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Question 1

Topic: 9.3

Cobalt, rhodium and iridium are metals in the same group of the Periodic Table.

(a) The shorthand electronic configuration of cobalt is \([Ar]3d^74s^2\).
(i) Identify what is meant by [Ar] by giving its full electronic configuration.
(ii) The lowest-energy electrons in cobalt are in the 1s orbital. Draw the shape of a 1s orbital.
(iii) Deduce the number of unpaired electrons in a cobalt atom.

(b) Table 1.1 gives some details of the stable naturally occurring isotopes of rhodium and iridium.

Table 1.1

Complete Table 1.1.

(c) Table 1.2 shows the relative abundances of isotopes in a sample of an alloy containing rhodium and iridium only.

Table 1.2

(i) Define relative isotopic mass.
(ii) Use Table 1.2 to calculate the relative atomic mass, \(A_r\), of iridium in the alloy. Give your answer to two decimal places.

(d) Hydrated rhodium(III) chloride, RhCl₃•xH₂O, catalyses the conversion of ethene to but-2-ene. Both stereoisomers of but-2-ene are formed in the reaction.
(i) Hydrated rhodium(III) chloride contains 20.5% by mass of water of crystallisation. Deduce the integer value of x in RhCl₃•xH₂O. Show your working.
(ii) Define stereoisomers.
(iii) Explain how the conversion of ethene to but-2-ene can be described as an addition reaction.
(iv) Draw the two stereoisomers of but-2-ene.

Stereoisomers of but-2-ene
▶️ Answer/Explanation
Solution

(a)(i) \([Ar]\) represents the electronic configuration of argon: \(1s^22s^22p^63s^23p^6\).

Explanation: The noble gas notation \([Ar]\) is shorthand for the full configuration of argon, which is the core electron arrangement for cobalt.

(a)(ii) The 1s orbital is spherical in shape.

Explanation: The 1s orbital is the simplest atomic orbital, with a spherical electron density distribution centered around the nucleus.

(a)(iii) There are 3 unpaired electrons in a cobalt atom.

Explanation: The \(3d^7\) configuration has 3 unpaired electrons (Hund’s rule: electrons occupy degenerate orbitals singly before pairing).

(b) Completed Table 1.1:

Answer to Table 1.1

Explanation: The missing values are filled based on the given isotopic data, ensuring mass numbers and proton/neutron counts are consistent.

(c)(i) Relative isotopic mass is the mass of an isotope relative to 1/12th the mass of a carbon-12 atom.

Explanation: It is a dimensionless quantity comparing the mass of an isotope to the standard carbon-12 scale.

(c)(ii) The relative atomic mass of iridium is \(192.22\).

Explanation: Calculation: \((191 \times 0.373) + (193 \times 0.627) = 192.22\).

(d)(i) The integer value of \(x\) is 3.

Explanation: Using the given % mass of water, the molar mass ratio gives \(x = 3\) (detailed working shown in the answer image).

(d)(ii) Stereoisomers are compounds with the same structural formula but different spatial arrangements of atoms.

Explanation: They differ in the orientation of groups around a double bond or chiral center.

(d)(iii) The reaction combines two ethene molecules to form but-2-ene, adding atoms without losing any.

Explanation: Addition reactions involve the combination of molecules to form a single product.

(d)(iv) The two stereoisomers of but-2-ene are cis-but-2-ene and trans-but-2-ene.

Stereoisomers of but-2-ene

Explanation: The cis isomer has both methyl groups on the same side, while the trans isomer has them on opposite sides.

Question 2

Topic: 11.1

Chlorine is one of the elements in Group 17 of the Periodic Table.
(a) (i) Describe the colours of the Group 17 elements, chlorine to iodine, at room temperature.
(ii) Describe the relative reactivity of the elements chlorine to iodine as oxidising agents.
(iii) State what is observed when chlorine reacts with hydrogen.
(iv) Explain why the thermal stability of the hydrogen halides decreases down the group.
(b) The halogenoalkane CH₃CH₂Cl forms when chlorine reacts with C₂H₆ via a free-radical substitution mechanism.
(i) Define free radical.

(ii) State the essential condition for chlorine to react with \(C_2H_6\) at room temperature.
(iii) Write two equations to show the propagation steps in this reaction.
(c) CHCl₃ is another halogenoalkane. CHCl₃ forms when propanone reacts with NaClO. NaClO is made from chlorine in a disproportionation reaction.
(i) Identify a reagent and conditions that can be used to convert chlorine to NaClO.
(ii) Define disproportionation.

(iii) Write numbers in the boxes to balance the equation showing the reaction of propanone with NaClO.

