9701_m24_qp

Question 1

Topic: 11.4

Bismuth is an element in Group 15 of the Periodic Table.
(a) Bismuth has metallic bonding. Draw a labelled diagram to show the metallic bonding in bismuth.

(b) Bismuth reduces water to form bismuth oxide, Bi₂O₃. A colourless gas that ignites with a squeaky pop also forms.
(i) Construct an equation for the reduction of water by bismuth.
(ii) Bi₂O₃ is a yellow insoluble solid that melts at 1090K. The molten compound conducts electricity. Deduce the structure and bonding of Bi₂O₃. Explain your answer.

(c) Bi₂O₃ can be used to form NaBiO₃, as shown in equation 1.
equation 1 Na₂O + Bi₂O₃ + O₂ → 2NaBiO₃
(i) Deduce the oxidation number of Bi in Bi₂O₃ and in NaBiO₃.
oxidation number of Bi:
in Bi₂O₃ ……………….. in NaBiO₃ ……………………..
(ii) Identify the reducing agent in equation 1.

(d) \(NaBiO_3\) is an oxidising agent with similar properties to KMnO₄. Fig. 1.1 shows an example of the use of NaBiO₃ as an oxidising agent.

(i) Explain the term oxidising agent.
(ii) Compound X forms when methylbut-2-ene reacts with \(KMnO_4\). State the essential conditions for this reaction.
(iii) Complete Table 1.1 to show what is observed when compounds Y and Z react separately with the named reagents.

(iv) Construct an equation for the reaction of Z with \(NaBH_4\). Use [H] to represent an atom of hydrogen from the reducing agent.

(e) NaBiO₃ can be used to determine the concentration of Mn²⁺(aq). The ionic equation for the reaction is shown in equation 2.
equation 2 \(2Mn^{2+} + 5BiO_3^– + 14H^+ \to 2MnO_4^– + 5Bi^{3+} + 7H_2O\)
A student uses the following procedure in an experiment.
• Add 100.0 cm³ of a saturated solution of Mn²⁺(aq) to a volumetric flask.
• Add distilled water to the flask to make a 1.00 dm³ diluted solution.
• Titrate a 25.00 cm³ sample of the diluted solution with 0.100 moldm⁻³ NaBiO₃ (aq).
The 25.00 cm³ sample of the diluted solution of Mn²⁺(aq) reacts completely with exactly 21.50 cm³ of 0.100 moldm⁻³ NaBiO₃(aq). Calculate the concentration, in mol/dm³, of Mn²⁺(aq) in the saturated solution. Show your working.

▶️ Answer/Explanation

Solution

(a)

Explanation: The diagram shows metallic bonding in bismuth, consisting of a lattice of positive ions (Bi³⁺) surrounded by a “sea” of delocalized electrons. These free-moving electrons are responsible for conductivity and malleability.

(b)(i) \(2Bi + 3H_2O \to Bi_2O_3 + 3H_2\)

Explanation: Bismuth reacts with water, reducing it to hydrogen gas (which ignites with a squeaky pop) and forming bismuth oxide (Bi₂O₃).

(b)(ii) Giant ionic structure.

Explanation: Bi₂O₃ has a high melting point (1090K) and conducts electricity when molten, indicating a giant ionic lattice with strong electrostatic forces between Bi³⁺ and O²⁻ ions.

(c)(i) Oxidation number of Bi: in Bi₂O₃ +3, in NaBiO₃ +5.

Explanation: In Bi₂O₃, oxygen is -2, so Bi is +3. In NaBiO₃, Na is +1 and O is -2, so Bi is +5 to balance the charges.

(c)(ii) Reducing agent: Bi₂O₃.

Explanation: Bi₂O₃ is oxidized (Bi changes from +3 to +5), so it acts as the reducing agent.

(d)(i) An oxidising agent accepts electrons from another substance, causing oxidation.

Explanation: Oxidising agents gain electrons (are reduced) while oxidizing other species.

(d)(ii) Cold and dilute KMnO₄.

Explanation: These conditions ensure diol formation from alkenes without further oxidation.

(d)(iii)

Explanation: Y (ketone) gives no reaction with Tollens’ reagent but forms an orange precipitate with 2,4-DNPH. Z (aldehyde) gives a silver mirror with Tollens’ and an orange precipitate with 2,4-DNPH.

(d)(iv) CH₃CHO + 2[H] → CH₃CH₂OH

Explanation: NaBH₄ reduces aldehydes to primary alcohols by adding hydrogen (represented as [H]).

(e) Concentration of Mn²⁺ = 0.344 mol/dm³.

Explanation: Moles of NaBiO₃ = 0.0215 × 0.100 = 0.00215 mol. From the equation, moles of Mn²⁺ = (2/5) × 0.00215 × 40 = 0.0344 mol. Concentration = 0.0344 / 0.100 = 0.344 mol/dm³.

