You're viewing free preview questions. Upgrade to access more Chemistry HL questions.Upgrade

Reactivity: What Are the Mechanisms of Chemical Change? — Free Chemistry HL Practice Questions

1FoundationMCQAcid-base reactions: Bronsted-Lowry theory1 markPaper 1~2 min
In the reaction between aqueous ammonia and ethanoic acid, ammonia acts a Brønsted–Lowry base. Which species acts as the Brønsted–Lowry acid in this reaction, and what is the conjugate acid formed?
diagram

Solutions

2MasteryMCQAcid-base reactions: Bronsted-Lowry theory1 markPaper 1~2 min
In the reaction H2SO3(aq)+H2O(l)⇌HSO3−(aq)+H3O+(aq)\text{H}_2\text{SO}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HSO}_3^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}, a student claims that H2O\text{H}_2\text{O} acts a Brønsted–Lowry base. Which statement best justifies this claim?
diagram

Solutions

3FoundationMCQAcid-base reactions: Bronsted-Lowry theory1 markPaper 1~2 min
According to Brønsted-Lowry theory, when propanoic acid (CH3CH2COOH\text{CH}_3\text{CH}_2\text{COOH}) acts an acid in aqueous solution, which species its conjugate base?
diagram

Solutions

4FoundationMCQAcid-base reactions: Bronsted-Lowry theory1 markPaper 1~2 min
In the reaction between hydrogen chloride gas and water, HCl+H2O→H3O++Cl−\text{HCl} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^-, what is the role of water according to Brønsted–Lowry theory?
diagram

Solutions

5FoundationMCQOxidizing and reducing agents1 markPaper 1~2 min
In the blast furnace extraction of iron, iron(III) oxide is reduced to molten iron according to the equation: Fe2O3(s)+3CO(g)→2Fe(l)+3CO2(g)\text{Fe}_2\text{O}_3\text{(s)} + 3\text{CO(g)} \rightarrow 2\text{Fe(l)} + 3\text{CO}_2\text{(g)} Which species undergoes oxidation in this reaction?
diagram

Solutions

6MasteryMCQRedox reactions and electron transfer1 markPaper 1~2 min
In a galvanic cell, a magnesium electrode in 1.0 mol dm−3 Mg(NO3)2(aq)1.0\ \text{mol dm}^{-3}\ \text{Mg(NO}_3)_2\text{(aq)} is connected by a salt bridge to a copper electrode in 1.0 mol dm−3 Cu(NO3)2(aq)1.0\ \text{mol dm}^{-3}\ \text{Cu(NO}_3)_2\text{(aq)}. The standard electrode potentials are E∘(Mg2+/Mg)=−2.37 VE^\circ(\text{Mg}^{2+}/\text{Mg}) = -2.37\ \text{V} and E∘(Cu2+/Cu)=+0.34 VE^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}. Which row correctly identifies the direction of electron flow in the external circuit and the species oxidised?
diagram

Solutions

7FoundationMCQOxidizing and reducing agents1 markPaper 1~2 min
In a standard Daniell cell, a zinc electrode in 1.0 mol dm−31.0\ \text{mol dm}^{-3} ZnSO4(aq)\text{ZnSO}_4(\text{aq}) and a copper electrode in 1.0 mol dm−31.0\ \text{mol dm}^{-3} CuSO4(aq)\text{CuSO}_4(\text{aq}) are connected by a salt bridge. The measured cell potential is +1.10 V+1.10\ \text{V}. Which statement correctly identifies the reducing agent and its role in this cell?
diagram

Solutions

8FoundationMCQOxidizing and reducing agents1 markPaper 1~2 min
In a titration, 25.0 cm325.0 \ \text{cm}^3 of 0.0200 mol dm−30.0200 \ \text{mol dm}^{-3} acidified KMnO4\text{KMnO}_4 reacts exactly with 25.0 cm325.0 \ \text{cm}^3 of FeSO4\text{FeSO}_4 solution. The MnO4−\text{MnO}_4^- ion is reduced to Mn2+\text{Mn}^{2+} and Fe2+\text{Fe}^{2+} is oxidised to Fe3+\text{Fe}^{3+}. What is the concentration, in mol dm−3\text{mol dm}^{-3}, of the FeSO4\text{FeSO}_4 solution?
diagram

Solutions

9FoundationMCQPolar vs non-polar covalent bonds in molecular compounds1 markPaper 1~2 min
A molecule of CCl4\text{CCl}_4 has polar C–Cl bonds yet a zero net dipole moment. Which statement correctly explains this?
diagram

Solutions

10MasteryMCQCovalent bonds and electron sharing1 markPaper 1~2 min
In ethene (C2H4\text{C}_2\text{H}_4), the carbon–carbon double bond consists of one σ\sigma bond and one π\pi bond. During electrophilic addition of bromine (Br2\text{Br}_2), the π\pi bond breaks preferentially. Which statement correctly explains why the π\pi bond is weaker and more reactive than the σ\sigma bond?
diagram

Solutions

11FoundationMCQPolar vs non-polar covalent bonds in molecular compounds1 markPaper 1~2 min
Elements X, Y, and Z have Pauling electronegativity values of 0.90.9, 3.03.0, and 3.53.5 respectively. Which bond formed between a pair of these elements has the smallest electronegativity difference and is therefore closest to non-polar covalent?
diagram

Solutions

12FoundationMCQPolar vs non-polar covalent bonds in molecular compounds1 markPaper 1~2 min
Oxygen gas and nitrogen gas are both homonuclear diatomic molecules, yet they differ in bond order: O2\text{O}_2 has a bond order of 2 and N2\text{N}_2 has a bond order of 3. Which statement correctly describes the polarity and bond order of these molecules?
diagram

Solutions

13FoundationMCQCoordinate covalent bonding1 markPaper 1~2 min
When ammonia reacts with a hydrogen ion to form the ammonium ion, NH3+H+→NH4+\text{NH}_3 + \text{H}^+ \rightarrow \text{NH}_4^+, which statement correctly describes the bond formed and the formal charge on nitrogen in NH4+\text{NH}_4^+?
diagram

Solutions

14MasteryMCQLewis acids and bases1 markPaper 1~2 min
Boron trifluoride, BF3\text{BF}_3, reacts with diethyl ether, (C2H5)2O(\text{C}_2\text{H}_5)_2\text{O}, to form a stable adduct in which the oxygen atom donates a lone pair to the boron atom. Which change in hybridisation of the boron atom occurs when this adduct forms?
diagram

Solutions

15FoundationMCQCoordinate covalent bonding1 markPaper 1~2 min
A hydride ion, H−\text{H}^-, donates its lone pair to boron trifluoride, BF3\text{BF}_3, forming the tetrahydridoborate ion, BH4−\text{BH}_4^-. What is the change in hybridisation of the boron atom and the molecular geometry of the species as BF3\text{BF}_3 becomes BH4−\text{BH}_4^-?
diagram

Solutions

16FoundationMCQCoordinate covalent bonding1 markPaper 1~2 min
In the complex ion [Cu(NH3)4]2+[\text{Cu(NH}_3)_4]^{2+}, which statement correctly describes the Cu–N bond and identifies the Lewis acid in this interaction?
diagram