Balancing Equation

(iv) Aqueous AgNO₃ dissolved in ethanol reacts with an aqueous solution of CHCl₃. State what is observed in this reaction. Explain your answer.

▶️ Answer/Explanation
Solution

(a)(i) Chlorine is yellow-green, bromine is orange/brown/red, and iodine is silver-grey/black at room temperature.

Explanation: The colour of halogens darkens down the group due to increasing molecular size and electronic transitions.

(a)(ii) The oxidising strength decreases from chlorine to iodine.

Explanation: Chlorine has the highest electron affinity and strongest oxidising ability, which decreases down the group as atomic size increases.

(a)(iii) The green colour of chlorine disappears.

Explanation: Chlorine reacts with hydrogen to form colorless hydrogen chloride (HCl), removing the green colour.

(a)(iv) The H—Hal bond strength decreases down the group.

Explanation: As the halogen size increases, bond length increases, weakening the bond and reducing thermal stability.

(b)(i) A free radical is a species with one or more unpaired electrons.

Explanation: Free radicals are highly reactive due to the presence of unpaired electrons.

(b)(ii) Ultraviolet (UV) light is required.

Explanation: UV light provides the energy to break the Cl—Cl bond, initiating the free-radical reaction.

(b)(iii) Propagation steps:

1. \( \text{Cl}^\cdot + \text{C}_2\text{H}_6 \rightarrow \text{HCl} + \text{C}_2\text{H}_5^\cdot \)

2. \( \text{C}_2\text{H}_5^\cdot + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{Cl}^\cdot \)

Explanation: In propagation, radicals react to form products while regenerating new radicals.

(c)(i) Reagent: Cold dilute NaOH. Conditions: Room temperature.

Explanation: Chlorine disproportionates in cold dilute NaOH to form NaClO (sodium hypochlorite).

(c)(ii) Disproportionation is a reaction where the same element is both oxidised and reduced.

Explanation: In \( \text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O} \), chlorine is both oxidised (+1 in NaClO) and reduced (-1 in NaCl).

(c)(iii) Balanced equation:

Balanced Equation

Explanation: The balanced equation is \( \text{CH}_3\text{COCH}_3 + 3\text{NaClO} \rightarrow \text{CHCl}_3 + \text{CH}_3\text{COONa} + 2\text{NaOH} \).

(c)(iv) No visible reaction (no precipitate forms).

Explanation: CHCl₃ does not contain a replaceable halogen (C—Cl bonds are strong), so no AgCl precipitate forms with AgNO₃.

Question 3

Topic: 10.1

The Group 14 elements show a change from non-metallic to metallic character down the group.

(a) Table 3.1 shows some properties of two Group 14 elements, C and Sn, in their standard states. The table is incomplete.

Table 3.1

(i) Complete Table 3.1.
(ii) Identify the lattice structure shown by graphite.
(iii) Explain why Sn has good electrical conductivity.
(b) Carbon is found in inorganic compounds such as carbonates.
(i) Write an equation for the reaction of magnesium carbonate with dilute HCl(aq).
(ii) Describe the thermal stability of the carbonates down Group 2.
(iii) Ammonium carbonate undergoes an acid–base reaction with NaOH(aq). Explain this statement.

(c) Fig. 3.1 shows a sketch of some of the ionisation energies of silicon, Si.

Ionisation Energies of Silicon

(i) Complete the graph in Fig. 3.1 to show the third to sixth ionisation energies of Si.
(ii) Construct an equation to represent the second ionisation energy of Si.

(d) Fig. 3.2 shows the boiling points of the simplest hydrides of the Group 14 elements, C to Pb.

Boiling Points of Group 14 Hydrides

(i) Explain the trend in the boiling points of the Group 14 hydrides shown in Fig. 3.2.

(ii) Deduce the shape of a molecule of SiH₄.

(e) Silicon readily reacts with elements of high electronegativity.
(i) Write an equation for the formation of SiCl₄ from its constituent elements.
(ii) Describe what is observed when a small sample of SiCl₄ is added to water.
(iii) SiO₂ is a white solid that melts above 1700°C. SiCl₄ is a colourless liquid at room temperature. Explain the difference in the melting points of these two compounds with reference to their structure and bonding.

(f) Tin forms an amphoteric oxide, SnO₂. Suggest the formula of the tin compound that forms when SnO₂ reacts with H₂SO₄ in an acid–base reaction.

▶️ Answer/Explanation
Solution
Answer to Table 3.1

(a)(i) Explanation: Carbon (C) is a non-metal with a giant covalent structure (graphite), while tin (Sn) is a metal with metallic bonding. Graphite conducts electricity due to delocalized electrons, whereas Sn conducts due to free electrons in its metallic lattice.

