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Structure: Models of Bonding and Structure — Free Chemistry SL Practice Questions

1FoundationMCQProperties of ionic compounds1 markPaper 1~2 min
Both sodium chloride (NaCl) and magnesium oxide (MgO) adopt the same rock-salt crystal structure. Which statement correctly explains why MgO has a significantly higher melting point than NaCl?
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2MasteryMCQProperties of ionic compounds1 markPaper 1~2 min
Four solid compounds are tested for electrical conductivity. Compound X conducts in the solid state only. Compound Y conducts when molten but not in the solid state. Compound Z does not conduct in either the solid or molten state. Compound W conducts in both the solid and molten states. Which row correctly identifies the bonding in each compound?
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3FoundationMCQProperties of ionic compounds1 markPaper 1~2 min
Solid sodium chloride does not conduct electricity, but molten sodium chloride does. Which statement best explains this difference?
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4FoundationMCQProperties of ionic compounds1 markPaper 1~2 min
A white crystalline solid dissolves in water to give a solution that conducts electricity. The solid has a melting point above 800 °C, and the molten liquid conducts electricity. When struck with a hammer, the solid shatters. Which type of bonding is present in this solid?
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5MasteryMCQProperties of ionic compounds1 markPaper 1~2 min
Sodium chloride (NaCl) and magnesium oxide (MgO) both adopt the same rock-salt crystal structure. The lattice enthalpy of MgO is approximately four times greater than that of NaCl. Which statement best explains this difference?
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6FoundationMCQCovalent bonding and electron sharing1 markPaper 1~2 min
In a Cl2\text{Cl}_2 molecule, each chlorine atom forms one single covalent bond. How many electrons are counted in the valence shell of each chlorine atom, and what principle does this illustrate?
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7MasteryMCQCovalent bonding and electron sharing1 markPaper 1~2 min
Phosphorus pentachloride, PCl5\text{PCl}_5, has a Lewis structure in which phosphorus forms five single covalent bonds, giving it 10 electrons in its valence shell. Which statement best explains why this Lewis structure is acceptable?
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8FoundationMCQCovalent bonding and electron sharing1 markPaper 1~2 min
In ethene (C2H4\text{C}_2\text{H}_4), the carbon-carbon double bond consists of one σ\sigma bond and one π\pi bond. Which statement correctly describes the π\pi bond?
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9FoundationMCQCovalent bonding and electron sharing1 markPaper 1~2 min
Which of the following correctly states the number of electrons shared in the bonds of N2\text{N}_2 (triple bond) and O2\text{O}_2 (double bond)?
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10MasteryMCQCovalent bonding and electron sharing1 markPaper 1~2 min
Ammonia (NH3\text{NH}_3) reacts with boron trifluoride (BF3\text{BF}_3) to form the adduct H3NBF3\text{H}_3\text{N} \rightarrow \text{BF}_3. The nitrogen atom donates its lone pair into the empty orbital on boron. Which term correctly describes the N–B bond formed, and what distinguishes it from an ordinary covalent bond?
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11FoundationMCQMetallic bonding and electron sea model1 markPaper 1~2 min
Which statement best describes the electron sea model of metallic bonding?
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12MasteryMCQMetallic bonding and electron sea model1 markPaper 1~2 min
