A Level Chemistry 9701 Paper 23, Oct/Nov 2023

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Question paper

9701/23 · Oct/Nov 2023

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1

The elements phosphorus, sulfur and chlorine are in Period 3 of the Periodic Table. Table 1.1 shows some properties of the elements P to Cl. The first ionisation energy of S is not shown.

Figure 1 of 1 for question 1
a(i)

Complete Table 1.1 to show the number of electrons in the 3p subshell and the total number of unpaired electrons in an atom of P, S and Cl.

[2]
a(ii)

Construct an equation to represent the first ionisation energy of P.

[1]
a(iii)

Three possible values for the first ionisation energy of S are given. 1000 kJ mol−11160 kJ mol−11320 kJ mol−1 Circle the correct value. Explain your choice by comparing your chosen value to those of P and Cl.

[4]
b(i)

P3−, S2− and Cl− have the same number of electrons. Give the full electronic configuration of P3−.

[1]
b(ii)

State the trend in ionic radius shown by P3−, S2− and Cl−. Explain your answer.

[2]
c

A student does three tests on separate samples of NaCl(aq). Complete Table 1.2 with the observations the student makes in each test.

[3]
d(i)

POCl3 shows similar chemical properties to PCl5. POCl3 has a melting point of 1 ∘C and a boiling point of 106 ∘C. POCl3 reacts vigorously with water, forming misty fumes and an acidic solution. Explain how the information in (d) suggests the structure and bonding of POCl3 is simple covalent.

[2]
d(ii)

Construct an equation for the reaction of POCl3 with water. POCl3+.................................→.....................................................................

[1]
d(iii)

POCl3 contains a double covalent bond between P and O. Complete the dot-and-cross diagram, in Fig. 1.1, to show the bonding in POCl3. Show outer shell electrons only.

[2]
e(i)

POCl3(g) forms when PCl3(g) reacts with O2(g). 2PCl3(g)+O2(g)→2POCl3(g) Table 1.3 gives some relevant data. Define enthalpy change of formation, ΔHf.

[2]
e(ii)

Calculate the bond energy of P=O in POCl3 using the data in Table 1.3. Show your working.

[2]
2

Barium hydroxide, Ba(OH)2, is a strong base used in inorganic and organic reactions. Fig. 2.1 shows a reaction scheme involving Ba(OH)2.

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

State the variation in solubilities of group 2 hydroxides.

[1]
a(ii)

State what is observed in reaction 1.

[1]
a(iii)

Suggest a reactant for reaction 2.

[1]
a(iv)

Identify A.

[1]
a(v)

Ba(OH)2 is made by the reaction of Ba with water. Write an equation for this reaction.

[1]
b(i)

The mineral barytocalcite contains both BaCO3 and CaCO3. Both compounds decompose on heating. State which compound decomposes first when barytocalcite is heated. Explain your answer.

[1]
b(ii)

Construct an equation for the complete thermal decomposition of barytocalcite. The formula of barytocalcite is BaCa(CO3)2. BaCa(CO3)2 ................................................................................................

[1]
c

Ba(OH)2 is used to hydrolyse organic compounds. Fig. 2.2 shows the reaction of B with Ba(OH)2, followed by acidification. Draw the structures of the organic products of the process shown in Fig. 2.2.

[3]
3

Potassium chlorate, KClO3, is widely used as an oxidising agent and to make O2(g).

Figure 1 of 2 for question 3
Figure 2 of 2 for question 3
a

Define oxidising agent.

[1]
b

KClO3(s) decomposes when heated. MnO2(s) catalyses the exothermic decomposition reaction. Complete and label the diagram in Fig. 3.1 to show the effect of MnO2(s) on the decomposition of KClO3(s).

[2]
c

When KClO3 is heated without a catalyst, KClO4 and KCl form. 4KClO3→3KClO4+KCl Explain why this reaction is described as a disproportionation reaction.

[1]
d(i)

Molten KClO3 reacts with glucose, C6H12O6. 4KClO3+C6H12O6→6CO2+6H2O+4KCl KClO3 melts at 630 K. At this temperature, both CO2 and H2O are gases. Use the ideal gas equation to calculate the volume, in m3, of one mole of gas at 630 K and 1.00×105 Pa. Show your working. Give your answer to 3 significant figures.

