Answers Chapter 1 Exercises 1 (a) CuCO3 → CuO + CO2 (b) 2Mg + O2 → 2MgO (c) H2SO4 + 2NaOH → Na2SO4 + 2H2O (d) N2 + 3H2 → 2NH3 (e) CH4 + 2O2 → CO2 + 2H2O 2 (a) 2K + 2H2O → 2KOH + H2 (b) C2H5OH + 3O2 → 2CO2 + 3H2O (c) Cl2 + 2KI → 2KCl + I2 (d) 4CrO3 → 2Cr2O3 + 3O2 (e) Fe2O3 + 3C → 3CO + 2Fe 3 (a) 2C4H10 + 13O2 → 8CO2 + 10H2O (b) 4NH3 + 5O2 → 4NO + 6H2O (c) 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O (d) 6H2O2 + 2N2H4 → 2N2 + 10H2O + O2 (e) 4C2H7N + 15O2 → 8CO2 + 14H2O + 2N2 4 (a) Sand and water: heterogeneous (b) Smoke: heterogeneous
6 X has diffused more quickly, so it must be a lighter gas. Its particles have greater velocity than the particles of Y at the same temperature. (Note though that they will both have the same value for average kinetic energy.) 7 From the kinetic molecular theory we would expect a solid to be more dense than its liquid, and therefore that ice would sink in water. 8 Bubbles will be present through the volume of the liquid. A brown gas is visible above the brown liquid. As the two states are at the same temperature, the particles have the same average kinetic energy and are moving at the same speed. The inter-particle distances in the gas are significantly larger than those in the liquid. 9 At certain conditions of low temperature and low humidity, snow changes directly to water vapour by sublimation, without going through the liquid phase.
(f) Steel: homogeneous
10 Steam will condense on the skin, releasing energy as it forms liquid at the same temperature (E–D on Figure 1.4). This is additional to the energy released when both the boiling water and the condensed steam cool on the surface of the skin.
5 (a) 2KNO3(s) → 2KNO2(s) + O2(g)
(b) CaCO3(s) + H2SO4(aq) → CaSO4(s) + CO2(g) + H2O(l)
12 AW 01_EX_12
(c) Sugar and water: homogeneous (d) Salt and iron filings: heterogeneous (e) Ethanol and water: homogeneous
(c) 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) (d) Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq) (e) 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(l) In the answers, exercises and text there is inconsistency about whether the water formed in combustion reactions is gas or liquid. At room temperature it will ultimately finish up as a liquid, as shown here.
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13 These calculations have used L = 6.02 × 1023
This is the same as the number of moles of carbon atoms used.
Therefore the number of carbon atoms used = moles of chalk × (6.02 × 1023 mol–1) =
(a) 7.22 × 10 (b) 3.01 × 10 22
(c) 1.20 × 1023 14 0.53 mol H
15 0.250 mol
6.02 × 1023 y 100.09
16 (a) 262.87 g mol−1 (b) 176.14 g mol−1
36 (a) 2.50 mol (b) 5.63 mol
(c) 164.10 g mol−1 (d) 248.22 g mol−1
(c) 665.5 g
17 189.1 g
37 (a) 2C4H10 + 13O2 → 8CO2 + 10H2O
18 1.5 mol
(b) 1.59 g
19 0.0074 mol Cl−
38 4.355 kg
20 1.83 × 1024 C atoms
39 (a) CaCO3 → CaO + CO2
21 171 g (integer value because no calculator) 22 10.0 g H2O 23 2.0 mol N2 > 3.0 mol NH3 > 25.0 mol H2 > 1.0 mol N2H4 24 (a) CH (b) CH2O (c) C12H22O11 (d) C4H9
(b) 92.8% (c) CaCO3 is the only source of CO2; all the CaCO3 undergoes complete decomposition; all CO2 released is captured; heating does not cause any change in the mass of the other minerals present. 40 (a) 85.2 g (b) 1.3 g H2
(e) C4H7 (f) CH2O
41 5.23 g C2H4Cl2
42 253.2 g theoretical CaSO3; 77.9%
43 3.16 g ester
44 107 g of C6H6 needed
45 (a) 2.40 mol (b) 0.0110 mol (c) 44 mol
29 6.94 Li
46 (a) 35.65 dm3 (b) 5.7 dm3
30 CdS 31 empirical formula CH; molecular formula C6H6 32
empirical formula H2PO3; molecular formula H4P2O6
33 C10H16N5P3O13 for both empirical and molecular formulas
35 Let y = mass of chalk in grams. moles of chalk used =
yg y 100.09 g mol–1
48 0.138 mol Br2 and 0.156 mol Cl2, so more molecules of Cl2 49 0.113 dm3 50 0.28 dm3
47 0.652 dm3
mass used × Mr(CaCO3)
51 90 kPa 52 16 °C 53 3.0 dm3 54 2.8 dm3
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Answers 55 M = 133 g mol−1 so gas is Xe
73 52% NH3 by mass; assuming no side reactions occur and gases behave as ideal gases
56 90.4 g mol−1
74 3.20 × 105 kg
75 0.225 mol dm−3 (or round to 0.23 mol dm−3)
58 311 dm3 59 empirical formula and molecular formula = SO3 60 At higher altitude the external pressure is less. As the air in the tyre expands on heating (due to friction with the road surface), the internal pressure increases. 61 (a) Particles are in constant random motion and collide with each other and with the walls of the container in perfectly elastic collisions. The kinetic energy of the particles increases with temperature. There are no inter-particle forces and the volume of the particles is negligible relative to the volume of the gas. (b) At low temperature, the particles have lower kinetic energy, which favours the formation of inter-particle forces and reduces gas pressure.
PV Cl > Cl+
22 Sodium floats on the surface; it melts into a sphere; there is fizzing/effervescence/bubbles; sound is produced; solution gets hot; white smoke is produced.
(b) Structure and bonding
pH of solution
MgO(s) +H2O(l) → Mg(OH)2(aq)
SiO2 neutral – oxide (quartz) is insoluble
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
P4O10(s) + 6H2O(l) → 4H3PO4(aq)
SO2(l) + H2O(l) → H2SO3(aq)
23 D 24 The reactivities of the alkali metals increase but those of the halogens decrease. 25 C
(d) (i) Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l)
(ii) Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4(aq)
33 The oxides of Na and Mg are basic; the oxide of Al is amphoteric; the oxides of Si to Cl are acidic. Ar forms no oxide. Na2O + H2O → 2NaOH
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SO3 + H2O → H2SO4
(c) Pt or Pd
47 (a) Homogeneous catalysts are in the same state of matter as the reactants; heterogeneous catalysts are in a different state from the reactants.
Ti3+ Ni2+ Zn2+ 35 D
46 (a) Ni (b) V2O5
(b) They provide a surface for the reactant molecules to come together with the correct orientation.
38 (a) 1s22s22p63s23p63d104s2 (b) 1s22s22p63s23p63d10 (c) The element does not form ions with partially filled d orbitals. 39 Calcium has one oxidation state: +2 (typical of Group 2). Chromium has common oxidation states of +2, +3 and +6. Calcium(II) and chromium(VI) have noble gas electron configurations, which are typically stable. However, the extremely high charge density of chromium(VI) makes it unstable and other oxidation states are more common. The chromium(II) oxidation state has lost its outer 4s electron and one 3d electron. Chromium(III) forms when the atom loses its 4s electron and two 3d electrons.
(c) They can be easily removed by filtration from the reaction mixture and re-used. 48 D
50 Chromium has the electron configuration [Ar]3d54s1; it has six unpaired electrons, which is the maximum number for the series. Zn has the [Ar]3d10 configuration with no unpaired electrons.
(b) (i) Fe3O4: +2.67
(ii) MnO4−: +7
51 In a complex the d sub-level splits into two energy levels due to the presence of the ligand’s lone pair of electrons. The energy difference between the two sets of d orbitals depends on the oxidation state of the central metal, the number of ligands and the identity of the ligand. Electron transitions between d orbitals result from the absorption of energy from the visible region of the electromagnetic spectrum. The wavelength (colour) of light absorbed depends on the size of the splitting between the two sets of d orbitals.
(iii) CrO42−: +6
(iv) [FeCN6]4−: +2
44 (a) Zn
45 (a) +2 (b) The N atoms adopt a square planar arrangement. (c) The planar structure allows oxygen molecules easy access to the iron ion, which can accept a lone pair of electrons from an oxygen moledule and form a coordinate bond. This bond is not strong, so it the process is easily reversible. This allows the complex to absorb oxygen where oxygen is in high concentrations (i.e. in the lungs) but release oxygen in tissues with low oxygen concentrations.
As the two complexes both contain a cobat ion in the +2 oxidation state the difference in colour is due to the identity of the ligands (H2O vs Cl–1) and the coordination number (6 in [Co(H2O)6]2+ and 4 in [CoCl4]2–), which changes from H2O to Cl–.
52 (a) difference in nuclear charge of metal (ion) (b) difference in oxidation number (c) difference in ligand 53 Fe2+ has configuration [Ar]3d6 and Zn2+ is [Ar]3d10. Colour in transition metal complexes is due to the splitting of the d subshell into two sets of d orbitals with different energy levels;
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Answers the absorption of visible light results in electrons being excited from the lower energy set to the higher energy set and the colour observed is complementary to the colour (wavelength) of light absorbed. Light can only be absorbed if the d orbitals are partially filled and the higher energy set has an empty or partially filled orbital that can accept an electron from the lower energy set. Fe2+ has partially filled d orbitals and so electronic transitions can occur from the lower energy set to the higher energy set with the absorption of visible light and it appears colour in solution. In Zn2+ all of the d orbitals are fully occupied so an electronic transition cannot occur from the lower energy set to the higher energy set so it is unable to absorb visible light and Zn2+ is not coloured in solution. Fe2+ not in its highest oxidation state and so can be oxidized by removal of d electron; Zn2+ in its highest oxidation state and so can’t be oxidized (and so can’t act as reducing agent). 54 λmax= 525 nm. The colour absorbed is green; the colour transmitted is red. 55
(a) [Ar] 3d6
(b) The splitting would be greater for [Fe(CN)6]4−. 56 (a) [Fe(H2O)6]3+ is yellow and [Cr(H2O)6]3+ is green; the colours they show are complementary to the colours of light they absorb; colour is caused by transitions between the two sets of d orbitals in the complex; the different metals in the two complexes cause the d orbitals to split differently as they have different nuclear charges and this results in different wavelengths (colours) of light being absorbed. (b) The oxidation state of the central ion is different in the two complexes and this affects the size of the d orbital splitting due to the different number of electrons present in d orbitals. Fe2+ has the electron configuration [Ar]3d6 and Fe3+ has the electron configuration [Ar]3d5.