Copper wire carries a large current without melting or breaking. Which statement correctly explains both the electrical conductivity and the ductility of copper in terms of its bonding and structure?
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13FoundationMCQMetallic bonding and electron sea model1 markPaper 1~2 min
In the metallic bonding model, the strength of the bond depends on the charge density of the metal cations and the number of delocalised electrons. Sodium (Group 1) melts at 98 °C, while magnesium (Group 2) melts at 650 °C. Which statement best explains why magnesium has a significantly higher melting point than sodium?
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14FoundationMCQMetallic bonding and electron sea model1 markPaper 1~2 min
According to the electron sea model, which statement best explains why metals are both malleable and ductile?
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15MasteryMCQMetallic bonding and electron sea model1 markPaper 1~2 min
A student tests the electrical conductivity of three unlabelled solid samples: sodium chloride (NaCl), graphite (C), and magnesium (Mg). One sample conducts only when molten, one conducts only when solid, and one conducts in both states. Which row correctly identifies magnesium?
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16FoundationMCQSolids, liquids, and gases: Structural differences1 markPaper 1~2 min
A student compares the three states of matter. Which row correctly describes the typical particle separation and motion in each state?
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17MasteryMCQSolids, liquids, and gases: Structural differences1 markPaper 1~2 min
Three liquid compounds — butane (C4H10\text{C}_4\text{H}_{10}), propanal (C3H6O\text{C}_3\text{H}_6\text{O}), and propan-1-ol (C3H8O\text{C}_3\text{H}_8\text{O}) — are compared at room temperature. Which option correctly ranks these liquids in order of increasing boiling point?
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18MasteryMCQSolids, liquids, and gases: Structural differences1 markPaper 1~2 min
A sealed syringe contains gaseous ammonia, NH3\text{NH}_3, at constant temperature. The plunger is pushed in, halving the volume. Which statement correctly describes the ammonia particles after compression?
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19FoundationMCQSolids, liquids, and gases: Structural differences1 markPaper 1~2 min
Three samples of the same pure molecular substance are at different temperatures: Sample X is a solid, Sample Y is a liquid, and Sample Z is a gas. Which statement correctly describes the relative effect of intermolecular forces on particle arrangement across the three samples?
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20MasteryMCQSolids, liquids, and gases: Structural differences1 markPaper 1~2 min
Which statement best explains the relative melting points of iodine (I2\text{I}_2, 114 °C114\ °\text{C}), sodium chloride (NaCl, 801 °C801\ °\text{C}), and silicon dioxide (SiO2\text{SiO}_2, 1710 °C1710\ °\text{C})?
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21ChallengeSAQ-LProperties of ionic compounds7 marksPaper 2~11 min
Sodium chloride (NaCl) and magnesium oxide (MgO) are both ionic compounds with the same face-centred cubic lattice arrangement. The melting point of NaCl is 801C801\,^\circ\text{C} and that of MgO is 2852C2852\,^\circ\text{C}. The lattice enthalpies are: NaCl =788kJ mol1= -788\,\text{kJ mol}^{-1}; MgO =3795kJ mol1= -3795\,\text{kJ mol}^{-1}. Ionic radii: Na+=102pm\text{Na}^+ = 102\,\text{pm}, Cl=181pm\text{Cl}^- = 181\,\text{pm}, Mg2+=72pm\text{Mg}^{2+} = 72\,\text{pm}, O2=140pm\text{O}^{2-} = 140\,\text{pm}.
(a)
State the charge on each ion in MgO and in NaCl. [1 mark]
(b)
Calculate the product of ionic charges for MgO and for NaCl. [1 mark]
(c)
Using your answer to (b), the ionic radii given, and the concept of electrostatic force, explain why the melting point of MgO is significantly higher than that of NaCl. [3 marks]
