise definition may include “infinite dilution” to make clear that no further ion–ion interactions are present.

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

The hydration of $\text{Ca}^{2+}$ is much more exothermic than that of $\text{K}^+$ .

$\text{K}^+$ has a larger ionic radius and a smaller charge ( $+1$ ) than $\text{Ca}^{2+}$ ( $+2$ ).
$\text{Ca}^{2+}$ therefore has a higher charge density and forms stronger ion–dipole attractions with water, releasing more energy.

Final answer

K+ has a larger radius and smaller charge; Ca2+ has stronger attraction for water, so its hydration enthalpy is more negative.

Detailed explanation

Background Concept

The enthalpy of hydration results from ion–dipole attractions between an ion and surrounding water molecules. The strength of the attraction depends on the charge density of the ion, which can be thought of as charge divided by size. A higher charge and a smaller radius produce stronger attractions, and therefore a more exothermic (more negative) hydration enthalpy.

Understanding the Question

The table gives $\Delta H_{\text{hyd}}$ values: $\text{K}^+$ is $-322\ \text{kJ mol}^{-1}$ and $\text{Ca}^{2+}$ is $-1650\ \text{kJ mol}^{-1}$ . The question asks why the $\text{Ca}^{2+}$ value is so much more negative. Both ions are isoelectronic, each having 18 electrons, but they differ in nuclear charge and therefore in ionic radius.

Approach

State the two relevant factual comparisons: ionic radius and charge. Then link these to charge density and ion–dipole attraction, and finally connect stronger attraction to a more exothermic hydration.

Step-by-Step Reasoning

$\text{K}^+$ has the same electron configuration as $\text{Ca}^{2+}$ , but $\text{Ca}^{2+}$ has a greater nuclear charge pulling the electrons in, so $\text{Ca}^{2+}$ has the smaller ionic radius.
$\text{K}^+$ has a charge of $+1$ , while $\text{Ca}^{2+}$ has a charge of $+2$ .
Overall, $\text{K}^+$ has a larger radius and a smaller charge, so it has a lower charge density.
$\text{Ca}^{2+}$ has a higher charge density, so it attracts the partial negative charge on the oxygen of water molecules more strongly.
Stronger ion–dipole attractions release more energy when the ion is hydrated, making the hydration enthalpy of $\text{Ca}^{2+}$ much more negative.

Key Takeaways

For ions, both charge and radius matter. High charge and small radius give high charge density, stronger ion–dipole attraction, and more exothermic hydration.

Common Mistakes

Saying only that $\text{Ca}^{2+}$ has a higher charge without mentioning radius. Both points are needed for the two marks.
Saying $\text{Ca}^{2+}$ has “more electrons”; both ions are isoelectronic.
Comparing lattice energy instead of hydration: the question is specifically about hydration by water.
Using “bigger enthalpy” without clarifying “more negative / more exothermic.”

Things to Be Careful About

Use precise wording: “ $\text{K}^+$ has a larger radius and smaller charge,” and “ $\text{Ca}^{2+}$ has stronger attraction for water.” Hydration enthalpies are negative, so “greater magnitude” means “more exothermic,” not a more positive value.

Techniques used

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

(iii)

Define lattice energy.

1M Similar

DifficultyEasy

Worked solution

Answer

Lattice energy is the energy change when one mole of an ionic solid is formed from its gaseous ions.

Final answer

Energy change when one mole of an ionic solid is formed from its gaseous ions.

Detailed explanation

Background Concept

Ionic solids are made of oppositely charged ions held together by electrostatic forces. The lattice energy is a measure of the strength of these attractions. It is defined as the energy change when one mole of an ionic solid is formed from its separate gaseous ions.

For calcium fluoride, the process represented is:

\text{Ca}^{2+}(\text{g}) + 2\text{F}^-(\text{g}) \rightarrow \text{CaF}_2(\text{s})

This process is exothermic, so the lattice energy is negative.

Understanding the Question

This part only asks for the definition. The surrounding table is not needed; the answer is a single sentence with the correct key terms.

Approach

Recall the definition: “energy change when one mole of an ionic solid forms from its gaseous ions.” Make sure the words “one mole,” “ionic solid,” and “gaseous ions” all appear.

Step-by-Step Reasoning

The definition must be understood in the formation direction: separate gaseous ions coming together to make the lattice. The quantity is quoted per mole of ionic solid, so for a compound such as $\text{CaF}_2$ it refers to the formation of one mole of $\text{CaF}_2$ (s), not to one mole of ions.

Key Takeaways

Lattice energy is a formation enthalpy: gaseous ions → solid lattice. It is exothermic and highly negative because of strong electrostatic attraction between oppositely charged ions.

Common Mistakes

Saying “breaking the lattice” instead of “forming the lattice.” The breaking process is the opposite of lattice energy.
Using “one mole of ions” rather than “one mole of ionic solid.”
Confusing lattice energy with lattice enthalpy of atomisation or hydration.

Things to Be Careful About

State clearly that the starting particles are gaseous ions. Do not say “gaseous atoms” or “molecules.” The sign convention used by CIE is formation of the solid from gaseous ions, so the lattice energy is negative.

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 $\text{CaF}_2$ .

$\Delta H_{\text{sol}}$ of $\text{CaF}_2$ = .............................. $\text{kJ mol}^{-1}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For each mole of $\text{CaF}_2$ dissolved, one $\text{Ca}^{2+}$ ion and two $\text{F}^-$ ions are hydrated:

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

\Delta H_{\text{sol}} = (-1650) + 2(-506) - (-2602)

\Delta H_{\text{sol}} = -2662 + 2602 = -60\ \text{kJ mol}^{-1}

Answer

$-60\ \text{kJ mol}^{-1}$

Final answer

-60 kJ mol^-1

Detailed explanation

Background Concept

Dissolving an ionic solid can be treated as two steps:

Break the lattice into gaseous ions. Lattice energy is defined as the energy change when gaseous ions form the solid, so the reverse process is endothermic and has the opposite sign.
Hydrate the gaseous ions. This releases energy and is the sum of the hydration enthalpies of the individual ions.

By Hess's law, the overall enthalpy change of solution is:

\Delta H_{\text{sol}} = \text{sum of hydration enthalpies} - \Delta H_{\text{latt}}

where $\Delta H_{\text{latt}}$ is the negative lattice formation energy.

Understanding the Question

We are given $\Delta H_{\text{latt}}$ for $\text{CaF}_2$ as $-2602\ \text{kJ mol}^{-1}$ , and the hydration enthalpies of $\text{Ca}^{2+}$ and $\text{F}^-$ from Table 4.1. We need the enthalpy change when $\text{CaF}_2$ (s) dissolves in water. Because one formula unit contains two fluoride ions, the hydration enthalpy of fluoride must be counted twice.

Approach

Use the energy cycle:

\text{CaF}_2(\text{s}) \rightarrow \text{Ca}^{2+}(\text{g}) + 2\text{F}^-(\text{g}) \rightarrow \text{Ca}^{2+}(\text{aq}) + 2\text{F}^-(\text{aq})

The first step uses the reverse of lattice formation, $+2602\ \text{kJ mol}^{-1}$ . The second step uses the total hydration energy, $-1650 + 2(-506)\ \text{kJ mol}^{-1}$ .

Step-by-Step Reasoning

Total hydration energy for the ions in one mole of $\text{CaF}_2$ :

-1650 + 2(-506) = -1650 - 1012 = -2662\ \text{kJ mol}^{-1}

Reverse of lattice formation:

-\Delta H_{\text{latt}} = -(-2602) = +2602\ \text{kJ mol}^{-1}

Add the two contributions:

\Delta H_{\text{sol}} = +2602 - 2662 = -60\ \text{kJ mol}^{-1}

The negative value shows that the solution process is exothermic for this ionic substance under the standard convention used.

Key Takeaways

Use Hess's law to connect lattice energy, hydration enthalpies, and enthalpy of solution. Always multiply hydration enthalpies by the stoichiometric number of ions in the formula unit.

Common Mistakes

Forgetting the factor of 2 for fluoride ions.
Using $-2602$ directly instead of $+2602$ when reversing lattice formation.
Adding hydration energies to the lattice energy instead of subtracting the lattice formation energy.
Omitting the units or the negative sign in the final answer.

Things to Be Careful About

The sign convention of lattice energy matters. Because CIE defines lattice energy as formation from gaseous ions, it is negative, so the reverse process contributes positive energy to solution. State symbols should be included if equations are written.

Techniques used

use the energy cycle relating hydration and lattice energiesaccount for two fluoride ions per formula unitcalculate the enthalpy change of solution

(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.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$ and rearranging:

\Delta S^\ominus = \frac{\Delta H^\ominus - \Delta G^\ominus}{T}

Substitute $T = 298\ \text{K}$ , $\Delta H^\ominus = -1214\ \text{kJ mol}^{-1}$ and $\Delta G^\ominus = -1162\ \text{kJ mol}^{-1}$ :

\Delta S^\ominus = \frac{-1214 - (-1162)}{298}\ \text{kJ K}^{-1}\text{ mol}^{-1}

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

Convert to $\text{J K}^{-1}\text{ mol}^{-1}$ :

= \frac{-52000}{298} = -174.5\ \text{J K}^{-1}\text{ mol}^{-1}

Answer

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

Final answer

-174.5 J K^-1 mol^-1

Detailed explanation

Background Concept

The Gibbs free energy change is related to enthalpy and entropy by:

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

Here $T$ is the absolute temperature in kelvin. $\Delta H$ is usually given in kilojoules per mole, while $\Delta S$ is usually quoted in joules per kelvin per mole. Therefore it is essential to convert units before doing the calculation, or convert the final answer to the units requested.

Understanding the Question

The reaction forming $\text{CaF}_2$ (s) is given with standard enthalpy and free energy values at 298 K:

\Delta H^\ominus = -1214\ \text{kJ mol}^{-1}

\Delta G^\ominus = -1162\ \text{kJ mol}^{-1}

The question asks for the standard entropy change, $\Delta S^\ominus$ , in $\text{J K}^{-1}\text{ mol}^{-1}$ . We need to rearrange the Gibbs equation to isolate $\Delta S$ .

Approach

Rearrange $\Delta G = \Delta H - T\Delta S$ to:

\Delta S = \frac{\Delta H - \Delta G}{T}

Substitute the given values with $T = 298\ \text{K}$ , and then convert the numerator from kilojoules to joules so that the entropy answer is in $\text{J K}^{-1}\text{ mol}^{-1}$ .