Challenge yourself 1 Ytterbium, yttrium, terbium, erbium 2 Two liquids, 11 gases 3 ‘Metalloid’ refers to the properties of certain elements in relation to the periodic table. ‘Semiconductor’ refers to the physical properties of materials (including alloys, compounds). There is a partial overlap between the two sets. 4 1s22s22p63s23p64s23d104p65s24d105p64f76s2 or [Xe]4f76s2 5 The entropy change is positive as there are more particles on the right-hand side. 6 The broad absorption spectrum of the complex ions should be contrasted with the sharp lines of atomic spectra (discussed in Chapter 2). Both phenomena are due to electronic transitions, but the spectrum of a complex ion is affected by the surrounding ligands as the complex ion also has vibrational and rotational energy levels. This allows the complex ion to absorb a wider range of frequencies due to the large number of vibrational and rotational excited states also available. Because the absorption of complex ions is measured in solution, interactions with the solvent further increase the number of energy levels present in the complex ion and the number of associated frequencies it can absorb, resulting in the broad absorption bands observed.
The isolated gaseous ions do not have vibrational or rotational energy levels available to them and will only absorb energy of the exact wavelength required to move an electron from a lower energy to a higher energy atomic orbital, generating discrete line spectra.
Practice questions 1 C
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13 (a) the amount of energy required to remove one (mole of) electron(s) 
from (one mole of) an atom(s) in the gaseous state
(b) greater positive charge on nucleus / greater number of protons / greater core charge 
greater attraction by Mg nucleus for electrons (in the same shell) / smaller atomic radius
14 Na2O(s) + H2O(l) → 2NaOH(aq)
SO3(l) + H2O(l) → H2SO4(aq)
State symbols are not needed.
Na2O is basic and SO3 is acidic
15 (a) solution becomes yellow/orange/brown/ darker
chlorine is more reactive than iodine (and displaces it from solution) / OWTTE
Allow correct equation Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(s) for second mark or stating that iodine/I2 is formed.
Na+ has greater net positive charge/same number of protons pulling smaller number of electrons (b) Si4+: 10 e− in 2 (filled) energy levels / electron arrangement 2.8 / OWTTE P3−: 18 e− in 3 (filled) energy levels / electron arrangement 2.8.8, thus larger / OWTTE OR Si4+ has 2 energy levels whereas P3− has 3 / P3− has one more (filled) energy level Si4+ has 10 e− whereas P3− has 18 e− / Si4+ has fewer electrons / P3+ has more electrons 18 (a) in the solid state ions are in fixed positions / there are no moveable ions / OWTTE
Do not accept answer that refers to atoms or molecules.
(b) 2O2− → O2 + 4e− / O2− → 12O2 + 2e−
(b) no colour change / nothing happens as fluorine is more reactive than chlorine / OWTTE 16 (a) atomic number / Z
17 (a) Na: 11 p, 11/2.8.1 e− and Na+: 11 p, 10/2.8 e− OR Na+ has 2 shells/energy levels, Na has 3 / OWTTE
(c) (i) basic
Accept nuclear charge / number of protons.
(b) Across period 3: increasing number of protons / atomic number / Z / nuclear charge
Accept e instead of e−.
(ii) Na2O + H2O → 2NaOH / Na2O + H2O → 2Na+ + 2OH−
Do not accept
(atomic) radius/size decreases / same shell/ energy level / similar shielding/screening (from inner electrons) 
19 (a) first ionization energy: M(g) → M+(g) + e−/e OR the (minimum) energy (in kJ mol−1) to remove one electron from a gaseous atom OR the energy required to remove one mole of electrons from one mole of gaseous atoms
No mark for shielding/screening or shielding/ screening increases.
periodicity: repeating pattern of (physical and chemical) properties 
Noble gases: do not form bonds (easily) / have a full/stable octet/shell/energy level / cannot attract more electrons 
Do not accept ‘inert’ or ‘unreactive’ without reference to limited ability/inability to form bonds or attract electrons.
Two of: the outer energy level/shell is full; the increased charge on the nucleus; great(est) attraction for electrons 
(c) 17 p in Cl nucleus attract the outer level more than 11 p in Na nucleus / greater nuclear charge attracts outer level more 
13 Z08_CHE_SB_IBDIP_9755_ANS.indd 13
Allow converse for Na. Do not accept ‘has larger nucleus’.
(d) S has one proton less/smaller nuclear charge so outer level held less strongly / OWTTE 2−
Allow converse for chloride. Do not accept ‘has larger nucleus’.
(e) the radii of the metal atoms increase (from Li → Cs) (so the forces of attraction are less between them) / OWTTE
Only penalize once if roman numerals are used or if written as 2+ or 3+.
23 (a) [Fe(CN)6]4− +2
(b) [FeCl4] +3
Award [1 max] if answers given as 2+ and 3+, 2 and 3 or II and III. 24 electron transitions between (split, partially filled) d orbitals
the forces of attraction between halogen molecules are van der Waals forces 
these forces increase with increasing mass/ number of electrons 
water molecules replaced by ammonia molecules
20 (a) complex (ion) / the charge is delocalized over all that is contained in the brackets 
ammonia (ligands) increase the splitting between the d orbitals/larger energy difference
absorption moves to shorter wavelength/higher frequency/towards blue end of spectrum  [max 3]
(b) colour is due to energy being absorbed when electrons are promoted within the split d orbitals OR the colour observed is the complementary colour to the energy absorbed / OWTTE
Accept either answer for the first mark.
changing the ligand / coordination number / geometry 
changes the amount the d orbitals are split/ energy difference between the d orbitals / OWTTE
21 NH3 is a Lewis base and Cu2+ is a Lewis acid 
each NH3/ligand donates an electron pair (to Cu2+)
NH3 replace H2O ligands around Cu2+ ion/ around central ion
forming coordinate (covalent)/dative covalent bond [max 3]
22 (a) +2 (b) +3
absorption depends on energy difference between the split d orbitals
 (b) splitting increases/greater
NH3 has greater electron/charge density / NH3 higher in spectrochemical series / NH3 stronger base 
Allow converse argument for H2O. Do not award second mark for stating that NH3 is a stronger ligand or has a smaller size. If ‘decreases’ is given for the first part of the answer then no mark can be scored in the second part.
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Chapter 4 Exercises δ
1 lead nitrate, Pb(NO3)2
12 (a) H
barium hydroxide, Ba(OH)2
potassium hydrogencarbonate, KHCO3
magnesium carbonate, MgCO3
copper sulfate, CuSO4
calcium phosphate, Ca3(PO4)2
13 (a) C 2.6 H 2.2
ammonium chloride, NH4Cl
(b) Si 1.9 Li 1.0
(c) Na2SO4 (d) CuBr2
3 (a) tin(II) phosphate
(d) barium sulfate (e) mercury sulfide 4 (a) Sn2+ (b) Ti4+ (c) Mn2+ (d) Ba2+ (e) Hg+ 5 A3B2
difference = 0.9
Si 1.9 Cl 3.2 difference = 1.3, more polar
(c) N 3.0 Cl 3.2 difference = 0.2
N 3.0 Mg 1.3 difference = 1.7, more polar
(b) titanium(IV) sulfate
(c) manganese(II) hydrogencarbonate
difference = 0.4
C 2.6 Cl 3.2 difference = 0.6, more polar
2 (a) KBr (b) ZnO (e) Cr2(SO4)3 (f) AlH3
O O F (d)
(c) Cl (e) H
O C O Br (b)
H 14 (a)
F F (b) F C F Cl
Cl H H (c) (d) Cl P Cl H C C H H H H (e) H C
H C H (f) C H
15 (a) 16 (b) 24
6 Mg 12: electron configuration [Ne]3s2
(c) 32 (d) 8
Br 35: electron configuration [Ar]3d104s24p5
(e) 20 (f) 26
The magnesium atom loses its two electrons from the 3s orbital to form Mg2+. Two bromine atoms each gain one electron into their 4p subshell to form Br −. The ions attract each other by electrostatic forces and form a lattice with the empirical formula MgBr2.
(c) O N
9 Test the melting point: ionic solids have high melting points.
Test the solubility: ionic compounds usually dissolve in water but not in hexane.
Test the conductivity: ionic compounds in aqueous solution are good conductors.
H H O H H O H 16 H+ O 17 (a) O N O (b) N
O (e) O S O
(d) O O O
H H (f) N N H H
18 (a) 105° bond angle, shape is bent (b) 109.5° bond angle, shape is tetrahedral (c) 180° bond angle, shape is linear
15 05/09/2014 10:01
Answers (d) 107° bond angle, shape is trigonal pyramidal
(c) London (dispersion) forces
(e) 120° bond angle, triangular planar
(d) dipole–dipole, London (dispersion) forces
(f) 107° bond angle, trigonal pyramidal
30 (a) C2H6 (b) H2S
(g) 105° bond angle, shape is bent 19 (a) 120° bond angle, shape is trigonal planar (b) 120° bond angle, shape is trigonal planar (c) 180° bond angle, shape is linear (d) 120° bond angle, shape is bent (e) 105° bond angle, shape is bent
(c) Cl2 (d) HCl 31 B 32 (a) malleability, thermal conductivity, thermal stability (b) light, strong, forms alloys
(f) 107° bond angle, shape is trigonal pyramidal
(c) thermal conductivity, thermal stability, noncorrosive
20 (a) 4 (b) 3 or 4
(d) light, strong, non-corrosive
33 (i) anodizing: increasing the thickness of the surface oxide layer helps resist corrosion
(e) 3 21 (a) polar (b) non-polar (c) polar (d) non-polar
(ii) alloying: mixing Al with other metals such as Mg and Cu increases hardness and strength
(e) non-polar (f) polar
34 (a) linear, 180°
(g) non-polar (h) non-polar
(b) triangular pyramidal, 107°
(c) bent, 105°
22 cis isomer has a net dipole moment
(d) tetrahedral, 109.5° (e) octahedral, 90°
23 CH3OH < CO32− < CO2 < CO
(f) seesaw, 117°
24 NO3 has three resonance structures, HNO3 has two resonance structures; N–O bonds shorter in HNO3
35 (a) 6 (b) 6
25 Similarities: strong, high melting points, insoluble in water, non-conductors of electricity, good thermal conductors.