(d)
The ratio of the lattice enthalpies (37957884.8)\left(\dfrac{3795}{788} \approx 4.8\right) is greater than the ratio predicted from charge products alone (41=4)\left(\dfrac{4}{1} = 4\right). Deduce one reason, other than ionic charge, why the lattice enthalpy of MgO is more negative than the charge-product ratio predicts, and explain the origin of this additional contribution. [2 marks]
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22MasterySAQ-SLattice structure in ionic solids5 marksPaper 2~8 min
Calcium fluoride (CaF2\text{CaF}_2) is an ionic compound in which Ca2+\text{Ca}^{2+} ions form a face-centred cubic (FCC) lattice and F\text{F}^- ions occupy all tetrahedral holes.
(a)
State the meaning of the term coordination number in an ionic lattice. [1 mark]
(b)
The coordination number of Ca2+\text{Ca}^{2+} in CaF2\text{CaF}_2 is 8. State the coordination number of F\text{F}^- in CaF2\text{CaF}_2[1 mark]
(c)
The distance between the centres of a Ca2+\text{Ca}^{2+} ion and a neighbouring F\text{F}^- ion is 236pm236\,\text{pm}. In the FCC unit cell of CaF2\text{CaF}_2, each tetrahedral hole lies at a body-diagonal distance of 34a\frac{\sqrt{3}}{4}a from the nearest corner, where aa is the edge length, so that dCa–F=34ad_{\text{Ca–F}} = \frac{\sqrt{3}}{4}a. Calculate the volume of the unit cell in pm3\text{pm}^3[3 marks]
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23ChallengeSAQ-LLattice structure in ionic solids7 marksPaper 2~11 min
Sodium chloride (NaCl) and caesium chloride (CsCl) are ionic compounds that adopt different lattice structures. In NaCl, each ion is surrounded by six oppositely charged ions (coordination number 6); in CsCl, each ion is surrounded by eight oppositely charged ions (coordination number 8). Ionic radii: Na+=102pm\text{Na}^+ = 102\,\text{pm}, Cs+=167pm\text{Cs}^+ = 167\,\text{pm}, Cl=181pm\text{Cl}^- = 181\,\text{pm}. Critical radius ratio ranges: coordination number 4 requires 0.225r+r<0.4140.225 \leq \frac{r_+}{r_-} < 0.414; coordination number 6 requires 0.414r+r<0.7320.414 \leq \frac{r_+}{r_-} < 0.732; coordination number 8 requires r+r0.732\frac{r_+}{r_-} \geq 0.732.
(a)
Calculate the radius ratio r+r\dfrac{r_+}{r_-} for NaCl and for CsCl, and state the coordination number predicted for each compound. [2 marks]
(b)
Explain why NaCl adopts a face-centred cubic (FCC) lattice while CsCl adopts a simple cubic lattice, with reference to the relative sizes of the ions and the stability of each arrangement. [3 marks]
(c)
Zinc sulfide (ZnS, zinc blende) has ionic radii Zn2+=88pm\text{Zn}^{2+} = 88\,\text{pm} and S2=184pm\text{S}^{2-} = 184\,\text{pm}, and an experimentally observed coordination number of 4. Evaluate the ability of the ionic radius-ratio model to account for this structure. [2 marks]
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24MasterySAQ-SLattice structure in ionic solids5 marksPaper 2~8 min
Zinc sulfide (ZnS) can crystallize in two different lattice structures: the zinc blende (sphalerite) structure and the wurtzite structure. In both structures, the coordination number of Zn2+\text{Zn}^{2+} is 4.
(a)
State the coordination number of S2\text{S}^{2-} in both structures. [1 mark]
(b)
In the zinc blende structure, the S2\text{S}^{2-} ions form a face-centred cubic (FCC) arrangement and touch along the face diagonal of the unit cell. The ionic radius of S2\text{S}^{2-} is 184pm184\,\text{pm}. (i) Calculate the length of the face diagonal of the unit cell. [1]
(ii) Hence determine the edge length aa of the unit cell. (The face diagonal of a cube is d=a2d = a\sqrt{2}.) [1 mark]
(c)
Explain why ZnS has a lower melting point than MgO, even though both compounds contain ions of the same charge magnitude. [2 marks]
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25MasterySAQ-SLattice structure in ionic solids5 marksPaper 2~8 min