Solutions

17ChallengeSAQ-LStrong vs weak acids and bases8 marksPaper 2~12 min
A student investigates the acid-base properties of two organic compounds, benzoic acid (C6H5COOH\text{C}_6\text{H}_5\text{COOH}) and phenol (C6H5OH\text{C}_6\text{H}_5\text{OH}). Benzoic acid has a pKa\text{p}K_a of 4.204.20 at 298 K298\,\text{K}; phenol has a pKa\text{p}K_a of 9.959.95 at 298 K298\,\text{K}. The student prepares separate 0.100 mol dm−30.100\,\text{mol}\,\text{dm}^{-3} aqueous solutions of each compound. The following expressions may be used: pH=−log⁡[H3O+],pKa=−log⁡Ka,Ka=[H3O+][A−][HA],Kw=1.00×10−14 mol2 dm−6\text{pH} = -\log[\text{H}_3\text{O}^+], \quad \text{p}K_a = -\log K_a, \quad K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}, \quad K_w = 1.00 \times 10^{-14}\,\text{mol}^2\,\text{dm}^{-6}
(a)
State the expression for pKa\text{p}K_a in terms of KaK_a. [1 mark]
(b)
(i) The pH of the 0.100 mol dm−30.100\,\text{mol}\,\text{dm}^{-3} benzoic acid solution is 2.602.60. Calculate the concentration of hydronium ions, [H3O+][\text{H}_3\text{O}^+], in this solution. [1]
(ii) Using your answer to (b)(i), determine the percentage dissociation of benzoic acid in this solution. [1 mark]
(c)
(i) The student adds 10.0 cm310.0\,\text{cm}^3 of 0.100 mol dm−30.100\,\text{mol}\,\text{dm}^{-3} sodium hydroxide solution to 25.0 cm325.0\,\text{cm}^3 of the 0.100 mol dm−30.100\,\text{mol}\,\text{dm}^{-3} benzoic acid solution (Ka=6.31×10−5 mol dm−3K_a = 6.31 \times 10^{-5}\,\text{mol}\,\text{dm}^{-3}). Calculate the pH of the resulting solution. [2]
(ii) State and explain whether the solution formed in (c)(i) is acidic, basic, or neutral. [1 mark]
(d)
The student repeats the procedure in part (c) using 25.0 cm325.0\,\text{cm}^3 of 0.100 mol dm−30.100\,\text{mol}\,\text{dm}^{-3} phenol instead of benzoic acid, adding the same volume of sodium hydroxide solution. Calculate the pH of the resulting phenol buffer solution and evaluate whether the two buffer solutions formed in (c)(i) and (d) would be effective over the same pH range. [2 marks]

Solutions

18ChallengeSAQ-LStrong vs weak acids and bases8 marksPaper 2~12 min

Data

Ka(CH3COOH)=1.8×10−5 mol dm−3K_a(\text{CH}_3\text{COOH}) = 1.8 \times 10^{-5}\,\text{mol dm}^{-3}; Kw=1.00×10−14 mol2 dm−6K_w = 1.00 \times 10^{-14}\,\text{mol}^2\,\text{dm}^{-6} at 298 K298\,\text{K}
A chemistry teacher demonstrates the difference between strong and weak acids by measuring the electrical conductivity of two 0.500 mol dm−30.500\,\text{mol dm}^{-3} acid solutions at 298 K298\,\text{K}: hydrochloric acid (HCl) and ethanoic acid (CH3_3COOH). The conductivity of the HCl solution is 42.5 mS cm−142.5\,\text{mS cm}^{-1}, while the conductivity of the ethanoic acid solution is 0.85 mS cm−10.85\,\text{mS cm}^{-1}.
(a)
Explain why the conductivity of the hydrochloric acid solution is significantly higher than that of the ethanoic acid solution, despite both having the same concentration. [2 marks]
(b)
The teacher dilutes both solutions by a factor of 10 to 0.0500 mol dm−30.0500\,\text{mol dm}^{-3}. For the HCl solution, the conductivity drops to approximately 4.25 mS cm−14.25\,\text{mS cm}^{-1}. Predict, with reasoning, whether the conductivity of the diluted ethanoic acid solution will be greater than, less than, or equal to 0.085 mS cm−10.085\,\text{mS cm}^{-1}. [3 marks]
(c)
The KaK_a of ethanoic acid is 1.8×10−5 mol dm−31.8 \times 10^{-5}\,\text{mol dm}^{-3} at 298 K298\,\text{K}. A student adds concentrated HCl to the 0.500 mol dm−30.500\,\text{mol dm}^{-3} ethanoic acid solution until [Cl−]=0.100 mol dm−3[\text{Cl}^-] = 0.100\,\text{mol dm}^{-3}. Using the common ion effect, determine the equilibrium [H3O+][\text{H}_3\text{O}^+] in the resulting solution and hence evaluate whether the overall conductivity of the solution increases or decreases compared to the original ethanoic acid solution. [3 marks]

Solutions

19ChallengeSAQ-LStrong vs weak acids and bases8 marksPaper 2~12 min

Data

pH=−log⁡[H3O+]\text{pH} = -\log[\text{H}_3\text{O}^+], pKa=−log⁡KapK_a = -\log K_a, Ka=[H3O+][A−][HA]K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}, Kw=1.00×10−14 mol2 dm−6K_w = 1.00 \times 10^{-14}\,\text{mol}^2\,\text{dm}^{-6} at 298 K298\,\text{K}
A forensic chemist analyses a sample suspected to contain a mixture of nitric acid (HNO3\text{HNO}_3) and propanoic acid (CH3CH2COOH\text{CH}_3\text{CH}_2\text{COOH}, pKa=4.88pK_a = 4.88 at 298 K298\,\text{K}). A 25.0 cm325.0\,\text{cm}^3 sample of the mixture is titrated with 0.200 mol dm−30.200\,\text{mol}\,\text{dm}^{-3} sodium hydroxide solution. The titration curve shows two distinct equivalence points: the first at 12.5 cm312.5\,\text{cm}^3 of NaOH added, and the second at 25.0 cm325.0\,\text{cm}^3 of NaOH added.
(a)
Identify which acid is neutralised at each equivalence point. Justify your answer using the volume data. [2 marks]
(b)
Calculate the concentration of nitric acid in the original mixture. [2 marks]
(c)
The concentration of propanoic acid in the original mixture is 0.100 mol dm−30.100\,\text{mol}\,\text{dm}^{-3}. Calculate the pH of the solution at the half-equivalence point of the propanoic acid titration. [2 marks]
(d)
Predict, with justification, whether methyl orange indicator (colour change range pH 3.1\text{pH}\ 3.1–4.44.4) would be suitable for detecting either equivalence point in this titration. [2 marks]

Solutions

20ChallengeSAQ-LStrong vs weak acids and bases9 marksPaper 2~14 min

Data

Kw=1.0×10−14 mol2 dm−6K_w = 1.0 \times 10^{-14}\,\text{mol}^2\,\text{dm}^{-6} at 298 K298\,\text{K}
A student investigates the acid strength of three unknown monoprotic acids, HA, HB, and HC. They prepare 0.100 mol dm−30.100\,\text{mol dm}^{-3} aqueous solutions of each acid and measure the pH at 298 K298\,\text{K}. The results are: - Acid HA: pH=1.00\text{pH} = 1.00 - Acid HB: pH=4.00\text{pH} = 4.00 - Acid HC: pH=2.87\text{pH} = 2.87
(a)
Identify which acid is a strong acid and state the reason for your choice. [2 marks]
(b)
Calculate the acid dissociation constant, KaK_a, for acid HB, assuming the equilibrium concentration of undissociated acid equals its initial concentration. [2 marks]
(c)
Explain why the pH at the equivalence point is greater than 7.007.00 when 25.0 cm325.0\,\text{cm}^3 of 0.100 mol dm−30.100\,\text{mol dm}^{-3} acid HC is titrated against 0.100 mol dm−30.100\,\text{mol dm}^{-3} sodium hydroxide solution. [2 marks]
(d)
Evaluate the suggestion that using 1.00 mol dm−31.00\,\text{mol dm}^{-3} sodium hydroxide solution instead of 0.100 mol dm−30.100\,\text{mol dm}^{-3} would make the titration of acid HC more accurate. [3 marks]