Differences: diamond is stronger and more lustrous; silicon can be doped to be an electrical conductor.
26 Graphite and graphene have delocalized electrons that are mobile and so conduct electrical charge. In diamond all electrons are held in covalent bonds and are not mobile. 27 A metal
B giant molecular
C polar molecular
D non-polar molecular
E ionic compound
28 A 29 (a) London (dispersion) forces (b) H bonds, dipole–dipole, London (dispersion) forces
(c) 6 (d) 5 (e) 2 or 5 36 (a) 90° (b) 107° (c) 90° 37 (a) polar (b) polar (c) non-polar (d) polar (e) non-polar (f) non-polar 38 (a) polar (b) non-polar (c) non-polar (d) non-polar (e) non-polar (f) polar 39 In BF3 all the atoms have formal charge of 0.
B: FC = 3 – 3 = 0
each F: FC = 7 – (1 + 6) = 0
Zero formal charge represents the most stable, preferred structure, so this is favoured despite violating the octet rule by having fewer than 8 electrons around Be.
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40 (i) S: FC = 6 – 4 = +2 each O: FC = 6 – (1 + 6) = –1 2 O O S O
angles of 109.5° give the puckered shape. In C6H6 (benzene) the carbon atoms are all sp2 hybridized, forming a planar triangular arrangement with bond angles of 120°.
O (ii) S: FC = 6 – 6 = 0 each O: FC = 6 – (1 + 6) = –1 each O: FC = 6 – (2 + 4) = 0 2 O O S O O
The structure with 12 electrons round the S atom has FC = 0 and so is the preferred structure.
41 O–O bonds in O3 are weaker than in O2, due to lower bond order, therefore dissociation occurs with lower energy light (longer wavelength). 42 O3 breakdown is catalyzed by NOx and CFCs in the atmosphere, e.g. CCl2F2(g) → CClF2•(g) + Cl•(g)
CFCs break down in upper atmosphere.
Cl•(g) + O3(g) → O2(g) + ClO•(g)
Chlorine radical reacts with ozone and another radical is produced.
ClO•(g) + O•(g) → O2(g) + Cl•(g)
Chlorine radical is regenerated and so acts as a catalyst for ozone destruction.
43 Electrons in a sigma bond are most concentrated in the bond axis, the region between the nuclei. Electrons in a pi bond are concentrated in two regions, above and below the plane of the bond axis. Cl–Cl in Cl2 44 (b) H–F in HF (c) (d) C–H in CH4 (e) C–H in C2H4 (f) C–H in C2H2 (g) C–Cl in C2H3Cl 45 (a) sp2 (b) sp3 (c) sp2 (d) sp (e) sp2 46 In C6H12 (cyclohexane) the carbon atoms are sp3 hybridized, each forming a tetrahedral arrangement with two neighbouring carbon atoms and two hydrogen atoms. The bond
Challenge yourself 1 Aluminium oxide is less ionic than MgO due to a smaller difference in electronegativity. It has some partially covalent character, which means the comparison with more ionic oxides is not fully valid. 2 F2 has lower bond enthalpy than expected from its atomic radius due to repulsion. The bond length is so short that the lone pairs in the two atoms repel each other, weakening the bond. 3 when bonded to F, e.g. in OF2 4 Run each solution out from separate burettes, and see whether the stream of liquid is deflected in the presence of a charged rod. Only the polar solution will show deflection.
Test solubility with ionic and covalent solutes. The polar solution will be a better solvent for polar/ionic solutes; the non-polar solution for covalent/non-polar solutes.
5 The high thermal conductivity of diamond is because of its strong covalent bonds. When heated the bonds becoming vibrationally excited, and as they are all connected heat energy could be readily transferred through the network from one bond to the next. Silicon is similarly a good thermal conductor – which is why computer chips need to be cooled. 6 Diamonds are kinetically stable with respect to graphite, as the conversion has a very high activation energy (see Chapter 6). So the reaction generally occurs too slowly to be observed. 7 It is difficult to know the number of valence electrons a transition metal has. Treating bonds from ligands as pure covalent molecules results in transition metals in complex ions with large negative formal charges. The formal charge model may not be useful for complex ions, as the values obtained do not make much sense.
17 Z08_CHE_SB_IBDIP_9755_ANS.indd 17
Answers 8 In the polar winter small amounts of water vapour freeze into ice crystals in the atmosphere, which provide a reactive surface. Reactions on the crystals’ surfaces produce species such as Cl2• which later dissociate into Cl•, which breaks down ozone.
Practice questions 1 C
16 Hydrogen bonding in butan-1-ol; stronger than  dipole–dipole attractions in butanal.
Accept converse argument. Do not penalize ‘dipole–dipole bonding’ instead of ‘dipole– dipole attractions’.
Lewis structure: Br
Allow x’s, dots or lines to represent electrons.
Shape: trigonal/triangular pyramidal
F Allow x’s, dots or lines to represent electrons.
Penalize missing lone pairs on terminal atoms once only for the two Lewis structures.
Penalize missing lone pairs on terminal atoms once only for the two Lewis structures. (ii)
Bond angle: less than 109.5°
Bond angle: 90°
Allow any angle less than 109.5° but greater than or equal to 100° (experimental value is 101°).
Ignore extra correct bond angles (e.g. 90° and 180° scores but not 90° and 120°).
(iii) Polarity: polar
(iii) Polarity: non-polar
Explanation: net dipole (moment) / polar PBr bonds and molecule not symmetrical/ bond dipoles do not cancel / asymmetric distribution of electron cloud /
Explanation: no net dipole (moment) / polar SF bonds but molecule symmetrical/ bond dipoles cancel / symmetric distribution of electron cloud / OWTTE
/ OWTTE 
18 Z08_CHE_SB_IBDIP_9755_ANS.indd 18
Do not allow ECF in this question from incorrect Lewis structure. Allow  max for stating that PBr is polar and SF is non-polar without giving a reason or if explanations are incorrect.
(iv) in cis two carboxylic acid groups close together so on heating cyclic anhydride forms (with elimination of water) / OWTTE Allow converse argument for trans.
Allow polar bonds do not cancel for PBr3 and polar bonds cancel for SF6.
(c) O of OH sp3 and O of C=O sp2
Do not allow asymmetric molecule as reason for PBr3 or symmetric molecule for SF6 as reason alone.
18 (a) Award [2 max] for three of the following features:
(b) (i) σ bond: end-on/axial overlap with electron density between the two carbon atoms/nuclei / end-on/axial overlap of orbitals so shared electrons are between atoms / OWTTE
π bond: sideways/parallel overlap of p orbitals with electron density above and below internuclear axis/σ bond / sideways/ parallel overlap of p orbitals so shared electrons are above and below internuclear axis/σ bond / OWTTE Marks can be scored from a suitable diagram. Award [1 max] for stating end-on/axial overlap for σ and sideways/parallel overlap for π only i.e. without mentioning electron density OR stating electron density between the two atoms/nuclei for σ and above and below internuclear axis for π. p p
(ii) 11 σ and 3 π (iii) (strong) intermolecular hydrogen bonding in trans but (strong) intramolecular hydrogen bonding in cis so attraction between different molecules is less (hence lower melting point)  Allow between molecules for intermolecular and within molecules for intramolecular.
Oxygens must be identified.
Bonding Graphite and C60 fullerene: covalent bonds and van der Waals’/London/dispersion forces Diamond: covalent bonds (and van der Waals’/London/dispersion forces) Delocalized electrons Graphite and C60 fullerene: delocalized electrons Diamond: no delocalized electrons Structure Diamond: network/giant structure / macromolecular / three-dimensional structure and Graphite: layered structure / two-dimensional structure / planar C60 fullerene: consists of molecules / spheres made of atoms arranged in hexagons/ pentagons Bond angles Graphite: 120° and Diamond: 109° C60 fullerene: bond angles between 109–120°
Allow Graphite: sp2 and Diamond: sp3. Allow C60 fullerene: sp2 and sp3.
Number of atoms each carbon is bonded to Graphite and C60 fullerene: each C atom attached to 3 others Diamond: each C atom attached to 4 atoms / tetrahedral arrangement of C (atoms) [6 max]
(b) (i) network/giant structure / macromolecular
each Si bonded covalently to 4 oxygen atoms and each O atom bonded covalently to 2 Si atoms / single covalent bonds
19 Z08_CHE_SB_IBDIP_9755_ANS.indd 19
Answers Award [1 max] for answers such as network-covalent, giant-covalent or macromolecular-covalent.
Both M1 and M2 can be scored by a suitable diagram. (ii) Silicon dioxide: strong/covalent bonds in network/giant structure/macromolecule
Carbon dioxide: weak/van der Waals’/ dispersion/London forces between molecules
(c) triple (covalent) bond
one electron pair donated by oxygen to carbon atom / dative (covalent)/coordinate (covalent) bond 
Ethanol contains a hydrogen atom bonded directly to an electronegative oxygen atom / hydrogen bonding can occur between two ethanol molecules / intermolecular hydrogen bonding in ethanol; the forces of attraction between molecules are stronger in ethanol than in methoxymethane / hydrogen bonding stronger than van der Waals’/dipole-dipole attractions. max 
Award  max if covalent bonds breaking during boiling is mentioned in the answer.
Penalize only once if no reference given to intermolecular nature of hydrogen bonding or van der Waals.
Award [1 max] for representation of C≡O.
20 (a) (bond formed by) sideways overlap of p orbitals.
Award  if CO shown with dative covalent bond.
(b) C1 is sp3 and C2 is sp2.
Lewis structure S
Name of shape
2 lone pairs on S required for the mark F (e)
Seesaw/ distorted tetrahedral
F 1 lone pair on S required for the mark F
SF6 19 Methoxymethane is very weakly polar/weak van der Waals’/dipole–dipole forces exist between methoxymethane molecules.
