Bonding and Structure - Worksheets, Questions and Revision

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A-Level · Chemistry

AC2 Bonding and Structure

AQA 7405 · Calculator allowed · about 160 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question is about ionic bonding and the properties of ionic compounds, using sodium chloride, NaCl, as an example.

(a)Define the term ionic bonding.(1)

(b)Describe the structure formed by sodium chloride.(2)

(c)Explain why sodium chloride has a high melting point.(3)

(d)Explain why solid sodium chloride does not conduct electricity, but molten sodium chloride does.(3)

(e)Ionic radii: Na+ = 102 pm, Cl- = 181 pm, Mg2+ = 72 pm, O2- = 140 pm. Predict, giving a reason, whether magnesium oxide, MgO, will have a higher or lower melting point than sodium chloride.(2)

(Total for Question 1 is 11 marks)

This question is about covalent bonding, including dative covalent (coordinate) bonding.

(a)Define the term covalent bond.(1)

(b)Describe the covalent bonding present in a molecule of methane, CH4.(2)

(c)Define the term dative covalent (coordinate) bond.(1)

(d)Ammonia, NH3, reacts with a hydrogen ion, H+, to form the ammonium ion, NH4+. Explain how this bond forms, identifying which atom donates the electron pair.(3)

(e)State the shape of the ammonium ion, NH4+, and its bond angle.(2)

(Total for Question 2 is 9 marks)

This question is about electronegativity and bond polarity.

(a)Define the term electronegativity.(1)

(b)State and explain the general trend in electronegativity across Period 3, from sodium to chlorine.(3)

(c)Using these electronegativity values, H = 2.1, Cl = 3.0, F = 4.0, I = 2.5, rank the hydrogen halides HF, HCl and HI in order of increasing bond polarity, showing your working.(3)

(d)A student states: "Carbon dioxide, CO2, has polar C=O bonds, so it must be a polar molecule." Evaluate this statement.(3)

(Total for Question 3 is 10 marks)

This question is about predicting the shapes of molecules and ions using electron-pair repulsion theory.

(a)State the electron-pair repulsion theory used to predict the shapes of molecules and ions.(2)

(b)Predict the shape and bond angle of a molecule of boron trifluoride, BF3, explaining your answer.(3)

(c)Predict the shape and bond angle of a molecule of ammonia, NH3, explaining why its bond angle differs from that in methane, CH4 (109.5 degrees).(3)

(d)Predict the shape and bond angle of a molecule of water, H2O, explaining the effect of having two lone pairs.(3)

(Total for Question 4 is 11 marks)

This question is about metallic bonding and the properties of metals and alloys.

(a)Describe the structure and bonding present in a typical metal.(2)

(b)Explain why metals are good conductors of electricity in both the solid and molten states.(3)

(c)Explain why metals are malleable.(2)

(d)Explain, in terms of structure, why alloys are generally harder than the pure metals from which they are made.(3)

(Total for Question 5 is 10 marks)

This question is about intermolecular forces: London (dispersion) forces and permanent dipole-dipole forces.

(a)State what is meant by a London (dispersion) force.(2)

(b)Explain why the boiling points of the noble gases (He, Ne, Ar, Kr, Xe) increase down the group.(3)

(c)Hydrogen chloride, HCl (Mr = 36.5), has a higher boiling point than argon, Ar (Mr = 40), even though Ar has a slightly greater Mr. Explain why.(3)

(d)Given that the boiling point of argon is -186 C and the boiling point of HCl is -85 C, use this data to support your answer to part (c).(2)

(Total for Question 6 is 10 marks)

This question is about hydrogen bonding and its effect on the properties of water.

(a)State the three conditions that must be met for hydrogen bonding to occur between molecules.(3)

(b)Describe how a hydrogen bond forms between two water molecules.(2)

(c)Explain, in terms of hydrogen bonding, why ice is less dense than liquid water.(3)

(d)The density of ice is 0.917 g/cm3 and the density of liquid water is 1.000 g/cm3 at 0 C. Calculate the percentage increase in volume when a fixed mass of water freezes to form ice.(4)

(Total for Question 7 is 12 marks)

This question is about the giant covalent structures of diamond, graphite and graphene, all forms of carbon.