Step-by-Step Reasoning

Write the rearranged equation:

\Delta S = \frac{\Delta H - \Delta G}{T}

Subtract $\Delta G$ from $\Delta H$ :

-1214 - (-1162) = -1214 + 1162 = -52\ \text{kJ mol}^{-1}

Convert to joules:

-52\ \text{kJ mol}^{-1} = -52000\ \text{J mol}^{-1}

Divide by temperature:

\Delta S = \frac{-52000}{298} = -174.5\ \text{J K}^{-1}\text{ mol}^{-1}

The negative sign is chemically reasonable: a solid is formed from calcium solids and fluorine gas, so gas is consumed and entropy decreases.

Key Takeaways

Rearrange the Gibbs equation carefully to solve for $\Delta S$ .
Always match units: $\Delta H$ and $\Delta G$ are in kilojoules, but the requested $\Delta S$ is in joules per kelvin per mole.
A negative $\Delta S$ is consistent with a decrease in the number of gas moles.

Common Mistakes

Forgetting to convert $-52\ \text{kJ}$ into $-52000\ \text{J}$ , leading to $-0.1745$ instead of $-174.5$ .
Using $T$ in degrees Celsius rather than kelvin.
Rearranging incorrectly as $\Delta S = \frac{\Delta G - \Delta H}{T}$ , which gives the wrong sign.
Quoting the answer in $\text{kJ K}^{-1}\text{ mol}^{-1}$ without converting.

Things to Be Careful About

The mark scheme expects $-174.5\ \text{J K}^{-1}\text{ mol}^{-1}$ . Include the negative sign and the correct unit. When writing $\Delta S$ with the Gibbs equation, remember that $T\Delta S$ has units of energy, so $\Delta S$ must ultimately be in energy per kelvin per mole.

Techniques used

rearrange the Gibbs free energy equationconvert kilojoules to joulessubstitute ΔH and ΔG values at 298 K

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

(ii)

Explain why transition elements have variable oxidation states.

1M Similar

(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

(ii)

Write an equation for the reaction taking place.

1M Similar

(c)

3M

(i)

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

Describe any further observations made.

1M Similar

(ii)

Write an equation for the reaction taking place.

1M Similar

(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 V.

2M Similar

(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 $E_{\text{cell}}$ .

$E_{\text{cell}}$ = .............................. V

1M Similar

(iii)

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

1M Similar

(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

(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 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.

current = .............................. A

3M Similar

Q6MediumTransition Elements

(a)

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

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 transition metal ion surrounded by ligands. Ligands are ions or molecules that donate a lone pair of electrons to the metal ion to form dative (coordinate) bonds. The overall charge of a complex ion is the algebraic 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. Complex ion A has an $\text{Fe}^{3+}$ centre with six $\text{CN}^-$ ligands. Complex ion B has an $\text{Fe}^{2+}$ centre with six $\text{CN}^-$ ligands. The formula must include the central metal, the ligands in brackets, and the overall charge as a superscript.

Approach

To find the overall charge, multiply the number of ligands by their individual charge and add this to the charge of the central metal ion. Then write the formula using standard coordination chemistry notation: $[\text{Metal}(\text{Ligand})_n]^{\text{charge}}$ .

Step-by-Step Reasoning

For complex ion A:

Central metal ion: $\text{Fe}^{3+}$ (charge = $+3$ )
Ligands: six $\text{CN}^-$ (total ligand charge = $6 \times -1 = -6$ )
Overall charge = $+3 + (-6) = -3$
Formula: $[\text{Fe}(\text{CN})_6]^{3-}$

For complex ion B:

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

Key Takeaways

The overall charge of a complex ion is simply the sum of the oxidation state of the central metal and the charges of all attached ligands. The formula is written with the metal first, followed by the ligands in parentheses, enclosed in square brackets, with the net charge outside.

Common Mistakes

Forgetting to include the overall charge in the formula.
Calculating the charge incorrectly (e.g., adding $+3$ and $-6$ to get $+9$ instead of $-3$ ).
Writing the charge inside the bracket or without the sign.

Things to Be Careful About

Ensure the charge is written as a superscript with the number first and sign second (e.g., $3-$ not $-3$ ). The formula for the cyanide ligand is $\text{CN}^-$ , not $\text{NC}^-$ , although in some structures the bond is written through carbon, the formula remains $\text{CN}$ .

Techniques used

calculate overall charge of a complex ionwrite complex ion formulae

(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 (and different numbers of d electrons), so the d-orbital splitting energy ( $\Delta E$ ) is different.
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/wavelengths of visible light.

Detailed explanation

Background Concept

Transition metal complexes are coloured because of electronic transitions between the split d-orbitals. When ligands approach a central metal ion, the degeneracy of the d-orbitals is lifted, creating two sets of orbitals at different energy levels (e.g., $t_{2g}$ and $e_g$ in an octahedral field). The energy difference between these sets is called the crystal field splitting energy, $\Delta E$ . Electrons can be excited from the lower energy d-orbitals to the higher energy ones by absorbing photons of visible light. The energy of the absorbed photon matches $\Delta E$ ( $E = hf = hc/\lambda$ ). The colour observed is the complementary colour to the one absorbed.

Understanding the Question

The question asks why solutions of complex ion A ( $[\text{Fe}(\text{CN})_6]^{3-}$ , containing $\text{Fe}^{3+}$ ) and complex ion B ( $[\text{Fe}(\text{CN})_6]^{4-}$ , containing $\text{Fe}^{2+}$ ) have different colours. Both have the same ligands ( $\text{CN}^-$ ) and the same geometry (octahedral). The only difference is the oxidation state (charge) of the central iron ion.

Approach

To explain the colour difference, we must connect the difference in metal ion charge to the difference in $\Delta E$ , and then connect $\Delta E$ to the frequency/wavelength of light absorbed.

Step-by-Step Reasoning

Different charges: Complex ion A contains $\text{Fe}^{3+}$ and complex ion B contains $\text{Fe}^{2+}$ . The $\text{Fe}^{3+}$ ion has a higher charge density and fewer d-electrons than $\text{Fe}^{2+}$ .
Different $\Delta E$ : The higher charge on the $\text{Fe}^{3+}$ ion exerts a stronger electrostatic attraction on the ligand electrons, leading to a greater repulsion and a larger d-orbital splitting energy ( $\Delta E$ ) compared to $\text{Fe}^{2+}$ . (Alternatively, the different number of d-electrons affects electron-electron repulsions within the d-subshell, also altering $\Delta E$ ).
Different light absorption: Since $\Delta E = hf = hc/\lambda$ , a different splitting energy means the complexes will absorb photons of different frequencies ( $f$ ) or wavelengths ( $\lambda$ ) in the visible spectrum.
Different colours: Because they absorb different parts of the visible spectrum, the transmitted (observed) colours are different.

Key Takeaways

The colour of a transition metal complex depends on the d-orbital splitting energy ( $\Delta E$ ). $\Delta E$ is influenced by the charge of the metal ion (higher charge = larger $\Delta E$ ), the type of ligand, and the geometry. A change in metal ion charge changes $\Delta E$ , which changes the wavelength of light absorbed, resulting in a different observed colour.

Common Mistakes

Saying "they absorb different colours" without explaining why (i.e., linking it to $\Delta E$ and frequency/wavelength).
Stating that the number of ligands is different (they are both hexacyano complexes).
Forgetting to mention frequency or wavelength; just saying "different energy" is not specific enough to the visible spectrum.

Things to Be Careful About

The mark scheme specifically looks for "different $\Delta E$ " and "different frequency/wavelength of visible light". Ensure you explicitly state that the splitting energy is different due to the different metal ion charges.

Techniques used

relate d-orbital splitting to light absorptionrelate frequency/wavelength to colour

(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

DifficultyEasy

Worked solution

Answer

Dative covalent (or coordinate) bonding.

Final answer

Dative covalent (coordinate) bonding

Detailed explanation

Background Concept

In a complex ion, the central metal ion acts as a Lewis acid (electron pair acceptor) and the ligands act as Lewis bases (electron pair donors). The bond formed between the ligand and the metal ion is called a dative covalent bond (or coordinate bond). Unlike a normal covalent bond where each atom contributes one electron to the shared pair, in a dative covalent bond, both electrons in the shared pair come from the same atom (the ligand).

Understanding the Question

The question states that the carbon atom of each $\text{CN}^-$ ligand bonds to the $\text{Fe}^{3+}$ ion in complex ion A. It asks for the type of bonding involved.

Approach

This is a direct recall question about the nature of the bond between a ligand and a central metal ion in a coordination complex.

Step-by-Step Reasoning

The $\text{CN}^-$ ligand has a lone pair of electrons on the carbon atom. The $\text{Fe}^{3+}$ ion has empty d-orbitals that can accept these lone pairs. When the lone pair from carbon is shared with the iron ion, a bond is formed where both electrons originate from the carbon. This is the definition of a dative covalent (or coordinate) bond.

Key Takeaways

All bonds between ligands and central metal ions in complex ions are dative covalent (coordinate) bonds. The ligand donates the electron pair, and the metal ion accepts it into an empty orbital.

Common Mistakes

Writing "ionic bonding" (the bond is covalent in nature, albeit with significant ionic character depending on the ligand).
Writing "metallic bonding" or "covalent bonding" without specifying "dative" or "coordinate". "Covalent" alone is often rejected because it implies equal sharing or contribution from both atoms.

Things to Be Careful About

Use the precise terminology: "dative covalent" or "coordinate". "Coordinate bond" or "dative bond" are also acceptable. Avoid just saying "covalent".

Techniques used

identify dative/coordinate bonding in complexes

(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

Worked solution

Answer

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 carbon monoxide ( $\text{Ni}(\text{CO})_4$ ): Tetrahedral geometry. Bond angle: $109.5^\circ$ .
Pd $^{2+}$ with chloride ( $[\text{PdCl}_4]^{2-}$ ): Square planar geometry. Bond angle: $90^\circ$ .

Final answer

See diagram for 3D geometries and bond angles.

Detailed explanation

Background Concept

The geometry of a complex ion is determined by its coordination number (the number of ligand donor atoms bonded to the central metal ion) and the electronic configuration of the metal ion.

Coordination number 2: Linear geometry, bond angle $180^\circ$ . Example: $[\text{Ag}(\text{NH}_3)_2]^+$ .
Coordination number 4: Can be tetrahedral ( $109.5^\circ$ ) or square planar ( $90^\circ$ ). Tetrahedral is common for $d^{10}$ ions like $\text{Zn}^{2+}$ or neutral atoms like $\text{Ni}$ in $\text{Ni}(\text{CO})_4$ . Square planar is common for $d^8$ ions like $\text{Pd}^{2+}$ , $\text{Pt}^{2+}$ , and $\text{Ni}^{2+}$ with strong field ligands.
Coordination number 6: Octahedral geometry, bond angles $90^\circ$ and $180^\circ$ . Example: $[\text{Fe}(\text{CN})_6]^{3-}$ .