Solutions

21ChallengeSAQ-LRedox reactions and electron transfer7 marksPaper 2~11 min
Vanadium is a transition metal used in high-strength steel alloys. A student investigates the stepwise reduction of ammonium vanadate(V), NH4VO3\text{NH}_4\text{VO}_3, using zinc metal in acidic aqueous solution at 298 K298\,\text{K}. Standard electrode potentials (acidic solution, 298 K298\,\text{K}): VO2+(aq)+2H+(aq)+e−→VO2+(aq)+H2O(l)E∘=+1.00 V\text{VO}_2^+(aq) + 2\text{H}^+(aq) + e^- \rightarrow \text{VO}^{2+}(aq) + \text{H}_2\text{O}(l) \quad E^\circ = +1.00\,\text{V} VO2+(aq)+2H+(aq)+e−→V3+(aq)+H2O(l)E∘=+0.34 V\text{VO}^{2+}(aq) + 2\text{H}^+(aq) + e^- \rightarrow \text{V}^{3+}(aq) + \text{H}_2\text{O}(l) \quad E^\circ = +0.34\,\text{V} V3+(aq)+e−→V2+(aq)E∘=−0.26 V\text{V}^{3+}(aq) + e^- \rightarrow \text{V}^{2+}(aq) \quad E^\circ = -0.26\,\text{V} Zn2+(aq)+2e−→Zn(s)E∘=−0.76 V\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \quad E^\circ = -0.76\,\text{V} Colours of vanadium species in aqueous solution: VO2+\text{VO}_2^+ (yellow), VO2+\text{VO}^{2+} (blue), V3+\text{V}^{3+} (green), V2+\text{V}^{2+} (violet).
(a)
State the oxidation state of vanadium in VO2+\text{VO}_2^+ and in VO2+\text{VO}^{2+}. [2 marks]
(b)
Calculate Ecell∘E^\circ_\text{cell} for the reduction of VO2+\text{VO}_2^+ to VO2+\text{VO}^{2+} using zinc metal as the reductant. [2 marks]
(c)
Excess zinc metal is added to a yellow solution containing VO2+\text{VO}_2^+ ions. Explain, with reference to E∘E^\circ values, the sequence of colour changes observed and identify the final vanadium-containing species present. [3 marks]
diagram

Solutions

22MasterySAQ-SOxidizing and reducing agents6 marksPaper 2~9 min
A student investigates the rusting of iron by placing an iron nail in a dish of water containing a small amount of sodium chloride and a few drops of a redox indicator that turns blue in the presence of Fe2+\text{Fe}^{2+} ions. After one hour, a blue colour appears near the bottom of the nail but not at the top.
(a)
State the species that is oxidised at the site where the blue colour appears. [1 mark]
(b)
Write the balanced half-equation for the reduction reaction occurring at the site where the blue colour does not appear. [2 marks]
(c)
The student wraps a strip of magnesium metal tightly around the nail and repeats the experiment. The blue colour does not appear. Explain why magnesium prevents rusting, using the concept of relative reducing strength. [3 marks]
diagram

Solutions

23ChallengeSAQ-LRedox reactions and electron transfer7 marksPaper 2~11 min
A student studies the redox chemistry of hydrogen peroxide, H2O2\text{H}_2\text{O}_2, in environmental remediation. The student investigates the reaction of hydrogen peroxide with aqueous potassium iodide, KI, in acidic solution: H2O2(aq)+2I−(aq)+2H+(aq)→I2(aq)+2H2O(l)\text{H}_2\text{O}_2\text{(aq)} + 2\text{I}^-\text{(aq)} + 2\text{H}^+\text{(aq)} \rightarrow \text{I}_2\text{(aq)} + 2\text{H}_2\text{O(l)} Standard electrode potentials, E∘E^\circ, at 298 K298\,\text{K}: H2O2(aq)+2H+(aq)+2e−→2H2O(l)E∘=+1.78 V\text{H}_2\text{O}_2\text{(aq)} + 2\text{H}^+\text{(aq)} + 2e^- \rightarrow 2\text{H}_2\text{O(l)} \quad E^\circ = +1.78\,\text{V} O2(g)+2H+(aq)+2e−→H2O2(aq)E∘=+0.68 V\text{O}_2\text{(g)} + 2\text{H}^+\text{(aq)} + 2e^- \rightarrow \text{H}_2\text{O}_2\text{(aq)} \quad E^\circ = +0.68\,\text{V} I2(aq)+2e−→2I−(aq)E∘=+0.54 V\text{I}_2\text{(aq)} + 2e^- \rightarrow 2\text{I}^-\text{(aq)} \quad E^\circ = +0.54\,\text{V} MnO4−(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(l)E∘=+1.51 V\text{MnO}_4^-\text{(aq)} + 8\text{H}^+\text{(aq)} + 5e^- \rightarrow \text{Mn}^{2+}\text{(aq)} + 4\text{H}_2\text{O(l)} \quad E^\circ = +1.51\,\text{V}
(a)
State the half-equation for the reduction reaction and the half-equation for the oxidation reaction occurring in the overall reaction above. [2 marks]
(b)
Calculate the standard cell potential, Ecell∘E^\circ_\text{cell}, for the reaction between H2O2\text{H}_2\text{O}_2 and I−\text{I}^- in acidic solution. [2 marks]
(c)
H2O2\text{H}_2\text{O}_2 can also act as a reducing agent in the presence of MnO4−\text{MnO}_4^- in acidic solution. Write the half-equation for the oxidation of H2O2\text{H}_2\text{O}_2, calculate Ecell∘E^\circ_\text{cell} for this reaction, and deduce whether the reaction is thermodynamically feasible under standard conditions. [3 marks]
diagram