Accept alternatives to van der Waals’ such as London and dispersion forces
Accept square bipyrimidal 
Penalise missing lone pairs on fluorine atoms once in correct structures only.
20 Z08_CHE_SB_IBDIP_9755_ANS.indd 20
For Lewis structures candidates are not expected to draw exact shapes of molecules.
2 23 (a) (i)
Do not allow ECF for wrong Lewis structures.
22 (a) (i) Cl
Cl P Cl Cl
trigonal pyramid in the range of 100–108° N H / (ii) N H H H
(iii) F F
(b) (i) sigma bonds are formed by end on/axial overlap of orbitals with electron density between the two atoms/nuclei pi bonds are formed by sideways overlap of parallel p orbitals with electron density above and below internuclear axis/σ bond 
Accept suitably annotated diagrams (ii) 8 sigma/σ 1 pi/π (iii) 109°/109.5° 120°
(iv) sp hybridization
1 sigma and 2 pi
sigma bond formed by overlap between the two sp hybrid orbitals (on each of the two carbon atoms) / pi bonds formed by overlap between remaining p orbitals (on each of the two carbon atoms) / diagram showing 2 sp hybrid orbitals and 2 p orbitals 
square planar 90°  Penalize once only if electron pairs are missed off outer atoms
Must include minus sign for the mark. bent/V-shaped in the range of 100–106° 
octahedral/octahedron/square bipyramidal 90° / 90° and 180° + (ii) O N O
Allow dots, crosses or lines in Lewis structures.
Penalize missing charge, missing bracket once only in (i) and (ii).
Lone pairs required for BOTH (i) and (ii).
(b) NO2: O
Award  for correct representation of the bent shape and  for showing the net dipole moment, or explaining it in words (unsymmetrical distribution of charge). CO2: O
Award  for correct representation of the linear shape and for showing the two equal but opposite dipoles or explaining it in words (symmetrical distribution of charge).  For both species, allow either arrow or arrow with bar for representation of dipole moment.
Allow correct partial charges instead of the representation of the vector dipole moment.
Ignore incorrect bonds.
Lone pairs not needed.
(c) Structure: network/giant lattice / macromolecular / repeating tetrahedral units
Answers Bonding: (single) covalent (bonds)
It is not necessary to identify which part refers to structure and bonding specifically.
(iv) > k1 so) step 1 rate-determining / rds / slow step; two molecules of NO2 involved in step 1 consistent with rate expression / rate of overall reaction must equal rate of step 1 which is rate = k1[NO2]2 / OWTTE 
25 (a) 2AB2 → A2 + 2B2
(d) Ea = −R × m; measurement of gradient from two points on line
22 Rate = k[NO2][CO] 23 Yes, it fits the kinetic data and the overall stoichiometry.
(b) Rate = k [AB2]2 (c) mol−1 dm3 s−1 26 C
29 134 kJ mol−1
Challenge yourself 1 Collecting a gas over warm water will cause its temperature and therefore its volume to increase. 2 If the partially made/broken bonds are treated as containing only one electron, we can calculate
Accept a gradient in range –2.14 × 104 K to –2.27 × 104 K.
correct answer for Ea;
correct units kJ mol−1 / J mol−1 corresponding to answer
Allow kJ or J.
A typical answer: Ea = 1.85 × 102 kJ mol−1.
Allow answers for Ea in range 1.75 × 102 kJ mol−1 to 1.91 × 102 kJ mol−1.
Award  for correct final answer with some working shown. Award [2 max] for correct final answer without any working shown.
30 Z08_CHE_SB_IBDIP_9755_ANS.indd 30
12 (a) the concentration (of nitrogen(II) oxide)
(b) mol−2 dm6 s−1 / dm6 mol−2 s−1
Number of particles
Award  if reference made to equilibrium. 
Accept (mol dm ) s . −1
13 (a) k increases with increase in T / k decreases with decrease in T 
−30000 (K) = −Ea/R
Allow value in range −28800 to −31300 (K).
Ea = (30000 × 8.31=) 2.49 × 105 J mol−1/ 249 kJ mol−1
axes correctly labelled x = energy/ velocity/speed, y = number/% of molecules/particles/probability
graph showing correct curve for Maxwell-Boltzmann distribution
If two curves are drawn, first and second marks can still be scored, but not third.
Curve(s) must begin at origin and not go up at high energy.
two activation energies shown with Ecat shown lower 
Award the mark for the final point if shown on an enthalpy level diagram.
Allow value in range 240–260 kJ mol−1.
Allow  for correct final answer. (c) 0.9 × 0.200 = 0.180 (mol dm ); −3
rate = (0.244 × (0.180)2 =) 7.91 × 10−3 mol dm−3 s−1
(ii) catalyst provides an alternative pathway of lower energy / OWTTE
Award  for correct final answer. Award [1 max] for either 9.76 × 10−3 mol dm−3 s−1 or 9.76 × 10−5 mol dm−3 s−1.
14 (a) to maintain a constant volume / OWTTE 
(b) (i) [H+] order 1, [CH3COCH3] order 1, [I2] order 0;
15 (a) XY + Z → X + YZ
(rate = ) k[H+][CH3COCH3]
Award  for correct rate expression.
Allow expressions including [I2]0.
(ii) neither were correct / Alex was right about propanone and wrong about iodine / Hannah was right about propanone and hydrogen ions but wrong about iodine / OWTTE
Accept catalyst lowers activation energy (of reaction).
transition state 2
Ea (no catalyst)
Do not allow answers giving just the Arrhenius equation or involving ln k relationships.
(b) gradient = −Ea/R;
Ea (with catalyst)
transition state 4
3 intermediate Ea 1 Reactants
(c) [CH3COCH3] = 0.100 mol dm and [H+] = 0.100 mol dm−3 4.96 × 10−6 =4 .96 × 10−4 mol−1 dm−3 k= (0.100 × 0.100) s−1
5 Products ∆H extent of reaction
Award  for all 5 correct,  for 4 correct,  for 3 correct.
(c) 1 X–Y + Z, W
2 X–Y–W 4 W–Y–Z
Ignore calculation of [I2].
5 W, Y–Z + X
Award  for all 5 correct,  for 4 correct,  for 3 correct.
No ECF here for incorrect units.
31 Z08_CHE_SB_IBDIP_9755_ANS.indd 31
Answers (d) 1st step is rate-determining step (highest energy of transition state);
rate equation = k[W][XY]
(e) reaction is catalysed by W, which is not chemically changed at the end of the reaction
(f) see graph in (b) above
Chapter 7 Exercises 1 A
4 (a) Kc =
[NO2] [NO2] [H2O] (b) Kc = 2 [NO] [O2] [NH3]4[O2]7 2
[CH3OH][Cl−] (c) Kc = [CH3Cl][OH−] 5 (a) N2O4(g)
6 (a) 3F2(g) + Cl2(g) [ClF3]2 Kc = [F2]3[Cl2]
(b) Amount of CO will decrease CO(g) + 3H2(g)
(b) 2NO(g) N2(g) + O2(g) [N2][O2] Kc = [NO]2 (c) CH4(g) + H2O(g) [CO][H2]3 Kc = [CH4][H2O]
(d) Shift to the right (e) Shift to the right 15 (a) Amount of CO will decrease
(b) CH4(g) + H2O(g)
(c) This is equivalent to an increase in pressure, so shifts to the left
CO(g) + 3H2(g)
(c) Amount of CO will increase (d) No change in CO 16 C
18 The Haber process is exothermic in the forward direction. Therefore, increasing temperature will decrease the value of Kc. This represents a decrease in the reaction yield. 19
7 (a) Mostly reactants (b) Mostly reactants
[H2][I2] (0.11) = = 2.0 × 10−2 [HI]2 (0.78)2 2
[HOCl]2 8 (a) > K; not at equilibrium; reaction [H2O][Cl2O] proceeds to the left (b) At equilibrium [HOCl]2 (c) > K; not at equilibrium; reaction [H2O][Cl2O] proceeds to the left 9 (a) 7.73 × 104 (b) 3.60 × 10−3 (c) 6.00 × 10−2 11 D
(c) Mostly products
13 (a) Shift to the left (b) Shift to the right (c) No shift in equilibrium
(b) At the higher temperature, the value of Kc is higher, so the reaction must be endothermic. N2(g)
1.6 − x
Equilibrium: 1.6 − x
As Kc is very small, 1.6 – x » 1.6 [NO]2 (2x)2 = = 1.7 × 10−3 Kc = [N2][O2] (1.6)2
x = 0.03298, so 2x = 0.066
[NO]eqm = 0.066 mol dm−3
14 (a) Shift to the left (b) Shift to the right
32 Z08_CHE_SB_IBDIP_9755_ANS.indd 32
CO(g) + H2O(g) = H2(g) + CO2(g) 3 The concentration of a pure solid or pure liquid is a constant, effectively its density, which is Initial: 4.0 6.4 0.0 0.0 independent of its amount. These constant Change: −3.2 −3.2 +3.2 +3.2 values therefore do not form part of the Equilibrium: 0.8 3.2 3.2 3.2 equilibrium expression, but are included in the [H2][CO2] (3.2)2 value of K. = = 4.0 Kc = [CO][H2 O] (0.8)(3.2) 4 The value for Kc at 298 K for the reaction (b) Put the values into the equilibrium N2(g) + O2(g) 2NO(g) is extremely low, so the expression to determine Q: equilibrium mixture lies to the left with almost no production of NO. But at higher temperatures, (3.0)2 Q= 2 = 0.56 such as in vehicle exhaust fumes, the reaction (4.0) shifts to the right and a higher concentration This is not equal to the value of Kc so the of NO is produced. This gas is easily oxidized reaction is not at equilibrium. As the value in the air, producing the brown gas NO2 which of this mixture is lower than Kc, the reaction is responsible for the brownish haze: 2NO(g) + will move to the right before equilibrium is O2(g) → 2NO2(g). established. 5 The atom economies of the Haber process 22 C and the Contact process reactions described 23 (a) 0 (b) Negative are both 100% as there is only one product. In other words, there is no waste. But this does not (c) Positive mean that all reactants are converted to product, 24 (a) 79.8 kJ so the stoichiometric yield is less than 100%. (b) Increasing temperature has increased the It is the goal of these industries to maximize value of K so it must be an endothermic yield and efficiency by choosing the optimum reaction. conditions, taking equilibrium and kinetic considerations into account. 21
Challenge yourself 1 Earth receives energy from the Sun and disperses energy, largely as heat. But exchange of matter is minimal – the only exceptions to Earth being a closed system are matter received from space such as asteroids and space dust, and matter lost to space such as spacecraft. 2 The different values of Kc indicate different stabilities of the hydrogen halides. The bonding in HCl is the strongest and in HI the weakest. This is largely because of the size of the atoms. As I has a larger atomic radius than Cl, in HI the bonding pair is further from the nucleus than the bonding pair in HCl, and so experiences a weaker pull. The HI bond breaks more easily and so the dissociation reaction is favoured.