Data

- Ionic radius of Na+\text{Na}^+: 102pm102\,\text{pm} - Ionic radius of Cs+\text{Cs}^+: 167pm167\,\text{pm} - Ionic radius of Cl\text{Cl}^-: 181pm181\,\text{pm} - Radius ratio ranges: 0.4140.4140.7320.732 → coordination number (CN) 6; >0.732> 0.732 → coordination number (CN) 8
Sodium chloride (NaCl) and caesium chloride (CsCl) are both ionic compounds but adopt different crystal lattice structures.
(a)
State the coordination number of Na+\text{Na}^+ in the NaCl lattice. [1 mark]
(b)
Calculate the radius ratio (r+r)\left(\dfrac{r_+}{r_-}\right) for each of NaCl and CsCl. [2 marks]
(c)
Using your answers from (b), explain why NaCl and CsCl adopt different coordination numbers, and deduce one structural difference between the two lattices that results from this difference in coordination number. [2 marks]
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26MasterySAQ-SPolar vs non-polar covalent bonds5 marksPaper 2~8 min
A student investigates the solubility of tetrachloromethane (CCl4\text{CCl}_4) and trichloromethane (CHCl3\text{CHCl}_3) in water. The student predicts that CCl4\text{CCl}_4 will be insoluble in water, while CHCl3\text{CHCl}_3 will show some solubility. Electronegativity values: χ(H)=2.20\chi(\text{H}) = 2.20, χ(C)=2.55\chi(\text{C}) = 2.55, χ(Cl)=3.16\chi(\text{Cl}) = 3.16, χ(O)=3.44\chi(\text{O}) = 3.44
(a)
State why the C-Cl\text{C{-}Cl} bond is polar, and explain why the CCl4\text{CCl}_4 molecule has no net dipole moment. [2 marks]
(b)
Deduce why CHCl3\text{CHCl}_3 has a net dipole moment, whereas CCl4\text{CCl}_4 does not. [2 marks]
(c)
Explain why CHCl3\text{CHCl}_3 shows greater solubility in water than CCl4\text{CCl}_4[1 mark]
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27MasterySAQ-SPolar vs non-polar covalent bonds5 marksPaper 2~8 min
Dinitrogen monoxide (N2O\text{N}_2\text{O}) is a linear molecule with the arrangement N–N–O. The measured bond lengths are: N–N bond =112pm= 112\,\text{pm} and N–O bond =119pm= 119\,\text{pm}. Electronegativity values: N =3.04= 3.04, O =3.44= 3.44. Compare: typical N–N single bond 145pm\approx 145\,\text{pm}, N=N double bond 125pm\approx 125\,\text{pm}, N≡N triple bond 110pm\approx 110\,\text{pm}.
(a)
State the polarity of the N–N bond and the N–O bond in N2O\text{N}_2\text{O}[1 mark]
(b)
Explain whether the N2O\text{N}_2\text{O} molecule as a whole is polar or non-polar, with reference to its molecular geometry and bond dipoles. [2 marks]
(c)
The bond lengths in N2O\text{N}_2\text{O} differ from those of a pure single or triple N–N bond a pure double N–O bond. Deduce what this indicates about the bonding in N2O\text{N}_2\text{O} and explain how this affects the magnitude of the molecular dipole moment. [2 marks]
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28ChallengeSAQ-LPolar vs non-polar covalent bonds7 marksPaper 2~11 min
Silicon tetrafluoride, SiF4\text{SiF}_4, and sulfur tetrafluoride, SF4\text{SF}_4, are both covalent molecules containing four fluorine atoms. Their molecular polarities differ significantly. Electronegativity values (Pauling scale): Si=1.90\text{Si} = 1.90, S=2.58\text{S} = 2.58, F=3.98\text{F} = 3.98 Boiling points: SiF4=86C\text{SiF}_4 = -86\,^\circ\text{C}; SF4=38C\text{SF}_4 = -38\,^\circ\text{C}
(a)
Calculate the electronegativity difference for the Si–F bond and for the S–F bond. [2 marks]
(b)
Explain, using VSEPR theory, why SiF4\text{SiF}_4 is non-polar and SF4\text{SF}_4 is polar. [3 marks]
(c)
Evaluate whether molecular polarity alone is sufficient to account for the difference in boiling points of SiF4\text{SiF}_4 and SF4\text{SF}_4[2 marks]
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29MasterySAQ-SPolar vs non-polar covalent bonds5 marksPaper 2~8 min
Carbon dioxide (CO2\text{CO}_2) and sulfur dioxide (SO2\text{SO}_2) are both oxides of non-metals. A student measures their dipole moments: CO2=0D\text{CO}_2 = 0\,\text{D}; SO2=1.63D\text{SO}_2 = 1.63\,\text{D}. Electronegativity values: C=2.55\text{C} = 2.55, S=2.58\text{S} = 2.58, O=3.44\text{O} = 3.44
(a)
State the polarity of the CO\text{C}{-}\text{O} bond and explain why the overall CO2\text{CO}_2 molecule has a dipole moment of 0D0\,\text{D}[2 marks]
(b)
Using VSEPR theory, deduce the shape of SO2\text{SO}_2 and explain why its dipole moment is non-zero, despite the SO\text{S}{-}\text{O} bonds having a similar electronegativity difference to the CO\text{C}{-}\text{O} bonds. [3 marks]
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30MasterySAQ-SPolar vs non-polar covalent bonds5 marksPaper 2~8 min
A student compares two simple covalent hydrides: methane (CH4\text{CH}_4) and ammonia (NH3\text{NH}_3). Their boiling points are 161.5C-161.5\,^\circ\text{C} and 33.3C-33.3\,^\circ\text{C} respectively. Electronegativity values: C=2.55\text{C} = 2.55, H=2.20\text{H} = 2.20, N=3.04\text{N} = 3.04
(a)
(i) State the electronegativity difference for the C-H\text{C{-}H} bond and for the N-H\text{N{-}H} bond, and hence state the polarity of each bond. [1]
(ii) The bond angle in CH4\text{CH}_4 is 109.5109.5^\circ and the molecule is tetrahedral; NH3\text{NH}_3 is trigonal pyramidal with a bond angle of 107107^\circ. State the overall polarity of each molecule and justify your answer in terms of molecular geometry. [1 mark]
(b)
Explain how the difference in molecular polarity between CH4\text{CH}_4 and NH3\text{NH}_3 leads to different types of intermolecular forces, and account for the large difference in their boiling points. [3 marks]
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31ChallengeSAQ-LMetallic bonding and electron sea model7 marksPaper 2~11 min

