A Level Chemistry 9701 Paper 21, Oct/Nov 2024

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9701/21 · Oct/Nov 2024

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1

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

a(i)

The shorthand electronic configuration of cobalt is [Ar]3d74s2. Identify what is meant by [Ar] by giving its full electronic configuration.

[1]
a(ii)

The lowest-energy electrons in cobalt are in the 1s orbital. Draw the shape of a 1s orbital.

[1]
a(iii)

Deduce the number of unpaired electrons in a cobalt atom.

[1]
b

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

isotopenumber of protonsnumber of neutronstotal number of electron shells
45103Rh58
77191Ir6
77193Ir6

Complete Table 1.1.

[3]
c(i)

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

isotoperelative isotopic massrelative abundance in alloy
45103Rh102.9150.00
77191Ir190.9615.18
77193Ir192.9634.82

Define relative isotopic mass.

[2]
c(ii)

Use Table 1.2 to calculate the relative atomic mass, Ar, of iridium in the alloy. Give your answer to two decimal places. relative atomic mass of iridium = ..............................

[2]
d(i)

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. 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. x = ..............................

[2]
d(ii)

Define stereoisomers.

[1]
d(iii)

Explain how the conversion of ethene to but-2-ene can be described as an addition reaction.

[1]
d(iv)

Draw the two stereoisomers of but-2-ene.

[2]
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.

[1]
a(ii)

Describe the relative reactivity of the elements chlorine to iodine as oxidising agents.

[1]
a(iii)

State what is observed when chlorine reacts with hydrogen.

[1]
a(iv)

Explain why the thermal stability of the hydrogen halides decreases down the group.

[1]
b(i)

The halogenoalkane CH3CH2Cl forms when chlorine reacts with C2H6 via a free-radical substitution mechanism. Define free radical.

[1]
b(ii)

State the essential condition for chlorine to react with C2H6 at room temperature.

[1]
b(iii)

Write two equations to show the propagation steps in this reaction. 1 ........................................................................................................................................ 2 ........................................................................................................................................

[2]
c(i)

CHCl3 is another halogenoalkane. CHCl3 forms when propanone reacts with NaClO. NaClO is made from chlorine in a disproportionation reaction. Identify a reagent and conditions that can be used to convert chlorine to NaClO.

[1]
c(ii)

Define disproportionation.

[1]
c(iii)

Write numbers in the boxes to balance the equation showing the reaction of propanone with NaClO. CH3COCH3+  □  NaClO→  □  CHCl3+  □  CH3COONa+  □  NaOH

[1]
c(iv)

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

[2]
3

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

Figure 1 of 2 for question 3
Figure 2 of 2 for question 3
a(i)

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

C (graphite)Sn
state and appearance in standard stategrey shiny solidsilvery solid
electrical conductivitygood
type of bondingmetallic
type of structuregiant

Complete Table 3.1.

[3]
a(ii)

Identify the lattice structure shown by graphite.

[1]
a(iii)

Explain why Sn has good electrical conductivity.

[1]
b(i)

Carbon is found in inorganic compounds such as carbonates. Write an equation for the reaction of magnesium carbonate with dilute HCl(aq).

[1]
b(ii)

Describe the thermal stability of the carbonates down Group 2.

[1]
b(iii)

Ammonium carbonate undergoes an acid–base reaction with NaOH(aq). Explain this statement.

[2]
c(i)

Fig. 3.1 shows a sketch of some of the ionisation energies of silicon, Si. Complete the graph in Fig. 3.1 to show the third to sixth ionisation energies of Si.

[2]
c(ii)

Construct an equation to represent the second ionisation energy of Si.

[1]
d(i)

Fig. 3.2 shows the boiling points of the simplest hydrides of the Group 14 elements, C to Pb. Explain the trend in the boiling points of the Group 14 hydrides shown in Fig. 3.2.

[2]
d(ii)

Deduce the shape of a molecule of SiH4.

[1]
e(i)

Silicon readily reacts with elements of high electronegativity. Write an equation for the formation of SiCl4 from its constituent elements.

[1]
e(ii)

Describe what is observed when a small sample of SiCl4 is added to water.

[1]
e(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.

[2]
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.

[1]
4

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

Figure 1 of 4 for question 4
Figure 2 of 4 for question 4
Figure 3 of 4 for question 4
Figure 4 of 4 for question 4
a(i)

Reaction 1 is a nucleophilic addition reaction. 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.

[3]
a(ii)

Explain why A does not show optical isomerism.

[1]
b

Suggest the reagents and conditions for reaction 2.

[1]
c(i)

Reaction 3 is a reduction reaction. Construct an equation to represent reaction 3. Use [H] to represent one atom of hydrogen from the reducing agent.

[1]
c(ii)

Name C.

[1]
d

State what is observed in reaction 4.

[1]
e

Explain why Fehling’s reagent does not react with propanone.

[1]
f(i)

Compounds A, B and C can be distinguished using infrared spectroscopy. Fig. 4.3 shows the infrared spectrum of one of the compounds. Table 4.1 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.

[1]
f(ii)

Identify which of compounds A, B or C produces the infrared spectrum in Fig. 4.3. Explain your answer. compound .............................. explanation ........................................................................................................................ ........................................................................................................................................... ...........................................................................................................................................