(a)Describe the structure of diamond.(2)

(b)Explain why diamond has a very high melting point.(2)

(c)Describe the structure of graphite and explain why graphite conducts electricity while diamond does not.(4)

(d)Explain why graphite is used as a lubricant.(2)

(e)Graphene is a single, isolated layer of graphite, one atom thick. Suggest one property of graphene that would make it useful in electronics, referring to its structure.(2)

(Total for Question 8 is 12 marks)

A student, Priya, investigates the structure and bonding of four unknown solid substances, P, Q, R and S, by measuring their melting points and testing their electrical conductivity and solubility. This tests key required practical skills: safe use of heating apparatus, and using experimental data to classify substances by structure and bonding.
Table 1: physical properties of four unknown solid substances

Substance	Melting point / C	Conducts as a solid?	Conducts when molten?	Soluble in water?	Soluble in hexane?
P	801	No	Yes	Yes	No
Q	1085 (malleable solid)	Yes	Yes	No	No
R	180 (decomposes above 250 C, releasing a toxic gas)	No	No	No	Yes
S	1610	No	No	No	No

(a)Describe how Priya should use an electrothermal (electric) melting point apparatus to determine the melting point of a solid sample safely and accurately.(3)

(b)Substance R decomposes above 250 C, releasing a toxic gas. State one safety precaution Priya should take when heating substance R.(1)

(c)Using the data in Table 1, identify the type of structure and bonding present in each of substances P, Q, R and S, justifying each conclusion using at least two pieces of data.(8)

(d)Priya's determined melting point for P was 802 C. The accepted literature value is 801 C. Calculate the percentage error in Priya's result.(2)

(e)Explain why it is necessary to test the electrical conductivity of both the solid and the molten states of a substance in order to distinguish an ionic compound from a metal.(3)

(f)Suggest one improvement Priya could make to her method for determining melting point that would increase the reliability of her results.(2)

(g)A second student, Jamal, repeats the conductivity test on substance S and obtains an anomalous result showing conduction. Suggest a possible reason for this anomaly.(2)

(Total for Question 9 is 21 marks)

This question uses ionic radii to compare the strength of ionic bonding in sodium chloride, NaCl, and magnesium oxide, MgO. Ionic radii: Na+ = 102 pm, Cl- = 181 pm, Mg2+ = 72 pm, O2- = 140 pm. The electrostatic force of attraction between two ions is proportional to (Q1 x Q2) / r², where Q1 and Q2 are the sizes of the ionic charges and r is the distance between the ion centres (the sum of their ionic radii).

(a)State two factors that affect the strength of ionic bonding (the magnitude of lattice enthalpy).(2)

(b)Show that the value of (Q1 x Q2) / r for MgO is more than five times greater than the value for NaCl, where r is the sum of the two ionic radii (in pm).(4)

(c)Explain how the calculation in part (b) accounts for the fact that magnesium oxide has a higher melting point than sodium chloride.(2)

(d)Going down Group 2, from MgO to CaO to SrO to BaO, explain how you would expect the lattice enthalpy and melting point of the Group 2 oxide to change, and why.(2)

(Total for Question 10 is 10 marks)

Table 2 shows the boiling points of the hydrides of Group 14 and Group 16.
Table 2: boiling points of Group 14 and Group 16 hydrides

Group 14 hydride	Mr	Boiling point / C
CH4	16	-162
SiH4	32	-112
GeH4	77	-88
SnH4	123	-52

Group 16 hydride	Mr	Boiling point / C
H2O	18	100
H2S	34	-60
H2Se	81	-41
H2Te	130	-2

Explain, using ideas about intermolecular forces, why the boiling points of the Group 14 hydrides increase steadily down the group, but the boiling point of H2O is anomalously high compared with the trend shown by the other Group 16 hydrides.

(Total for Question 11 is 6 marks)

This question is about ionic polarisation and the bonding in aluminium chloride, AlCl3.

(a)Explain, using the idea of ionic polarisation, why aluminium chloride, AlCl3, has significant covalent character rather than being purely ionic.(4)

(b)Predict, with a reason, whether aluminium fluoride, AlF3, or aluminium chloride, AlCl3, has the greater ionic character.(2)

(c)State the type of structure adopted by solid aluminium chloride as a result of its covalent character.(1)

(d)Explain how the dative covalent bonds form within the Al2Cl6 dimer.(2)

(Total for Question 12 is 9 marks)

Boron trifluoride, BF3, reacts with ammonia, NH3, to form the adduct F3B-NH3, in which a new dative covalent bond forms from the nitrogen atom to the boron atom.