Data

- Electrical resistivity of pure aluminium =2.65×108Ωm= 2.65 \times 10^{-8}\,\Omega\,\text{m} at 298K298\,\text{K} - Electrical resistivity of Duralumin =3.45×108Ωm= 3.45 \times 10^{-8}\,\Omega\,\text{m} at 298K298\,\text{K} - Melting point of pure aluminium =933K= 933\,\text{K} - Melting point of Duralumin 775K\approx 775\,\text{K} - Atomic radii: Al=143pm\text{Al} = 143\,\text{pm}; Cu=128pm\text{Cu} = 128\,\text{pm}; Mg=160pm\text{Mg} = 160\,\text{pm}; Mn=127pm\text{Mn} = 127\,\text{pm} *Note: Resistivity is not in the Data Booklet; use the values above as experimental data. Percentage change =neworiginaloriginal×100%= \dfrac{\text{new} - \text{original}}{\text{original}} \times 100\%*
The aerospace alloy Duralumin (approximately 94%94\% aluminium, 4%4\% copper, 1%1\% magnesium, 1%1\% manganese) is stronger than pure aluminium yet maintains good electrical conductivity.
(a)
Calculate the percentage increase in electrical resistivity when aluminium is alloyed to form Duralumin. [1 mark]
(b)
Using the electron sea model and the atomic radii data, explain why Duralumin has a higher electrical resistivity than pure aluminium. [2 marks]
(c)
Using the electron sea model and the atomic radii data, explain why Duralumin has a lower melting point than pure aluminium. [2 marks]
(d)
Evaluate the electron sea model as an explanation for the properties of Duralumin. Identify one specific limitation and suggest how a more sophisticated model would address it. [2 marks]
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32ChallengeSAQ-LMetallic bonding and electron sea model7 marksPaper 2~11 min

