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Question 1

Topic: 9.1

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.

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.

(i) Define relative isotopic mass.
(ii) Use Table 1.2 to calculate the relative atomic mass, Ar, 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.

▶️ Answer/Explanation
Solution

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

Explanation: The shorthand notation \([Ar]\) stands for the noble gas argon’s full configuration, which serves as the core electrons for cobalt.

(a)(ii) The shape of a 1s orbital is a sphere.

Explanation: The 1s orbital is spherically symmetric, representing the region where the electron is most likely to be found.

(a)(iii) Cobalt has 3 unpaired electrons.

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

(b)

Explanation: The table is completed using the given isotopic data, calculating missing values such as nucleon numbers and percentages where necessary.

(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: Calculated as \((191 \times 0.373) + (193 \times 0.627) = 192.22\) (weighted average of isotopic masses).

(d)(i) \(x = 3\).

Explanation: Given 20.5% water by mass, solving \(\frac{18x}{209.5 + 18x} = 0.205\) yields \(x \approx 3\).

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

Explanation: They differ in the 3D orientation of groups around a double bond or ring.

(d)(iii) The reaction combines two ethene molecules (\(C_2H_4\)) to form but-2-ene (\(C_4H_8\)), adding atoms without losing any.

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

(d)(iv)

Explanation: The two stereoisomers of but-2-ene are the cis (same side) and trans (opposite sides) forms.

Question 2

Topic: 11.2

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 C₂H₆ 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 halogens exhibit distinct colors due to their molecular structure and electronic transitions. Chlorine is pale yellow-green, bromine is reddish-brown, and iodine is dark grey or purple-black.

(a)(ii) Oxidising strength decreases from chlorine to iodine.

Explanation: Chlorine is the strongest oxidising agent in Group 17 due to its high electronegativity and small atomic size, while iodine is the weakest as it has lower electronegativity and larger atomic size.

(a)(iii) The green color disappears.

Explanation: Chlorine reacts with hydrogen to form colorless hydrogen chloride (HCl), causing the green color of chlorine to vanish.

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

Explanation: As the halogen atom size increases, the bond with hydrogen becomes weaker, reducing thermal stability.

(b)(i) A species with one or more unpaired electrons.

Explanation: Free radicals are highly reactive intermediates formed during bond homolysis, containing unpaired electrons.

(b)(ii) Ultraviolet light.

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

(b)(iii) \( C_2H_6 + Cl^\bullet \to C_2H_5^\bullet + HCl \)
\( C_2H_5^\bullet + Cl_2 \to C_2H_5Cl + Cl^\bullet \)

Explanation: The propagation steps involve chlorine radicals reacting with ethane to form ethyl radicals, which then react with more chlorine to produce chloroethane.

(c)(i) Cold NaOH(aq).

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

(c)(ii) A reaction where a species is both oxidised and reduced simultaneously.

Explanation: In disproportionation, the same element undergoes both oxidation and reduction in the same reaction.

(c)(iii) \( CH_3COCH_3 + 3NaClO \to 1CHCl_3 + 1CH_3COONa + 2NaOH \)

Explanation: Propanone reacts with sodium hypochlorite in a haloform reaction, producing chloroform and sodium acetate.

(c)(iv) White precipitate forms. CHCl₃ is hydrolysed, releasing chloride ions which react with Ag⁺ to form AgCl.

Explanation: The hydrolysis of CHCl₃ releases Cl⁻ ions, which react with AgNO₃ to form insoluble silver chloride (white precipitate).

Question 3

Topic: 11.1

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.

Fig. 3.1

(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.

Fig. 3.2

(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 \(SiH_4\).

(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: The table is completed by noting that carbon (C) has a giant covalent structure (graphite) and is a non-metal, while tin (Sn) has a metallic structure and is a metal. Graphite conducts electricity due to delocalized electrons, whereas Sn conducts due to its metallic bonding with free electrons.

(a)(ii) Hexagonal layer lattice.

Explanation: Graphite has a hexagonal layered structure where carbon atoms form sheets with weak van der Waals forces between layers.

(a)(iii) Tin has delocalized electrons in its metallic lattice.