(a)Suggest, with reasoning, the shape around the boron atom in the adduct F3B-NH3, given that BF3 itself is trigonal planar.(3)

(b)Explain why BF3 must have an incomplete octet (an empty orbital) on boron in order to accept the dative bond from nitrogen.(2)

(c)In the adduct F3B-NH3, the nitrogen atom's lone pair is now used in the dative bond to boron. Suggest how this affects the ability of F3B-NH3 molecules to hydrogen bond to one another, and hence compare the expected volatility of the adduct with that of a simple molecular compound of similar Mr that has no hydrogen bonding capability.(2)

(d)If ammonia, NH3, were replaced with trimethylamine, N(CH3)3, in this reaction, a dative bond to boron would still form. Suggest why the melting point of the resulting adduct, F3B-N(CH3)3, might differ from that of F3B-NH3, and in which direction you would expect this change.(2)

(Total for Question 13 is 9 marks)

Mark scheme · AC2 Bonding and Structure

Question 1

(a) B1 the electrostatic force of attraction between oppositely charged ions oe
(a) Answer: The electrostatic force of attraction between oppositely charged ions.
(b) B1 a giant ionic lattice (giant structure) oe
(b) B1 a regular, repeating three-dimensional arrangement of alternating Na+ and Cl- ions oe
(b) Answer: A giant ionic lattice: a regular, repeating 3-D arrangement of alternating Na+ and Cl- ions.
(c) M1 NaCl has a giant ionic lattice held together by strong electrostatic forces of attraction between oppositely charged ions oe
(c) A1 these forces act in all directions throughout the lattice (many strong ionic bonds must be broken) oe
(c) A1 a large amount of (thermal) energy is needed to overcome these strong forces, so the melting point is high oe
(c) Answer: NaCl has a giant ionic lattice with strong electrostatic attractions acting throughout the structure; a large amount of energy is needed to overcome these, giving a high melting point.
(d) B1 in the solid, the ions are held in fixed positions in the lattice (only vibrate about a fixed point) and are not free to move oe
(d) B1 with no mobile charge carriers, solid NaCl cannot conduct electricity oe
(d) B1 when molten, the lattice breaks down and the ions become free to move, so they can carry charge and conduct electricity oe
(d) Answer: In the solid, ions are fixed in the lattice and cannot move to carry charge. When molten, ions are free to move, so molten NaCl conducts.
(e) B1 predicts MgO has a higher melting point than NaCl
(e) B1 correct reason: MgO ions have greater charge (2+ and 2-, compared with 1+ and 1- in NaCl) and/or smaller ionic radii (ions closer together), so the electrostatic force of attraction between the ions is stronger oe
(e) Answer: MgO has a higher melting point than NaCl, because its ions have greater charge and smaller radii, giving stronger electrostatic attraction.

Question 2

(a) B1 a shared pair of electrons, attracted to the nuclei of both atoms oe
(a) Answer: A shared pair of electrons, attracted to the nuclei of both bonded atoms.
(b) B1 carbon shares one electron with each of four hydrogen atoms, forming four single covalent bonds oe
(b) B1 each shared (bonding) pair consists of one electron from carbon and one electron from the hydrogen atom, giving carbon a share in 8 electrons (a full outer shell/octet) oe
(b) Answer: Carbon forms four single covalent bonds, each a shared pair (one electron from C, one from H), giving carbon a share of 8 outer electrons.
(c) B1 a shared pair of electrons in which both electrons are provided (donated) by only one of the two bonded atoms oe
(c) Answer: A covalent bond in which both shared electrons come from only one of the two atoms.
(d) B1 the nitrogen atom in NH3 has a lone pair of electrons oe
(d) B1 the H+ ion has an empty (1s) orbital and no electrons of its own to contribute oe
(d) B1 nitrogen donates its lone pair into the empty orbital on H+, forming a new dative covalent bond oe
(d) Answer: The nitrogen lone pair in NH3 is donated into the empty orbital of H+, forming a new N-H dative covalent bond.
(e) B1 tetrahedral
(e) B1 109.5 degrees
(e) Answer: Tetrahedral, with a bond angle of 109.5 degrees.