Practice questions 1 C
9 D 10 (a) (i) (K =) [SO3]2/[O2][SO2]2
(ii) yield (of SO3) increases / equilibrium moves to right / more SO3 formed;
3 gaseous molecules → 2 gaseous molecules / decrease in volume of gaseous molecules / fewer gaseous molecules on right hand side
Do not allow ECF.
(iii) yield (of SO3) decreases;
33 Z08_CHE_SB_IBDIP_9755_ANS.indd 33
Answers (c) value of Kc increases;
forward reaction is exothermic / reverse/ backwards reaction is endothermic / equilibrium shifts to absorb (some of) the heat 
[SO2Cl2] increases; decrease in temperature favours (forward) reaction which is exothermic 
Do not accept exothermic reaction or Le Châtelier’s principle.
Do not allow ECF. (d) no effect on the value of Kc / depends only on temperature;
Do not allow ECF. (iv) rates of both forward and reverse reactions increase equally; no effect on position of equilibrium; no effect on value of Kc 
[SO2Cl2] decreases; increase in volume favours the reverse reaction which has more gaseous moles  Do not allow ECF.
(b) 2NO(g) + 2H2(g) → N2(g) + 2H2O(g) NO(g)
Initial/ mol dm−3
Change/ mol dm−3
Equilibrium/ mol dm−3
(e) no effect; catalyst increases the rate of forward and reverse reactions (equally) / catalyst decreases activation energies (equally) 12
increasing temperature favours endothermic/ reverse reaction; as yield decreases with increasing temperature [2 max]
[N2] at equilibrium = 0.019 (mol dm−3); [H2O] at equilibrium = 0.138 (mol dm−3);
(b) yield increases / equilibrium moves to the right / more ammonia;
Kc = [N2][H2O]2/[NO]2[H2]2 = (0.019)(0.138)2/ (0.062)2(0.013)2 = 5.6 × 102 
increase in pressure favours the reaction which has fewer moles of gaseous products  [NH3]2  (c) (Kc =) [N2][H2]3
Award  for final correct answer. Accept any value also in range 557–560. Do not penalize significant figures. (a) (Kc =)
(d) [N2]: (at equilibrium = 1.00 − 0.031 =) 0.969 (mol dm−3);
Ignore state symbols.
[H2]: (at equilibrium = 3.00 − 3(0.031) =) 2.91 (mol dm−3); (0.062)2  Kc = = 1.6(1) × 10−4 (0.969)(2.91)3
Square brackets [ ] required for the equilibrium expression.
(b) 7.84 × 10−3 mol of SO2 and 7.84 × 10−3 mol of Cl2;
Award  for 10.34. Award  for the correct final answer.
7.84 × 10−3 mol dm−3 of SO2, 7.84 × 10−3 mol dm−3 of Cl2 and 7.65 × 10−4 mol dm−3 of SO2Cl2; 12.5
(a) exothermic Accept either of the following for the second mark.
[H2] at equilibrium = 0.013 (mol dm−3);
Award  for Kc = 1.4 × 10−4 (e) no effect 13
(a) reactants and products in same phase/state; rate of forward reaction = rate of reverse reaction;
34 Z08_CHE_SB_IBDIP_9755_ANS.indd 34
concentrations of reactants and products remain constant / macroscopic properties remain constant [2 max]
(f) no effect (on the value of the equilibrium constant)
Do not accept concentrations are equal. [HI]2 [H2][I2] (c) no change to position of equilibrium (b) (Kc =)
as it speeds up forward and reverse reaction / concentrations of reactants and products do not change / position of equilibrium does not change / no change in yield
(d) the reaction is exothermic / heat is given out / ∆H is negative 
14 (a) ∆G° = 0
(e) amount of H2 remaining at equilibrium 1.80 = 0.70 mol = 1.60 − 2 amount of I2 remaining at equilibrium 1.80 = 0.10 mol = 1.0 − 2 (1.80 / 4.0)2 1.802 Kc = / (0.70 / 4.00) × (0.10 / 4.00) 0.70 × 0.10 (1.80)2 Kc = = 46.3  0.70 × 0.10
Award  for correct final answer.
(b) (i) HI originally placed = 2.0 mol dm−3
(ii) I2 at equilibrium = 0.218 mol dm−3; HI at equilibrium = 1.56 mol dm−3
(b) ΔG = −70 kJ mol−1 The reaction has a very high value for K, so will go essentially to completion – from the equilibrium yield, this reaction is likely to give a high production of methanol. However, kinetic data are not available, so the rate cannot be deduced.
15 (a) [H2(g)] = 4.0 mol dm−3 [I2(g)] = 1.0 mol dm−3
[HI] = 4.0 mol dm−3
Chapter 8 Exercises 1 (a) HSO3− (b) CH3NH3+ (c) C2H5COOH (d) HNO3 (e) HF (f) H2SO4 2 (a) H2PO4− (b) CH3COO− (c) HSO3− (d) SO42− (e) O2− (f) Br− 3 (a) CH3COOH (acid)/CH3COO− (base) NH3 (base)/NH4+ (acid) (b) CO32− (base)/HCO3− (acid)
4 HPO42−(aq) + H2O(l) behaviour)
PO43−(aq) + H3O+(aq) (acid
HPO42−(aq) + H2O(l) (base behaviour)
H2PO4−(aq) + OH−(aq)
5 (a) H2SO4(aq) + CuO(s) → CuSO4(aq) + H2O(l) (b) HNO3(aq) + NaHCO3(s) → NaNO3(aq) + H2O(l) + CO2(g) (c) H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3H2O(l)
H3O (acid)/H2O (base)
(d) 6CH3COOH(aq) + 2Al(s) → 2Al(CH3COO)3(aq) + 3H2(g)
(c) NH4+ (acid)/NH3 (base)
NO2 (base)/HNO2 (acid)
35 Z08_CHE_SB_IBDIP_9755_ANS.indd 35
Answers 8 (a) nitric acid + sodium carbonate / sodium hydrogencarbonate / sodium hydroxide 2HNO3(aq) + Na2CO3(aq) → 2NaNO3(aq) + H2O(l) + CO2(g) (b) hydrochloric acid + ammonia solution HCl(aq) + NH4OH(aq) → NH4Cl(aq) + H2O(l)
(c) copper(II) oxide + sulfuric acid H2SO4(aq) + CuO(s) → CuSO4(aq) + H2O(l) (d) methanoic acid + potassium hydroxide HCOOH(aq) + KOH(aq) → KCOOH(aq) + H2O(l)
[OH−] = 5.9 × 10−8 mol dm−3 acidic 24 (a) 0.40 (b) 10.57 (c) 10.00 25 B [C2H5NH3+][OH−] [C2H5NH2] [H SO ][OH−] (b) Kb = 2 4 − [HSO4 ] [HCO3−][OH−] (c) Kb = [CO32−] 26 (a) Kb =
9 pH increases by 1 unit
27 HNO2 < H3PO4 < H2SO3
10 pH = 4.72
28 Strong acids and bases are fully dissociated, so it is not useful to think of them in terms of an equilibrium mixture. The pH of their solutions can be derived directly from their concentration.
11 [H+] = 1.0 × 10−9 mol dm−3, [OH−] = 1.0 × 10−5 mol dm−3 12 (a) [OH−] = 2.9 × 10−6; basic (b) [H+] = 1.0 × 10−12; basic (c) [H+] = 1.0 × 10−4; acidic (d) [OH−] = 1.2 × 10−10; acidic 13 pH = 2.0 14 (a) pH = 6.9 (b) pH = 2 (c) pH = 4.8
30 Kb = 5.6 × 10−4 31 [H+] = 1.0 × 10−4 mol dm−3; [OH−] = 1.0 × 10−10 mol dm−3 32 pH = 3.22; [H+] = 6.0 × 10−4 mol dm−3 33 A 34 HF is the stronger acid.
15 pH = 13.17 16 B
35 pKb CN− = 4.79; pKb F− = 10.83
18 (a) H2CO3 (b) HCOOH 19 (a) Lewis acid Zn2+; Lewis base NH3
CN− is the stronger base. 36 (a) pKb CH3COO− = 9.24
(c) Lewis acid Mg2+; Lewis base H2O
(b) Methanoic acid is a stronger acid than ethanoic acid from its lower pKa. Therefore, its conjugate base is weaker.
20 D, CH4 because it does not possess a lone pair.
21 C, there is no exchange of H+.
39 (i) Because it has a higher concentration of the acid and its conjugate base.
(b) Lewis acid BeCl2; Lewis base Cl−
22 [H+] = [OH−] = 1.55 × 10−7
pH = pOH = 6.81
40 (a) equal to 7 (b) less than 7
pH + pOH = pKw = 13.62
(c) less than 7 (d) greater than 7
41 B; salt of strong base and weak acid
23 pOH = 7.23
42 (a) less than 7 (b) greater than 7
[H ] = 1.7 × 10 mol dm +
(c) equal to 7
36 Z08_CHE_SB_IBDIP_9755_ANS.indd 36
Answers 43 D (III only)
(c) Particulates act as catalysts in the production of secondary pollutants.