When drawing 3D geometries, solid wedges represent bonds coming out of the plane towards the viewer, dashed wedges represent bonds going into the plane away from the viewer, and plain lines represent bonds in the plane of the paper.

Understanding the Question

The question asks to complete Fig. 6.1 by drawing the correct 3D geometry for four different complex ions and labelling one bond angle for each. The central atoms and ligands are given, but the geometries and angles must be drawn and labelled.

Approach

Identify the coordination number and expected geometry for each complex ion based on the central metal/ion and ligands. Draw the structure using appropriate 3D bond representations (solid/dashed wedges) and label the characteristic bond angle.

Step-by-Step Reasoning

Complex ion A ( $[\text{Fe}(\text{CN})_6]^{3-}$ ):
- Central ion: $\text{Fe}^{3+}$
- Ligands: six $\text{CN}^-$ (coordination number 6)
- Geometry: Octahedral
- Bond angle: $90^\circ$ (between adjacent ligands)
- Drawing: Draw Fe in the centre. Draw two vertical bonds (one solid up, one plain down), two bonds in the plane (plain left and right), and two bonds in 3D (solid wedge forward, dashed wedge backward). Label $90^\circ$ between any two adjacent bonds.
Ag $^+$ with ammonia ( $[\text{Ag}(\text{NH}_3)_2]^+$ ):
- Central ion: $\text{Ag}^+$
- Ligands: two $\text{NH}_3$ (coordination number 2)
- Geometry: Linear
- Bond angle: $180^\circ$
- Drawing: Draw Ag in the centre. Draw two plain lines extending horizontally in opposite directions. Attach $\text{H}_3\text{N}$ to the left and $\text{NH}_3$ to the right. Label $180^\circ$ above the Ag atom.
Ni with carbon monoxide ( $\text{Ni}(\text{CO})_4$ ):
- Central atom: $\text{Ni}$ (neutral, $d^{10}$ configuration)
- Ligands: four $\text{CO}$ (coordination number 4)
- Geometry: Tetrahedral
- Bond angle: $109.5^\circ$
- Drawing: Draw Ni in the centre. Draw one plain bond upwards, one plain bond to the left, one solid wedge to the right-front, and one dashed wedge to the right-back. Attach CO to each. Label $109.5^\circ$ between two adjacent bonds (e.g., between the left plain bond and the solid wedge).
Pd $^{2+}$ with chloride ( $[\text{PdCl}_4]^{2-}$ ):
- Central ion: $\text{Pd}^{2+}$ ( $d^8$ configuration)
- Ligands: four $\text{Cl}^-$ (coordination number 4)
- Geometry: Square planar
- Bond angle: $90^\circ$
- Drawing: Draw Pd in the centre. Draw four plain bonds extending up, down, left, and right in a cross shape. Attach Cl to each end. Label $90^\circ$ between any two adjacent bonds (e.g., between the up and right bonds).

Key Takeaways

Coordination number dictates geometry: 2 is linear ( $180^\circ$ ), 4 is tetrahedral ( $109.5^\circ$ ) or square planar ( $90^\circ$ ), and 6 is octahedral ( $90^\circ$ ). $d^8$ ions like $\text{Pd}^{2+}$ and $\text{Pt}^{2+}$ almost always form square planar complexes with coordination number 4. Neutral $d^{10}$ complexes like $\text{Ni}(\text{CO})_4$ are tetrahedral. Always use 3D bond notation (wedges) when drawing tetrahedral and octahedral geometries to show depth.

Common Mistakes

Drawing $[\text{PdCl}_4]^{2-}$ as tetrahedral instead of square planar. $\text{Pd}^{2+}$ is a $4d^8$ ion and forms square planar complexes.
Drawing $\text{Ni}(\text{CO})_4$ as square planar. Neutral $\text{Ni}$ is $d^{10}$ and forms tetrahedral complexes.
Forgetting to use solid and dashed wedges for 3D representation in octahedral and tetrahedral drawings.
Labelling the wrong angle (e.g., labelling $109.5^\circ$ for a square planar complex, or $90^\circ$ for a tetrahedral complex).
Not labelling any bond angle, or labelling an angle that doesn't exist in that geometry.

Things to Be Careful About

Ensure the bond angles match the geometry: octahedral ( $90^\circ$ ), linear ( $180^\circ$ ), tetrahedral ( $109.5^\circ$ ), square planar ( $90^\circ$ ).
In the octahedral drawing, make sure to show the 3D bonds clearly (one forward, one backward) to distinguish it from a flat 2D cross.
The question asks to label one bond angle on each complex ion. Labeling more is fine, but at least one must be correct and clearly marked.

Techniques used

draw 3D geometries of complex ionslabel bond angles

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

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

Fig. 7.1

(a)

Give the systematic name of E.

1M Similar

(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

(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 would be no difference between the two spectra because compound E has no exchangeable protons (no $\text{-OH}$ , $\text{-NH}$ , or $\text{-COOH}$ groups) that can be replaced by deuterium.

Final answer

No difference; no exchangeable protons in E

Detailed explanation

Background Concept

When a sample is shaken with $\text{D}_2\text{O}$ , any protons attached to electronegative atoms (such as O, N, or S) that are labile/exchangeable will be replaced by deuterium. These protons disappear from the $^1\text{H}$ NMR spectrum because deuterium does not resonate in the proton window. This is a diagnostic test for identifying $\text{-OH}$ , $\text{-NH}$ , or $\text{-COOH}$ groups.

Understanding the Question

Compound E is 1,3-diethylbenzene — a pure hydrocarbon containing only $\text{C-H}$ bonds. The question asks what happens when its proton NMR spectrum is taken before and after shaking with $\text{D}_2\text{O}$ .

Approach

Examine the structure of compound E for any protons that could be exchanged with deuterium. Since E is a hydrocarbon with only $\text{C-H}$ bonds (alkyl and aromatic), no exchange occurs.

Step-by-Step Reasoning

$\text{D}_2\text{O}$ exchange removes peaks due to protons on electronegative atoms (O, N, S).
Compound E contains only carbon and hydrogen — all protons are bonded to carbon.
$\text{C-H}$ protons are not labile and do not exchange with $\text{D}_2\text{O}$ .
Therefore, the spectrum before and after shaking with $\text{D}_2\text{O}$ would be identical.

Key Takeaways

$\text{D}_2\text{O}$ exchange is specific to protons on heteroatoms (O, N, S), not $\text{C-H}$ protons.
A hydrocarbon will show no change in its NMR spectrum upon $\text{D}_2\text{O}$ treatment.

Common Mistakes

Suggesting that aromatic protons would be exchanged — they are not labile.
Confusing $\text{D}_2\text{O}$ exchange with deuteration of the ring (which requires different conditions).

Things to Be Careful About

The answer must state both that there is no difference AND the reason (no exchangeable protons).

Techniques used

predict the effect of D2O exchange on a proton NMR spectrumidentify exchangeable protons in a molecule

(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	[crossed out]	[crossed out]

Table 7.2 shows some proton NMR chemical shift values.

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 responsible for the peak	group responsible for the peak	splitting pattern
1.3	6	$\text{CH}_3$	triplet
2.7	4	$\text{CH}_2$	quartet
7.1	4

Working

Compound D is 1,2-diethylbenzene: a benzene ring with two $\text{-CH}_2\text{CH}_3$ groups at positions 1 and 2.

$\delta = 1.3$ ppm: The two $\text{CH}_3$ groups are equivalent (symmetry of the molecule). Each $\text{CH}_3$ has 3 H, so total = 6 H. The $\text{CH}_3$ is adjacent to a $\text{CH}_2$ (2 neighbouring H), so splitting = $2 + 1 = 3$ (triplet).
$\delta = 2.7$ ppm: The two $\text{CH}_2$ groups are equivalent. Each $\text{CH}_2$ has 2 H, so total = 4 H. The $\text{CH}_2$ is adjacent to a $\text{CH}_3$ (3 neighbouring H), so splitting = $3 + 1 = 4$ (quartet). The chemical shift is consistent with alkyl protons next to an aromatic ring (2.3–3.0 ppm range).
$\delta = 7.1$ ppm: Four aromatic protons remain on the ring (positions 3, 4, 5, 6). These are in the aromatic region (6.0–9.0 ppm).

Final answer

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

Detailed explanation

Background Concept

In proton NMR, each distinct chemical environment of hydrogen gives a separate signal. The number of protons in each environment is determined by the integration (area under the peak). The splitting pattern follows the $n+1$ rule: a proton signal is split into $n+1$ peaks where $n$ is the number of equivalent protons on adjacent carbon atoms. Chemical shift values indicate the electronic environment of the proton.

Understanding the Question

Compound D is 1,2-diethylbenzene. The spectrum shows three signals at $\delta = 1.3$ , $2.7$ , and $7.1$ ppm. The table requires the number of protons, the group responsible, and the splitting pattern for each peak. The $\delta = 7.1$ row has the group and splitting crossed out, so only the number of protons is needed there.

Approach

Draw the structure of compound D and identify all proton environments.
Use symmetry to determine which protons are equivalent.
Apply the $n+1$ rule for splitting.
Match chemical shifts to expected ranges using Table 7.2.

Step-by-Step Reasoning

Structure of D: 1,2-diethylbenzene has a plane of symmetry through the ring between C1-C2 and C4-C5 (bisecting the angle between the two ethyl groups). This makes the two ethyl groups equivalent to each other.

Peak at $\delta = 1.3$ ppm:

This is in the alkane range (0.9–1.7 ppm), consistent with $\text{CH}_3$ groups.
Both $\text{CH}_3$ groups are equivalent due to symmetry: $2 \times 3 = 6$ protons.
Each $\text{CH}_3$ is bonded to a $\text{CH}_2$ (2 adjacent H), so splitting = $2 + 1 = 3$ → triplet.

Peak at $\delta = 2.7$ ppm:

This is in the range for alkyl next to aromatic ring (2.3–3.0 ppm), consistent with $\text{CH}_2$ groups attached to the benzene ring.
Both $\text{CH}_2$ groups are equivalent: $2 \times 2 = 4$ protons.
Each $\text{CH}_2$ is bonded to a $\text{CH}_3$ (3 adjacent H), so splitting = $3 + 1 = 4$ → quartet.

Peak at $\delta = 7.1$ ppm:

This is in the aromatic region (6.0–9.0 ppm).
The benzene ring has 6 positions; 2 are occupied by ethyl groups, leaving 4 aromatic H.
Number of protons = 4.
(Group and splitting not required — crossed out in the question.)

Key Takeaways

Symmetry makes equivalent groups give a single signal.
The $n+1$ rule applies to protons on adjacent carbons only.
Chemical shift ranges help identify the type of proton environment.

