A Level Chemistry 9701 Paper 22, Oct/Nov 2024

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9701/22 · Oct/Nov 2024

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1

Vanadium, niobium and tantalum are metals in the same group of the Periodic Table.

Figure 1 of 1 for question 1
a(i)

The shorthand electronic configuration of vanadium in the ground state is [Ar]3d34s2. State what is meant by the term ground state.

[1]
a(ii)

Show the electronic configuration of vanadium using electrons in boxes notation.

[1]
a(iii)

Deduce the total number of electrons in the p sub-shells of a vanadium atom.

[1]
b(i)

Pelopium was the suggested name for a new element discovered in a mineral. Pelopium was later found to be a mixture of niobium, Nb, and tantalum, Ta. Only one naturally occurring isotope exists for each of Nb and Ta. Complete Table 1.1. Table 1.1

isotoperelative isotopic massnumber of protonsnumber of neutrons
4193Nb92.91
73181Ta180.95
[2]
b(ii)

Define relative isotopic mass.

[2]
b(iii)

A sample of pelopium contains 90.9% by mass 4193Nb and 9.1% by mass 73181Ta. Calculate the theoretical relative atomic mass of pelopium based on these data and Table 1.1. Give your answer to two decimal places. Show your working.

[2]
2

Oxygen is a Group 16 element.

Figure 1 of 1 for question 2
a(i)

Write equations for the following reactions. • sodium and oxygen • sulfur and oxygen

[2]
a(ii)

Draw a dot-and-cross diagram to show the species present in Al2O3. Draw outer electrons only.

[1]
a(iii)

The maximum oxidation state of the Period 3 elements in their oxides varies across the period. State and explain the variation.

[2]
b(i)

H2O reacts with both inorganic and organic compounds. Complete Table 2.1 to give details of the reactions of some Period 3 oxides with H2O. Table 2.1

Period 3 oxideproduct of reaction with H2OpH of solution formed
Mg(OH)2
P4O10
[2]
b(ii)

Write an equation for the reaction of CH3CN with H2O in acidic conditions. CH3CN+……H2O+……H+…………………………………………………………………………………………………

[1]
b(iii)

Draw the structures of the two alcohols formed in the reaction shown in equation 1. equation 1 C3H6(g)+H2O(g)→H3PO4C3H8O(g)

[2]
b(iv)

Explain why alcohols are less acidic than water.

[2]
c(i)

Fig. 2.1 shows the boiling points of H2O and other Group 16 hydrides. Explain the trend in the boiling points of the Group 16 hydrides H2S to H2Te.

[2]
c(ii)

Explain why the boiling point of H2O is much higher than that of H2S.

[1]
3

Nitrogen and phosphorus are elements in Group 15 of the Periodic Table.

Figure 1 of 1 for question 3
a(i)

Nitrogen is found in inorganic compounds such as nitrogen oxides (NOx), nitrates and nitric acid. Identify one natural and one man-made occurrence of nitrogen oxides in the atmosphere. natural ............................................................................................................................... man-made .........................................................................................................................

[2]
a(ii)

Write an equation to describe the role of NO2 in the direct formation of acid rain.

[1]
a(iii)

Peroxyacetyl nitrate, PAN, is a component of photochemical smog. Describe how PAN forms from NO2.

[1]
a(iv)

Nitric acid reacts with basic oxides to form nitrates. Write an equation for the reaction of nitric acid with calcium oxide.

[1]
a(v)

Describe what is seen when solid calcium nitrate is heated strongly.

[1]
b(i)

A common test for nitrates is the reaction with NaOH and Al. Equation 1 shows the reaction. equation 1 3NO3−+8Al+5OH−+18H2O→3NH3+8[Al(OH)4]− Deduce the oxidation state of nitrogen in NO3−.

[1]
b(ii)

Identify the species that is oxidised in equation 1.

[1]
b(iii)

NH3 is a basic gas. Describe how NH3 is able to act as a base.

[1]
b(iv)

Suggest the shape of the [Al(OH)4]− ion.

[1]
c(i)

Fig. 3.1 shows a sketch of some of the ionisation energies of phosphorus, P. Construct an equation to represent the third ionisation energy of P.

[1]
c(ii)

Complete the graph in Fig. 3.1 to show the third to sixth ionisation energies of P.

[2]
d

Complete Table 3.1 to show the properties of nitrogen and phosphorus in their standard states. Table 3.1

nitrogenphosphorus
state and appearance of standard statecolourless gaswhite solid
electrical conductivitypoor
type of bonding
type of structuresimple
[2]
e

A form of solid nitrogen has a lattice structure similar to solid iodine. Identify the type of lattice structure of solid nitrogen.