[1]

Mark scheme

9701/21 · Oct/Nov 2024

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1a(i) · 1 mark

  • 1 mark

    1s2 2s2 2p6 3s2 3p6

1a(ii) · 1 mark

  • 1 mark

    circle / spherical shape drawn

1a(iii) · 1 mark

  • 1 mark

    3 (three)

1b · 3 marks

  • 1 mark

    proton number for Rh = 45 and electron shells = 5

  • 1 mark

    proton number for Ir = 77 (both)

  • 1 mark

    neutrons: 114 and 116

1c(i) · 2 marks

  • 1 mark

    mass of an atom of an isotope

  • 1 mark

    compared to the mass of the unified atomic mass unit (amu ) OR on a scale on which a carbon-12 atom has a mass of exactly 12 units OR divided by 1/12 mass of a carbon-12 atom

1c(ii) · 2 marks

  • 1 mark

    (190.96 × 15.18 + 192.96 × 34.82) / 50.00

  • 1 mark

    = 192.35

1d(i) · 2 marks

  • M1

    79.5 / 209.4 : 20.5 / 18.0 = 0.380 : 1.14 = 1 : 3 OR 18x / (102.9 + 3(35.5) + 18x) = 0.205 x = 3 OR (102.9 + 3(35.5)) / (209.4 + 18x) = 0.795 x = 3

  • M2

    x = 3

1d(ii) · 1 mark

  • 1 mark

    molecules with the same structural formula and same molecular formula with different 3D / different spatial arrangement of atoms / groups

1d(iii) · 1 mark

  • 1 mark

    a single product is made

1d(iv) · 2 marks

  • 1 mark

    structure of trans-isomer

  • 1 mark

    structure of cis-isomer

2a(i) · 1 mark

  • 1 mark

    chlorine = yellow-green, bromine = orange / brown / red, iodine = silver grey / black

2a(ii) · 1 mark

  • 1 mark

    oxidising strength decreases from chlorine to iodine

2a(iii) · 1 mark

  • 1 mark

    green colour disappears

2a(iv) · 1 mark

  • 1 mark

    H—Hal covalent bond strength decreases down the group

2b(i) · 1 mark

  • 1 mark

    a species with one or more unpaired electrons

2b(ii) · 1 mark

  • 1 mark

    ultraviolet

2b(iii) · 2 marks

  • 1 mark

    C2H6 + Cl● → C2H5● + HCl

  • 1 mark

    C2H5● + Cl2 → C2H5Cl + Cl●

2c(i) · 1 mark

  • 1 mark

    cold NaOH(aq)

2c(ii) · 1 mark

  • 1 mark

    a reaction where a species / same species is both oxidised and reduced simultaneously / in the same reaction

2c(iii) · 1 mark

  • 1 mark

    CH3COCH3 + 3NaClO → 1CHCl3 + 1CH3COONa + 2NaOH

2c(iv) · 2 marks

  • M1

    white precipitate/white ppt forms

  • M2

    CHCl3 is hydrolysed / chloride ions are released OR Cl– / chloride ions which are released react with Ag+ OR silver chloride / AgCl is formed

3a(i) · 3 marks

  • 1 mark

    electrical conductivity for C (graphite): good / conductor

  • 1 mark

    type of bonding for C (graphite): covalent

  • 1 mark

    type of structure for Sn: giant

3a(ii) · 1 mark

  • 1 mark

    giant molecular

3a(iii) · 1 mark

  • 1 mark

    its delocalised electrons are free to move

3b(i) · 1 mark

  • 1 mark

    MgCO3 + 2HCl → MgCl2 + CO2 + H2O

3b(ii) · 1 mark

  • 1 mark

    thermal stability increases down the group

3b(iii) · 2 marks

  • M1

    NH4+ / ammonium ion / ammonium cation is a proton / H+ donor

  • M2

    OH– is a proton / H+ acceptor OR the OH– / NaOH is a stronger base than NH3

3c(i) · 2 marks

  • M1

    general increase

  • M2

    increase in IE between IE4 and IE5 is noticeably the largest increase between IE’s shown

3c(ii) · 1 mark

  • 1 mark

    Si+(g) → Si2+(g) + e–

3d(i) · 2 marks

  • M1

    molecules CH4 to PbH4 have greater number of electrons / number of electrons increase

  • M2

    greater / stronger instantaneous dipole–induced forces / London forces / dispersion forces OR more energy required to overcome the instantaneous dipole–induced forces / London forces / dispersion forces

3d(ii) · 1 mark

  • 1 mark

    tetrahedral

3e(i) · 1 mark

  • 1 mark

    Si + 2Cl2 → SiCl4

3e(ii) · 1 mark

  • 1 mark

    effervescence / misty / steamy fumes

3e(iii) · 2 marks

  • M1

    SiO2 is giant covalent AND SiCl4 is simple covalent / simple molecular / molecular

  • M2

    so need less energy to overcome IMF’s in SiCl4, compared to the bonds in SiO2 ora

3f · 1 mark

  • 1 mark

    Sn(SO4)2

4a(i) · 3 marks

  • 1 mark

    curly arrow from lone pair on C of CN– to carbonyl C AND correct dipole on C=O

  • 1 mark

    curly arrow from C=O bond to O AND correct intermediate (the lone pair(s) not needed for this bullet point)

  • 1 mark

    curly arrow from lone pair on —O to (source of) H+

4a(ii) · 1 mark

  • 1 mark

    A’s central carbon atom is not bonded to four different atoms/groups of atoms

4b · 1 mark

  • 1 mark

    H2SO4(aq) or HCl(aq)

4c(i) · 1 mark

  • 1 mark

    CH3COCH3 + 2[H] → CH3CH(OH)CH3

4c(ii) · 1 mark

  • 1 mark

    propan-2-ol

4d · 1 mark

  • 1 mark

    red / orange / yellow ppt

4e · 1 mark

  • 1 mark

    Fehling’s solution cannot oxidise ketones OR ketones are not easily oxidised

4f(i) · 1 mark

  • 1 mark

    All three have a C—H OR CH bond

4f(ii) · 1 mark

  • 1 mark

    compound A AND absorption at 2200–2250 cm–1 indicates C≡N

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