Question 2

Topic: 11.2

Chlorine, Cl₂, reacts with many elements and compounds to form chlorides. Table 2.1 shows information about some chlorides of Period 3 elements.

(a) Complete Table 2.1.
(b) When Cl₂ reacts with cold NaOH(aq), Cl₂ is both oxidised and reduced. The products are NaCl, water, and G.
(i) State the type of redox reaction in which the same species is both oxidised and reduced.
(ii) Identify G.
(iii) Write an equation for the reaction between Cl₂ and hot NaOH(aq).
(iv) Describe fully what is observed when AgNO₃ (aq) is added to the aqueous solution of the chloride of sodium, followed by dilute NH₃ (aq).

(c) An excess of Cl₂ reacts with phosphorus to form PCl₅.
(i)PCl₅ is a simple molecule in the gas phase. It also exists in a solid form as two ions, PCl₄⁺ and PCl₆⁻. Complete Table 2.2 to identify the shapes of each of these species.

(ii) PCl₅ reacts with J to form H₃PO₄. Identify J and state the type of reaction.

(d) Cl₂ reacts readily with propene to form K, 1,2-dichloropropane. K can be used to form L.

(i) Complete Fig. 2.2 to show the mechanism for the reaction of Cl₂ with propene in reaction 1. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.

(ii) Identify the reagent and conditions for reaction 2.
(iii) Draw one repeat unit of the addition polymer that forms from L.

▶️ Answer/Explanation

Solution

(a)

Explanation: The table is completed by identifying the oxidation states and bonding types for the chlorides of Period 3 elements. Sodium chloride (NaCl) has ionic bonding and oxidation state +1, while phosphorus pentachloride (PCl₅) has covalent bonding and oxidation state +5.

(b)(i) Disproportionation.

Explanation: Disproportionation occurs when the same species is both oxidized and reduced. Here, Cl₂ is oxidized to NaClO and reduced to NaCl.

(b)(ii) NaClO / sodium chlorate(I).

Explanation: The product G is sodium chlorate(I) (NaClO), formed alongside NaCl and H₂O in the disproportionation reaction of Cl₂ with cold NaOH.

(b)(iii) 3Cl₂ + 6NaOH → NaClO₃ + 5NaCl + 3H₂O.

Explanation: With hot NaOH, Cl₂ undergoes further disproportionation to form sodium chlorate(V) (NaClO₃), NaCl, and water.

(b)(iv) White precipitate forms, which dissolves in dilute NH₃(aq).

Explanation: AgNO₃ reacts with NaCl to form a white AgCl precipitate, soluble in NH₃(aq) due to the formation of [Ag(NH₃)₂]⁺.

(c)(i) ● Trigonal bipyramidal ● Octahedral.

Explanation: PCl₄⁺ has a trigonal bipyramidal shape (5 regions of electron density), while PCl₆⁻ has an octahedral shape (6 regions).

(c)(ii) H₂O / water, hydrolysis.

Explanation: PCl₅ reacts with water (J) in a hydrolysis reaction to form phosphoric acid (H₃PO₄).

(d)(i)

Explanation: The mechanism involves electrophilic addition of Cl₂ to propene, with curly arrows showing the movement of electrons and the formation of a carbocation intermediate.

(d)(ii) NaOH in ethanol AND heat.

Explanation: Reaction 2 is an elimination reaction, requiring NaOH in ethanol under heat to form L (propene).

(d)(iii)

Explanation: The repeat unit of the addition polymer (polypropene) is derived from the monomer propene (L).

Question 3

Topic: 10.1

Nitrogen, \(N_2\), is generally an unreactive molecule but it does react under certain conditions.
(a) Give two reasons to explain the lack of reactivity of nitrogen

(b) \(N_2\) can react with oxygen in an internal combustion engine to form a mixture of NO and \(NO_2\). Fig. 3.1 shows a reaction scheme involving \(N_2\).

(i) Write an equation to show the formation of a mixture of NO and NO₂ in reaction 1.
(ii) Give the formulae of the products of reaction 2.
(iii) State one environmental consequence of reaction 3.

(c) The Haber process involves the reaction of N₂ and H₂ to form ammonia, NH₃. A catalyst is used, which allows the process to be carried out at a lower temperature and pressure.

(i) Use the information in (c) to complete Table 3.1.

(ii) Explain how the presence of a catalyst affects the reaction.
(iii) State and explain the effect, if any, on the rate of the Haber process as the pressure is lowered.

(d) The N₂F₂ molecule has a double covalent bond between its nitrogen atoms. This consists of a σ and a π bond.
(i) Complete Fig. 3.2 to show the dot-and-cross diagram for N₂F₂. Show outer electrons only.