Question 3

(a) B1 the power (ability) of an atom to attract the bonding (shared) pair of electrons in a covalent bond oe
(a) Answer: The ability of an atom to attract the bonding pair of electrons in a covalent bond.
(b) M1 electronegativity increases across Period 3 from Na to Cl
(b) A1 nuclear charge (number of protons) increases across the period while the number of occupied shells (and shielding by inner shells) stays approximately constant oe
(b) A1 atomic radius decreases, so there is a greater attraction for the bonding pair of electrons oe
(b) Answer: Electronegativity increases from Na to Cl because nuclear charge increases with roughly constant shielding, so atomic radius decreases and the attraction for bonding electrons increases.
(c) M1 calculates the electronegativity difference for each: HF = 4.0-2.1 = 1.9; HCl = 3.0-2.1 = 0.9; HI = 2.5-2.1 = 0.4
(c) A1 correct order of increasing polarity: HI < HCl < HF
(c) A1 correct justification: a greater electronegativity difference means a more polar bond oe
(c) Answer: Order of increasing bond polarity: HI < HCl < HF.
(d) M1 the statement is not fully correct; the shape of the molecule must also be considered oe
(d) M1 CO2 is a linear, symmetrical molecule oe
(d) A1 the two C=O bond dipoles are equal in size and point in exactly opposite directions, so they cancel out, giving no overall (net) molecular dipole: CO2 is non-polar oe
(d) Answer: The statement is incorrect. CO2 has polar bonds, but because the molecule is linear and symmetrical the bond dipoles cancel, so CO2 itself is a non-polar molecule.

Question 4

(a) B1 electron pairs (bonding and lone pairs) around a central atom repel one another and arrange themselves as far apart as possible, to minimise repulsion oe
(a) B1 the order of repulsion strength is: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair oe
(a) Answer: Electron pairs around a central atom repel each other and arrange as far apart as possible; lone pair-lone pair repulsion is strongest, then lone pair-bonding pair, then bonding pair-bonding pair.
(b) B1 boron has three bonding pairs and no lone pairs of electrons around it
(b) B1 shape is trigonal planar
(b) B1 bond angle is 120 degrees, since the three bonding pairs repel equally and spread out as far apart as possible oe
(b) Answer: Trigonal planar, bond angle 120 degrees.
(c) B1 shape is trigonal pyramidal
(c) B1 bond angle is 107 degrees
(c) B1 the lone pair on nitrogen repels the three bonding pairs more strongly than a bonding pair repels another bonding pair, pushing the N-H bonds closer together and reducing the angle from 109.5 to 107 degrees oe
(c) Answer: Trigonal pyramidal, bond angle 107 degrees; the extra repulsion from the lone pair compresses the angle below the tetrahedral 109.5 degrees.
(d) B1 shape is non-linear/bent (V-shaped)
(d) B1 bond angle is 104.5 degrees
(d) B1 two lone pairs on oxygen cause even greater repulsion than one lone pair, pushing the two O-H bonding pairs even closer together, reducing the angle further than in NH3 (107 to 104.5 degrees) oe
(d) Answer: Non-linear (bent), bond angle 104.5 degrees; two lone pairs cause greater repulsion than one, compressing the angle further than in NH3.

Question 5

(a) B1 a giant metallic lattice of positive metal ions arranged in a regular, close-packed structure oe
(a) B1 surrounded by a "sea" of delocalised electrons that are free to move throughout the whole structure oe
(a) Answer: A giant lattice of positive metal ions in a regular arrangement, surrounded by a sea of delocalised electrons.
(b) B1 the delocalised electrons are free to move throughout the metallic structure and can carry charge oe
(b) B1 metallic bonding (and the delocalised electrons) is present in both the solid and molten states, since only the regular arrangement of ions is lost on melting, not the bonding itself oe
(b) B1 so mobile charge carriers (delocalised electrons) are available to conduct electricity in both states oe
(b) Answer: Delocalised electrons are free to move and carry charge in both solid and molten metal, since metallic bonding persists on melting.
(c) B1 the layers of positive metal ions can slide over one another when a force is applied oe
(c) B1 because metallic bonding is non-directional, the delocalised electrons continue to attract and hold the ions together as the layers move, so the metal bends/changes shape without breaking oe
(c) Answer: Layers of metal ions can slide over each other under force, while the non-directional sea of delocalised electrons keeps holding the structure together, so metals bend rather than shatter.
(d) M1 an alloy contains atoms of one or more other elements, which are usually a different size from the atoms of the pure metal oe
(d) M1 this distorts the regular arrangement of layers of ions in the lattice oe
(d) A1 the layers can no longer slide over each other as easily, making the alloy harder/stronger than the pure metal oe
(d) Answer: Alloys contain atoms of different sizes, which distort the regular layered structure and prevent layers sliding easily, making alloys harder than pure metals.