Solutions

24ChallengeSAQ-LRedox reactions and electron transfer11 marksPaper 2~17 min

Data

Q=ItF=96500 C mol−1n=QFc=nVQ = It \qquad F = 96500\,\text{C mol}^{-1} \qquad n = \frac{Q}{F} \qquad c = \frac{n}{V}
This question is about the redox chemistry of vanadium and its use in a vanadium redox flow battery (VRFB). Data — standard electrode potentials, E⊖E^\ominus: VO2+(aq)+2H+(aq)+e−→VO2+(aq)+H2O(l)E⊖=+1.00 VVO_2^+(aq) + 2H^+(aq) + e^- \rightarrow VO^{2+}(aq) + H_2O(l) \quad E^\ominus = +1.00\text{ V} VO2+(aq)+2H+(aq)+e−→V3+(aq)+H2O(l)E⊖=+0.34 VVO^{2+}(aq) + 2H^+(aq) + e^- \rightarrow V^{3+}(aq) + H_2O(l) \quad E^\ominus = +0.34\text{ V} V3+(aq)+e−→V2+(aq)E⊖=−0.26 VV^{3+}(aq) + e^- \rightarrow V^{2+}(aq) \quad E^\ominus = -0.26\text{ V} A VRFB operates by pumping solutions of vanadium ions in sulfuric acid through two half-cells separated by a proton-exchange membrane. During discharge, VO2+VO_2^+ is reduced at the positive electrode and V2+V^{2+} is oxidised at the negative electrode.
(a)
State the half-equation for the reaction occurring at the negative electrode during discharge. [1 mark]
(b)
Calculate the standard cell potential, Ecell⊖E^\ominus_{\text{cell}}, for the spontaneous reaction during discharge of the VRFB. [3 marks]
(c)
The positive electrode half-cell initially contains 1.00 dm31.00\,\text{dm}^3 of 2.00 mol dm−32.00\,\text{mol dm}^{-3} VO2+VO_2^+ in 4.00 mol dm−34.00\,\text{mol dm}^{-3} H2SO4H_2SO_4. During a test discharge, a current of 5.00 A5.00\,\text{A} flows for 9650 s9650\,\text{s}. Calculate the concentration of VO2+VO_2^+ remaining in the positive half-cell at the end of the discharge. Assume the volume remains constant and that all charge is used to reduce VO2+VO_2^+. [3 marks]
(d)
The standard reduction potential for O2(g)+4H+(aq)+4e−→2H2O(l)O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l) is +1.23 V+1.23\,\text{V}. Explain why V2+(aq)V^{2+}(aq) is thermodynamically unstable in the presence of air, and evaluate whether this instability represents a significant practical limitation for VRFB operation when the negative-electrode reservoir is sealed under an inert atmosphere. [4 marks]
diagram

Solutions

25MasterySAQ-SCovalent bonds and electron sharing5 marksPaper 2~8 min
Ethene (C2H4\text{C}_2\text{H}_4) is an unsaturated hydrocarbon used in the manufacture of polymers. Each carbon atom in ethene forms three sigma bonds and one pi bond.
(a)
State the number of sigma (σ\sigma) bonds and pi (π\pi) bonds present in the carbon–carbon double bond of ethene. [1 mark]
(b)
Use the bond enthalpy data below to calculate the enthalpy change, in kJ mol−1\text{kJ mol}^{-1}, for the hydrogenation of ethene: C2H4(g)+H2(g)→C2H6(g)\text{C}_2\text{H}_4(\text{g}) + \text{H}_2(\text{g}) \rightarrow \text{C}_2\text{H}_6(\text{g}) | Bond enthalpy / kJ mol−1\text{kJ mol}^{-1} | C=C — +612+612 C–C — +348+348 H–H — +436+436 [2 marks]
(c)
Explain, in terms of orbital overlap, why the pi bond in the C=C double bond is weaker than the sigma bond. [2 marks]
diagram

Solutions

26MasterySAQ-SCovalent bonds and electron sharing5 marksPaper 2~8 min
Nitrogen trifluoride (NF3\text{NF}_3) is used in semiconductor manufacture. It has a trigonal pyramidal geometry, a bond angle of 102.5°102.5°, and a dipole moment of 0.24 D0.24\,\text{D}. Ammonia (NH3\text{NH}_3) also has trigonal pyramidal geometry, a bond angle of 107.3°107.3°, and a dipole moment of 1.47 D1.47\,\text{D}. Both molecules have three N–X single covalent bonds and one lone pair on nitrogen. Electronegativity values: N=3.0\text{N} = 3.0, H=2.1\text{H} = 2.1, F=4.0\text{F} = 4.0
(a)
State the direction of the bond dipole in the N–F bond and in the N–H bond, and calculate the electronegativity difference for each bond. [2 marks]
(b)
Explain why the dipole moment of NF3\text{NF}_3 (0.24 D0.24\,\text{D}) is much smaller than that of NH3\text{NH}_3 (1.47 D1.47\,\text{D}), despite both molecules having the same geometry. [3 marks]
diagram

Solutions

27ChallengeSAQ-LCovalent bonds and electron sharing7 marksPaper 2~11 min

Data

- Average bond enthalpy: Si–O=466 kJ mol−1\text{Si–O} = 466\,\text{kJ mol}^{-1} - Average bond enthalpy: C=O (in CO2\text{CO}_2) =799 kJ mol−1= 799\,\text{kJ mol}^{-1} Bond enthalpy is the energy required to break one mole of bonds in the gaseous state.
Silicon dioxide (SiO2\text{SiO}_2) and carbon dioxide (CO2\text{CO}_2) are both oxides of Group 14 elements, yet they have very different structures and properties. CO2\text{CO}_2 sublimes at −78 ∘C-78\,^\circ\text{C}, while SiO2\text{SiO}_2 (quartz) melts above 1600 ∘C1600\,^\circ\text{C}.
(a)
State the number of bonds formed by silicon in SiO2\text{SiO}_2 and by carbon in CO2\text{CO}_2, giving the bond type in each case. [1 mark]
(b)
Calculate the total bond enthalpy required to completely atomise one mole of SiO2\text{SiO}_2 and one mole of CO2\text{CO}_2. [2 marks]
(c)
Explain why SiO2\text{SiO}_2 has a much higher melting point than CO2\text{CO}_2, in terms of structure and the forces that must be overcome on melting. [2 marks]
(d)
A student claims: *"SiO2\text{SiO}_2 has a higher melting point than CO2\text{CO}_2 because the Si–O bond is stronger than the C=O bond."* Using your answer to (b) and your knowledge of structure, evaluate this claim. [2 marks]
diagram

Solutions

28MasterySAQ-SCovalent bonds and electron sharing5 marksPaper 2~8 min
Sulfur hexafluoride (SF6\text{SF}_6) is a gas used as an electrical insulator in high-voltage equipment. It has an octahedral molecular geometry. Each S–F bond has an electronegativity difference of 1.51.5, but the molecule has a net dipole moment of zero. The S–F bond enthalpy is 327 kJ mol−1327\,\text{kJ mol}^{-1} and the bond length is 156 pm156\,\text{pm}. (Electronegativity: S =2.5= 2.5, F =4.0= 4.0)
(a)
State the direction of the bond dipole in the S–F bond. [1 mark]
(b)
Explain why SF6\text{SF}_6 has a net dipole moment of zero despite having highly polar S–F bonds. [2 marks]
(c)
The S–F bond enthalpy of 327 kJ mol−1327\,\text{kJ mol}^{-1} is significantly higher than a typical S–F single bond value of approximately 285 kJ mol−1285\,\text{kJ mol}^{-1}. Deduce, with reasoning, what this difference suggests about the bond order in SF6\text{SF}_6. [2 marks]
diagram