Common Mistakes

Writing 3 H instead of 6 H for the $\text{CH}_3$ peak (forgetting that both ethyl groups are equivalent).
Confusing the splitting: $\text{CH}_3$ gives a triplet (adjacent to $\text{CH}_2$ with 2H), not a quartet.
Writing 2 aromatic protons instead of 4.

Things to Be Careful About

The number of protons must account for symmetry — multiply by the number of equivalent groups.
Splitting is determined by neighbouring protons on adjacent carbons, not within the same group.

Techniques used

assign proton environments from a molecular structuredetermine splitting patterns using the n+1 rulecorrelate chemical shifts with proton environments

(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.

Working

Compound D (1,2-diethylbenzene):

Plane of symmetry bisects the C1-C2 bond and the C4-C5 bond.
Equivalent carbons: both $\text{CH}_3$ (1 signal), both $\text{CH}_2$ (1 signal), C1=C2 (1 signal), C4=C5 (1 signal), C3=C6 (1 signal).
Total: 5 peaks.

Compound E (1,3-diethylbenzene):

Plane of symmetry passes through C2 and C5.
Equivalent carbons: both $\text{CH}_3$ (1 signal), both $\text{CH}_2$ (1 signal), C1=C3 (1 signal), C2 (1 signal), C4=C6 (1 signal), C5 (1 signal).
Total: 6 peaks.

Final answer

D: 5 peaks; E: 6 peaks

Detailed explanation

Background Concept

In $^{13}\text{C}$ NMR, each chemically distinct carbon environment produces one signal. Symmetry reduces the number of signals: carbons that are related by a symmetry operation (mirror plane, rotation axis) are equivalent and give a single peak. The number of peaks therefore equals the number of unique carbon environments in the molecule.

Understanding the Question

Compounds D (1,2-diethylbenzene) and E (1,3-diethylbenzene) are positional isomers. The question asks how many $^{13}\text{C}$ peaks each produces, which depends on the symmetry of each molecule.

Approach

Identify the symmetry elements in each molecule.
Group carbons into equivalent sets based on symmetry.
Count the number of unique sets = number of peaks.

Step-by-Step Reasoning

Compound D (1,2-diethylbenzene):
The molecule has a mirror plane that bisects the C1-C2 bond and passes through the midpoint of the C4-C5 bond. This makes:

The two $\text{CH}_3$ carbons equivalent → 1 signal
The two $\text{CH}_2$ carbons equivalent → 1 signal
C1 and C2 (ring carbons bearing substituents) equivalent → 1 signal
C4 and C5 (para to each other, ortho to one substituent) equivalent → 1 signal
C3 and C6 (meta to one substituent, para to the other) equivalent → 1 signal

Total: 5 peaks

Compound E (1,3-diethylbenzene):
The molecule has a mirror plane passing through C2 and C5 (the carbon between the two substituents and the carbon opposite to it). This makes:

The two $\text{CH}_3$ carbons equivalent → 1 signal
The two $\text{CH}_2$ carbons equivalent → 1 signal
C1 and C3 (bearing substituents) equivalent → 1 signal
C2 (between the two substituents) unique → 1 signal
C4 and C6 equivalent → 1 signal
C5 (opposite C2) unique → 1 signal

Total: 6 peaks

Key Takeaways

$^{13}\text{C}$ NMR counts unique carbon environments.
Symmetry analysis is the key to determining the number of peaks.
Positional isomers can have different numbers of peaks despite having the same molecular formula.

Common Mistakes

Forgetting that the $\text{CH}_3$ and $\text{CH}_2$ carbons are each equivalent by symmetry (counting them separately).
Misidentifying the symmetry plane in one of the isomers.
Confusing the number of peaks with the total number of carbon atoms (both have 10 carbons).

Things to Be Careful About

Each carbon environment must be checked for equivalence carefully.
The two isomers differ in their symmetry, leading to different peak counts (5 vs 6), which is how they are distinguished.

Techniques used

count distinct carbon environments using molecular symmetrycompare symmetry of positional isomers

(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

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

Strong oxidising agents such as alkaline $\text{KMnO}_4$ (followed by acidification) oxidise alkyl side chains on benzene rings completely to $\text{-COOH}$ groups, provided there is at least one hydrogen on the benzylic carbon. Each $\text{-CH}_2\text{CH}_3$ group is converted to $\text{-COOH}$ , with the terminal carbon being lost as $\text{CO}_2$ . This is a useful reaction for determining the position of substituents on a benzene ring.

Understanding the Question

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

Approach

Write the unbalanced equation with known formulae.
Balance C, H, and O atoms.
Determine the coefficient of $[\text{O}]$ needed.

Step-by-Step Reasoning

Unbalanced equation:

\text{C}_{10}\text{H}_{14} + x[\text{O}] \rightarrow \text{C}_8\text{H}_6\text{O}_4 + a\text{CO}_2 + b\text{H}_2\text{O}

Balance carbon:
Left: 10 C. Right: 8 (in F) + $a$ (in $\text{CO}_2$ ). So $a = 2$ .

Balance hydrogen:
Left: 14 H. Right: 6 (in F) + $2b$ (in $\text{H}_2\text{O}$ ). So $2b = 8$ , $b = 4$ .

Balance oxygen:
Right: 4 (in F) + $2 \times 2 = 4$ (in $\text{CO}_2$ ) + $4 \times 1 = 4$ (in $\text{H}_2\text{O}$ ) = 12 oxygen atoms total.
Left: $x$ oxygen atoms from $[\text{O}]$ .
So $x = 12$ .

Final balanced equation:

\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}

Key Takeaways

Each ethyl group loses one carbon (as $\text{CO}_2$ ) and gains an oxygen to become $\text{-COOH}$ .
Balancing with $[\text{O}]$ means counting oxygen atoms needed on the product side.
The molecular formula of phthalic acid is $\text{C}_8\text{H}_6\text{O}_4$ (benzene ring $\text{C}_6\text{H}_4$ + two $\text{COOH}$ groups).

Common Mistakes

Forgetting to include $\text{CO}_2$ as a product (the terminal carbons of the ethyl groups are lost).
Incorrectly balancing the oxygen atoms.
Writing the molecular formula of F incorrectly (e.g., $\text{C}_8\text{H}_4\text{O}_4$ instead of $\text{C}_8\text{H}_6\text{O}_4$ ).

Things to Be Careful About

The question specifies using $[\text{O}]$ , not $\text{O}_2$ or a full formula for the oxidising agent.
All atoms must balance: check C, H, and O independently.

Techniques used

balance a redox equation using [O] notationdetermine molecular formulae from structural information

(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)

Compound J ( $\text{C}_{10}\text{H}_8\text{O}_4$ ): cyclic diester (one molecule of G reacts with one molecule of ethane-1,2-diol, forming a ring)

Compound K ( $\text{C}_{18}\text{H}_{12}\text{Cl}_2\text{O}_6$ ): dimeric product (one molecule of ethane-1,2-diol bridges two molecules of G, with one acyl chloride group on each G remaining unreacted)

Working

G: F is benzene-1,2-dicarboxylic acid ( $\text{C}_8\text{H}_6\text{O}_4$ ). Reaction with excess $\text{SOCl}_2$ converts both $\text{-COOH}$ groups to $\text{-COCl}$ groups, giving $\text{C}_8\text{H}_4\text{Cl}_2\text{O}_2$ (phthaloyl chloride). This matches the given formula.

J: One $\text{-COCl}$ from each side of G reacts with the two $\text{-OH}$ groups of ethane-1,2-diol, forming a cyclic diester. Molecular formula: $\text{C}_8\text{H}_4\text{O}_2$ (from G, losing 2 Cl) + $\text{C}_2\text{H}_4\text{O}_2$ (from diol, losing 2 H) = $\text{C}_{10}\text{H}_8\text{O}_4$ . ✓

K: One molecule of ethane-1,2-diol reacts with one $\text{-COCl}$ from each of two molecules of G, forming a bridge. Each G retains one unreacted $\text{-COCl}$ . Molecular formula: $2 \times \text{C}_8\text{H}_4\text{ClO}$ (G losing 1 Cl each) + $\text{C}_2\text{H}_4\text{O}_2$ (diol losing 2 H) = $\text{C}_{18}\text{H}_{12}\text{Cl}_2\text{O}_6$ . ✓

Final answer

See structures: G = phthaloyl chloride, J = cyclic diester, K = dimeric diester with two remaining COCl groups

Detailed explanation

Background Concept

Acyl chlorides are formed when carboxylic acids react with $\text{SOCl}_2$ (thionyl chloride). Each $\text{-COOH}$ group is converted to $\text{-COCl}$ with loss of $\text{HCl}$ and $\text{SO}_2$ . Acyl chlorides are highly reactive towards nucleophiles including alcohols, forming esters via addition-elimination (nucleophilic acyl substitution). When a diacyl chloride reacts with a diol, multiple products are possible depending on stoichiometry: intramolecular cyclisation (if both acyl chloride groups react with the same diol molecule), or intermolecular bridging (if one acyl chloride from each of two diacyl chloride molecules reacts with one diol).

Understanding the Question

Compound F (phthalic acid) reacts with excess $\text{SOCl}_2$ to give G. G then reacts with ethane-1,2-diol to give a mixture including J and K. The molecular formulae are given, and structures must be deduced and drawn.

Approach

G: Recognise that $\text{SOCl}_2$ converts both $\text{-COOH}$ groups to $\text{-COCl}$ . Verify the formula matches.
J: A cyclic ester forms when both $\text{-COCl}$ groups of one G molecule react with both $\text{-OH}$ groups of one diol molecule. Verify the formula.
K: A dimeric product forms when one diol molecule bridges two G molecules (one $\text{-COCl}$ from each G reacts with one $\text{-OH}$ from the diol), leaving one $\text{-COCl}$ unreacted on each G. Verify the formula.

Step-by-Step Reasoning

Compound G ( $\text{C}_8\text{H}_4\text{Cl}_2\text{O}_2$ ):

F is $\text{C}_6\text{H}_4(\text{COOH})_2$ = $\text{C}_8\text{H}_6\text{O}_4$ .
Replacing both $\text{-OH}$ with $\text{-Cl}$ : $\text{C}_6\text{H}_4(\text{COCl})_2$ = $\text{C}_8\text{H}_4\text{Cl}_2\text{O}_2$ . ✓
This is phthaloyl chloride (benzene-1,2-dicarbonyl dichloride).