Question 6

(a) B1 a weak intermolecular force caused by temporary (instantaneous) dipoles, arising from the random movement of electrons in an electron cloud oe
(a) B1 the temporary dipole induces a dipole in a neighbouring molecule/atom, resulting in a weak force of attraction between them oe
(a) Answer: A weak attractive force caused by temporary dipoles (from random electron movement) inducing dipoles in neighbouring molecules.
(b) B1 the number of electrons (and atomic size/Mr) increases down the group oe
(b) B1 this increases the size of the electron cloud, so larger instantaneous and induced dipoles can form, increasing the strength of the London forces oe
(b) B1 more energy is needed to overcome these stronger forces, so the boiling point increases down the group oe
(b) Answer: Boiling point increases down the group because more electrons give larger, stronger London forces, requiring more energy to overcome.
(c) B1 HCl is a polar molecule (Cl is more electronegative than H), so it has a permanent dipole; Ar is non-polar oe
(c) B1 HCl molecules experience permanent dipole-dipole forces in addition to London forces, whereas only London forces act between Ar atoms oe
(c) B1 for molecules/atoms of similar Mr, the additional dipole-dipole forces are stronger overall than London forces alone, so more energy is needed to separate HCl molecules, giving a higher boiling point oe
(c) Answer: HCl has permanent dipole-dipole forces (from its polar bond) in addition to London forces, which are stronger overall than the London forces alone acting between non-polar Ar atoms, giving HCl a higher boiling point.
(d) M1 calculates the difference in boiling point: -85 - (-186) = 101 C
(d) A1 concludes that despite Ar having the greater Mr (and so stronger London forces alone), HCl boils 101 C higher, showing that permanent dipole-dipole forces make a significant additional contribution to intermolecular attraction in HCl oe
(d) Answer: HCl boils 101 C higher than Ar, showing the extra dipole-dipole forces in HCl contribute significantly more attraction than London forces alone.

Question 7

(a) B1 a hydrogen atom must be covalently bonded directly to a small, highly electronegative atom: nitrogen, oxygen or fluorine (N, O or F) oe
(a) B1 the electronegative atom (N, O or F) must have at least one lone pair of electrons available oe
(a) B1 the hydrogen bond forms between the delta+ hydrogen on one molecule and a lone pair on the N, O or F atom of a neighbouring molecule oe
(a) Answer: H must be bonded directly to N, O or F; that atom must have a lone pair; the hydrogen bond forms between the delta+ H and a lone pair on N/O/F of a neighbouring molecule.
(b) B1 the delta+ hydrogen atom of one water molecule is attracted to a lone pair on the delta- oxygen atom of a neighbouring water molecule oe
(b) B1 the hydrogen bond acts as a roughly linear, weak attraction (often shown as a dotted/dashed line) between the H and the lone pair, O-H...O oe
(b) Answer: The delta+ H of one water molecule is attracted to a lone pair on the delta- O of a neighbouring molecule, forming a roughly linear O-H...O hydrogen bond.
(c) M1 in ice, hydrogen bonds hold water molecules in a rigid, open lattice, with each molecule hydrogen bonded to four neighbours oe
(c) M1 this rigid, open structure creates more empty space between the molecules than in the liquid oe
(c) A1 in liquid water, hydrogen bonds are constantly breaking and reforming, allowing molecules to pack more closely together, so liquid water is denser than ice oe
(c) Answer: Ice has a rigid, open hydrogen-bonded lattice with more space between molecules than liquid water, where molecules can pack more closely, making ice less dense.
(d) M1 uses volume = mass / density for a fixed mass (e.g. 1 g): volume of water = 1/1.000 = 1.000 cm3
(d) M1 volume of ice = 1/0.917 = 1.091 cm3 (awrt 1.09)
(d) M1 increase in volume = 1.091 - 1.000 = 0.091 cm3
(d) A1 percentage increase = (0.091/1.000) x 100 = 9.1% (accept awrt 9.0-9.1%) cao
(d) Answer: Approximately 9.1% increase in volume (accept 9.0-9.1%).