Solutions

29ChallengeSAQ-LCoordinate covalent bonding7 marksPaper 2~11 min

Data

- Electronegativity values: N=3.0\text{N} = 3.0, H=2.2\text{H} = 2.2, B=2.0\text{B} = 2.0 - Bond dissociation energies: N–H=391 kJ mol−1\text{N–H} = 391\,\text{kJ mol}^{-1}; B–H=389 kJ mol−1\text{B–H} = 389\,\text{kJ mol}^{-1}; N–B (coordinate)=310 kJ mol−1\text{N–B (coordinate)} = 310\,\text{kJ mol}^{-1} - Enthalpy of sublimation of NH3BH3(s)\text{NH}_3\text{BH}_3\text{(s)}: +85 kJ mol−1+85\,\text{kJ mol}^{-1}
Ammonium borane (NH3BH3\text{NH}_3\text{BH}_3) is a stable white crystalline solid at room temperature. It decomposes at around 110 ∘C110\,^\circ\text{C} rather than melting. The molecule contains a coordinate covalent (dative) bond between the nitrogen and boron atoms.
(a)
Calculate the enthalpy change, ΔH\Delta H, for the gas-phase reaction: NH3(g)+BH3(g)→NH3BH3(g)\text{NH}_3\text{(g)} + \text{BH}_3\text{(g)} \rightarrow \text{NH}_3\text{BH}_3\text{(g)} State the type of bond formed and identify which atom donates both electrons. [3 marks]
(b)
Using the electronegativity data provided, explain why the N–B coordinate bond has a lower bond dissociation energy than the N–H covalent bond. [2 marks]
(c)
Evaluate the claim: *"The existence of a stable solid such as NH3BH3\text{NH}_3\text{BH}_3 shows that coordinate covalent bonds are stronger than ordinary covalent bonds."* In your answer, use both the bond energy data and the enthalpy of sublimation to support your evaluation. [2 marks]
diagram

Solutions

30ChallengeSAQ-LCoordinate covalent bonding7 marksPaper 2~11 min
In the industrial synthesis of cisplatin, [Pt(NH3)2Cl2][\text{Pt(NH}_3)_2\text{Cl}_2], the square planar complex [PtCl4]2−[\text{PtCl}_4]^{2-} is first formed when platinum reacts with excess chloride ions. This complex then undergoes two sequential ligand substitution reactions with ammonia (NH3\text{NH}_3) to produce cisplatin, which binds to DNA guanine bases in cancer cells. - [PtCl4]2−[\text{PtCl}_4]^{2-} is square planar; overall charge =−2= -2; each Cl−\text{Cl}^- carries charge −1-1 - Electron configuration of Pt: [Xe] 4f14 5d9 6s1[\text{Xe}]\,4f^{14}\,5d^9\,6s^1 - NH3\text{NH}_3 is a stronger ligand than Cl−\text{Cl}^- in the spectrochemical series - Inside cells, [Cl−]≈4 mmol dm−3[\text{Cl}^-] \approx 4\,\text{mmol dm}^{-3}; in blood plasma, [Cl−]≈100 mmol dm−3[\text{Cl}^-] \approx 100\,\text{mmol dm}^{-3}
(a)
Deduce the oxidation state of platinum in [PtCl4]2−[\text{PtCl}_4]^{2-} and the number of d-electrons on the Pt2+\text{Pt}^{2+} ion. Show your reasoning. [3 marks]
(b)
Explain why NH3\text{NH}_3 is a more effective ligand than Cl−\text{Cl}^- for Pt2+\text{Pt}^{2+}, referring to the nature of the lone pair and the relative strength of the coordinate bond formed. [2 marks]
(c)
Evaluate how the difference in lability between the Pt–Cl and Pt–N coordinate bonds determines the selectivity of cisplatin as anticancer drug. In your answer, refer to the intracellular and extracellular chloride concentrations given. [2 marks]
diagram

Solutions

31ChallengeSAQ-LCoordinate covalent bonding7 marksPaper 2~11 min

Data

- Kf=5.0×109K_f = 5.0 \times 10^{9} for [Fe(en)3]2+[\text{Fe}(\text{en})_3]^{2+} at 298 K298\,\text{K} - Ionization energies of Fe: IE1=759 kJ mol−1\text{IE}_1 = 759\,\text{kJ mol}^{-1}; IE2=1562 kJ mol−1\text{IE}_2 = 1562\,\text{kJ mol}^{-1}; IE3=2957 kJ mol−1\text{IE}_3 = 2957\,\text{kJ mol}^{-1} - Fe has 26 electrons; R=8.31 J mol−1K−1R = 8.31\,\text{J mol}^{-1}\text{K}^{-1}; ΔG=−RTln⁡K\Delta G = -RT\ln K
The iron(II) ion, Fe2+\text{Fe}^{2+}, forms a hexaaqua complex ion, [Fe(H2O)6]2+[\text{Fe}(\text{H}_2\text{O})_6]^{2+}, in aqueous solution. Each water molecule donates a lone pair of electrons to the iron(II) ion, forming a coordinate covalent bond. Ethane-1,2-diamine (H2NCH2NH2\text{H}_2\text{NCH}_2\text{NH}_2, abbreviated as 'en') is a bidentate ligand. Each nitrogen atom has a lone pair and can form a coordinate covalent bond with a metal ion. When excess 'en' is added to a solution of [Fe(H2O)6]2+[\text{Fe}(\text{H}_2\text{O})_6]^{2+}, the water ligands are displaced, forming [Fe(en)3]2+[\text{Fe}(\text{en})_3]^{2+}.
(a)
State the total number of coordinate covalent bonds in one [Fe(en)3]2+[\text{Fe}(\text{en})_3]^{2+} ion. [1 mark]
(b)
Calculate ΔG\Delta G for the formation of [Fe(en)3]2+[\text{Fe}(\text{en})_3]^{2+} from [Fe(H2O)6]2+[\text{Fe}(\text{H}_2\text{O})_6]^{2+} and three 'en' ligands at 298 K298\,\text{K}. [3 marks]
(c)
Explain why [Fe(en)3]2+[\text{Fe}(\text{en})_3]^{2+} is more stable than [Fe(H2O)6]2+[\text{Fe}(\text{H}_2\text{O})_6]^{2+}, with reference to the chelate effect and the sign of ΔS\Delta S for the reaction. [2 marks]
(d)
Using the ionization energy data and the electron configuration of Fe, deduce why the iron centre is Fe2+\text{Fe}^{2+} rather than Fe3+\text{Fe}^{3+} in this complex. [1 mark]
diagram

Solutions

32ChallengeSAQ-LCoordinate covalent bonding10 marksPaper 2~15 min

Data

Quantity — Value Bond dissociation enthalpy, B–F in BF3\text{BF}_3 — 613 kJ mol−1613\,\text{kJ mol}^{-1} Bond dissociation enthalpy, B–F in F3B←NH3\text{F}_3\text{B} \leftarrow \text{NH}_3 — 572 kJ mol−1572\,\text{kJ mol}^{-1} Bond dissociation enthalpy, N–H in NH3\text{NH}_3 — 391 kJ mol−1391\,\text{kJ mol}^{-1} Bond dissociation enthalpy, N–H in F3B←NH3\text{F}_3\text{B} \leftarrow \text{NH}_3 — 386 kJ mol−1386\,\text{kJ mol}^{-1} Bond dissociation enthalpy, B←N\text{B} \leftarrow \text{N} coordinate bond — 156 kJ mol−1156\,\text{kJ mol}^{-1} ΔHf∘ [BF3(g)]\Delta H_f^\circ\,[\text{BF}_3(g)] — −1137 kJ mol−1-1137\,\text{kJ mol}^{-1} ΔHf∘ [NH3(g)]\Delta H_f^\circ\,[\text{NH}_3(g)] — −46.1 kJ mol−1-46.1\,\text{kJ mol}^{-1} ΔHf∘ [F3B←NH3(g)]\Delta H_f^\circ\,[\text{F}_3\text{B} \leftarrow \text{NH}_3(g)] — −1201 kJ mol−1-1201\,\text{kJ mol}^{-1}
A coordinate covalent bond forms when BF3\text{BF}_3 reacts with NH3\text{NH}_3 to give the adduct F3B←NH3\text{F}_3\text{B} \leftarrow \text{NH}_3.
(a)
Draw the Lewis (electron dot) structure of F3B←NH3\text{F}_3\text{B} \leftarrow \text{NH}_3, showing all bonding pairs, lone pairs, and the coordinate covalent bond. [2 marks]
(b)
Calculate the enthalpy change ΔH\Delta H for the reaction BF3(g)+NH3(g)→F3B←NH3(g)\text{BF}_3(g) + \text{NH}_3(g) \rightarrow \text{F}_3\text{B} \leftarrow \text{NH}_3(g) using the bond enthalpy data provided. [3 marks]
(c)
Explain why the B–F bond dissociation enthalpy decreases from 613 kJ mol−1613\,\text{kJ mol}^{-1} in BF3\text{BF}_3 to 572 kJ mol−1572\,\text{kJ mol}^{-1} in the adduct, with reference to hybridization and π\pi-back-bonding. [2 marks]
(d)
Use the standard enthalpies of formation given in the table to calculate ΔH∘\Delta H^\circ for the adduct formation reaction. Compare this value with your answer to (b) and evaluate which method gives a more reliable estimate of the reaction enthalpy, justifying your answer. [3 marks]
diagram