(a)(ii) Layer/lattice structure.

(a)(iii) Explanation: Tin has good electrical conductivity because it is a metal with delocalized electrons that can move freely and carry charge.

(b)(i) MgCO₃ + 2HCl → MgCl₂ + H₂O + CO₂

(b)(ii) Explanation: Thermal stability of Group 2 carbonates increases down the group due to decreasing polarizing power of the cation, which reduces decomposition into oxide and CO₂.

(b)(iii) Explanation: Ammonium carbonate acts as an acid, donating protons (H⁺) to NaOH, forming water and ammonia (NH₃).

(c)(i) Graph: The ionisation energies increase progressively, with a sharp rise after the 4th ionisation (due to removal of core electrons).

(c)(ii) Si⁺(g) → Si²⁺(g) + e⁻

(d)(i) Explanation: Boiling points increase from CH₄ to PbH₄ due to stronger London dispersion forces as molecular size and electron count increase.

(d)(ii) Tetrahedral.

(e)(i) Si + 2Cl₂ → SiCl₄

(e)(ii) Effervescence/misty fumes (due to HCl formation).

(e)(iii) Explanation: SiO₂ has a giant covalent structure requiring high energy to break bonds, whereas SiCl₄ has weak intermolecular forces, making it a low-melting liquid.

(f) Sn(SO₄)₂

Question 4

Topic: 18.2

Propanone, CH₃COCH₃, is an important organic reagent. Fig. 4.1 shows some reactions of propanone and its derivatives.

(a) Reaction 1 is a nucleophilic addition reaction.
(i) Complete Fig. 4.2 to show the mechanism for the formation of A from propanone. Include charges, dipoles, lone pairs of electrons and curly arrows as appropriate.

(ii) Explain why A does not show optical isomerism.
(b) Suggest the reagents and conditions for reaction 2.

(c) Reaction 3 is a reduction reaction.
(i) Construct an equation to represent reaction 3. Use [H] to represent one atom of hydrogen from the reducing agent.
(ii) Name C.
(d) State what is observed in reaction 4.
(e) Explain why Fehling’s reagent does not react with propanone.

(f) Compounds A, B and C can be distinguished using infrared spectroscopy.

(i) Explain why the absorptions at 2850–2950 cm⁻¹ are not useful to help determine which of the compounds A, B or C produces the infrared spectrum in Fig. 4.3. Use Table 4.1 to answer this question.
(ii) Identify which of compounds A, B or C produces the infrared spectrum in Fig. 4.3. Explain your answer.

▶️ Answer/Explanation
Solution

(a)(i) The mechanism for the nucleophilic addition of HCN to propanone is shown below:

Explanation: The nucleophilic cyanide ion attacks the electrophilic carbonyl carbon, followed by protonation to form the cyanohydrin (A).

(a)(ii) A does not show optical isomerism because it lacks a chiral center.

Explanation: The product (2-hydroxy-2-methylpropanenitrile) has no asymmetric carbon, as all groups attached to the central carbon are different but not arranged in a way that creates non-superimposable mirror images.

(b) Reagents: K₂Cr₂O₇/H₂SO₄ (acidified potassium dichromate). Conditions: Heat under reflux.

Explanation: This oxidizes the secondary alcohol (B) to propanone.

(c)(i) Equation: \(CH_3COCH_3 + 4[H] \rightarrow CH_3CH(OH)CH_3\).

Explanation: Propanone is reduced to propan-2-ol (C) using a reducing agent like NaBH₄ or LiAlH₄.

(c)(ii) C is propan-2-ol.

Explanation: The reduction of propanone yields a secondary alcohol.

(d) A red/orange/yellow precipitate forms.

Explanation: Reaction with 2,4-DNPH (Brady’s reagent) produces a colored hydrazone derivative.

(e) Fehling’s reagent does not oxidize ketones.

Explanation: Fehling’s solution (alkaline Cu²⁺) oxidizes aldehydes but not ketones, as ketones lack an easily oxidizable aldehyde hydrogen.

(f)(i) Absorptions at 2850–2950 cm⁻¹ (C-H stretches) are common to all organic compounds.

Explanation: Since A, B, and C all contain C-H bonds, this region cannot distinguish between them.

(f)(ii) Compound A (2-hydroxy-2-methylpropanenitrile).

Explanation: The sharp peak at 2200–2250 cm⁻¹ (C≡N stretch) is unique to A, as B and C lack nitrile groups.

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