Data

- Ionic radius of In3+=80pm\text{In}^{3+} = 80\,\text{pm} - Ionic radius of Sn4+=69pm\text{Sn}^{4+} = 69\,\text{pm} - Electrical conductivity of pure In2O3\text{In}_2\text{O}_3 at 298K=1.0×104Sm1298\,\text{K} = 1.0 \times 10^{-4}\,\text{S\,m}^{-1} - Electrical conductivity of ITO (90:1090:10 ratio) at 298K=1.0×105Sm1298\,\text{K} = 1.0 \times 10^{5}\,\text{S\,m}^{-1} - Melting point of In2O3=2183K\text{In}_2\text{O}_3 = 2183\,\text{K} - ITO does not melt; it decomposes above approximately 1800K1800\,\text{K} 1S=1Ω11\,\text{S} = 1\,\Omega^{-1}. No additional formulae are provided.
Indium tin oxide (ITO) is a ceramic material used in smartphone touchscreens. It is composed of approximately 90%90\% indium(III) oxide (In2O3\text{In}_2\text{O}_3) and 10%10\% tin(IV) oxide (SnO2\text{SnO}_2). When Sn4+\text{Sn}^{4+} ions substitute for In3+\text{In}^{3+} ions in the indium oxide lattice, the material becomes electrically conductive while remaining transparent.
(a)
State the charge imbalance introduced when one Sn4+\text{Sn}^{4+} ion substitutes for one In3+\text{In}^{3+} ion in the In2O3\text{In}_2\text{O}_3 lattice, and state what is released to maintain electrical neutrality. [1 mark]
(b)
Explain how this substitution leads to metallic conductivity in ITO, using the electron sea model. [2 marks]
(c)
Using the ionic radius data, explain why ITO decomposes over a range of temperatures rather than melting at a single sharp temperature as pure In2O3\text{In}_2\text{O}_3 does. [2 marks]
(d)
State one property of ITO that the electron sea model successfully explains and one property that it cannot adequately explain. Justify your answer for the property it cannot explain. [2 marks]
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33ChallengeSAQ-LElectrical conductivity of metals7 marksPaper 2~11 min
The electrical conductivity of metals is explained by the metallic bonding model, in which delocalized electrons move freely through a lattice of positive ions. Data for copper at 298K298\,\text{K}: - Electrical conductivity: σ=5.96×107Sm1\sigma = 5.96 \times 10^{7}\,\text{S\,m}^{-1} - Conductivity decreases by approximately 0.390.39 percent per kelvin increase in temperature - Number density of free electrons: n=8.49×1028m3n = 8.49 \times 10^{28}\,\text{m}^{-3} - Electron charge: e=1.60×1019Ce = 1.60 \times 10^{-19}\,\text{C} - Effective electron mass: me=9.11×1031kgm_e = 9.11 \times 10^{-31}\,\text{kg} σ=ne2τme\sigma = \frac{n e^2 \tau}{m_e}
(a)
Calculate the relaxation time τ\tau for copper at 298K298\,\text{K}[3 marks]
(b)
Explain, using the metallic bonding model, why the electrical conductivity of copper decreases as temperature increases. [2 marks]
(c)
A student claims: "Doubling the number density of free electrons in a metal will always double its electrical conductivity, regardless of temperature." Using the formula σ=ne2τme\sigma = \dfrac{n e^2 \tau}{m_e} and the temperature dependence of conductivity given in the data, evaluate this claim. [2 marks]
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34ChallengeSAQ-LElectrical conductivity of metals7 marksPaper 2~11 min