Solutions

33ChallengeLAQpH scale and calculations15 marksPaper 3~23 min

Data

- KaK_a of H2CO3\text{H}_2\text{CO}_3 (first dissociation) =4.3×10−7 mol dm−3= 4.3 \times 10^{-7}\,\text{mol}\,\text{dm}^{-3} at 25 °C25\,°\text{C} - Kw=1.0×10−14 mol2 dm−6K_w = 1.0 \times 10^{-14}\,\text{mol}^2\,\text{dm}^{-6} at 25 °C25\,°\text{C} - Concentration of dissolved CO2\text{CO}_2 in equilibrium with atmospheric CO2\text{CO}_2: [H2CO3]=1.2×10−5 mol dm−3[\text{H}_2\text{CO}_3] = 1.2 \times 10^{-5}\,\text{mol}\,\text{dm}^{-3}
The pH of natural rainwater is approximately 5.6 due to dissolved atmospheric carbon dioxide forming carbonic acid. In industrialised regions, acid rain with pH values as low as 4.0 is observed, primarily due to emissions of sulfur dioxide (SO2\text{SO}_2) and nitrogen oxides (NOx\text{NO}_x) from fossil fuel combustion. A 100 cm3100\,\text{cm}^3 sample of acid rain collected near an industrial area has a pH=4.00\text{pH} = 4.00. Assume all acidity arises solely from HNO3\text{HNO}_3, formed from NOx\text{NO}_x emissions.
(a)
State why HNO3\text{HNO}_3 can be assumed to be fully dissociated in aqueous solution, and write the equation for its dissociation. [2 marks]
(b)
Calculate the volume, in cm3\text{cm}^3, of 0.0100 mol dm−30.0100\,\text{mol}\,\text{dm}^{-3} NaOH solution required to completely neutralise the HNO3\text{HNO}_3 in the 100 cm3100\,\text{cm}^3 acid rain sample. [3 marks]
(c)
Explain why the pH of natural rainwater is approximately 5.6, rather than 7.0, even in the absence of anthropogenic pollutants. Include the relevant equilibrium equations and a calculation to verify the pH. [5 marks]
(d)
Evaluate the environmental impact of acid rain on aquatic ecosystems. In your answer, explain the concept of buffer capacity and, using relevant chemical equations, justify why lakes with granite (silicate) bedrock are more susceptible to acidification than those with limestone (CaCO3\text{CaCO}_3) bedrock. [5 marks]

Solutions

34ChallengeLAQpH scale and calculations15 marksPaper 3~23 min
A student investigates the acid-base properties of two unknown solutions, X and Y, using a pH meter and titration. Solution X is a colourless liquid with a pH=2.50\text{pH} = 2.50. Solution Y is also colourless with a pH=11.30\text{pH} = 11.30. Experiment 1: The student mixes 25.0 cm325.0\,\text{cm}^3 of solution X with 25.0 cm325.0\,\text{cm}^3 of distilled water. The pH of the resulting mixture is 2.802.80. Experiment 2: The student titrates 25.0 cm325.0\,\text{cm}^3 of solution X with solution Y (a strong base, concentration 0.100 mol dm−30.100\,\text{mol dm}^{-3}), using phenolphthalein indicator. The mean titre of solution Y required to reach the end point is 18.5 cm318.5\,\text{cm}^3. Data provided: - Kw=1.0×10−14K_w = 1.0 \times 10^{-14} at 25 °C25\,°\text{C} - pKa\text{p}K_a of phenolphthalein =9.3= 9.3 (colour transition range: pH 8.2\text{pH}\ 8.2–10.010.0) - pKa\text{p}K_a of ethanoic acid (CH3COOH)=4.76(\text{CH}_3\text{COOH}) = 4.76 - pKa\text{p}K_a of benzoic acid (C6H5COOH)=4.20(\text{C}_6\text{H}_5\text{COOH}) = 4.20 - pKa\text{p}K_a of carbonic acid (H2CO3, first dissociation)=6.35(\text{H}_2\text{CO}_3,\ \text{first dissociation}) = 6.35
(a)
Deduce whether solution X is a strong acid or a weak acid. Justify your answer using quantitative reasoning and the data from Experiment 1. [3 marks]
(b)
Calculate the concentration of solution X, in mol dm−3\text{mol dm}^{-3}, using the titration data from Experiment 2. [3 marks]
(c)
Using your answers to parts (a) and (b), calculate the experimental pKa\text{p}K_a of solution X and deduce which of the three acids listed is most likely to be solution X. [4 marks]
(d)
Calculate the pH at the equivalence point of the titration in Experiment 2, using your identified acid from part (c) and your concentration from part (b). Hence evaluate the suitability of phenolphthalein as an indicator for this titration. [5 marks]

