Topic: 3.3
The Pauling electronegativity values of elements can be used to predict the chemical properties of compounds. Use the information in Table 1.1 to answer the following questions.
(a) (i) Define electronegativity.
(ii) O and S are in Group 16. Explain the difference in the Pauling electronegativity values of O and S.
(b) (i) LiH is an ionic compound. Draw a dot‑and‑cross diagram of LiH. Include all electrons.
(ii) Suggest the shape of a molecule of \(H_2S\).
(c) (i) Write an equation that represents the first ionisation energy of H.
(ii) Explain why there is no information given in Table 1.1 for the second ionisation energy of H.
(iii) Give the full electronic configuration of \(S^{2+}\)(g).
(d) \(CO_2\) and \(SO_2\) are acidic gases.
(i) Write an equation for the reaction of \(SO_2\) with \(H_2O\).
(ii) Write an equation for the reaction of \(SO_2\) with NaOH.
(iii) Construct an equation for the reaction of \(CO_2\) with \(Mg(OH)_2\).
(e) (i) Complete Table 1.2 by placing a tick (✓) to show which of the compounds have molecules with an overall dipole moment.
(ii) At 150°C and 103kPa, all of the compounds listed in Table 1.2 are gases. Under these conditions, 0.284g of one of the compounds occupies a volume of \(127cm^3\). Use this information to calculate the \(M_r\) of the compound. Hence, identify the compound from those given in Table 1.2. Show your working.
▶️ Answer/Explanation
(a)(i) The power of an atom to attract the bonding pair of electrons towards itself in a covalent bond.
(a)(ii) Oxygen has a higher electronegativity than sulfur because its bonding electrons are closer to the nucleus (smaller atomic radius) and experience less shielding from inner shells, leading to a stronger effective nuclear charge and a greater attraction for electrons.
(b)(i)
Explanation: Lithium (Li) has one valence electron, and hydrogen (H) has one. In ionic LiH, lithium transfers its electron to hydrogen, forming \(Li^+\) and \(H^-\) ions.
(b)(ii) Non-linear / Bent / V-shaped
Explanation: \(H_2S\) has two bonding pairs and two lone pairs of electrons around the central sulfur atom, resulting in a bent molecular shape with a bond angle of approximately \(92^\circ\).
(c)(i) \(H(g) \rightarrow H^+(g) + e^-\)
Explanation: The first ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms.
(c)(ii) A hydrogen atom has only one electron, so it cannot have a second ionisation energy.
Explanation: The second ionisation energy would involve removing an electron from \(H^+\), which has no electrons to remove.
(c)(iii) \(1s^2 2s^2 2p^6 3s^2 3p^2\)
Explanation: A neutral sulfur atom (\(Z=16\)) has the configuration \(1s^2 2s^2 2p^6 3s^2 3p^4\). Removing two electrons gives \(S^{2+}\) with 14 electrons.
(d)(i) \(SO_2(g) + H_2O(l) \rightarrow H_2SO_3(aq)\)
Explanation: Sulfur dioxide dissolves in water to form sulfurous acid.
(d)(ii) \(SO_2(g) + 2NaOH(aq) \rightarrow Na_2SO_3(aq) + H_2O(l)\)
Explanation: Sulfur dioxide acts as an acidic oxide, reacting with a base to form a salt (sodium sulfite) and water.
(d)(iii) \(CO_2(g) + Mg(OH)_2(s) \rightarrow MgCO_3(s) + H_2O(l)\)
Explanation: Carbon dioxide reacts with the insoluble base magnesium hydroxide to form magnesium carbonate and water.
(e)(i)
Explanation: A molecule has an overall dipole moment if it is asymmetric and has polar bonds. \(CO_2\) is linear and symmetric (no dipole), \(SO_2\) is bent (has a dipole), \(H_2S\) is bent (has a dipole), \(CH_4\) is tetrahedral and symmetric (no dipole), \(BF_3\) is trigonal planar and symmetric (no dipole).
(e)(ii) Using the ideal gas equation \(PV = nRT\), converted to find molar mass \(M = \frac{mRT}{PV}\). Mass \(m = 0.284\) g, volume \(V = 0.127\) dm³, pressure \(P = 103\) kPa, temperature \(T = 423\) K, \(R = 8.314\) dm³ kPa mol⁻¹ K⁻¹. Substituting the values gives \(M = \frac{0.284 \times 8.314 \times 423}{103 \times 0.127} = 64.1\) g mol⁻¹. The compound with this \(M_r\) is sulfur dioxide, \(SO_2\).