[1]
f(i)

At very high temperatures, phosphorus can form P2 molecules. P2 contains a triple bond, P≡P. Describe the formation of the P≡P bond in terms of orbital overlap.

[2]
f(ii)

The bond energy of P≡P is 485 kJ mol−1. The bond energy of N≡N is 944 kJ mol−1. Compare the reactivity of P2 and N2. Explain your answer.

[1]
4

Bromoalkanes are used widely in industry, although there is increasing concern about their environmental impact. Fig. 4.1 shows a reaction scheme involving 1,2-dibromoethane.

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

Complete Fig. 4.2 to show the mechanism for the formation of 1,2-dibromoethane in reaction 1. Include charges, dipoles, lone pairs of electrons and curly arrows as appropriate.

[3]
b

The enthalpy change of reaction 1, ΔHr=−90.0 kJ mol−1. The enthalpy change of formation of ethene, ΔHf=+52.2 kJ mol−1. Calculate the enthalpy change of formation of 1,2-dibromoethane. ΔHf of 1,2-dibromoethane=…………………………………… kJ mol−1

[1]
c(i)

Complete Fig. 4.1 to: • draw the structure of compound A • name compound B.

[2]
c(ii)

Draw the structure of one repeat unit of polymer C in the box.

[1]
c(iii)

In reaction 5, compound B reacts with an excess of NaOH dissolved in ethanol. The products are HBr, H2O and an unsaturated hydrocarbon D. Suggest the identity of D.

[1]
d(i)

Compound E is the only isomer of 1,2-dibromoethane. Alkaline hydrolysis of E gives compound F. Identify the type of isomerism shown by E and 1,2-dibromoethane.

[1]
d(ii)

Name the homologous series that F belongs to.

[1]
d(iii)

Complete Table 4.1 to state what is observed when F reacts with the reagents listed. Table 4.1

reagentobservation with F
2,4-dinitrophenylhydrazine (2,4-DNPH reagent)
Tollens' reagent
alkaline I2(aq)
[3]
e(i)

Compound F reacts with reagent G to form compound H. The infrared spectrum of H is shown in Fig. 4.3. Table 4.2 gives characteristic infrared absorption ranges. H also shows a molecular ion peak at m/e=60 in its mass spectrum. Use the information in (e), Fig. 4.3 and Table 4.2 to deduce the structure of H. Explain your answer fully.

[3]
e(ii)

Suggest the role of reagent G.

[1]

Mark scheme

9701/22 · Oct/Nov 2024

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1a(i) · 1 mark

  • 1 mark

    lowest energy state owtte

1a(ii) · 1 mark

  • 1 mark

    [Ar] [↑][↑][↑][ ][ ] [↑↓] The 3 × d electrons may go in any of the d orbitals

1a(iii) · 1 mark

  • 1 mark

    12

1b(i) · 2 marks

  • 1 mark

    protons: 41 AND 73

  • 1 mark

    neutrons: 52 AND 108

1b(ii) · 2 marks

  • M1

    mass of an (atom of an) isotope Alternative route using mass of 1 mol throughout mass of one mol of an (atom of an) isotope

  • M2

    relative/compared to (the mass of) the unified atomic mass unit OR on a scale in which a carbon-12 atom / isotope has a mass of exactly 12 units OR divided by / compared to 112 of the mass of a carbon-12 atom / isotope Alternative route: relative/compared to 112 of the mass of 1 mol of C-12 atom / isotope OR when the mass of one mol C-12 atom / isotope is exactly 12(.000) g

1b(iii) · 2 marks

  • 1 mark

    92.91×0.909+180.95×0.091

  • 1 mark

    = 100.92

2a(i) · 2 marks

  • M1

    4Na+O2→2Na2O

  • M2

    S+O2→SO2

2a(ii) · 1 mark

  • 1 mark

    Al3+ [:O¨:]2−

2a(iii) · 2 marks

  • 1 mark

    it increases

  • 1 mark

    as number of valence electrons (which can be used in bonding / lost / shared / donated) increases

2b(i) · 2 marks

  • 1 mark

    MgO product is Mg(OH)2 AND 8⩽pH⩽12

  • 1 mark

    P4O10 product is H3PO4 AND 1⩽pH⩽4

2b(ii) · 1 mark

  • 1 mark

    CH3CN+2H2O+(1)H+→CH3COOH+NH4+

2b(iii) · 2 marks

  • 1 mark

    skeletal formula of propan-2-ol

  • 1 mark

    skeletal formula of butan-1-ol

2b(iv) · 2 marks

  • 1 mark

    alkyl groups are electron donating / have positive inductive effect.