Compound J ( $\text{C}_{10}\text{H}_8\text{O}_4$ ):

G ( $\text{C}_8\text{H}_4\text{Cl}_2\text{O}_2$ ) + diol ( $\text{C}_2\text{H}_6\text{O}_2$ ) → cyclic ester + 2HCl
Product: $\text{C}_8\text{H}_4\text{O}_2$ + $\text{C}_2\text{H}_4\text{O}_2$ = $\text{C}_{10}\text{H}_8\text{O}_4$ . ✓
Structure: a benzene ring with two adjacent $\text{-COO-}$ groups connected by a $\text{-CH}_2\text{-CH}_2\text{-}$ bridge, forming a 10-membered ring (benzene + two C=O + two O + two CH2).

Compound K ( $\text{C}_{18}\text{H}_{12}\text{Cl}_2\text{O}_6$ ):

2G ( $2 \times \text{C}_8\text{H}_4\text{Cl}_2\text{O}_2$ ) + diol ( $\text{C}_2\text{H}_6\text{O}_2$ ) → K + 2HCl
Product: $2(\text{C}_8\text{H}_4\text{ClO})$ + $\text{C}_2\text{H}_4\text{O}_2$ = $\text{C}_{18}\text{H}_{12}\text{Cl}_2\text{O}_6$ . ✓
Structure: two phthaloyl chloride units connected by an ethylene bridge ( $\text{-CH}_2\text{CH}_2\text{-}$ ) through ester linkages, with one remaining $\text{-COCl}$ group on each benzene ring.

Key Takeaways

$\text{SOCl}_2$ converts $\text{-COOH}$ to $\text{-COCl}$ .
Diacyl chlorides + diols can form cyclic esters (intramolecular) or bridged products (intermolecular).
Always verify molecular formulae match the proposed structure.

Common Mistakes

Drawing J as an open-chain diester rather than a cyclic structure.
Forgetting that K has two remaining $\text{-COCl}$ groups (one on each ring).
Incorrectly placing the ester linkages (the $\text{-COO-}$ must connect the ring carbon to the $\text{CH}_2$ of the diol).

Things to Be Careful About

The ester linkage direction: $\text{Ar-CO-O-CH}_2$ (not $\text{Ar-O-CO-CH}_2$ ).
In K, the two G units are connected by the diol, and each G still has one free $\text{-COCl}$ .
Check that all molecular formulae are consistent with the drawn structures.

Techniques used

deduce structure from molecular formulaidentify acyl chloride formation from carboxylic acid and SOCl2predict products of diacyl chloride reacting with diolconstruct cyclic ester structureconstruct dimeric ester structure with remaining acyl chloride groups

Q8MediumHalogen CompoundsHydrocarbons

Bromine reacts with methylbenzene in the dark in the presence of a suitable catalyst to form $\text{HBr}$ 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 the halogenation of arenes (such as benzene or methylbenzene), the halogen molecule ( $\text{Br}_2$ or $\text{Cl}_2$ ) is not electrophilic enough on its own to react with the stable delocalized $\pi$ system of the aromatic ring. A Lewis acid catalyst, typically a halogen compound of a Group 3 element like $\text{AlBr}_3$ or $\text{FeBr}_3$ , is required to polarize the halogen bond and generate a strong electrophile.

Understanding the Question

The question asks to complete the equation showing how the $\text{Br}^+$ electrophile is generated when bromine reacts with the catalyst. The catalyst is a Lewis acid that accepts an electron pair.

Approach

Bromine ( $\text{Br}_2$ ) reacts with aluminium bromide ( $\text{AlBr}_3$ ). The $\text{AlBr}_3$ acts as a Lewis acid, accepting a lone pair from one of the bromine atoms. This weakens the $\text{Br-Br}$ bond, leading to heterolytic fission and the formation of the bromonium ion ( $\text{Br}^+$ ) and the tetrabromoaluminate(III) ion ( $\text{AlBr}_4^-$ ).

Step-by-Step Reasoning

Identify the catalyst: For bromination of arenes, the standard catalyst is $\text{AlBr}_3$ (or $\text{FeBr}_3$ ).
Write the reactants: $\text{Br}_2 + \text{AlBr}_3$ .
Determine the products: The $\text{AlBr}_3$ accepts a bromide ion ( $\text{Br}^-$ ) from $\text{Br}_2$ , leaving behind $\text{Br}^+$ . The resulting anion is $\text{AlBr}_4^-$ .
Balance the equation: $\text{Br}_2 + \text{AlBr}_3 \rightarrow \text{Br}^+ + \text{AlBr}_4^-$ .

Key Takeaways

The Lewis acid catalyst is essential to create a sufficiently strong electrophile ( $\text{Br}^+$ ) for electrophilic aromatic substitution. The byproduct is the complex ion $\text{AlBr}_4^-$ .

Common Mistakes

Writing $\text{Fe}$ or $\text{FeBr}_3$ instead of $\text{AlBr}_3$ (both are acceptable in practice, but $\text{AlBr}_3$ is the standard expected answer unless specified; if $\text{FeBr}_3$ is used, the product is $\text{FeBr}_4^-$ ).
Forgetting the charge on the ions ( $\text{Br}^+$ and $\text{AlBr}_4^-$ ).
Writing $\text{AlBr}_2^-$ instead of $\text{AlBr}_4^-$ .

Things to Be Careful About

Ensure state symbols are not required unless specified, but charges must be correct. The equation must be balanced in terms of atoms and charge.

Techniques used

generate the electrophile in electrophilic aromatic substitution

(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:

Explanation:
The methyl group ( $-\text{CH}_3$ ) is an activating group and is ortho/para-directing. It increases the electron density at the 2, 4, and 6 positions of the ring through inductive electron donation and hyperconjugation. Therefore, the electrophile ( $\text{Br}^+$ ) preferentially attacks the ortho and para positions. The meta position (3) is not activated to the same extent, so the meta-isomer (3-bromo-1-methylbenzene) forms in the smallest amount.

Final answer

3-bromo-1-methylbenzene (meta-isomer); methyl group is ortho/para directing

Detailed explanation

Background Concept

When an alkyl group like methyl ( $-\text{CH}_3$ ) is attached to a benzene ring, it influences the distribution of incoming electrophiles. Alkyl groups are electron-donating via the inductive effect and hyperconjugation. This increases the electron density of the aromatic $\pi$ system, making the ring more reactive than benzene (activation). Furthermore, this electron density is not distributed evenly; it is concentrated most at the ortho (2, 6) and para (4) positions relative to the methyl group. Consequently, alkyl groups are ortho/para directors.

Understanding the Question

Methylbenzene reacts with $\text{Br}^+$ to form three isomeric products: 2-bromo-1-methylbenzene (ortho), 3-bromo-1-methylbenzene (meta), and 4-bromo-1-methylbenzene (para). The question asks to identify and draw the isomer that forms in the smallest amount and explain why.

Approach

Since the methyl group directs incoming electrophiles to the ortho and para positions, the meta position is the least favored. Therefore, the meta-isomer (3-bromo-1-methylbenzene) will be the minor product. We draw its structure and state that the methyl group is ortho/para directing.

Step-by-Step Reasoning

Identify the isomers: Bromination of methylbenzene yields ortho, meta, and para isomers.
Determine the minor product: The methyl group is an ortho/para director. Thus, ortho and para isomers are the major products, and the meta isomer is the minor product.
Draw the structure: The meta isomer is 3-bromo-1-methylbenzene. Draw a benzene ring with a $-\text{CH}_3$ group at position 1 and a $-\text{Br}$ atom at position 3.
Explain the reason: The methyl group donates electron density into the ring, particularly activating the ortho and para positions. The meta position is less activated, so attack there is less favorable, leading to a smaller yield.

Key Takeaways

Substituents on a benzene ring determine the regiochemistry of further electrophilic substitution. Electron-donating groups (like $-\text{CH}_3$ , $-\text{OH}$ ) are ortho/para directors, while electron-withdrawing groups (like $-\text{NO}_2$ , $-\text{COOH}$ ) are meta directors (with halogens being the exception: deactivating but ortho/para directing).

Common Mistakes

Drawing the ortho or para isomer instead of the meta isomer.
Stating that the methyl group is a "meta director" (it is not).
Saying "the methyl group blocks the meta position" (it doesn't block; it simply doesn't activate it as much as ortho/para).

Things to Be Careful About

Ensure the structural formula clearly shows the bromine and methyl groups in the 1,3-relationship (meta). The explanation must mention that the methyl group is ortho/para directing or increases electron density at ortho/para positions.

Techniques used

predict isomer distribution based on directing effectsdraw structural formula of aromatic isomer

(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 from the delocalized $\pi$ system (inside the hexagon) of methylbenzene to the $\text{Br}^+$ ion.
Intermediate: A cyclohexadienyl cation (arenium ion) where the carbon bonded to $\text{Br}$ is also bonded to $\text{H}$ , and the ring has a positive charge delocalized over the remaining five carbons.
Step 2: A curly arrow from the $\text{C-H}$ bond (at the carbon bearing the $\text{Br}$ ) into the ring to restore aromaticity. The products are the substituted arene (e.g., 2-bromo-1-methylbenzene) and $\text{H}^+$ .

(See diagram for full visual representation)

Final answer

See mechanism diagram

Detailed explanation

Background Concept

Electrophilic aromatic substitution (EAS) proceeds via a two-step mechanism:

Electrophilic attack: The electrophile ( $\text{E}^+$ ) attacks the electron-rich $\pi$ system of the aromatic ring, forming a non-aromatic carbocation intermediate (the sigma complex or arenium ion). The aromaticity is temporarily lost.
Deprotonation: A base (often the catalyst complex, e.g., $\text{AlBr}_4^-$ ) removes a proton ( $\text{H}^+$ ) from the carbon bearing the electrophile. The electrons from the $\text{C-H}$ bond return to the ring, restoring the stable delocalized $\pi$ system.

Understanding the Question

The question provides a template (Fig 8.1) showing methylbenzene + $\text{Br}^+$ $\rightarrow$ intermediate $\rightarrow$ product + $\text{H}^+$ . We need to complete the curly arrows and draw the intermediate structure. The mark scheme image shows attack at the ortho position.

Approach

Draw an arrow from the benzene ring's $\pi$ electrons to the $\text{Br}^+$ .
Draw the intermediate: the ring is no longer aromatic. The carbon that attacked the $\text{Br}^+$ is now $sp^3$ hybridized, bonded to both $\text{H}$ and $\text{Br}$ . The positive charge is delocalized around the remaining five carbons of the ring.
Draw an arrow from the $\text{C-H}$ bond on the $sp^3$ carbon into the ring to reform the double bond/aromatic system.
Draw the final product (e.g., ortho-bromotoluene) and $\text{H}^+$ .