Data

- Metal A (pure silver): electrical conductivity 6.30×107S m16.30 \times 10^7\,\text{S m}^{-1} at 298K298\,\text{K} - Metal B (silver–copper alloy, 10%10\% copper by mass): electrical conductivity 4.10×107S m14.10 \times 10^7\,\text{S m}^{-1} at 298K298\,\text{K} - Free electrons per atom: approximately 11 for both silver and copper - Atomic radius of silver: 165pm165\,\text{pm}; atomic radius of copper: 128pm128\,\text{pm} - Density of silver: 10.49g cm310.49\,\text{g cm}^{-3}; density of copper: 8.96g cm38.96\,\text{g cm}^{-3} - Molar mass of silver: 107.87g mol1107.87\,\text{g mol}^{-1}; molar mass of copper: 63.55g mol163.55\,\text{g mol}^{-1} - Avogadro's constant: 6.02×1023mol16.02 \times 10^{23}\,\text{mol}^{-1}
The metallic bonding model explains the high electrical conductivity of metals through the presence of a 'sea' of delocalized electrons. However, not all metals conduct electricity equally well.
(a)
Calculate the number of free electrons per unit volume, nn, for pure silver (Metal A). Show your working. [3 marks]
(b)
Explain why the electrical conductivity of Metal B is lower than that of Metal A, using the metallic bonding model. [2 marks]
(c)
A student proposes that the decrease in conductivity from Metal A to Metal B is solely due to a decrease in the number of free electrons per unit volume, nn, because copper atoms are smaller and contribute fewer free electrons per unit volume than silver atoms. Evaluate this proposal. In your answer, estimate whether nn changes significantly between Metal A and Metal B, and identify the dominant factor responsible for the conductivity decrease. [2 marks]
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35MasterySAQ-SMetallic bonding and electron sea model7 marksPaper 2~11 min
A student investigates the electrical conductivity of four metallic elements at 298K298\,\text{K}. Data are collected in the table below. Element — Conductivity — Density / g cm3\text{g cm}^{-3} — Melting point / °C W — moderate — 2.702.70660660 X — good — 1.741.74650650 Y — very good — 8.968.9610851085 Z — poor — 7.147.14420420
(a)
State the origin of mobile electrons in the electron sea model of metallic bonding. [1 mark]
(b)
Element Y is copper (Cu, Group 11) and Element Z is zinc (Zn, Group 12). Using the electron sea model and the data in the table, explain why Element Y has a higher melting point than Element Z. [2 marks]
(c)
Element W is aluminium (Al, Group 13). Using the electron sea model, predict and explain how the electrical conductivity of Element W changes as its temperature increases from 298K298\,\text{K} to 500K500\,\text{K}[2 marks]
(d)
A student claims: "Any solid that contains mobile charge carriers will show increased electrical conductivity at higher temperatures." Evaluate this claim by comparing the behaviour of Element W with that of a semiconductor such as silicon. [2 marks]

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36ChallengeSAQ-LSolids, liquids, and gases: Structural differences10 marksPaper 2~15 min
A student investigates the properties of three different solid substances at room temperature (298K298\,\text{K}) and standard pressure. - Substance X: Diamond — a giant covalent network solid - Substance Y: Solid iodine (I2\text{I}_2) — a simple molecular solid - Substance Z: Sodium chloride (NaCl) — an ionic lattice The student measures the electrical conductivity of each substance in the solid state and again after melting. Results are shown below. Substance — Conductivity (solid) — Conductivity (liquid) Diamond — Does not conduct Iodine — Does not conduct Sodium chloride — Does not conduct — Conducts
(a)
State the type of particle (atoms, ions, or molecules) present in each substance in the liquid state. [3 marks]
(b)
Explain how the bonding and structure of each substance accounts for its electrical conductivity, or lack thereof, in the liquid state. [3 marks]
(c)
Evaluate the following statement, using all three substances as evidence: "The boiling point of a substance is always higher than its melting point because intermolecular forces are stronger in the liquid state than in the solid state." [4 marks]
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37MasterySAQ-SPhysical properties of materials5 marksPaper 2~8 min
A materials scientist investigates two carbon-based materials for lightweight bicycle frames. Material X is a composite of long, aligned carbon fibres embedded in a polymer matrix. The fibres have strong covalent bonding along their length; only weak intermolecular forces act between the fibres and the polymer. Material Y is a single crystal of pure graphite, consisting of layers of carbon atoms arranged in hexagons. Within each layer, carbon atoms are joined by strong covalent bonds; the layers are held together only by weak London dispersion forces. A tensile strength test (pulling until fracture) gives the following results: Material — Tensile strength / MPa — Fracture surface X — 3500 — clean, straight Y — 25 — jagged, uneven The density of Material X is 1.60g cm31.60\,\text{g cm}^{-3} and the density of Material Y is 2.09g cm32.09\,\text{g cm}^{-3}.
(a)
State the type of bonding primarily responsible for the strength within each individual carbon fibre of Material X. [1 mark]
(b)
Explain why Material Y has a much lower tensile strength than Material X. [2 marks]
(c)
The scientist calculates the specific tensile strength of each material (tensile strength divided by density) to compare their suitability for a bicycle frame. Deduce, with a calculation, which material is more suitable for this application. [2 marks]
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38ChallengeSAQ-LPhysical properties of materials7 marksPaper 2~11 min
Shape memory alloys (SMAs) such as Nitinol (NiTi) exhibit a reversible martensitic phase transformation. In the low-temperature martensite phase, the alloy deforms easily by twinning (reorientation of crystal domains). On heating, it transforms to the austenite phase and recovers its original shape. A Nitinol wire of original length 10.0cm10.0\,\text{cm} is compressed in the martensite phase to 9.5cm9.5\,\text{cm}. On heating to the austenite phase, it recovers its original length, exerting a constant stress of 200MPa200\,\text{MPa} throughout recovery.
(a)
Calculate the work done per unit volume by the wire during shape recovery. [2 marks]
(b)
Explain, in terms of atomic arrangement, why the martensite phase is more easily deformed than the austenite phase. [2 marks]
(c)
Evaluate the validity of the constant-stress assumption used in part (a) for predicting the work output of a real SMA actuator, with reference to the stress–strain–temperature hysteresis loop. [3 marks]
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Solutions