Topic: 11.2
The Group 2 elements Mg to Ba are all silvery‑white reactive metals.
(a) (i) Draw a labelled diagram to show the bonding and structure of the Group 2 metals at room temperature.
(ii) Explain why Mg has a higher electrical conductivity than Na.
(b) Write an equation for the reaction of magnesium with cold water.
(c) Identify a single reagent that can be used to distinguish separate samples of dilute \(Mg(NO_3)_2(aq)\) and dilute \(Ba(NO_3)_2(aq)\). Explain your answer.
(d) (i) Describe what is observed when \(SrI_2(aq)\) reacts with concentrated sulfuric acid.
(ii) Compound X, an anhydrous Group 2 bromide, is dissolved in water and titrated against aqueous silver nitrate. A solution containing 0.250g of X requires \(33.65cm^3\) of \(0.0500moldm^{–3}\) \(AgNO_3(aq)\) for complete reaction. Identify X. Show your working.
▶️ Answer/Explanation
(a)(i)
Explanation: The diagram shows a giant metallic lattice structure. The key features are the regular arrangement of positive metal ions (cations) and the ‘sea’ of delocalised electrons surrounding them, which are responsible for bonding and conductivity.
(a)(ii) Mg has more delocalised e⁻ (than Na).
Explanation: Magnesium atoms contribute two valence electrons to the delocalised sea, while sodium atoms contribute only one. A higher number of charge carriers (delocalised electrons) per atom results in a higher electrical conductivity for magnesium.
(b) \(Mg + 2H_2O \rightarrow Mg(OH)_2 + H_2\)
Explanation: Magnesium reacts slowly with cold water to form the slightly soluble magnesium hydroxide and hydrogen gas. This is a typical reaction for a Group 2 metal, though magnesium is the least reactive in the group.
(c) Reagent = any named/formula of soluble sulfate OR H₂SO₄ OR any named/formula of soluble hydroxide.
BaSO₄ insoluble (& MgSO₄ soluble) OR (Ba(OH)₂ soluble &) Mg(OH)₂ insoluble.
Explanation: Sulfate ions form a white precipitate with Ba²⁺ but not with Mg²⁺. Alternatively, hydroxide ions form a precipitate with Mg²⁺ but not with Ba²⁺, as barium hydroxide is soluble.
(d)(i) white precipitate, yellow solid, effervescence / misty fumes, (dark) grey solid / purple gas, rotten egg smell.
Explanation: Concentrated sulfuric acid reacts with the iodide ion, oxidizing it. Observations include a white precipitate of strontium sulfate (SrSO₄), a yellow solid of sulfur (S), misty fumes of hydrogen iodide (HI), purple vapour of iodine (I₂), and the smell of hydrogen sulfide (H₂S) from further reduction.
(d)(ii) X is BaBr₂.
Explanation: Moles of AgNO₃ = \( \frac{33.65}{1000} \times 0.0500 = 1.6825 \times 10^{-3} \) mol.
The reaction is: \( MBr_2 + 2AgNO_3 \rightarrow 2AgBr + M(NO_3)_2 \), so moles of X = \( \frac{1}{2} \times \) moles of AgNO₃ = \( 8.4125 \times 10^{-4} \) mol.
\( M_r(X) = \frac{0.250}{8.4125 \times 10^{-4}} = 297.2 \) g/mol.
\( A_r \) of Group 2 element = \( 297.2 – 2(79.9) = 137.4 \). This atomic mass identifies the element as barium (Ba).
Topic: 18.2
Alkenes undergo an addition reaction with a 1:1 mixture of CO and H₂ to form aldehydes. Fig. 3.1 shows the reaction of propene with a 1:1 mixture of CO and H₂.
(a) (i) Define addition reaction.
(ii) Aldehydes A and B are structural isomers. State the type of structural isomerism shown by A and B.
(iii) Name A.
(iv) The complete reaction of propene with a 1:1 mixture of CO and H₂ produces A and B only. The product mixture contains 96% A and 4% B. Calculate the mass of A produced in this reaction when \(5.00×10^3 kg\) of propene is used.
(b) A and B show reactions typical of aliphatic aldehydes.
(i) A undergoes a nucleophilic addition reaction with a mixture of HCN and KCN, forming compound C. Complete the diagram to show the mechanism for this reaction. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate. Draw the structure of the organic intermediate.
(ii) Table 3.1 shows information about three experiments involving B. Complete Table 3.1.
(iii) B, C₄H₈O, is oxidised by acidified potassium manganate(VII). Complete the equation for this reaction. Use [O] to represent one atom of oxygen from the oxidising agent.