  • 1 mark

    strengthens the O—H bond / makes H+ less likely to be donated

2c(i) · 2 marks

  • 1 mark

    (molecules) H2S to H2Te / they have greater number of electrons

  • 1 mark

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

2c(ii) · 1 mark

  • 1 mark

    (only) H2O has hydrogen bonding AND hydrogen bonding (much) stronger than the other intermolecular forces OR H2O has hydrogen bonding AND because of higher electronegativity of O compared to S OR stronger van der waals’ forces because (only) H2O has hydrogen bonding OR hydrogen bond AND increased strength of permanent dipoles in H2O outweighs the increase in strength of id-id in the others

3a(i) · 2 marks

  • 1 mark

    (natural = )lightning

  • 1 mark

    (man-made = )internal combustion engines

3a(ii) · 1 mark

  • 1 mark

    2NO2+H2O→HNO2+HNO3 OR 4NO2+O2+2H2O→4HNO3

3a(iii) · 1 mark

  • 1 mark

    It / NO2 reacts with (unburned) hydrocarbons / VOCs

3a(iv) · 1 mark

  • 1 mark

    2HNO3+CaO→Ca(NO3)2+H2O

3a(v) · 1 mark

  • 1 mark

    brown fumes given off

3b(i) · 1 mark

  • 1 mark

    (+)5 / V

3b(ii) · 1 mark

  • 1 mark

    aluminium / Al

3b(iii) · 1 mark

  • 1 mark

    (NH3 is an) H+ acceptor

3b(iv) · 1 mark

  • 1 mark

    tetrahedral

3c(i) · 1 mark

  • 1 mark

    P2+(g)→P3+(g)+e(−)

3c(ii) · 2 marks

  • 1 mark

    general increase

  • 1 mark

    increase in IE between IE5 and IE6 is noticeably the largest increase between IEs shown

3d · 2 marks

  • 1 mark

    nitrogen electrical conductivity poor AND type of bonding covalent

  • 1 mark

    phosphorus type of bonding covalent AND type of structure simple

3e · 1 mark

  • 1 mark

    simple molecular (lattice structure)

3f(i) · 2 marks

  • M1

    one sigma / σ bond AND head-on (overlap of) p / sp (orbitals)

  • M2

    two pi / π bond(s) AND side-on (overlap of / involving) p (orbitals)

3f(ii) · 1 mark

  • 1 mark

    P≡P is much weaker so P2 is more reactive (than N2)

4a · 3 marks

  • 1 mark

    dipole on Br2 and curly arrow from C=C bond to a Brδ+

  • 1 mark

    curly arrow from Br–Br bond to the other Br and correct intermediate

  • 1 mark

    curly arrow from lone pair on Br− to C+

4b · 1 mark

  • 1 mark

    =−37.8 (kJ mol−1)

4c(i) · 2 marks

  • 1 mark

    HO-CH2-CH(OH)-CH3

  • 1 mark

    bromoethene

4c(ii) · 1 mark

  • 1 mark

    correct structure of poly(bromoethene)

4c(iii) · 1 mark

  • 1 mark

    C2H2

4d(i) · 1 mark

  • 1 mark

    structural / positional

4d(ii) · 1 mark

  • 1 mark

    aldehyde

4d(iii) · 3 marks

  • 1 mark

    2,4-DNPH: red / orange / yellow ppt

  • 1 mark

    Tollens’ reagent: silver mirror OR grey ppt OR black ppt

  • 1 mark

    alkaline I2(aq): yellow ppt

4e(i) · 3 marks

  • M1

    identify H = ethanoic acid structure

  • M2

    (broad) absorption within the range 3600–2500 cm–1 so O-H (bond) O-H (bond) is equivalent to OH bond

  • M3

    pt 1 and pt 2 •✓ OR pt 1 and pt 3 •✓ •pt 1 absorption within the range 1670–1750 cm–1 so C=O (bond) •pt 2 absorption within the range 1040–1300 cm–1 so C-O (bond) •pt 3 (M+ at m / e = 60 so it has) molecular mass / Mr = 60

4e(ii) · 1 mark

  • 1 mark

    oxidising agent

Questions

  1. 1
    Question 19 marks
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  2. 2
    Question 215 marks
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  3. 3
    Question 319 marks
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  4. 4
    Question 417 marks
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