Explanation: In metallic bonding, electrons are free to move, allowing Sn to conduct electricity efficiently.

(b)(i) \(MgCO_3 + 2HCl \rightarrow MgCl_2 + CO_2 + H_2O\).

Explanation: Magnesium carbonate reacts with hydrochloric acid to produce magnesium chloride, carbon dioxide, and water.

(b)(ii) Thermal stability increases down Group 2.

Explanation: Larger cations stabilize the carbonate ion more effectively due to lower polarizing power, reducing decomposition tendency.

(b)(iii) Ammonium ion (\(NH_4^+\)) acts as an acid, donating a proton to \(OH^-\).

Explanation: The reaction \(NH_4^+ + OH^- \rightarrow NH_3 + H_2O\) shows the Brønsted–Lowry acid–base behavior.

(c)(i) The graph shows increasing ionisation energies with successive removals of electrons.

Explanation: Higher ionisation energies are required as electrons are removed from increasingly positive ions.

(c)(ii) \(Si^+(g) \rightarrow Si^{2+}(g) + e^-\).

Explanation: The second ionisation energy represents removing an electron from a singly charged silicon ion.

(d)(i) Boiling points increase due to stronger London dispersion forces.

Explanation: Larger molecules (e.g., \(PbH_4\)) have more electrons, leading to stronger intermolecular forces and higher boiling points.

(d)(ii) Tetrahedral.

Explanation: \(SiH_4\) has four bonding pairs around silicon, adopting a tetrahedral shape (109.5° bond angles).

(e)(i) \(Si + 2Cl_2 \rightarrow SiCl_4\).

Explanation: Silicon reacts with chlorine to form silicon tetrachloride.

(e)(ii) Effervescence/steamy fumes.

Explanation: \(SiCl_4\) hydrolyzes in water, producing HCl gas (visible as fumes).

(e)(iii) SiO₂ has a giant covalent structure, while SiCl₄ is molecular.

Explanation: Breaking SiO₂’s strong covalent bonds requires much more energy than overcoming SiCl₄’s weak intermolecular forces.

(f) \(Sn(SO_4)_2\).

Explanation: SnO₂ reacts as a base with \(H_2SO_4\), forming tin(IV) sulfate.

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)

Explanation: The mechanism involves nucleophilic attack by CN⁻ on the carbonyl carbon of propanone, followed by protonation to form the cyanohydrin (A). Curly arrows show electron movement, and dipoles indicate partial charges.

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

Explanation: For optical isomerism, a molecule must have a chiral carbon (four different groups attached). In A, the central carbon is bonded to two identical CH₃ groups, making it achiral.

(b) Reagents: KCN and H₂SO₄ (dilute). Conditions: Room temperature.

Explanation: The reaction involves nucleophilic addition of HCN to propanone, which is catalyzed by dilute acid (H₂SO₄). KCN provides the CN⁻ nucleophile.

(c)(i) Equation: CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃

Explanation: Propanone is reduced to propan-2-ol (C) using a reducing agent like NaBH₄ or LiAlH₄, represented by [H].

(c)(ii) Name: Propan-2-ol

Explanation: The product (C) is a secondary alcohol with the OH group on the second carbon of propane.

(d) Observation: Red/orange/yellow precipitate forms.

Explanation: Reaction 4 involves oxidation of propanone (if it were an aldehyde) with Fehling’s reagent, but since propanone is a ketone, no reaction occurs. However, if it were an aldehyde, a colored precipitate of Cu₂O would form.

(e) Fehling’s reagent does not react with propanone because ketones are not easily oxidized.

Explanation: Fehling’s reagent oxidizes aldehydes but not ketones due to the lack of an easily oxidizable hydrogen on the carbonyl carbon in ketones.

(f)(i) The absorptions at 2850–2950 cm⁻¹ are not useful because all three compounds (A, B, C) contain C—H bonds.

Explanation: This region corresponds to C—H stretching, which is common in all organic compounds and does not help distinguish between A, B, or C.

(f)(ii) Compound: A

Explanation: The absorption at 2200–2250 cm⁻¹ indicates a C≡N group, which is present only in compound A (cyanohydrin).

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