Step-by-Step Reasoning

M1 (First arrow): Draw a curly arrow starting from inside the benzene ring (representing the $\pi$ electrons) and pointing to the $\text{Br}^+$ ion. This shows the formation of the new $\text{C-Br}$ bond.
M2 (Intermediate): Draw the intermediate structure. If attack is at the ortho position (carbon 2), that carbon now has four single bonds: to C1, C3, H, and Br. The ring has two double bonds (e.g., between C3-C4 and C5-C6) and a positive charge on C1 (or delocalized). The positive charge can be shown on the carbon adjacent to the $sp^3$ carbon.
M3 (Second arrow and products): Draw a curly arrow from the center of the $\text{C-H}$ bond (on the $sp^3$ carbon) pointing into the ring (towards the adjacent double bond or to restore aromaticity). This shows the loss of $\text{H}^+$ . Draw the final product: methylbenzene with $\text{Br}$ at the ortho position, and $\text{H}^+$ as the other product.

Key Takeaways

The mechanism of EAS involves a loss of aromaticity in the intermediate step, which is why the reaction requires a strong electrophile. The restoration of aromaticity is the driving force for the second step (loss of $\text{H}^+$ ).

Common Mistakes

Drawing the arrow from the $\text{Br}^+$ to the ring (electrons flow from electron-rich to electron-poor, so arrow must start at the ring).
Forgetting the positive charge on the intermediate or the final $\text{H}^+$ .
Drawing the intermediate with a full octet on all carbons and a positive charge on Br (incorrect; Br is neutral in the intermediate).
Not showing the $\text{H}$ atom on the carbon that loses the proton in the intermediate.

Things to Be Careful About

Curly arrows must show the movement of electron pairs, not atoms.
The intermediate is a carbocation; ensure the charge is clearly marked.
The final step produces $\text{H}^+$ (or $\text{HAlBr}_4$ if the base is $\text{AlBr}_4^-$ , but the template specifically asks for $\text{H}^+$ + ...).

Techniques used

draw electrophilic aromatic substitution mechanismuse curly arrows for electron movement

(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

DifficultyMedium-Easy

Worked solution

Answer

Chlorobutane reacts.
Observation: A white precipitate forms.

Final answer

Chlorobutane; white precipitate

Detailed explanation

Background Concept

Halogenoalkanes (alkyl halides) undergo nucleophilic substitution reactions, where the halogen is replaced by a nucleophile (like $\text{OH}^-$ or $\text{H}_2\text{O}$ ). When a halogenoalkane is warmed with aqueous silver nitrate ( $\text{AgNO}_3$ ), the halogen is hydrolyzed to release halide ions ( $\text{X}^-$ ). These ions react with silver ions ( $\text{Ag}^+$ ) to form an insoluble silver halide precipitate. Chlorides form a white precipitate ( $\text{AgCl}$ ), bromides form a cream precipitate ( $\text{AgBr}$ ), and iodides form a yellow precipitate ( $\text{AgI}$ ).

Arenes (halogenoarenes) like chlorobenzene do not undergo nucleophilic substitution under these mild conditions because the $\text{C-Cl}$ bond is strengthened by delocalization.

Understanding the Question

Two compounds, chlorobutane (a halogenoalkane) and chlorobenzene (a halogenoarene), are added to warm aqueous $\text{AgNO}_3$ . We need to identify which one reacts and describe the observation.

Approach

Chlorobutane is a typical halogenoalkane and will hydrolyze slowly in warm water/aqueous conditions to release $\text{Cl}^-$ ions. These ions will precipitate as $\text{AgCl}$ . Chlorobenzene is unreactive under these conditions. Therefore, chlorobutane reacts, producing a white precipitate.

Step-by-Step Reasoning

Identify the reactive compound: Chlorobutane ( $\text{C}_4\text{H}_9\text{Cl}$ ) is a halogenoalkane and can undergo hydrolysis. Chlorobenzene ( $\text{C}_6\text{H}_5\text{Cl}$ ) is a halogenoarene and is unreactive towards nucleophilic substitution under these conditions.
Describe the observation: The hydrolysis of chlorobutane releases chloride ions ( $\text{Cl}^-$ ). These react with $\text{Ag}^+$ to form silver chloride ( $\text{AgCl}$ ), which is a white precipitate.

Key Takeaways

Halogenoalkanes react with aqueous silver nitrate (often with ethanol as a co-solvent and warming) to give silver halide precipitates. Halogenoarenes do not react under these conditions due to the strength of the $\text{C-Cl}$ bond.

Common Mistakes

Saying chlorobenzene reacts (it does not; it requires extreme conditions like high temperature and pressure with concentrated $\text{NaOH}$ ).
Describing the precipitate color incorrectly (chloride = white, bromide = cream, iodide = yellow).
Not specifying that the precipitate is silver chloride.

Things to Be Careful About

Ensure the observation is specific: "white precipitate". Simply saying "a precipitate forms" may not earn full marks if the color is not specified.

Techniques used

identify reactive halogenoalkane in nucleophilic substitutiondescribe precipitation observations

(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} \text{ (s)}

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. Water acts as a weak nucleophile. The reaction is slow and is accelerated by warming. The products are an alcohol and hydrohalic acid (which dissociates into $\text{H}^+$ and $\text{X}^-$ ).

The precipitation of silver halides is a standard test for halide ions. The equation is a simple ionic combination: $\text{Ag}^+(\text{aq}) + \text{X}^-(\text{aq}) \rightarrow \text{AgX}(\text{s})$ .

Understanding the Question

We need two equations to explain the observation in (b)(i): the release of chloride ions from chlorobutane and their subsequent precipitation with silver ions.

Approach

Write the hydrolysis of chlorobutane. Since the reagent is aqueous $\text{AgNO}_3$ , water is the nucleophile. The equation is $\text{C}_4\text{H}_9\text{Cl} + \text{H}_2\text{O} \rightarrow \text{C}_4\text{H}_9\text{OH} + \text{H}^+ + \text{Cl}^-$ . (Using $\text{OH}^-$ is also acceptable if considering the basic conditions, but hydrolysis by water is more precise for neutral aqueous silver nitrate).
Write the ionic equation for the precipitation of silver chloride: $\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}$ .

Step-by-Step Reasoning

M1: Hydrolysis equation. Chlorobutane reacts with water to form 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: $\text{C}_4\text{H}_9\text{OH}$ represents butanol. State symbols are often not required for organic hydrolysis unless specified, but (aq) for ions and (s) for precipitate are good practice).
M2: Precipitation equation. The chloride ions react with silver ions from the silver nitrate. $\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}$ . (Include state symbol (s) for AgCl if possible, though the mark scheme doesn't strictly demand it in the text provided, it's good practice).

Key Takeaways

Hydrolysis of halogenoalkanes releases halide ions, which can be detected by silver nitrate solution. The two equations link the organic reaction to the inorganic observation.

Common Mistakes

Writing the hydrolysis equation with $\text{NaOH}$ instead of $\text{H}_2\text{O}$ or $\text{OH}^-$ . While $\text{OH}^-$ is acceptable, the reagent is aqueous $\text{AgNO}_3$ , so water is the primary nucleophile.
Forgetting to balance the hydrolysis equation (producing $\text{H}^+$ and $\text{Cl}^-$ ).
Writing molecular equations instead of ionic equations for the precipitation step (e.g., $\text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3$ ). The mark scheme accepts the ionic form $\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}$ .

Things to Be Careful About

Ensure the organic formula is correct: $\text{C}_4\text{H}_9\text{Cl}$ and $\text{C}_4\text{H}_9\text{OH}$ . The precipitation equation must show $\text{Ag}^+$ and $\text{Cl}^-$ combining.

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

Worked solution

Answer

In chlorobenzene, the lone pair of electrons on the chlorine atom is delocalized into the $\pi$ system (delocalized electron cloud) of the benzene ring. This gives the $\text{C-Cl}$ bond partial double bond character, making it stronger and harder to break. In chlorobutane, there is no delocalization, so the $\text{C-Cl}$ bond is a weaker single bond and is more easily broken (hydrolysed).

Final answer

Delocalization of Cl lone pair into benzene ring strengthens C-Cl bond

Detailed explanation

Background Concept

The reactivity of a carbon-halogen bond depends on its bond enthalpy (strength). In halogenoalkanes, the $\text{C-X}$ bond is a polar single bond, and the carbon is electrophilic, susceptible to nucleophilic attack.

In halogenoarenes (like chlorobenzene), the halogen atom is attached directly to an $sp^2$ hybridized carbon of the aromatic ring. The lone pair on the halogen atom can overlap with the $p$ -orbitals of the aromatic $\pi$ system. This is a resonance effect (delocalization). The lone pair is partially shared with the ring, giving the $\text{C-X}$ bond partial double bond character. Double bonds are shorter and stronger than single bonds, so the $\text{C-Cl}$ bond in chlorobenzene is stronger and less reactive towards nucleophilic substitution than in chlorobutane.

Understanding the Question

We need to explain why chlorobutane reacts with warm aqueous $\text{AgNO}_3$ (hydrolysis occurs) while chlorobenzene does not.

Approach

Discuss the electronic structure of chlorobenzene: delocalization of the chlorine lone pair into the benzene ring.
Explain the consequence: partial double bond character / stronger $\text{C-Cl}$ bond.
Contrast with chlorobutane: no delocalization, weaker single bond, easily broken.

Step-by-Step Reasoning

M1 (Delocalization): In chlorobenzene, the lone pair on the chlorine atom is delocalized into the delocalized $\pi$ system (or p-orbitals) of the benzene ring. This resonance stabilization involves the overlap of the chlorine $p$ -orbital with the ring $\pi$ system.
M2 (Bond strength): This delocalization gives the $\text{C-Cl}$ bond in chlorobenzene partial double bond character. As a result, the bond is stronger (higher bond enthalpy) and requires more energy to break. In chlorobutane, the $\text{C-Cl}$ bond is a normal single bond with no such delocalization, so it is weaker and more easily hydrolysed.

Key Takeaways

The reactivity of halogenoarenes is much lower than halogenoalkanes towards nucleophilic substitution due to the resonance interaction between the halogen lone pair and the aromatic ring, which strengthens the $\text{C-X}$ bond.

Common Mistakes

Saying "the benzene ring is stable, so it doesn't react" (too vague; needs to mention the specific effect on the $\text{C-Cl}$ bond).
Saying "the bond is ionic" (it's covalent).
Not mentioning delocalization or partial double bond character.
Confusing this with the reactivity of the ring towards electrophilic substitution (which is about the ring's electron density, not the $\text{C-Cl}$ bond strength).

Things to Be Careful About

Use precise terminology: "delocalized into the $\pi$ system", "partial double bond character", "stronger bond". Do not just say "the bond is stronger" without explaining why.