(iv) C is a chiral molecule. Circle any chiral centres in the structure of C shown in Fig. 3.2.
(c) When propene reacts with CO and an excess of H₂, an alkane and a mixture of alcohols are formed instead. The alcohols are isomers of each other. Suggest the molecular formulae of the alkane and the alcohols that are formed under these conditions.
(d) The reaction of ethene, C₂H₄, with a 1:1 mixture of CO and H₂ is shown in equation 1.
equation 1 C₂H₄(g) + CO(g) + H₂(g) \(\rightleftharpoons \) CH₃CH₂CHO(g)
At atmospheric pressure a cobalt‑based catalyst is used in this reaction.
(i) State and explain the effect of using a catalyst on this reaction.
(ii) Explain why the yield of CH₃CH₂CHO(g) increases when the overall pressure of the reaction mixture is increased.
(iii) Use the information in Table 3.2 to calculate the enthalpy change, \(ΔH_r\), of the reaction in equation 1.
equation 1 C₂H₄(g) + CO(g) + H₂(g) \(\rightleftharpoons \) CH₃CH₂CHO(g)
(iv) The reaction mixture is cooled to collect CH₃CH₂CHO as a liquid. Identify all types of van der Waals’ forces that are present between molecules of CH₃CH₂CHO.
▶️ Answer/Explanation
(a)(i) A reaction where two or more molecules combine to form a single product.
(a)(ii) Positional isomerism.
Explanation: A and B are aldehydes with the same carbon chain but the carbonyl group (-CHO) is attached at a different position.
(a)(iii) Butanal.
(a)(iv) 8230 kg (or 8229 kg).
Explanation: The mass of propene used is \(5.00 \times 10^3\) kg. The molar mass of propene (\(C_3H_6\)) is 42.0 g/mol. Moles of propene = \((5.00 \times 10^6 \, \text{g}) / 42.0 \, \text{g/mol} = 1.1905 \times 10^5\) mol. Since 96% forms A (butanal, \(C_4H_8O\), molar mass 72.0 g/mol), mass of A = \(0.96 \times 1.1905 \times 10^5 \, \text{mol} \times 72.0 \, \text{g/mol} = 8.229 \times 10^6\) g = 8230 kg.
(b)(i)
Explanation: The nucleophilic cyanide ion (CN⁻) attacks the electrophilic carbonyl carbon. A curly arrow shows this movement. Another curly arrow shows the movement of electrons from the C=O double bond to the oxygen atom, forming a tetrahedral intermediate.
(b)(ii) Row 1: acidified K₂Cr₂O₇
Row 2: Tollens’ reagent
Row 3: no reaction
Explanation: Aldehydes are oxidised by acidified potassium dichromate(VI) and reduce Tollens’ reagent. They do not react with iodine in sodium hydroxide (the iodoform test), which is for methyl ketones or alcohols with CH₃CH(OH)- group.
(b)(iii) C₄H₈O + [O] → C₄H₈O₂
Explanation: Aldehydes are oxidised to carboxylic acids. The aldehyde B gains one oxygen atom to form the carboxylic acid.
(b)(iv)
Explanation: A chiral centre is a carbon atom with four different substituents. In the structure of C, the carbon atom bonded to -OH, -CN, -H, and -CH₂CH₃ is chiral because all four groups are different.
(c) molecular formula of alkane: C₃H₈
molecular formula of alcohols: C₄H₁₀O
Explanation: With excess H₂, full hydrogenation occurs. Propene (C₃H₆) can be hydrogenated to propane (C₃H₈). The hydroformylation reaction with excess H₂ would produce butanols (C₄H₁₀O), which are positional isomers (butan-1-ol and butan-2-ol).
(d)(i) A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy.
(d)(ii) The yield increases because the reaction involves a decrease in the number of moles of gas (3 moles of reactants → 1 mole of product). According to Le Chatelier’s principle, increasing pressure favours the side with fewer moles.
(d)(iii) –128 kJ mol⁻¹
Explanation: Using Hess’s Law and the bond enthalpy data: \(ΔH_r = \sum (BE_{\text{bonds broken}}) – \sum (BE_{\text{bonds formed}})\). A simpler approach is to use the standard enthalpies of formation: \(ΔH_r = ΔH_f^\circ(\text{CH}_3\text{CH}_2\text{CHO}) – [ΔH_f^\circ(\text{C}_2\text{H}_4) + ΔH_f^\circ(\text{CO}) + ΔH_f^\circ(\text{H}_2)]\). Since \(ΔH_f^\circ(\text{H}_2) = 0\), \(ΔH_r = (-187) – [52 + (-111)] = -187 – (-59) = -128\) kJ mol⁻¹.