Solutions

35ChallengeLAQRedox reactions and electron transfer14 marksPaper 3~21 min
Potassium permanganate, KMnO4\text{KMnO}_4, is a strong oxidising agent used in redox titrations. In acidic solution, MnO4−(aq)\text{MnO}_4^-(aq) is reduced to Mn2+(aq)\text{Mn}^{2+}(aq); in neutral or alkaline solution it is reduced to MnO2(s)\text{MnO}_2(s). Standard reduction potentials at 298 K: MnO4−(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(l)E⊖=+1.51 V\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) \quad E^\ominus = +1.51\,\text{V} MnO4−(aq)+2H2O(l)+3e−→MnO2(s)+4OH−(aq)E⊖=+0.59 V\text{MnO}_4^-(aq) + 2\text{H}_2\text{O}(l) + 3e^- \rightarrow \text{MnO}_2(s) + 4\text{OH}^-(aq) \quad E^\ominus = +0.59\,\text{V} Fe3+(aq)+e−→Fe2+(aq)E⊖=+0.77 V\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \quad E^\ominus = +0.77\,\text{V} I2(s)+2e−→2I−(aq)E⊖=+0.54 V\text{I}_2(s) + 2e^- \rightarrow 2\text{I}^-(aq) \quad E^\ominus = +0.54\,\text{V} Cr2O72−(aq)+14H+(aq)+6e−→2Cr3+(aq)+7H2O(l)E⊖=+1.33 V\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) \quad E^\ominus = +1.33\,\text{V} Cl2(g)+2e−→2Cl−(aq)E⊖=+1.36 V\text{Cl}_2(g) + 2e^- \rightarrow 2\text{Cl}^-(aq) \quad E^\ominus = +1.36\,\text{V}
(a)
(i) State the balanced oxidation half-equation for Fe2+(aq)\text{Fe}^{2+}(aq) and the overall ionic equation for the reaction of Fe2+(aq)\text{Fe}^{2+}(aq) with acidified KMnO4(aq)\text{KMnO}_4(aq). [2]
(ii) Calculate the volume, in cm3\text{cm}^3, of 0.0200 mol dm−30.0200\,\text{mol\,dm}^{-3} KMnO4(aq)\text{KMnO}_4(aq) required to reach the endpoint when titrating 25.00 cm325.00\,\text{cm}^3 of 0.100 mol dm−30.100\,\text{mol\,dm}^{-3} Fe2+(aq)\text{Fe}^{2+}(aq) with acidified KMnO4(aq)\text{KMnO}_4(aq). [3 marks]
(b)
Explain why the volume of KMnO4(aq)\text{KMnO}_4(aq) required to reach the endpoint would differ if the titration in (a)(ii) were repeated using neutral KMnO4(aq)\text{KMnO}_4(aq) instead of acidified KMnO4(aq)\text{KMnO}_4(aq). Your answer must include the oxidation state of manganese in each product, the number of electrons transferred per mole of MnO4−\text{MnO}_4^- in each case, and a calculation of the new volume required. [4 marks]
(c)
Evaluate the suitability of acidified KMnO4\text{KMnO}_4 versus acidified K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 for the titration of Fe2+(aq)\text{Fe}^{2+}(aq) solutions that may contain Cl−(aq)\text{Cl}^-(aq). Your evaluation must address: the spontaneity of each reaction with Fe2+\text{Fe}^{2+}, the need for an indicator, and thermodynamic feasibility of oxidising Cl−\text{Cl}^- by each reagent. [5 marks]

Solutions

36ChallengeLAQRedox reactions and electron transfer15 marksPaper 3~23 min
The industrial production of chlorine gas via the electrolysis of brine (concentrated aqueous sodium chloride) is a major electrochemical process. The overall cell reaction is: 2NaCl(aq)+2H2O(l)→2NaOH(aq)+H2(g)+Cl2(g)2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g) + \text{Cl}_2(g) Standard reduction potentials at 298 K298\,\text{K}: Cl2(g)+2e−→2Cl−(aq)E⊖=+1.36 V\text{Cl}_2(g) + 2e^- \rightarrow 2\text{Cl}^-(aq) \qquad E^\ominus = +1.36\,\text{V} 2H2O(l)+2e−→H2(g)+2OH−(aq)E⊖=−0.83 V2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-(aq) \qquad E^\ominus = -0.83\,\text{V} O2(g)+4H+(aq)+4e−→2H2O(l)E⊖=+1.23 V\text{O}_2(g) + 4\text{H}^+(aq) + 4e^- \rightarrow 2\text{H}_2\text{O}(l) \qquad E^\ominus = +1.23\,\text{V} Na+(aq)+e−→Na(s)E⊖=−2.71 V\text{Na}^+(aq) + e^- \rightarrow \text{Na}(s) \qquad E^\ominus = -2.71\,\text{V}
(a)
(i) State the half-equation occurring at each electrode during the electrolysis of brine. [2]
(ii) Justify the choice of half-equation at each electrode using the standard reduction potentials provided. In your justification for the anode, explain why Cl−\text{Cl}^- is preferentially oxidised over water despite thermodynamic data. [3]
(iii) Calculate the minimum theoretical voltage required to drive the overall cell reaction. [2 marks]
(b)
The industrial cell operates at approximately 3.5 V3.5\,\text{V}, significantly higher than the value calculated in (a)(iii). Explain two reasons for this difference, referencing overpotential and ohmic losses. In your answer, state which electrode is most affected by overpotential and explain how this relates to the electrode choice conflict identified in (a)(ii). [4 marks]
(c)
In membrane-free electrolysis cells, Cl2\text{Cl}_2 produced at the anode contacts NaOH produced at the cathode. The following half-equations describe the resulting disproportionation reaction: Oxidation: Cl2(g)+4OH−(aq)→2ClO−(aq)+2H2O(l)+2e−\text{Oxidation: } \text{Cl}_2(g) + 4\text{OH}^-(aq) \rightarrow 2\text{ClO}^-(aq) + 2\text{H}_2\text{O}(l) + 2e^- Reduction: Cl2(g)+2e−→2Cl−(aq)\text{Reduction: } \text{Cl}_2(g) + 2e^- \rightarrow 2\text{Cl}^-(aq) (i) Deduce the overall ionic equation for this reaction and state the oxidation state of chlorine in Cl2\text{Cl}_2, ClO−\text{ClO}^-, and Cl−\text{Cl}^-. [2]
(ii) Evaluate the industrial consequences of this side reaction for both the yield of Cl2\text{Cl}_2 and the purity of the NaOH product. In your answer, consider whether the use of a membrane fully resolves both problems and justify your reasoning. [2 marks]

Solutions

37ChallengeLAQCovalent bonds and electron sharing15 marksPaper 3~23 min
The molecule dinitrogen tetroxide, N2O4\text{N}_2\text{O}_4, exists in equilibrium with nitrogen dioxide, NO2\text{NO}_2: N2O4(g)⇌2 NO2(g)\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g) The N–N bond in N2O4\text{N}_2\text{O}_4 is 175 pm175\,\text{pm} long, compared to a typical N–N single bond of 145 pm145\,\text{pm}. The molecule is planar.
(a)
Draw the Lewis (electron dot) structure of N2O4\text{N}_2\text{O}_4, including one resonance form. Show all lone pairs and any formal charges. [2 marks]
(b)
State the hybridization of each nitrogen atom in N2O4\text{N}_2\text{O}_4 and the bond order of the N–N bond. [2 marks]
(c)
Using the concepts of resonance, formal charge, and electrostatic effects, explain why the N–N bond in N2O4\text{N}_2\text{O}_4 is longer and weaker than a typical N–N single bond. Use the bond enthalpy data below to support your answer quantitatively. Bond enthalpy data: Bond — Enthalpy / kJ mol−1\text{kJ mol}^{-1} — Approximate length / pm N–N — 167 — 145 N–O — 201 — 140 N=O — 607 — 120 S–O — 265 — 166 S=O — 523 — 143 Electronegativity values: N =3.0= 3.0, O =3.5= 3.5, S =2.5= 2.5 [4 marks]
(d)
A hypothetical molecule N2O3S\text{N}_2\text{O}_3\text{S} is formed by replacing one terminal oxygen atom in N2O4\text{N}_2\text{O}_4 with a sulfur atom, giving a structure in which one N is bonded to two O atoms and the other N is bonded to one O and one S. (i) Predict whether the N–N bond length and bond dissociation enthalpy in N2O3S\text{N}_2\text{O}_3\text{S} are each greater than, less than, or equal to the corresponding values in N2O4\text{N}_2\text{O}_4. [2]
(ii) Justify your predictions in (d)(i) by referring to electronegativity values, the bond enthalpy data above, and the effect of atomic size on molecular planarity and π\pi conjugation. [5 marks]