Techniques used

explain reactivity difference using resonance and bond strength

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 (or C $_6$ H $_5$ NO $_2$ )
reagents: Sn and HCl (or Fe and HCl)
conditions: concentrated HCl and heat (or boil / reflux)

Final answer

nitrobenzene; Sn and HCl; concentrated HCl and heat

Detailed explanation

Background Concept

Aromatic amines such as phenylamine (aniline) are typically prepared by the reduction of nitroarenes. The nitro group (–NO $_2$ ) attached directly to the benzene ring can be reduced to an amino group (–NH $_2$ ) using a strong reducing agent in acidic conditions.

Understanding the Question

The question asks for the organic starting material to produce phenylamine via reduction, along with the specific reagents and conditions required for this transformation.

Approach

Recall the standard industrial and laboratory method for reducing a nitro group on an aromatic ring. The most common reagent system is tin (or iron) with concentrated hydrochloric acid, followed by heating.

Step-by-Step Reasoning

Organic compound: The precursor to phenylamine (C $_6$ H $_5$ NH $_2$ ) is nitrobenzene (C $_6$ H $_5$ NO $_2$ ). Reducing the nitro group yields the amino group.
Reagents: Tin (Sn) or iron (Fe) metal in the presence of hydrochloric acid (HCl) acts as the reducing system. The metal reduces the nitro group while the acid provides the necessary protons.
Conditions: The reaction requires concentrated HCl and heat (boiling or refluxing) to proceed to completion.

Key Takeaways

Nitroarenes are reduced to aromatic amines using Sn/HCl or Fe/HCl under reflux. This is a fundamental preparation method for phenylamine and its derivatives.

Common Mistakes

Writing only "H $_2$ and Ni/Pt catalyst" without specifying that catalytic hydrogenation of nitrobenzene is less commonly the expected answer in this specific syllabus context (though it can work, Sn/HCl is the standard mark-scheme answer).
Forgetting the "concentrated" qualifier for HCl or omitting the heat/reflux condition.
Writing "HCl" alone without the metal reducing agent.

Things to Be Careful About

Ensure the reagents are stated as a pair (metal + acid). Conditions must include both the concentration of the acid and the application of heat. Two correct points out of the three listed will earn full marks, but providing all three is safest.

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 (or CH $_3$ CH $_2$ CN / propylamide / C $_2$ H $_5$ CONH $_2$ )
reagent: LiAlH $_4$ (lithium tetrahydridoaluminate(III))

Final answer

propanenitrile; LiAlH $_4$

Detailed explanation

Background Concept

Aliphatic primary amines can be prepared by the reduction of nitriles (R–CN) or amides (R–CONH $_2$ ). Unlike nitro reduction, these functional groups require a strong hydride reducing agent because the C≡N or C=O bonds are not easily reduced by metal/acid systems.

Understanding the Question

Identify an organic compound that yields propylamine (CH $_3$ CH $_2$ CH $_2$ NH $_2$ ) upon reduction, and name the specific reducing reagent required.

Approach

Work backwards from the product: propylamine has a 3-carbon chain. A nitrile with 3 carbons is propanenitrile (CH $_3$ CH $_2$ CN). An amide with 3 carbons is propanamide (CH $_3$ CH $_2$ CONH $_2$ ). Both are reduced by lithium tetrahydridoaluminate(III) (LiAlH $_4$ ).

Step-by-Step Reasoning

Organic compound: Propanenitrile (CH $_3$ CH $_2$ CN) or propanamide (CH $_3$ CH $_2$ CONH $_2$ ). Reduction of the nitrile adds two hydrogen atoms across the triple bond to form the primary amine. Reduction of the amide replaces the C=O with CH $_2$ .
Reagent: LiAlH $_4$ is a powerful reducing agent capable of reducing nitriles to primary amines and amides to amines. It must be used in a dry ether solvent (like dry ethanol or tetrahydrofuran) followed by an aqueous acid workup, but stating "LiAlH $_4$ " is sufficient for the mark.

Key Takeaways

Nitriles and amides are reduced to amines using LiAlH $_4$ . This is the standard method for extending or maintaining the carbon chain while introducing an amine group.

Common Mistakes

Suggesting NaBH $_4$ as the reducing agent (it is not strong enough to reduce nitriles or amides).
Writing "H $_2$ /Ni" without specifying high pressure/temperature (catalytic hydrogenation can work but LiAlH $_4$ is the specific reagent expected here).
Incorrectly naming the organic compound (e.g., butanenitrile would give butylamine, not propylamine).

Things to Be Careful About

The mark scheme accepts either a nitrile or an amide as the organic compound, but the reagent must be LiAlH $_4$ . Ensure the carbon count in the organic compound matches the 3-carbon chain of propylamine.

Techniques used

identify reduction precursor for aliphatic aminestate reducing agent for nitriles or amides

(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) / bromine water
amine that gives a precipitate: phenylamine (or C $_6$ H $_5$ NH $_2$ )
structure of the compound that is precipitated:

Final answer

bromine(aq); phenylamine; see diagram for 2,4,6-tribromophenylamine

Detailed explanation

Background Concept

Phenylamine (aniline) has an –NH $_2$ group attached directly to the benzene ring. The lone pair on the nitrogen atom can partially delocalise into the ring's pi-system, making the ring highly activated towards electrophilic substitution. This activation is so strong that phenylamine reacts rapidly with bromine water at room temperature without a catalyst, substituting all three available ortho and para positions (2, 4, and 6) to form 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. They may react with bromine in other ways (e.g., oxidation or substitution on the alkyl chain under different conditions), but they do not produce an immediate white precipitate with bromine water under standard test conditions.

Understanding the Question

Identify a single test that produces a white precipitate with only one of the two amines (phenylamine or propylamine), name the amine that reacts, and draw the structure of the precipitated product.

Approach

Use the characteristic test for activated aromatic rings: bromine water. Phenylamine will decolourise bromine water and form a white precipitate. Draw the 2,4,6-tribromo derivative of phenylamine.

Step-by-Step Reasoning

Testing reagent: Bromine water (bromine(aq)).
Amine that gives a precipitate: Phenylamine (C $_6$ H $_5$ NH $_2$ ). The –NH $_2$ group strongly activates the ring, leading to rapid trisubstitution.
Structure of precipitated compound: 2,4,6-tribromophenylamine. The bromine atoms substitute at the two ortho positions (adjacent to –NH $_2$ ) and the para position (opposite to –NH $_2$ ). The –NH $_2$ group remains at position 1.

Key Takeaways

Phenylamine undergoes electrophilic substitution with bromine water to give 2,4,6-tribromophenylamine as a white precipitate. This is a definitive test to distinguish aromatic amines from aliphatic amines.

Common Mistakes

Suggesting the amine gives a precipitate with bromine water (propylamine does not).
Drawing only a mono- or di-brominated product (the reaction goes to trisubstitution due to the strong activating effect of –NH $_2$ ).
Forgetting to show the –NH $_2$ group in the product structure.

Things to Be Careful About

The structure must clearly show three bromine atoms at the 2, 4, and 6 positions relative to the amino group. A skeletal structure with a benzene ring, –NH $_2$ at the top, and Br atoms at the two adjacent vertices and the bottom vertex is correct. Ensure the formula matches C $_6$ H $_3$ Br $_3$ NH $_2$ .

Techniques used

identify electrophilic aromatic substitution test for activated benzene ringdraw substituted product structure

(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} \hspace{150pt} \text{most basic}

4M Similar

DifficultyMedium

Worked solution

Answer

phenylamine < ammonia < propylamine

Explanation:

A base accepts a proton (H $^+$ ) using the lone pair on the nitrogen atom.
In phenylamine, the lone pair on the nitrogen is delocalised into the pi-system of the benzene ring, 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 via the inductive effect, 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 $^+$ ). Two main electronic effects influence this availability:

Delocalisation: If the lone pair is delocalised into an adjacent pi-system (such as a benzene ring), it is less available for protonation, reducing 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 for protonation, thus increasing basicity.

Ammonia (NH $_3$ ) has no alkyl groups and no delocalisation, serving as the baseline for comparison.

Understanding the Question

Describe the relative basicities of phenylamine, ammonia, and propylamine, and explain the reasoning using electronic effects.

Approach

Compare the three compounds:

Phenylamine: lone pair delocalised into benzene ring → least basic.
Ammonia: no delocalisation, no alkyl groups → intermediate basicity.
Propylamine: electron-donating alkyl group increases electron density on N → most basic.

Step-by-Step Reasoning

Order of basicity: phenylamine < ammonia < propylamine.
General definition: A base is a proton acceptor; the nitrogen atom must have a lone pair available to accept a proton (H $^+$ ).
Phenylamine vs. ammonia: In phenylamine, the lone pair on the nitrogen atom is delocalised into the pi-system of the benzene ring. This delocalisation stabilises the lone pair and makes it less available to bond with a proton, so phenylamine is a weaker base than ammonia.
Propylamine vs. ammonia: In propylamine, the propyl group (–CH $_2$ CH $_2$ CH $_3$ ) is an electron-donating alkyl group. Through the positive inductive effect (+I), 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, so propylamine is a stronger base than ammonia.

Key Takeaways

Basicity of amines is governed by the availability of the nitrogen lone pair. Delocalisation into an aromatic ring decreases basicity, while electron-donating alkyl groups increase basicity via the inductive effect.

Common Mistakes

Stating that phenylamine is more basic than ammonia because it has a carbon group attached (forgetting the delocalisation effect).
Saying "alkyl groups donate electrons" without specifying the inductive effect or linking it to lone pair availability.
Forgetting to define a base as a proton acceptor using the lone pair.
Writing the order incorrectly (e.g., ammonia < phenylamine < propylamine).

Things to Be Careful About

The explanation must explicitly mention the lone pair on nitrogen and its availability to accept a proton.
For phenylamine, use the term delocalised into the pi-system or benzene ring.
For propylamine, mention the electron-donating nature of the alkyl group and the inductive effect.
Ensure the order matches the explanation: least basic to most basic.

Techniques used

compare basicity of amines using electronic effectsexplain delocalisation and inductive effects on lone pair availability

Q10MediumIntroduction to A Level Organic ChemistryCarboxylic 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 showing HO-C(=O)-C(=O)-OH with all atoms and bonds drawn.

Detailed explanation

Background Concept

A displayed (or full structural) formula shows every atom in the molecule and every bond between them. For carboxylic acids, this means drawing the C=O double bond, the C-O single bond, and the O-H bond in the hydroxyl group, as well as any C-H bonds on the carbon chain.

Understanding the Question

The question asks for the displayed formula of ethanedioic acid (commonly known as oxalic acid). Ethanedioic acid is a dicarboxylic acid with the molecular formula C $_2$ H $_2$ O $_4$ and the condensed structural formula HOOC-COOH. It consists of two carboxyl groups (-COOH) joined directly to each other.