[1]
d(ii)

5.00 g of C6H12O6 reacts completely with molten KClO3. Use your answer to (d)(i) to calculate the total volume of gas released at 630 K and 1.00×105 Pa in this reaction. (If you were unable to answer (d)(i), use 0.0463 m3 in this question. This is not the correct answer to (d)(i).)

[2]
e(i)

The structure of glucose, C6H12O6, is shown in Fig. 3.2. Complete Table 3.1 to identify the number of primary, secondary and tertiary alcohol groups present in the structure shown in Fig. 3.2.

[1]
e(ii)

Separate samples of aqueous glucose are tested with the reagents shown in Table 3.2. Complete Table 3.2 with the observation for each reaction. Write “no reaction” if applicable.

[3]
e(iii)

There are many structural isomers of C6H12O6. Define structural isomers.

[1]
4

Compounds C and D are alkenes with the same molecular formula, C5H10.

Figure 1 of 6 for question 4
Figure 2 of 6 for question 4
Figure 3 of 6 for question 4
Figure 4 of 6 for question 4
Figure 5 of 6 for question 4
Figure 6 of 6 for question 4
a(i)

Give the systematic name of D.

[1]
a(ii)

Explain why C and D do not show geometrical (cis/trans) isomerism.

[1]
a(iii)

Draw the structure of a molecule that is a positional isomer of C and D.

[1]
a(iv)

Give the structural formula of the compound formed when D reacts with H2(g) in the presence of a Pt catalyst.

[1]
a(v)

C can form an addition polymer. Draw the structure of one repeat unit of this addition polymer.

[1]
b

The mass spectrum of C shows a molecular ion peak at m/e=70. This peak has a relative intensity of 48.7. The relative intensity of the [M+1] peak is 2.7. Show that this information is consistent with the molecular formula of C.

[2]
c(i)

C and D both react with HBr. C reacts with HBr to form E. Complete the diagram in Fig. 4.2 to show the mechanism for this reaction. Draw the structure of the organic intermediate. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.

[3]
c(ii)

D reacts with HBr to produce F, a chiral bromoalkane. Draw the structure of F.

[1]
c(iii)

Explain why the reaction of HBr with C and D produces different major products.

[2]
d

C can be used to form H. One possible synthesis of H is shown in Fig. 4.5. Different portions of C are used in reactions 1 and 3. Some of the products are then combined to produce H. Fig. 4.5 does not show any of the inorganic products of the reactions. Complete Table 4.1 with the reagents and conditions required for each of the reactions shown in Fig. 4.5.

[3]

Mark scheme

9701/23 · Oct/Nov 2023

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1a(i) · 2 marks

  • 1 mark

    total # 3p e– 3 4 5

  • 1 mark

    total # unpaired e– 3 2 1

1a(ii) · 1 mark

  • 1 mark

    P(g) → P+(g) + e(–)

1a(iii) · 4 marks

  • 1 mark

    1000 (kJmol−1)

  • 1 mark

    (S) less / least / lowest nuclear attraction (to outer electrons) (than P OR Cl) OR nuclear attraction for Cl is stronger (compared to S (and or P))

  • 1 mark

    S less nuclear charge than Cl OR Cl has a greater nuclear charge than S

  • 1 mark

    (S has a greater nuclear charge (of outer electrons) than P BUT) S has two electrons in a (3)p orbital AND resulting in spin-pair repulsion

1b(i) · 1 mark

  • 1 mark

    1s2 2s2 2p6 3s2 3p6

1b(ii) · 2 marks

  • 1 mark

    decreases (from P3− to Cl−) and increased attraction (of outer e−) for nucleus owtte

  • 1 mark

    increased nuclear charge and same shielding

1c · 3 marks

  • 1 mark

    (Br2) no visible reaction OR colourless to orange / brown / yellow solution

  • 1 mark

    (conc. H2SO4) no visible reaction OR Solution remains colourless

  • 1 mark

    (dil. AgNO3) white precipitate forms

1d(i) · 2 marks

  • 1 mark

    (relatively low melting and boiling points suggests) weak intermolecular / VdW forces

  • 1 mark

    (vigorous reaction with water suggests) hydrolysis

1d(ii) · 1 mark

  • 1 mark

    POCl3 + 3H2O → H3PO4 + 3HCl

1d(iii) · 2 marks

  • 1 mark

    Correct bonding e−

  • 1 mark

    rest of electrons correct (i.e. to make 32 electrons in total)