(d) SO2(g): CaO(s) + SO2(g) → CaSO3(g)
9 (a) CuCl2(aq) + Ag(s)
oxidation Mg(s) → Mg2 (aq) 2e
(c) MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) (d) 2MnO4−(aq) + 16H+(aq) + 5C2O42−(aq) → 2Mn2+(aq) + 8H2O(l) + 10CO2(g)
BrO3−(aq) + 6H+ + 6I− → Br−(aq) + 3H2O(l) + 3I2(s)
E ⊖cell = E ⊖BrO3– − E ⊖I2 = +1.44 − (+0.54) = +0.90 V
18 Strongest oxidizing agent Cu2+; strongest reducing agent Mg. 19 (a) No reaction
(e) 6.16 × 10−3 (f) 6.16 × 10−3
(b) Reaction occurs
BrO3−(aq) + 6H+(aq) + 3Cd(s) → Br−(aq) + 3H2O(l) + 3Cd2+(aq)
12 (a) 0.117% (b) Solution changes from orange to green
E ⊖cell = E ⊖BrO3 − E ⊖Cd2+ = +1.44 − (−0.40) = 1.84 V (c) No reaction
13 (a) Zn / Zn2+ Fe / Fe2+ anode cathode
20 −270 kJ
Zn(s) → Zn2+(aq) + 2e−
21 (a) At anode: 2Br−(l) → Br2(l) + 2e−
Fe2+(aq) + 2e− → Fe(s)
At cathode: 2K+(l) + 2e− → 2K(l)
42 Z08_CHE_SB_IBDIP_9755_ANS.indd 42
Answers (b) At anode: 2F−(l) → F2(g) + 2e−
At cathode: Mg (l) + 2e → Mg(l) 2+
(c) At anode: S2−(l) → S(l) + 2e−
At cathode: Zn2+(l) + 2e− → Zn(l)
e− anode + oxidation
Cathode: Cu2+(aq) + 2e– → Cu(s)
Reaction at the anode would be different:
Cu(s) → Cu2+(aq) + 2e–
− cathode reduction inert electrodes Mg2+
(b) Reaction at the cathode would be the same with copper deposited on the copper electrode.
(b) Anode: 2Cl−(aq) → Cl2(g) + 2e−
Cathode: Mg2+(aq) + 2e− → Mg(s)
Overall: Mg2+(aq) + 2Cl−(aq) → Mg(s) + Cl2(g)
The copper electrode disintegrates as it is oxidized, releasing Cu2+ ions into the solution. The blue colour of the solution would not change as Cu2+ ions are produced and discharged at an equal rate.
26 During electrolysis of NaCl(aq) at the cathode H2O is reduced (rather than Na+, which is reduced in molten Na(l) and H2(g) is discharged:
2H2O(l) + 2e− → H2(g) + 20k–(aq)
24 Ions present: K+(aq), F−(aq)
27 AlCl3(l) → Al3+(l) + 3Cl−(l)
At anode: F−(aq) and H2O(l); H2O(l) will be oxidized. Reaction occurring is 2H2O(l) → 4H+(aq) + O2(g) + 4e– H+(aq) and O2(g) will be discharged at the anode.
At cathode: K+(aq) and H2O(l); H2O(l) will be reduced.
Reaction occurring is 2H2O(l) + 2c– → H2(g) + 2OH–(aq) OH–(aq) and H2(g) will be discharged at the cathode.
Products will be O2(g) and H2(g)
This is because H2O has a higher E ⊖ than K+ so is preferentially reduced at the cathode; H2O has a higher E ⊖ than F− so is preferentially oxidized at the anode (assuming the concentration of F− is not high enough to cause it to be discharged).
25 (a) At the anode, bubbles of gas emitted; at the cathode, pinky brown layer of copper metal deposited. The blue colour of the solution fades.
2H2O(l) → 4H+(aq) + O2(g) + 4e−
depending on the concentration of the solution.
Cathode: Cu2+(aq) + 2e− → Cu(s)
Blue colour fades as the concentration of Cu2+ ions in solution decreases.
1 mole 2 moles of Cl2 of electrons
3 moles 1 mole of electrons of Al
So the same quantity of electricity will produce Cl2 : Al 3 : 2
Therefore, yield of Al = 0.2 mol × 2/3 = 0.13 mol Al
Mass Al = 0.13 × M(Al) = 3.5 g
28 C 29 The mass of the silver anode will decrease as Ag is oxidized to Ag+ ions that are released into the solution. The mass of the cathode (spoon) will increase as a layer of Ag is deposited. Impurities may be visible collecting as a sludge at the bottom of the electrolyte as they fall from the decomposing anode.
Anode: 2Cl−(aq) → Cl2(g) + 2e−
2Cl−(l) → Cl2(g) + 2e−Al3+(l) + 3e− → Al(l)
1 H2O2 : H = +1, O= −1
Oxygen is halfway between 0 (element) and −2 (usual oxidation state in compounds), so can be oxidized (to 0) or reduced (to −2). It will more easily be reduced from −1 to −2 as it is a very electronegative element, and so acts mainly as an oxidizing agent.
43 Z08_CHE_SB_IBDIP_9755_ANS.indd 43
Answers 2 Cl2(aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H2O(l)
above hydrogen in the ECS / Fe is a better reducing agent than H2 / Fe is oxidized more readily than H2 
Cl changes from 0 to −1 (reduction)
Cl changes from 0 to +1 (oxidation)
(ii) −0.28 V
Both changes occur simultaneously.
(iii) Co2+ / cobalt(II) ion
(iv) 2Al + 3Fe2+ → 3Fe + 2Al3+
3 Iodine solution contains the triiodide ion, I3−, in which the central atom has five electron domains with two bonding and three non-bonding pairs in the equatorial plane. This gives a linear ion with a low charge density, which is able to slip into the coils of the hydrophobic interior of the amylose helix. 4 Solubility of gases decreases with increasing temperature as evaporation is higher. So the discharge of hot water will lower the dissolved O2 content. 5 Charge per e = 1.602189 × 10 −
Award  for correct reactants and products and  for correctly balanced; ignore states. Do not accept .
(v) To complete the electrical circuit / OWTTE;
Electrons per mole = 6.02 × 10 mol−1
Therefore, charge per mole = 1.602189 × 10−19 C × 6.02 × 1023 mol−1 = 96451.78 C mol−1
(b) (i) +2
Only penalize once if roman numerals are used or if written as 2+ or 3+.
(c) (i) electron flow
6 ∆G = −RT ln Kc and ∆G = −nFEcell RT ln Kc Therefore Ecell = nF Expressing in terms of log10 and combining all constants at 298 K: 0.0592 Ecell = log10 Kc n
by allowing the movement of ions
battery / source of electricity connected to two electrodes in the solution with positive and negative electrodes correctly labelled;
electrons / current flowing from the cell to the negative electrode;
labelled solution of sodium chloride
12 (a) (i) The voltage obtained when the half-cell is connected to the standard hydrogen electrode;
If the connecting wires to electrodes are immersed in the solution [1 max].
Under standard conditions of 298K and 1 mol dm−3 solutions;
Electrons flow (in the external circuit) from the half-cell to the hydrogen electrode / the metal in the half-cell is
(ii) Na+, H+/H3O+, Cl−, OH−
All four correct , any 3 correct .
(iii) hydrogen at (–)/cathode and oxygen at (+)/anode
2H+ + 2e− → H2 / 2H2O + 2e− → H2 + 2OH−
44 Z08_CHE_SB_IBDIP_9755_ANS.indd 44
Answers 4OH− → O2 + 2H2O + 4e− / 2H2O → O2 + 4H+ + 4e− 
Accept e instead of e–; if electrodes omitted or wrong way round [2 max].
(iv) ratio of H2 : O2 is 2 : 1
(d) (i) (–)/(cathode) 2H+ + 2e− → H2 / 2H2O + 2e− → H2 + 2OH−
(+)/(anode) 2Cl− → Cl2 + 2e−
Accept e instead of e–; if electrodes omitted or wrong way round [1 max].
Cathode / negative electrode: Na+ + e− → Na
Anode / positive electrode: 2Cl → Cl2 + 2e−/ Cl− → 12Cl2 + e−
Award [1 max] if the two electrodes are not labelled/labelled incorrectly for the two halfequations.
(ii) (–)/(cathode) Na+ + e− → Na
(+)/(anode) 2Br− → Br2 + 2e−
Accept e instead of e–; if electrodes omitted or wrong way round [1 max].
Electrolytic cell involves a non-spontaneous (redox) reaction and voltaic cell involves a spontaneous (redox) reaction. In an electrolytic cell, cathode is negative and anode is positive and vice versa for a voltaic cell / electrolytic cell, anode is positive and voltaic cell, anode is negative / electrolytic cell, cathode is negative and voltaic cell cathode is positive. Voltaic cell has two separate solutions and electrolytic cell has one solution / voltaic cell has salt bridge and electrolytic cell has no salt bridge. Electrolytic cell, oxidation occurs at the positive electrode/anode and voltaic cell, oxidation occurs at the negative electrode/ anode and vice versa. [2 max]
(b) (solid) ions in a lattice / ions cannot move
(molten) ions mobile / ions free to move
Overall cell reaction: Na+(l) + Cl−(l) → Na (l) + 12Cl2(g)
Award  for correct equation and  for correct state symbols.
13 (a) Electrolytic cell converts electrical energy to chemical energy and voltaic cell converts chemical energy to electrical energy / electrolytic cell uses electricity to carry out a (redox) chemical reaction and voltaic cell uses a (redox) chemical reaction to produce electricity / electrolytic cell requires a power supply and voltaic cell does not.
(c) Reduction occurs at the cathode / negative electrode and oxidation occurs at the anode / positive electrode
Allow NaCl(l) instead of Na+(l) and Cl–(l).
(d) Al does not corrode / rust; Al is less dense / better conductor / more malleable 
Accept Al is lighter (metal compared to Fe).
Accept converse argument.
(e) Cathode / negative electrode
Object to be plated
Allow a specific example here, e.g. spoon.
Accept inert metal / graphite.
Do not accept silver halides or their formulae.
Anode / positive electrode
Silver / Ag
Allow silver nitrate / AgNO3 / silver cyanide / any other suitable silver salt/solution.
Do not accept AgCl.
14 (a) 2Al(s) + 3Ni2+(aq) → 2Al3+(aq) + 3Ni(s)
Correct reactants and products, award .
Balancing award .
Ignore state symbols and equilibrium sign.
(b) (+) 1.40 (V)
(c) aluminium anode / negative electrode
nickel cathode / positive electrode
electron movement from Al to Ni
correct movement of cations and anions through salt bridge 
If electron movement shown correctly but not labelled, award the mark.