(d)(iv) Instantaneous dipole-induced dipole (id-id) forces and permanent dipole-permanent dipole (pd-pd) forces.
Explanation: All molecules experience id-id (London dispersion) forces. The carbonyl group in propanal has a permanent dipole moment, so pd-pd forces are also present.
Topic: 16.1
Fig. 4.1 shows some reactions of compound D, 2‑bromobutane.
(a) (i) State the reagent and conditions used to form E in reaction 1.
(ii) Draw the structure of one repeat unit of the addition polymer that forms from E.
(iii) E also forms when F is heated strongly in the presence of an Al₂O₃ catalyst. Write an equation for this reaction.
(b) (i) Predict what is observed in reaction 2.
(ii) Identify the yellow precipitate and the organic ion formed in reaction 3.
(c) (i) State the type of reaction that occurs in reaction 4.
(ii) Reaction 5 is similar to the reaction of LiAlH₄ with carboxylic acids to form alcohols. Suggest the role of LiAlH₄ in reaction 5.
(d) (i) Fig. 4.2 shows the infrared spectrum of one of the compounds D, E, F, G or H.
Use information from Table 4.1 (on page 14) to identify which of the compounds D, E, F, G or H produces the infrared spectrum in Fig. 4.2. Explain your answer.
(ii) In the mass spectrum of D, the relative abundance of the molecular ion peak is 3.4. Predict the relative abundance of the M+2 peak for D. Explain your answer.
▶️ Answer/Explanation
(a)(i) NaOH in alcohol / ethanol AND heat (under reflux)
Explanation: Reaction 1 is an elimination reaction where 2-bromobutane loses HBr to form the alkene E (but-2-ene). This requires a strong base (NaOH) dissolved in ethanol and heat to proceed.
(a)(ii)
Explanation: E is but-2-ene, an alkene. The repeat unit of its addition polymer, poly(but-2-ene), is derived from the alkene monomer by opening the double bond.
(a)(iii) \( \ce{C4H10O -> C4H8 + H2O} \) or \( \ce{CH3CH(OH)CH2CH3 -> CH3CH=CHCH3 + H2O} \)
Explanation: F is butan-2-ol. The dehydration of an alcohol to form an alkene occurs with concentrated sulfuric acid or by passing its vapors over a heated alumina (\( \ce{Al2O3} \)) catalyst.
(b)(i) cream(-coloured) / off-white precipitate (forms)
Explanation: Reaction 2 is the test for a bromide ion with silver nitrate. Silver bromide (\( \ce{AgBr} \)) is insoluble and forms a cream-colored precipitate.
(b)(ii)
yellow precipitate: \( \ce{CHI3} \) / iodoform / triiodomethane
organic ion: \( \ce{CH3CH2CO2-} \) / propanoate
Explanation: Reaction 3 is the iodoform test. A methyl ketone (G, butanone) is oxidized by iodine to form a yellow precipitate of iodoform (\( \ce{CHI3} \)) and the carboxylate ion propanoate (\( \ce{CH3CH2COO-} \)).
(c)(i) substitution
Explanation: Reaction 4 involves the cyanide ion (\( \ce{CN-} \)) replacing the bromine atom in 2-bromobutane to form the nitrile H (2-methylpropanenitrile), which is a nucleophilic substitution reaction.
(c)(ii) reducing agent
Explanation: Lithium aluminium hydride (\( \ce{LiAlH4} \)) is a strong reducing agent commonly used to reduce nitriles to primary amines, as it is for reducing carboxylic acids to alcohols.
(d)(i) G
(absorption at) 2200–2250 \( \ce{(cm^{-1})} \) AND C≡N
Explanation: The spectrum shows a strong, sharp absorption between 2200–2250 \( \ce{cm^{-1}} \), which is characteristic of the C≡N stretch in a nitrile group. Compound G is the only nitrile among the options.
(d)(ii) 3.4 AND relative abundance of \( ^{79}\ce{Br}:^{81}\ce{Br} \) ≅ 50:50 OR 1:1
Explanation: Bromine has two isotopes, \( ^{79}\ce{Br} \) and \( ^{81}\ce{Br} \), in an approximately 1:1 ratio. The M+2 peak corresponds to molecules containing the \( ^{81}\ce{Br} \) isotope. Since the molecular ion (M+) and M+2 peaks have almost equal abundance, the relative abundance of the M+2 peak is also 3.4.