(ii) Deduce the hybridisation of the N atoms in N₂F₂.
(iii) Draw a diagram of the π bond between the N atoms in N₂F₂ and describe how it forms.

▶️ Answer/Explanation

Solution

(a) strong triple bond / high triple bond enthalpy
non-polar (molecule)

Explanation: Nitrogen (\(N_2\)) is unreactive due to its strong triple covalent bond (high bond enthalpy) and its non-polar nature, which prevents easy interaction with other molecules.

(b)(i) \(N_2 + 1½O_2 \to NO + NO_2\)

Explanation: In an internal combustion engine, nitrogen reacts with oxygen to form a mixture of nitrogen monoxide (NO) and nitrogen dioxide (NO₂).

(b)(ii) HNO₂ and HNO₃

Explanation: The products of reaction 2 are nitrous acid (HNO₂) and nitric acid (HNO₃), formed when NO and NO₂ react with water.

(b)(iii) photochemical smog

Explanation: Reaction 3 leads to photochemical smog, as NO₂ contributes to the formation of ground-level ozone and other pollutants.

(c)(i) 0 for both \(\Delta H_f\) of \(N_2\) and \(H_2\)
–46 for \(\Delta H_f\) of NH₃

Explanation: The standard enthalpy of formation (\(\Delta H_f\)) for elements in their standard states (N₂, H₂) is 0, while for NH₃, it is –46 kJ/mol.

(c)(ii) increases rate by providing a different mechanism with a lower activation energy

Explanation: A catalyst lowers the activation energy, enabling the reaction to proceed faster at a lower temperature.

(c)(iii) rate is lowered
lower frequency of successful collisions / number of successful collisions per unit time

Explanation: Lowering the pressure reduces the rate of the Haber process because there are fewer collisions between reactant molecules per unit time.

(d)(i)

Explanation: The dot-and-cross diagram shows the outer electrons of N₂F₂, with a double bond (σ + π) between the nitrogen atoms and single bonds to fluorine.

(d)(ii) \(sp^2\)

Explanation: Each nitrogen atom in N₂F₂ is \(sp^2\) hybridized, forming three σ bonds (one N=N and two N-F) with one remaining p orbital for the π bond.

(d)(iii)

Explanation: The π bond forms by the sideways overlap of unhybridized p orbitals on the nitrogen atoms, allowing electron density above and below the bond axis.

Question 4

Topic: 22.2

Compound S is used in food flavourings. A possible synthesis of S is shown in Fig. 4.1.

(a) P, Q, R and S show stereoisomerism. Complete Table 4.1 by identifying with a tick the type of stereoisomerism that each molecule shows. The type of stereoisomerism shown by Q is given.

(b) (i) Give the structural formula of Q.
(ii) Name the mechanism in reaction 2.
(iii) Complete the equation for reaction 3. R is represented as \(C_4H_9CN\).
\(C_4H_9CN\) + ……………………..

(i) Use the data from Table 4.2 to show that U contains 7 carbon atoms. Show your working.

(ii) Fig. 4.2 shows the infrared spectrum of U.

Use Fig. 4.2 and Table 4.3 to identify the functional group present in U. Explain your answer fully.
(iii) T also has a single functional group. Use the information in (c)(i) and your answer to (c)(ii) to identify T and U. Draw the structures of T and U in the boxes.

▶️ Answer/Explanation

Solution

(a)

Explanation: P shows E/Z isomerism due to restricted rotation around a double bond with different substituents. Q shows optical isomerism (given). R shows E/Z isomerism (if applicable). S shows optical isomerism due to the presence of a chiral center.

(b)(i) Structural formula of Q:

Explanation: Q is an optically active compound with a chiral carbon, represented by the given structural formula.

(b)(ii) Mechanism in reaction 2: Nucleophilic substitution.

Explanation: The hydroxide ion (\(OH^-\)) replaces the halogen in an \(S_N2\) reaction.

(b)(iii) \(C_4H_9CN + 2H_2O + H^+ \to C_4H_9COOH + NH_4^+\)

Explanation: The nitrile (\(C_4H_9CN\)) undergoes acid hydrolysis to form a carboxylic acid (\(C_4H_9COOH\)) and ammonium ion (\(NH_4^+\)).

(c)(i) \(n = \frac{100}{1.1} \times \frac{0.55}{7.2} = 6.94\) (so 7 carbon atoms)

Explanation: Using the given data, the calculation confirms U has 7 carbon atoms.

(c)(ii) Functional group: ester (RCOOR’).

Explanation: The IR spectrum shows absorptions at 1040–1300 cm⁻¹ (C—O stretch) and 1710–1750 cm⁻¹ (C=O stretch), characteristic of esters.

(c)(iii) Structures of T and U:

Explanation: U is an ester formed from S (carboxylic acid) and T (alcohol). The structures are deduced from the given data and functional group analysis.

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