45 Z08_CHE_SB_IBDIP_9755_ANS.indd 45
Answers voltmeter e e− V −
cations salt bridge
Allow zero/nought for 0.
Nitrogen: +5 to +4 / decreases by 1 / −1 / 1−
(ii) nitric acid / HNO3 / NO3− / nitrate
(b) (i) 0.100 × 0.0285
2.85 × 10−3 (mol)
(ii) 2.85 × 10−3 (mol)
(iii) (63.55 × 2.85 × 10 ) = 0.181 g
Award  for correct final answer.  −3
15 (a) (i) Copper: 0 to +2 / increases by 2 / +2 / 2+
Allow 63.5. 0.181 × 100 = 39.7% (iv) 0.456 44.2 − 39.7 × 100 = 10 / 10.2% (v) 44.2
  
Allow 11.3%, i.e. percentage obtained in (iv) is used to divide instead of 44.2%.
Penalize missing + sign or incorrect notation such as 2+, 2+ or II, once only.
Chapter 10 Exercises 1 (a) carboxylic acid; butanoic acid
(b) halogenoalkane; 1,1-dichloropropane
(c) ketone; butanone
(d) ester; methyl ethanoate
(e) ether; methoxyethane
(f) ester; ethyl pentanoate
2 (a) CH3(CH2)4COOH
4 D Cl
7 Benzene is a cyclic molecule with a planar framework of single bonds between the six carbon atoms and six hydrogen atoms. The carbon atoms are also bonded to each other by a delocalized cloud of electrons which forms a symmetrical region of electron density above
46 Z08_CHE_SB_IBDIP_9755_ANS.indd 46
Answers (b) C2H5COOH(aq) + C4H9OH(aq) → C2H5COOC4H9(aq) + H2O(l)
and below the plane of the ring. This is a very stable arrangement, so benzene has much lower energy than would be expected.
8 (a) Similar molar mass will mean molecules have approximately equal London (dispersion) forces and so differences in boiling point can be attributed to differences in dipole–dipole or hydrogen bonding.
(b) methanal; orange → green
(c) no reaction; no colour change
(b) Solubility in hexane will increase with increasing chain length as the non-polar part of the molecule makes a larger contribution to its structure.
9 (a) C5H14(l) + 6O2(g) → 5CO(g) + 7H2O(l)
(b) 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)
(c) C3H4(g) + O2(g) → 3C(s) + 2H2O(l)
10 Bromine + ethane
initiation UV light
2Br• bromine radicals
Br• + C2H6 → C2H5• + HBr C2H5• + Br2 → C2H5Br + Br• C2H5Br + Br• → C2H4Br• + HBr C2H4Br• + Br2 → C2H4Br2 + Br·
Br• + Br• → Br2 C2H5• + Br• → C2H5Br C2H5 + C2H5 → C4H10 •
Overall, these reactions show how a mixture of products is formed.
11 (a) CH3CH2CH2CH3 butane
(b) CH3CH2CH(OH)CH3 butan-2-ol
(c) CH3CH2CHBrCH3 2-bromobutane
12 (a) No observable change.
(b) Burns with very smoky flame.
(c) The bromine water changes from brown to colourless.
13 (a) C2H5OH(l) + 3O2 → 2CO2(g) + 3H2O(l) 2C3H7OH(l) + 9O2(g) → 6CO2(g) + 8H2O(l)
14 (a) butanone; orange → green
15 Nucleophilic substitution involves an electronrich species (e.g. OH−) attacking an electrondeficient carbon atom (e.g. in chloroethane), leading to substitution of the halogen functional group by the nucleophile. C2H5Cl + OH− → C2H5OH + Cl− 16 Benzene has a very stable structure as a result of its symmetrical ring of delocalized electrons. Addition reactions would involve breaking this ring and therefore decreasing its stability. Substitution reactions in which one or more hydrogen atoms of the ring are replaced by other atoms or groups preserves the aromatic ring structure and therefore its stability. 17 (a) CH3CH2CH2CH2Br primary CH3CH2CHBrCH3 secondary C(CH3)3Br tertiary
(b) The tertiary halogenoalkane reacts by an SN1 mechanism.
S = substitution; N = nucleophilic; 1 = unimolecular
(c) RBr → R+ + Br−
18 C 19 (a) The carbon–halogen bond breaks more easily in the iodo- and bromo- derivatives than in the chloro- derivatives, so these compounds more readily undergo substitution reactions.
(b) The substitution reaction of OH for Cl occurs in both these compounds, displacing Cl− and forming the white precipitate of AgCl, which darkens on exposure to air. The tertiary halogenoalkane C(CH3)3Cl isomer reacts more quickly than the primary isomer CH3CH2CH2CH2Cl because it undergoes an SN1 mechanism, which is faster.
47 Z08_CHE_SB_IBDIP_9755_ANS.indd 47
Answers 20 Alkenes have a double bond which is an electrondense region and so is susceptible to attack by electrophiles which are themselves electron deficient. They undergo addition reactions because they are unsaturated; one of the bonds in the double bond breaks and incoming groups can add to the two carbon atoms.
When bromine approaches but-2-ene, it is polarized by the electron density in the double bond. Electrons in the bromine–bromine bond are repelled away from the double bond, leading to the heterolytic fission of the bromine molecule. The Br+ product now attaches itself to one of the carbon atoms as the carbon–carbon bond breaks. This produces an unstable carbocation which then rapidly reacts with the Br − ion. The product is 2,3-dibromobutane.
Application of Markovnikov’s rule enables us to predict that the electrophile H+ will add to the terminal carbon-forming a secondary carbocation, as this is stabilized by the positive inductive effect of the alkyl groups. Br− will then add to carbon 2, forming 2-bromobutane.
22 ICl is polarized: I Cl owing to the greater electronegativity of Cl than I. So when it undergoes heterolytic fission it will form I+ and Cl−. By application of Markovnikov’s rule, the I+ will attach to the terminal carbon, while Cl− will add to carbon 2. The product is therefore 1-iodo-2-chloropropane. δ+
24 (a) Use LiAlH4 in dry ether and heat. The acid is reduced first to the aldehyde and then to the alcohol.
The stronger acid H2SO4 protonates the HNO3, leading to production of the nitronium ion NO2+. This is a strong electrophile which reacts with the π electrons of the benzene ring, substituting for H.
23 Concentrated H2SO4 and concentrated HNO3.
21 But-1-ene + HBr → 2-bromobutane
(b) Nitrobenzene is heated under reflux with tin and concentrated HCl, and the product is reacted with NaOH.
(c) Ethanal is heated with NaBH4(aq).
25 Start with ethanol. Take one portion and oxidize it using acidified potassium(VI) dichromate solution and heat under reflux to allow the reaction to go to completion. C2H5OH
The product is ethanoic acid.
React the ethanoic acid product with another portion of the ethanol by warming it in the presence of some concentrated H2SO4. The esterification reaction yields ethyl ethanoate.
CH3COOH + C2H5OH → CH3COOC2H5
48 Z08_CHE_SB_IBDIP_9755_ANS.indd 48
Answers 26 React the 1-chlorobutane with NaOH in warm aqueous solution to convert it into butan-1-ol. C4H9Cl + NaOH → C4H9OH + NaCl
Oxidize the butan-1-ol using acidified potassium(VI) dichromate solution and heat under reflux to allow the reaction to go to completion. [+O]
27 C 28 3-methylhexane: CH3CH2CH(CH3)CH2CH2CH3 H C2H5 29 (a) H
E-pent-2-ene (b) H
3 Heterolytic describes breaking of the bond, producing two different products. The products are ions, and the reaction mechanism involves attraction of the electron density of the C=C double bond to the positive ion. 4 The repeating unit in polystyrene is —CH(C6H5)–CH2— 5 The cyanide ion, CN−, and ammonia, NH3, are nucleophiles that react with halogenoalkanes in substitution reactions. They act as ligands with transition metal ions, forming complexes such as [Cu(NH3)4(H2O)2]2+ and [Cu(CN)4]3−. They act as Lewis bases by donating a lone pair of electrons. For example: NH3 + BCl3 → NH3BCl3
Incomplete combustion: 2C2H6 + 5O2 → 4CO + 6H2O
C: −3 → +2
2 Complete combustion: 2C2H6 + 7O2 → 4CO2 + 6H2O
1 All four C atoms in the molecule are sp3 hybridized because they form four single bonds which are tetrahedrally arranged. The nitrogen atom is also sp3 hybridized, as its four electron domains are also tetrahedrally arranged. Note that here the hybridization also includes the lone pair on the nitrogen atom.
C: −3 → +4
6 The order of the reaction with respect to each reactant can be deduced from experiments in which the concentration of each reactant in turn is changed, and the initial rate of the reaction then measured. If, for example, the concentration of halogenoalkane is doubled while the concentration of OH− remains constant, and the rate is found to have doubled, then it indicates that the reaction is first order with respect to halogenoalkane. Examples of this type of experiment and the interpretation of the results are given in Chapter 6.
49 Z08_CHE_SB_IBDIP_9755_ANS.indd 49
Answers 7 With bromine water, the water can also take part in the second part of the reaction because of its lone pairs. The carbocation is attacked by water in competition with Br−, and the major product is 2-bromoethanol, CH2BrCH2OH. The bromine water is decolorized from brown. The relative concentration of bromoethanol and 1,2-dibromoethane depends on the strength of the bromine water used. 8 The –NH2 group in phenylamine is electron donating due to conjugation of the lone pair of electrons on N with the ring electrons. As a result, the electron density of the ring is increased, making it more susceptible to electrophilic attack. In contrast, the –NO2 group in nitrobenzene is electron withdrawing due to the electronegativity of the nitrogen and oxygen atoms. d2 O
than the trans isomer, and its density is less. cis-Butenedioic acid is a stronger acid because when H+ is lost, the cis anion is more stable than the trans form.
Practice questions 1 C
15 (a) A: 1-bromobutane
Penalize incorrect punctuation, e.g. commas for hyphens, only once. Accept 2-bromomethylpropane and 1-bromomethylpropane for C and D respectively.
Also, the electrons in its double bond conjugate with the π electrons in the ring, causing the electron density of the ring to be decreased, making it less susceptible to electrophilic attack.