Question 8

(a) B1 a giant covalent (macromolecular) structure/lattice oe
(a) B1 each carbon atom is covalently bonded to four other carbon atoms in a tetrahedral arrangement, extending throughout the whole structure oe
(a) Answer: A giant covalent lattice in which each carbon atom is bonded to four others tetrahedrally, throughout the whole structure.
(b) B1 diamond has a giant covalent structure containing many strong covalent bonds extending throughout the lattice oe
(b) B1 a very large amount of energy is needed to break these many strong covalent bonds, giving a very high melting point oe
(b) Answer: Diamond's giant covalent lattice has many strong covalent bonds; a very large amount of energy is needed to break them, giving a very high melting point.
(c) B1 graphite consists of layers of carbon atoms, each covalently bonded to only three other carbon atoms, arranged in hexagonal rings oe
(c) B1 there are strong covalent bonds within each layer, but only weak London (van der Waals) forces between layers oe
(c) B1 each carbon atom in graphite has one delocalised electron (not used in bonding), which is free to move within a layer, allowing graphite to conduct electricity oe
(c) B1 in diamond, all four outer electrons of every carbon atom are used in (localised) covalent bonds, so there are no delocalised electrons/free charge carriers, and diamond does not conduct oe
(c) Answer: Graphite has layers of carbon atoms each bonded to 3 others (hexagonal rings), with one delocalised electron per atom free to move within each layer, allowing conduction; diamond has all 4 outer electrons localised in covalent bonds, so it cannot conduct.
(d) B1 the layers in graphite are held together only by weak London (van der Waals) forces oe
(d) B1 these weak forces allow the layers to slide over one another easily (and flake off), reducing friction between surfaces oe
(d) Answer: Weak forces between graphite's layers let them slide over each other easily, reducing friction, so graphite acts as a lubricant.
(e) B1 identifies a valid property, e.g. very high electrical conductivity (or extremely high strength/thinness/flexibility) oe
(e) B1 links the property to its structure/usefulness, e.g. every delocalised electron is at the surface (none are "wasted" between internal layers as in graphite), giving very low resistance/fast electron movement, useful in very thin, flexible electronic components oe
(e) Answer: Graphene has very high electrical conductivity, since all its delocalised electrons are at the exposed single layer, useful for extremely thin, flexible conducting components.