Approach

Identify the carbon chain length (two carbons) and the functional groups (two carboxyl groups). Draw the C-C single bond, then attach the C=O and C-O-H groups to each carbon. Ensure all hydrogen atoms are shown.

Step-by-Step Reasoning

Start with the two carbon atoms bonded to each other: C-C.
To each carbon, attach a double bond to an oxygen atom: C=O.
To each carbon, also attach a single bond to an oxygen atom, which is in turn bonded to a hydrogen atom: C-O-H.
The final structure is HO-C(=O)-C(=O)-OH, with all bonds and atoms explicitly drawn. No hydrogen atoms are attached directly to the carbon chain in ethanedioic acid.

Key Takeaways

Displayed formulas must show every single bond and every atom, including hydrogens in functional groups like -OH. This is distinct from a structural formula (which groups atoms like OH) or a skeletal formula (which hides carbons and hydrogens on the chain).

Common Mistakes

Drawing a skeletal or condensed formula instead of a displayed formula (e.g., writing HOOC-COOH or drawing lines without explicit O and H atoms).
Forgetting the hydrogen atom on the hydroxyl oxygen (writing C-O instead of C-O-H).
Adding hydrogen atoms to the carbon atoms (the carbons in ethanedioic acid have no C-H bonds; their four bonds are used for C-C, C=O, and C-O).

Things to Be Careful About

Ensure the C=O bonds are drawn as double lines and the C-O and O-H bonds as single lines. The geometry around each carbon is trigonal planar, but for a displayed formula, a 2D representation showing all connections is sufficient as long as all bonds are explicit.

Techniques used

draw displayed formulashow all bonds and atoms explicitly

(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

reagent 1: Tollens' reagent (or Fehling's solution)
reagent 2: acidified potassium manganate(VII) (or acidified MnO $_4^-$ )

	observation when treated with reagent 1	observation when treated with reagent 2
propanoic acid	no change	no change
methanoic acid	silver mirror (or red precipitate)	colourless
ethanedioic acid	no change	colourless

Final answer

reagent 1: Tollens' reagent; reagent 2: acidified KMnO4; observations: propanoic (no change, no change), methanoic (silver mirror/red ppt, colourless), ethanedioic (no change, colourless)

Detailed explanation

Background Concept

Methanoic acid (HCOOH) is unique among carboxylic acids because it contains a hydrogen atom attached directly to the carbonyl carbon, giving it an aldehyde-like structure (H-C=O). This allows it to act as a reducing agent and be oxidized to carbon dioxide and water. Ethanedioic acid (HOOC-COOH) is also easily oxidized by strong oxidizing agents like acidified potassium manganate(VII) to produce CO $_2$ and H $_2$ O. Propanoic acid (CH $_3$ CH $_2$ COOH) lacks these structural features and is resistant to oxidation by these reagents.

Understanding the Question

The question asks to distinguish three weak acids using two chemical tests, neither of which is an acid-base indicator. We need to identify two reagents that give different, observable results for at least some of these acids, and complete a table of observations.

Approach

Identify the unique structural feature of methanoic acid (aldehyde-like H) and choose a mild oxidizing agent that tests for it (Tollens' or Fehling's).
Identify the oxidizability of ethanedioic acid and choose a strong oxidizing agent that will react with it but not propanoic acid (acidified MnO $_4^-$ ).
Predict the observations for each acid with each reagent.

Step-by-Step Reasoning

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

Tollens' reagent (ammoniacal silver nitrate) is a mild oxidizing agent. It oxidizes aldehydes to carboxylic acids, reducing Ag $^+$ to metallic silver.
Methanoic acid has an aldehyde-like group, so it reduces Tollens' reagent to give a silver mirror (or with Fehling's, a red precipitate of Cu $_2$ O).
Propanoic acid and ethanedioic acid do not have this aldehyde-like group (ethanedioic acid is not oxidized by mild reagents like Tollens'), so there is no change.

Reagent 2: Acidified potassium manganate(VII) (KMnO $_4$ )

Acidified MnO $_4^-$ is a strong oxidizing agent (purple solution).
Ethanedioic acid is oxidized by acidified MnO $_4^-$ to CO $_2$ and H $_2$ O, reducing MnO $_4^-$ to Mn $^{2+}$ (which is pale pink/essentially colourless in dilute solution). Thus, the purple colour disappears.
Methanoic acid is also oxidized by acidified MnO $_4^-$ to CO $_2$ and H $_2$ O, so the solution also turns colourless.
Propanoic acid is not oxidized, so the solution remains purple (or no change in colour).

Completing the table:

Propanoic acid: no change with Tollens'; no change (purple remains) with acidified MnO $_4^-$ .
Methanoic acid: silver mirror/red ppt with Tollens'; colourless with acidified MnO $_4^-$ .
Ethanedioic acid: no change with Tollens'; colourless with acidified MnO $_4^-$ .

Key Takeaways

Methanoic acid's dual nature (carboxylic acid + aldehyde) makes it uniquely reactive towards mild oxidizing agents. Ethanedioic acid's structure makes it susceptible to strong oxidizing agents like acidified manganate(VII). Propanoic acid is relatively inert to both.

Common Mistakes

Suggesting an acid-base indicator (e.g., universal indicator or pH paper) as a test. The question explicitly excludes these.
Confusing the observations: Tollens' gives a silver mirror (not a black precipitate, which is copper with Fehling's if misremembered, though Fehling's gives red Cu $_2$ O). Acidified MnO $_4^-$ goes from purple to colourless, not green (green is MnO $_4^{2-}$ in alkaline conditions).
Forgetting that methanoic acid also decolorizes acidified MnO $_4^-$ .

Things to Be Careful About

Ensure reagent names are precise: "Tollens' reagent" or "Fehling's solution", not just "silver nitrate" (which must be ammoniacal for Tollens').
For acidified manganate, specify "acidified" or "acidified potassium manganate(VII)"; neutral or alkaline conditions give different products/colours.
Observations must be specific: "colourless" for the decolorization of MnO $_4^-$ , not just "reacts" or "bubbles" (though CO $_2$ is produced, the primary visual mark is the loss of purple colour).

Techniques used

identify reducing properties of methanoic acididentify oxidizability of ethanedioic acidpredict observations for redox 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-Easy

Worked solution

The chlorine atoms are electron-withdrawing (or electronegative). This weakens the O–H bond in the carboxyl group, or stabilises the conjugate base (carboxylate ion) by dispersing/delocalising the negative charge.

Final answer

Chlorine atoms are electron-withdrawing, weakening the O-H bond and stabilising the conjugate base.

Detailed explanation

Background Concept

The strength of a carboxylic acid depends on the ease with which it donates a proton (H $^+$ ) from the carboxyl group (-COOH). This is governed by two related factors:

Polarity of the O-H bond: A more polarized (weaker) O-H bond makes it easier for the H $^+$ to leave.
Stability of the conjugate base (carboxylate ion, -COO $^-$ ): Once the proton is lost, the negative charge on the carboxylate ion is delocalized over the two oxygen atoms. Electron-withdrawing groups attached to the carbon chain pull electron density away from the carboxylate group, helping to disperse and stabilize this negative charge.

Electron-withdrawing groups (like halogens) exert an inductive effect. They pull electron density through the sigma bonds towards themselves. The closer and more numerous the electron-withdrawing groups, the stronger the effect.

Understanding the Question

We are comparing propanoic acid (CH $_3$ CH $_2$ COOH) with 2,2-dichloropropanoic acid (CH $_3$ CCl $_2$ COOH). The latter has two chlorine atoms attached to the alpha-carbon (C2). We must explain why the dichloro derivative is the stronger acid, referencing the structure of both compounds.

Approach

Identify the structural difference: the presence of two highly electronegative chlorine atoms on C2 in 2,2-dichloropropanoic acid.
Explain the inductive effect of these chlorine atoms.
Link this effect to either the weakening of the O-H bond or the stabilization of the resulting carboxylate anion.

Step-by-Step Reasoning

Structure comparison: Propanoic acid has an ethyl group (CH $_3$ CH $_2$ -) attached to the carboxyl group. 2,2-Dichloropropanoic acid has a -CCl $_2$ CH $_3$ group. The key difference is the two Cl atoms on C2.
Inductive effect: Chlorine is highly electronegative. The two C-Cl bonds are polarized, with electron density pulled towards the chlorine atoms. Through the sigma bond framework, this creates an electron-withdrawing inductive effect (-I effect) that pulls electron density away from the carboxyl group.
Effect on acidity: This withdrawal of electron density has two equivalent ways to be described for marking purposes:
- It weakens the O–H bond in the -COOH group by pulling electron density away from the oxygen, making the bond more polar and easier to break heterolytically.
- Alternatively (and often preferred in modern mark schemes), it stabilises the conjugate base (the 2,2-dichloropropanoate ion, CCl $_2$ CH $_3$ COO $^-$ ). The negative charge on the carboxylate oxygen is dispersed/delocalized more effectively because the electron-withdrawing chlorines pull some of that charge density away from the ion.
Either explanation (weakened O-H bond OR stabilized conjugate base) combined with the identification of the electron-withdrawing nature of chlorine earns the marks.

Key Takeaways

Electron-withdrawing substituents (like halogens) increase the acidity of carboxylic acids via the inductive effect. The effect is distance-dependent (stronger when closer to the -COOH group) and additive (more Cl atoms = stronger acid). Always link the structural feature (Cl atoms) to the electronic effect (electron-withdrawing) and then to the consequence (weaker O-H bond or more stable conjugate base).

Common Mistakes

Saying "chlorine pulls electrons" without specifying it is an inductive effect or that chlorine is electronegative/electron-withdrawing.
Only stating one point (e.g., just "it stabilizes the ion") without linking it to the structure (mentioning the Cl atoms or the inductive effect).
Confusing inductive effects with resonance/delocalization effects (halogens on an alkyl chain act via induction, not resonance).
Forgetting to reference the structure: the mark scheme requires mentioning the chlorine atoms or their electronegativity.

Things to Be Careful About

The mark scheme explicitly accepts either "weakening O–H bond" OR "stabilising conjugate base". You do not need to write both, but writing both is fine. However, you must pair it with the reason (electron-withdrawing Cl atoms).
Do not say "chlorine is a good leaving group"; this is irrelevant to acid strength (which is about losing H $^+$ , not Cl $^-$ ).
Ensure you mention the conjugate base or the carboxylate ion if using the stabilization argument, not just "the acid".

Techniques used

apply inductive electron-withdrawing effectexplain bond weakening and conjugate base stabilization

Chemistry 9701/43 — May/June 2025

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

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

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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

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Background Concept

Understanding the Question