9 In both square planar and octahedral compounds, geometric isomers can arise due to groups having the possibility of being in adjacent (cis) or in across (trans) positions. In tetrahedral compounds, all positions are adjacent to each other, so these isomers are not possible. 10
(b) (i) C / 2-bromo-2-methylpropane; unimolecular nucleophilic substitution 
(ii) RBr → R+ + Br−
+ Br H CH2CH2CH3
Allow use of 2-bromo-2-methylpropane instead of RBr.
(iii) A / 1-bromobutane / D / 1-bromo-2methylpropane
cis-butenedioic acid melting point 139 °C
The cis isomer, maleic acid, has a lower melting point as it forms fewer intermolecular bonds. The cis isomer is much more soluble in water
+ Br − H CH(CH3)2
50 Z08_CHE_SB_IBDIP_9755_ANS.indd 50
curly arrow going from lone pair/negative charge on O in OH– to C
Do not allow curly arrow originating on H in OH−.
curly arrow showing Br leaving
Accept curly arrow either going from bond between C and Br to Br in 1-bromobutane or in the transition state.
representation of transition state showing negative charge, square brackets and partial bonds
Do not penalize if HO and Br are not at 180° to each other.
Do not award fourth mark if OH–C bond is represented.
(c) (b) (i) no change as [OH ] does not appear in the rate equation / in the rate determining step –
(b) (iii) rate doubles as the rate is proportional to [OH–] / OH– appears in the ratedetermining / slow step / first order with respect to OH–  Award  if correctly predicts no rate change for SN1 and doubling of rate for SN2 without suitable explanation.
(d) rate of 1-bromobutane is faster; C–Br bond is weaker / breaks more easily than C–Cl bond
(e) 2-bromobutane / B; (plane-) polarized light shone through; enantiomers rotate plane of plane-polarized light to left or right / opposite directions (by same amount)
No mark for change to clear, or for decolorized with no reference to original colour.
(b) Chloroethene: H
No mark if the lone pairs are missing on Cl. Accept lines, dots or crosses for e– pairs.
(c) (hydration of ethene for the manufacture of) ethanol / C2H4 + H2O → C2H5OH; (synthesis of) CH3COOH / ethanoic / acetic acid; (synthesis of) ethylene glycol / 1,2-ethanediol / ethane-1,2-diol; (synthesis of) drugs / pesticides; (hydrogenation of unsaturated oils in the manufacture of) margarine [2 max]
Accept other commercial applications.
17 (a) A: H
H B: H
H C: H
Do not accept ‘similar’ in place of ‘identical’.
16 (a) Colour change from yellow / orange / rust colour / red / brown to colourless 
n and square brackets are not required. Continuation bonds must be shown.
Accept ‘turn’ instead of ‘rotate’ but not ‘bend’/‘reflect’. Physical properties identical (apart from effect on plane-polarized light); chemical properties are identical (except with other chiral compounds) 
51 05/09/2014 10:01
H of A and D. M2 (for B) and M5 (for E) may also be scored for substitution product if primary chloroalkane used. Penalize missing hydrogens once only in Q7.
 for reactants and  for products.
(concentrated) sulfuric acid / H2SO4
Do not accept just H+ or acid.
18 (a) (the solution changes) from orange to green
(b) CH3CH2COOH + CH3OH → CH3CH2COOCH3 + H2O
Accept condensed formulas.
Award [1 max] if A and D are other way round (and nothing else correct). Award [2 max] if A and D are other way round but one substitution product B or E is correct based on initial choice of A and D. Award [3 max] if A and D are other way round but both substitution products B and E are correct based on initial choice
Do not accept 6, 6+ or the use of Roman numerals unless they have already been penalized in 2(a).
(c) Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O
(d) CH3CH2OH → CH3CHO + 2H+ + 2e–
Cr2O72– + 3CH3CH2OH + 8H+ → 2Cr3+ + 3CH3CHO + 7H2O For second equation award  for correct reactants and products and  for correct balancing.
(e) H+ is a reactant / OWTTE
(f) ethanoic acid / CH3COOH / acid
Accept acetic acid.
Chapter 11 Exercises 1
The smallest division is 1 so the uncertainty is ±0.5.
2 The missing diamond has a mass of between 9.87 and 9.97 g.
The found diamond has a mass between 9.9 and 10.3 g.
As the ranges overlap, it could be the missing diamond.
2.22 × 10 cm 3 (a) 4 × 10 g (b) −2
(c) 3.0 × 10−2 g
(d) 3 × 10 or 3.0 × 10 °C (unspecified) 4 (a) 4 (b) unspecified (c) 3 (d) 4 5 A
12 The average value = 49.0 s
The uncertainty in the measurements is given as ±0.1 s but the results show that there is
52 Z08_CHE_SB_IBDIP_9755_ANS.indd 52
Answers additional uncertainty, suggesting that the value could be anywhere between 48.8 and 49.2 s. So the value could be quoted as 49.0 s ± 0.2 s. 13 D
20 Number of moles = concentration × volume/1000
= 1.00 × 10.0/1000 = 0.0100 mol
% uncertainty in concentration = (0.05/1.00) × 100 = 5%
% uncertainty in volume = (0.1/10.0) × 100 = 1%
% uncertainty in number of moles = 5% + 1% = 6%
Absolute uncertainty in number of moles = (6/100) × 0.0100 = 0.0006
Number of moles = 0.0100 ± 0.0006 mol
29 A (the spectrum on the left) corresponds to CH3CH2CHO
B (the spectrum on the right) corresponds to CH3COCH3
Both have a molecular ion corresponding to 58.
A has peaks corresponding to 29 (CH3CH2+) and 28 (loss of CH3CH2).
B has a peak corresponding to 43 (loss of CH3).
Mass / charge
C3H7+ (loss of CH3)
21 (a) ∆T = 43.2 − 21.2°C = 22.0°C
absolute uncertainty = (±)0.2°C
(b) % uncertainty = 0.2/22.0 × 100% ≈ 1%
(b) CH3CH2CH2CH3 31
Corresponding saturated non-cyclic molecule
(f) experimental value for ∆H = −184 (±) 2 kJ mol−1
The literature value is outside this range.
The random errors involved in reading the thermometer do not account for this difference.
(c) ∆H = −4.18 × 22.0/0.500 = −184 kJ mol−1 (d) 1% (e) absolute uncertainty = 1/100 × 184 = (±) 2 kJ mol−1
There are systematic errors. The assumptions on which the calculation is based are not strictly valid. Some of the heat of reaction passes into the surroundings and the other uncertainties in the measurements cannot be ignored. It should also be noted that the standard value for ΔH refers to standard conditions of 298 K and 100 kPa.
25 Concentration of chromium (from graph for absorbance of 0.215) = 3.34 µg dm−3
35 (a) Empirical formula CH2O. Molecular formula C2H4O2. (b) IHD = 1
symmetric stretch IR active
asymmetric stretch IR active
symmetric bend IR active
53 Z08_CHE_SB_IBDIP_9755_ANS.indd 53
Answers 38 The polarity (of bond or molecule) changes as the bonds are bent or stretched. 39 Hex-1-ene shows an absorption in the range 1610–1680 cm−1 due to the presence of the C=C bond.
49 X-ray crystallography. 50 Monochromatic means all the X-rays have the same wavelength.
40 C–H bond 41 CH3OCH3 42 C
51 Hydrogen atoms have a low electron density.
44 (a) 2 (b) 1 (c) 1
45 The H atoms are in three different environments. There are three peaks in the 1H NMR spectrum.
52 The atoms must have a regular arrangement if an ordered diffraction pattern is to be produced. 53 (a) C6H5CH3
46 (a) CH3COCH2CH3 Type of Chemical No. of H Splitting hydrogen atom shift / ppm atoms pattern
Compound CH3CHO CH3COCH3
(c) The saturated non-cyclic compound is C7H16
Chemical shift / ppm
(b) The peak at 8.0 ppm corresponds to R–COOH. There is no splitting as there are no hydrogen atoms bonded to neighbouring carbon atoms.
IHD = 12(16 − 8) = 4 (the IHD of a benzene ring = 4)
1 Y1ave = Y2ave = 3
The peak at 1.3 ppm corresponds to a CH3 group. The peak is split into a triplet because there is a neighbouring CH2 group.
(−2×(−2)) + (−1×(−1)) + 0 + (1 × 1) + (2 × 2) (−2)2 + (−1)2 + 02 + 12 + 22 =1
2 Y1ave = Y2ave = 3
48 (a) Possible structures: CH3CH2COOH, CH3COOCH3, HCOOCH2CH3.
(b) Hydrogen atoms do not appear because of their low electron density
Challenge yourself Splitting pattern
No. of H atoms
Number of peaks
The angle of diffraction depends on the wavelength. If the X-rays have different wavelengths, different diffraction angles/pattern would be obtained. It would be impossible to match the angles with the wavelengths.
(−2 × 2) + (−1 × 1) + 0 + (1×(−1)) + (2×(−2)) (−2)2 + (−1)2 + 02 + 12 + 22 = −1
3 Y1ave = Y2ave = 3
(−2×(−2)) + (−1 × 2) + 0 + (1 × 1) + (2 × (−1)) 22 + 12 + 02 + 12 + 22 4−2+1−2 = 0.10 = 10
4 Saturated hydrocarbons have the general formula CnH2n+2
The peak at 4.3 ppm corresponds to the R–CH2–COO group. The peak is split into a quartet as there is a neighbouring CH3 group.
H atoms needed = 2n + 2 − p
Molecular structure: CH3CH2COOH
H2 molecules needed = IHD = 12(2n + 2 − p) For CnHpOq:
54 Z08_CHE_SB_IBDIP_9755_ANS.indd 54
Oxygen forms two covalent bonds. Comparing ethane, C2H6: C–H, to ethanol, C2H5OH: C–O–H, we see that the presence of O has no impact on the IHD:
only this compound would give a peak in the 9.4–10 ppm region / OWTTE [2 max]
(b) 2.5 ppm peak
IHD = 12(2n + 2 − p)
CH3–CO–CH3 also has hydrogen atoms on a carbon next to the >C=O group 
(c) (i) 1700–1750 cm−1 (>C=O)
Nitrogen forms three covalent bonds. Comparing C–H to C–N–H, we see that the presence of one N increases the IHD by 1:
(ii) 1610–1680 cm−1 (>C=C