Question 9

(a) B1 grinds the solid to a fine powder and packs a small amount into a (sealed) capillary tube oe
(a) B1 heats the sample slowly and at a controlled/reduced rate (e.g. no more than about 1-2 C per minute) as the temperature approaches the expected melting point, observing the sample closely oe
(a) B1 records the temperature range over which the solid starts to melt and finishes melting (the melting point), and repeats with a fresh sample to check reproducibility oe
(a) Answer: Grind the solid, pack into a sealed capillary tube, heat slowly (especially near the expected melting point) while observing, and record the melting range; repeat for reproducibility.
(b) B1 carry out the heating in a fume cupboard (or use a similarly effective means of avoiding inhalation of the toxic gas) oe
(b) Answer: Heat substance R in a fume cupboard.
(c) B1 P: identifies (giant) ionic lattice
(c) B1 P: justifies using at least two data points, e.g. conducts only when molten (ions become mobile) and is soluble in water/has a high melting point oe
(c) B1 Q: identifies giant metallic lattice
(c) B1 Q: justifies using at least two data points, e.g. conducts as both a solid and a liquid (delocalised electrons present in both states) and is malleable oe
(c) B1 R: identifies simple molecular structure
(c) B1 R: justifies using at least two data points, e.g. low melting point/decomposes rather than surviving to a high temperature (weak intermolecular forces) and does not conduct/soluble in non-polar hexane oe
(c) B1 S: identifies giant covalent (macromolecular) structure
(c) B1 S: justifies using at least two data points, e.g. very high melting point (strong covalent bonds throughout the lattice) and does not conduct/insoluble in both solvents oe
(c) Answer: P: giant ionic. Q: giant metallic. R: simple molecular. S: giant covalent (macromolecular).
(d) M1 percentage error = (|802-801| / 801) x 100
(d) A1 = 0.12% (accept awrt 0.1-0.13%) cao
(d) Answer: 0.12% (accept awrt 0.1-0.13%)
(e) B1 an ionic solid does not conduct electricity because its ions are held in fixed positions in the lattice and are not free to move oe
(e) B1 an ionic compound only conducts when molten, because the lattice breaks down and the ions become free to move and carry charge oe
(e) B1 a metal conducts electricity in both the solid and molten states, because its delocalised electrons are always free to move, regardless of state; testing both states therefore distinguishes the two bonding types oe
(e) Answer: Ionic solids do not conduct (ions fixed) but conduct when molten (ions mobile); metals conduct in both states (delocalised electrons always mobile), so testing both states distinguishes ionic compounds from metals.
(f) B1 identifies a valid improvement, e.g. repeat the determination using at least two further fresh samples and calculate a mean melting point, or reduce the heating rate further as the expected melting point is approached oe
(f) B1 correct reasoning: repeating and averaging reduces the effect of random error/anomalous readings, or a slower heating rate gives a narrower, more precise melting range oe
(f) Answer: Repeat the determination with fresh samples and calculate a mean, to reduce the effect of random error.
(g) B1 identifies a valid reason, e.g. the sample of S was contaminated with a conducting impurity (such as a trace of graphite, a metal, or an ionic salt) oe
(g) B1 or the probes did not make good/consistent electrical contact with the solid surface, giving an unreliable reading oe
(g) Answer: The sample may have been contaminated with a conducting impurity, or the probes made poor contact with the solid surface.

Question 10

(a) B1 the size of the ionic charge (a greater charge gives stronger electrostatic attraction) oe
(a) B1 the ionic radius (a smaller radius means ions are closer together, giving stronger attraction) oe
(a) Answer: Ionic charge and ionic radius.
(b) M1 sum of ionic radii: NaCl, r = 102 + 181 = 283 pm; MgO, r = 72 + 140 = 212 pm
(b) M1 charge product: NaCl, Q1 x Q2 = 1 x 1 = 1; MgO, Q1 x Q2 = 2 x 2 = 4
(b) M1 value for NaCl = 1/283 = 3.53 x 10^-3 (pm^-1); value for MgO = 4/212 = 1.887 x 10^-2 (pm^-1) (allow awrt)
(b) A1 ratio = 1.887x10^-2 / 3.53x10^-3 = 5.3 (accept 5.3-5.4), confirming the value for MgO is more than five times greater than for NaCl, cso
(b) Answer: Ratio approx 5.3: the MgO value is about 5.3 times greater than the NaCl value, confirming the statement.
(c) M1 the larger calculated value for MgO (ft from (b)) shows a stronger electrostatic force of attraction between its ions, meaning MgO has a greater (more exothermic) lattice enthalpy oe
(c) A1 more energy is required to overcome these stronger ionic bonds, so a higher temperature is needed to melt MgO than NaCl oe
(c) Answer: The larger value for MgO shows stronger ionic attraction (greater lattice enthalpy), so more energy (a higher temperature) is needed to melt MgO, giving it a higher melting point than NaCl.
(d) B1 the ionic radius of the Group 2 cation increases down the group, while its charge stays the same (2+) oe
(d) B1 the ions become further apart, so electrostatic attraction weakens: lattice enthalpy becomes less negative (decreases in magnitude) and melting point decreases down the group oe
(d) Answer: Lattice enthalpy (in magnitude) and melting point both decrease going down Group 2, because increasing cation radius (same charge) weakens the electrostatic attraction between ions.