39ChallengeSAQ-LPhysical properties of materials10 marksPaper 2~15 min

Data

- Density of UHMWPE: 0.93g cm30.93\,\text{g cm}^{-3} - Density of carbon fibre: 1.80g cm31.80\,\text{g cm}^{-3} - Young's modulus of UHMWPE: 0.8GPa0.8\,\text{GPa} - Young's modulus of carbon fibre: 230GPa230\,\text{GPa} - Young's modulus of cortical bone: 15GPa15\,\text{GPa} - UHMWPE wear rate: 45mg per million cycles45\,\text{mg per million cycles} - CF-UHMWPE wear rate: 12mg per million cycles12\,\text{mg per million cycles} - Carbon fibre volume fraction in composite: 20%20\%
Ultra-high molecular weight polyethylene (UHMWPE) is a polymer used as a bearing surface in artificial hip replacements. A composite material, UHMWPE reinforced with short carbon fibres (CF-UHMWPE), has been developed to improve wear resistance.
(a)
Calculate the density of CF-UHMWPE, assuming no voids. [2 marks]
(b)
The rule of mixtures gives the Young's modulus EE of a fibre-reinforced composite as: Ecomposite=VfEf+VmEmE_{\text{composite}} = V_f E_f + V_m E_m where VfV_f and VmV_m are the volume fractions of fibre and matrix, and EfE_f and EmE_m are their respective moduli. Calculate the Young's modulus of CF-UHMWPE. [2 marks]
(c)
Explain, at the molecular and structural level, why the addition of carbon fibres increases the stiffness of the composite compared to pure UHMWPE. [2 marks]
(d)
Calculate the percentage reduction in wear rate when CF-UHMWPE replaces UHMWPE. [1] (e) Evaluate whether CF-UHMWPE is more suitable than UHMWPE as a hip-replacement bearing surface. In your answer, refer to: - the Young's moduli of the composite, pure polymer, and cortical bone - the wear data from (d) - one structural limitation of short-fibre composites under cyclic loading [3 marks]
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Solutions

40MasterySAQ-SNanotechnology and applications6 marksPaper 2~9 min

Data

In the diameter range 10100nm10\text{–}100\,\text{nm}, the SPR peak wavelength of spherical gold nanoparticles increases by approximately 1.5nm1.5\,\text{nm} for every 1nm1\,\text{nm} increase in diameter.
Gold nanoparticles exhibit surface plasmon resonance (SPR), in which conduction electrons oscillate collectively in response to incident light, causing strong absorption at a characteristic peak wavelength. For spherical gold nanoparticles suspended in water, the SPR peak shifts to longer wavelengths as particle diameter increases. This property is exploited in lateral flow immunoassays used in medical diagnostics.
(a)
Describe the change in colour of a gold nanoparticle suspension as the average particle diameter increases from 10nm10\,\text{nm} to 80nm80\,\text{nm}[2 marks]
(b)
A researcher measures the SPR peak of a 15nm15\,\text{nm} diameter gold nanoparticle sample at 520nm520\,\text{nm}. Calculate the predicted SPR peak wavelength for a sample with an average diameter of 60nm60\,\text{nm}[2 marks]
(c)
The linear relationship given in the data is a simplification. Suggest one reason why the actual SPR peak shift may deviate from this linear model at larger particle diameters, and explain the chemical or physical basis for your suggestion. [2 marks]
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Solutions