1e(i) · 2 marks

  • 1 mark

    enthalpy change when one mole of a compound / substance is formed

  • 1 mark

    from its constituent elements in their standard states

1e(ii) · 2 marks

  • 1 mark

    2(–592) = 2(–289) + (496) – 2(P=O) OR (2) P=O = 2(–289) + (496) + 2(592)

  • 1 mark

    EP= O = ½(1102) = +551 (kJ mol−1)

2a(i) · 1 mark

  • 1 mark

    increases down the group

2a(ii) · 1 mark

  • 1 mark

    white solid disappears

2a(iii) · 1 mark

  • 1 mark

    CO2 / carbon dioxide

2a(iv) · 1 mark

  • 1 mark

    Ba(CH3COO)2 / barium ethanoate

2a(v) · 1 mark

  • 1 mark

    Ba + 2H2O → Ba(OH)2 + H2

2b(i) · 1 mark

  • 1 mark

    CaCO3 AND thermal stability of carbonates increases down the group

2b(ii) · 1 mark

  • 1 mark

    BaCa(CO3)2 → BaO + CaO + 2CO2

2c · 3 marks

  • 1 mark

    CH3OH

  • 1 mark

    one COOH group AND correct carbon framework (5C’s) (each C has the correct valency)

  • 1 mark

    both COOH groups

3a · 1 mark

  • 1 mark

    compound / molecule / substance that oxidises another / is reduced

3b · 2 marks

  • 1 mark

    uncatalysed hump

  • 1 mark

    catalysed hump lower than uncatalysed

3c · 1 mark

  • 1 mark

    Cl / chlorine is both oxidised and reduced

3d(i) · 1 mark

  • 1 mark

    pV = nRT ∴ V = 8.31  630  100 000 = 0.0524 (m3)

3d(ii) · 2 marks

  • 1 mark

    moles of gas = 12  5.00 / 180

  • 1 mark

    volume of gas = M1  (d)(i) = 0.0175 (m3)

3e(i) · 1 mark

  • 1 mark

    1 4 0

3e(ii) · 3 marks

  • 1 mark

    Solution turns from purple to colourless

  • 1 mark

    Orange / red precipitate forms

  • 1 mark

    no reaction

3e(iii) · 1 mark

  • 1 mark

    molecules with the same molecular formula but different structural formulae

4a(i) · 1 mark

  • 1 mark

    3-methylbut-1-ene

4a(ii) · 1 mark

  • 1 mark

    one end / one C of the C=C bond has the same groups attached

4a(iii) · 1 mark

  • 1 mark

    Correct skeletal formula of 3-methylbut-1-ene

4a(iv) · 1 mark

  • 1 mark

    (CH3)2CHCH2CH3

4a(v) · 1 mark

  • 1 mark

    Correct 3D representation

4b · 2 marks

  • 1 mark

    2.7 / 48.7  100 / 1.1 = 5(.04)

  • 1 mark

    Mr = 5  12 + 10  1 = 70

4c(i) · 3 marks

  • 1 mark

    correct dipole on H-Br AND curly arrow from C=C bond to Hδ+ AND curly arrow from H—Br bond to Brδ–

  • 1 mark

    correct (Markovnikoff) intermediate

  • 1 mark

    curly arrow from lone pair on Br− to C+ of intermediate

4c(ii) · 1 mark

  • 1 mark

    Skeletal structure of 2-bromo-3-methylbutane

4c(iii) · 2 marks

  • 1 mark

    C has a tertiary carbocation and a 2ndry carbocation

  • 1 mark

    D has a secondary carbocation and a primary carbocation

  • 0 marks

    alkyl groups have a +ve / positive / +I / inductive effect

  • 0 marks

    more alkyl groups / more inductive effects give a more stable intermediate / carbocation / C+

  • 0 marks

    In C tertiary carbocation is more stable than the secondary carbocation

4d · 3 marks

  • 1 mark

    (reaction 2) NaOH(aq)

  • 1 mark

    (reaction 3) hot acidified concentrated KMnO4

  • 1 mark

    (reaction 4) conc H2SO4 catalyst

Questions

  1. 1
    Question 122 marks
    Mark scheme ›
  2. 2
    Question 210 marks
    Mark scheme ›
  3. 3
    Question 312 marks
    Mark scheme ›
  4. 4
    Question 416 marks
    Mark scheme ›

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