Solutions

38ChallengeLAQElectronegativity and bond polarity15 marksPaper 3~23 min
A research chemist investigates halogen-substituted methanes to understand how bond properties influence nucleophilic substitution reactivity. She prepares three compounds: CH3F\text{CH}_3\text{F}, CH3Cl\text{CH}_3\text{Cl}, and CH3I\text{CH}_3\text{I}. Pauling electronegativity values: H=2.20\text{H} = 2.20, C=2.55\text{C} = 2.55, F=3.98\text{F} = 3.98, Cl=3.16\text{Cl} = 3.16, I=2.66\text{I} = 2.66. Mean bond enthalpies: C–F=485 kJ mol−1\text{C–F} = 485\,\text{kJ mol}^{-1}, C–Cl=339 kJ mol−1\text{C–Cl} = 339\,\text{kJ mol}^{-1}, C–I=218 kJ mol−1\text{C–I} = 218\,\text{kJ mol}^{-1}.
(a)
State which of the three halogens has the highest electronegativity. [1 mark]
(b)
Calculate the electronegativity difference, ΔEN\Delta\text{EN}, for the C–X bond in each of CH3F\text{CH}_3\text{F}, CH3Cl\text{CH}_3\text{Cl}, and CH3I\text{CH}_3\text{I}. [3 marks]
(c)
Using your answers from (b) and the bond enthalpy data above, explain why CH3F\text{CH}_3\text{F} has the most polar C–X bond yet undergoes SN2S_\text{N}2 reaction with aqueous hydroxide most slowly. In your answer, refer to the role of bond strength and leaving group ability. [4 marks]
(d)
The observed SN2S_\text{N}2 reactivity order with aqueous hydroxide is CH3I>CH3Cl>CH3F\text{CH}_3\text{I} > \text{CH}_3\text{Cl} > \text{CH}_3\text{F}. Using the electronegativity differences from (b) and the bond enthalpy data, evaluate the relative importance of bond polarity and bond strength in determining this reactivity order. Your answer should identify which factor is dominant and justify this conclusion with reference to the numerical data. [7 marks]

Solutions

39ChallengeLAQLewis acids and bases15 marksPaper 3~23 min
Aluminium chloride, AlCl3\text{AlCl}_3, is a white solid that sublimes at 178 ∘C178\,^\circ\text{C} and exists a dimer, Al2Cl6\text{Al}_2\text{Cl}_6, in the vapour phase. It is widely used as a catalyst in the Friedel–Crafts acylation of benzene, where it activates an acyl chloride, RCOCl, by forming a complex.
(a)
State, with a reason, whether AlCl3\text{AlCl}_3 acts a Lewis acid or a Lewis base when it forms the dimer Al2Cl6\text{Al}_2\text{Cl}_6. [2 marks]
(b)
Describe the bonding and geometry in the dimer Al2Cl6\text{Al}_2\text{Cl}_6, identifying the type of bond formed at the bridging positions and the resulting coordination geometry around each aluminium atom. [2 marks]
(c)
The Friedel–Crafts acylation of benzene with ethanoyl chloride, CH3COCl\text{CH}_3\text{COCl}, in the presence of AlCl3\text{AlCl}_3 proceeds via the formation of an intermediate complex [CH3CO]+⋯[AlCl4]−[\text{CH}_3\text{CO}]^+\cdots[\text{AlCl}_4]^-. (i) Explain, using Lewis acid–base theory, the role of AlCl3\text{AlCl}_3 in the formation of this complex. Include the direction of electron-pair movement in your explanation. [4]
(ii) Evaluate why AlCl3\text{AlCl}_3 is considered a catalyst in this reaction, even though it forms a new chemical species, [AlCl4]−[\text{AlCl}_4]^-, during the process. [3 marks]
(d)
Boron trifluoride, BF3\text{BF}_3, is also a Lewis acid but does not dimerise under comparable conditions. Using the electronic structures of BF3\text{BF}_3 and AlCl3\text{AlCl}_3, and considering both the strength of the relevant bonds and the ability of the halide ligands to donate lone pairs, construct a reasoned argument to explain why dimerisation is thermodynamically favourable for AlCl3\text{AlCl}_3 but not for BF3\text{BF}_3. Bond enthalpies: Al–Cl=420 kJ mol−1,B–F=613 kJ mol−1\text{Bond enthalpies: } \text{Al–Cl} = 420\,\text{kJ mol}^{-1},\quad \text{B–F} = 613\,\text{kJ mol}^{-1} [4 marks]

Solutions

40ChallengeLAQLewis acids and bases15 marksPaper 3~23 min

Data

- Electron configuration of Fe2+\text{Fe}^{2+}: [Ar] 3d6[\text{Ar}]\,3d^6 - Electron configuration of Au+\text{Au}^+: [Xe] 4f14 5d10[\text{Xe}]\,4f^{14}\,5d^{10} - Kf([Fe(CN)6]4−)=1.0×1035K_f\bigl([\text{Fe}(\text{CN})_6]^{4-}\bigr) = 1.0 \times 10^{35} - Kf([Au(CN)2]−)=2.0×1038K_f\bigl([\text{Au}(\text{CN})_2]^{-}\bigr) = 2.0 \times 10^{38}
The cyanide ion, CN−\text{CN}^-, is a classic Lewis base that forms stable complexes with many metal ions. In the extraction of gold from low-grade ores, crushed rock is treated with a dilute aqueous solution of sodium cyanide in the presence of oxygen. The gold dissolves according to: 4Au(s)+8CN−(aq)+O2(g)+2H2O(l)→4[Au(CN)2]−(aq)+4OH−(aq)4\text{Au}(s) + 8\text{CN}^-(aq) + \text{O}_2(g) + 2\text{H}_2\text{O}(l) \rightarrow 4[\text{Au}(\text{CN})_2]^-(aq) + 4\text{OH}^-(aq) Relevant
(a)
State the Lewis acid and Lewis base in the complex ion [Au(CN)2]−[\text{Au}(\text{CN})_2]^-, and explain your reasoning in terms of electron-pair donation and acceptance. [3 marks]
(b)
- (i) State the shape and bond angle of [Au(CN)2]−[\text{Au}(\text{CN})_2]^-. [1] -
(ii) The Au+\text{Au}^+ ion has a 5d105d^{10} electron configuration. Explain, using hybridisation, why [Au(CN)2]−[\text{Au}(\text{CN})_2]^- adopts this geometry. [3 marks]
(c)
An environmental chemist investigates the toxicity of free cyanide in water. The complex [Fe(CN)6]4−[\text{Fe}(\text{CN})_6]^{4-} is much less toxic than free CN−\text{CN}^-. - (i) The formation constant of [Au(CN)2]−[\text{Au}(\text{CN})_2]^- is significantly larger than that of [Fe(CN)6]4−[\text{Fe}(\text{CN})_6]^{4-}, even though Fe2+\text{Fe}^{2+} carries a higher formal charge than Au+\text{Au}^+. Explain this observation with reference to the nature of the metal–ligand bonding in each complex. [4] -
(ii) A proposal is made to detoxify water containing free CN−\text{CN}^- by adding excess Fe2+\text{Fe}^{2+} ions to form [Fe(CN)6]4−[\text{Fe}(\text{CN})_6]^{4-}. Write the equation for the formation of [Fe(CN)6]4−[\text{Fe}(\text{CN})_6]^{4-} from Fe2+\text{Fe}^{2+} and CN−\text{CN}^-, and calculate the equilibrium constant for the reverse reaction (dissociation of [Fe(CN)6]4−[\text{Fe}(\text{CN})_6]^{4-}). Hence evaluate whether the proposal is chemically feasible. [4 marks]

Solutions