Question 11

Level 3 (5-6): A full and coherent explanation addressing both patterns: correctly identifies that the Group 14 hydrides are non-polar, symmetrical (tetrahedral) molecules held together only by London (dispersion) forces, and correctly explains that these forces strengthen down the group as the number of electrons (and Mr) increases; and correctly identifies that H2O (unlike H2S, H2Se, H2Te) can form hydrogen bonds, because its H atoms are bonded directly to the small, highly electronegative O atom which has lone pairs available, and explains that hydrogen bonds are much stronger than the London and permanent dipole-dipole forces present in the other Group 16 hydrides, requiring significantly more energy to overcome and giving H2O an anomalously high boiling point.
Level 2 (3-4): The explanation addresses most of the required chemistry for both patterns but is incomplete in at least one respect (for example, does not fully explain why H2O can hydrogen bond and H2S cannot, or omits reference to the number of electrons/Mr in the Group 14 trend), or contains minor inaccuracies; both the Group 14 trend and the Group 16 anomaly are attempted.
Level 1 (1-2): A basic or partial explanation with limited use of correct terminology; may address only one of the two patterns (for example, only explains the Group 14 trend, or only states "hydrogen bonding" without further detail linking it to boiling point).
Level 0 (0): No relevant chemistry, or the answer does not relate to the question.
Indicative content:
- The Group 14 hydrides (CH4, SiH4, GeH4, SnH4) are non-polar, symmetrical, tetrahedral molecules with no permanent dipole, so the only intermolecular force present between them is London (dispersion) forces.
- The number of electrons (and Mr) increases down Group 14 from CH4 to SnH4.
- A greater number of electrons gives a larger electron cloud, allowing larger instantaneous and induced dipoles to form, so the strength of the London forces increases down the group.
- Stronger London forces require more energy to overcome, so the boiling point of the Group 14 hydrides increases steadily down the group.
- The Group 16 hydrides H2S, H2Se and H2Te also show an increasing boiling point down the group, for the same reason (increasing number of electrons strengthens London forces); these three are also weakly polar, adding weak permanent dipole-dipole forces.
- H2O is different because its hydrogen atoms are bonded directly to oxygen, which is small and highly electronegative and has two lone pairs, satisfying the conditions needed for hydrogen bonding.
- Hydrogen bonds are much stronger than permanent dipole-dipole forces and London forces.
- Significantly more energy is needed to overcome the additional hydrogen bonding between H2O molecules, giving H2O a much higher boiling point than the trend (based on Mr/London forces alone) would predict.
- H2S, H2Se and H2Te cannot form hydrogen bonds because S, Se and Te are larger and less electronegative than O, so their H-S, H-Se and H-Te bonds are not polar enough to meet the conditions for hydrogen bonding.

Question 12

(a) M1 Al3+ is a small ion with a high (3+) charge, so it has a high charge density (high polarising power) oe
(a) M1 Cl- is a relatively large ion, so its electron cloud is polarisable (easily distorted) oe
(a) M1 the small, highly charged Al3+ ion distorts (polarises) the electron cloud of the Cl- ion, pulling electron density towards itself oe
(a) A1 this distortion means electron density is shared between the ions, giving the Al-Cl bond significant covalent character oe
(a) Answer: Al3+ has high charge density and strongly polarises the large, polarisable Cl- ion, distorting its electron cloud and sharing electron density, giving the bond significant covalent character.
(b) B1 AlF3 has the greater ionic character
(b) B1 correct reason: F- is smaller and less polarisable than Cl- (its electron cloud is held more tightly), so Al3+ causes less distortion, and the bonding remains closer to purely ionic oe
(b) Answer: AlF3 has the greater ionic character, because F- is smaller and less polarisable than Cl-, so it is less distorted by Al3+.
(c) B1 a dimer, Al2Cl6 (two AlCl3 units joined together) oe
(c) Answer: It exists as the dimer Al2Cl6.
(d) M1 a chlorine atom on one AlCl3 unit uses one of its lone pairs of electrons to form a new bond, donated to the aluminium atom of the neighbouring AlCl3 unit, which has an empty orbital (only 6 electrons around Al in AlCl3, an incomplete octet) oe
(d) A1 two such dative bonds form (from two bridging chlorine atoms), joining the two AlCl3 units into the dimer Al2Cl6 oe
(d) Answer: A lone pair from a Cl atom on one AlCl3 unit is donated into the empty orbital on the Al atom of the other unit; two such dative bonds (bridging chlorines) form the Al2Cl6 dimer.