1 (a) Cr is +6 (7 × −2 = −14, but ion is 2−, so 2 × Cr = +12 or +6 each).
(b) I is +5 (3 × −2 = −6, but the ion is 1−, so I = +5).
2 3MnO42− + 4H+ → 2MnO4− + MnO2 + 2H2O
Note that the change in oxidation number going from MnO42− to MnO4− is +1 and that from MnO42− to MnO2 it is −2, so there must be three MnO42− ions, two of which are oxidised to MnO4− and one reduced to MnO2.
3 The oxidation number of manganese in MnO4− is +7 and in Mn2+ it is +2. That of carbon in C2O42− is +3 each or +6 in total, and in CO2 it is +4. The oxidation number of manganese decreases by 5 and that of each of the two carbon atoms increases by 1, so there must be two MnO4− ions and five C2O42− ions so that the total change of each is 10. In order to balance charge and oxygen atoms, there must be 16H+ on the left and 8H2O on the right. Thus the equation is:
2MnO4− + 5C2O42− + 16H+ → 2Mn2+ + 10CO2 + 8H2O
4 The electrode consists of a platinum plate dipping into a solution that is 1 mol dm−3 in MnO4−
ions, H+ ions and Mn2+ ions, all at a temperature of 25°C.
5 Item 64. Each nitrogen atom changes from +4 to +2, so the total change is 4 and hence there must be 4e− on the left of the half-equation.
N2O4(g) + 4H+(aq) + 4e− → (or ) 2NO(g) + 2H2O(l) E = +1.03 V
Item 67. Each iodine atom changes from +5 to 0, so the total change and the number of electrons is 10.
2IO3−(aq) + 12H+(aq) + 10e− → (or ) I2(aq) + 6H2O(l) E = +1.19 V
E cell = +1.19 − (+1.03) = +0.16 V, so iodate(V) ions will oxidise nitrogen(II) oxide because E cell is positive.
6 Either: using the data on page 17 ½Cl2 + e− Cl−
E = +1.36V
HClO + H+ + e− ½Cl2 + H2O E = +1.59 V
Reversing the first equation and adding it to the second gives the overall equation.
The value is positive, so the reaction is feasible.
Or: using items 70 and 76
Cl2(aq), 2Cl−(aq) E = + 1.36 V
[2HClO(aq) + 2H+(aq)], [Cl2(aq) + 2H2O(l)] E = +1.59 V
Using the anticlockwise rule, HClO, in acid solution, (on the left of the lower data) will oxidise Cl− ions (on the right of the upper data) to Cl2 with E cell = +1.59 − (+1.36) = +0.23 V.
The value is positive, so the reaction is feasible.
As the change in oxidation number is 1 per chlorine atom in both, you need 2HClO for each Cl2 so that both half-equations will have 2e−, so the overall equation is:
2HClO(aq) + 2H+(aq) + 2Cl−(aq) → 2Cl2(aq) + 2H2O (This can now be divided by 2.)
7 Either: using item 33
The E value given is less positive than that for VO2+ to VO2+ and for VO2+ to V3+ but more positive than that for V3+ to V2+, so H2S will reduce VO2+ to VO2+ and then to V3+ but no further.
Or: using the half-equation from page 17 and reversing it H2S S + 2H+ + 2e−
E = −(+0.14) = −0.14 V
Adding this to the 5→4 half-equation in the table on page 14 which has E = 1.00 V gives a positive E cell and so reduction occurs.
Likewise adding it to the 4→3 half-equation also gives a positive E cell, so reduction will go to V3+.
However, adding it to the 3→2 half-equation gives a negative E cell and so H2S will reduce VO2+ ions to V3+ but no further.
8 amount of thiosulfate in titre
= 0.100 mol dm−3 × 0.02645 dm3 = 0.002645 mol
amount of iodine released = ½ × 0.002645
= 197 g mol−1
% of ethanedioic acid in rhubarb leaves =
= 1.20%
10 Not all the ethanedioic acid was extracted from the leaves.
There might be other substances in the leaves that will react with manganate(VII) ions.
11 The overall reactions are the same, so the E cell values will be the same.
12 (a) Cr3+ is [Ar], 3d3 and (b) Mn2+ is [Ar], 3d5 (you can add 4s0 to both if you wish).
13 (a) As there are no ligands in anhydrous copper(II) sulfate, the d-orbitals are not split. This means that light cannot be absorbed.
(b) The electronic configuration of copper(I) ions is [Ar], 3d10. Even though the five d-orbitals are split in energy, all five are full and so no d–d transitions can take place.
14 The ring that would have to form between the two nitrogen atoms and the central metal ion would only contain four atoms. The resulting strain would make its formation energetically impossible.
15 Using method 1 on page 10:
E cell = E (oxidising agent) − E (reducing agent) = E (Cr3+/Cr2+) − E (Zn2+/Zn) = (−0.41) − (−0.76) = +0.35 V. This is a positive number and so zinc will reduce Cr3+ ions to Cr2+.
Using the anticlockwise rule:
Zn2+ + 2e− Zn E = −0.76 V Cr3+ + e− Cr2+
E = −0.41 V
Going anticlockwise from bottom left: Cr3+ will be reduced by Zn and E cell = −0.41 − (−0.76) = + 0.35 V. This is a positive number so the reaction will happen.
16 Using method 1 on page 10:
E cell = E (oxidising agent) − E (reducing agent) = E (Fe3+/Fe2+) − E (Cr2O72−/Cr3+) = +0.77
− (1.33) = −0.56V. This is a negative number, so Fe3+ ions will not oxidise Cr3+ ions in acid solution.
Or using the anticlockwise rule (more difficult this time):
Fe3+ + e− Fe2+
E = +0.77 V
Cr2O72− + 14H+ + 6e− 2Cr3+ + 7H2O E = +1.33 V
Going anticlockwise from bottom left: Cr2O72− ions would react with Fe2+ ions with E cell = +0.56 V, so the opposite reaction of Cr3+ being oxidised by Fe3+ will not happen.
17 It is not. Although chlorine is both oxidised and reduced, the two chlorine atoms are in different species. This type of reaction is sometimes called reverse disproportionation.
18 The bond energy calculation uses average bond enthalpies, so is less accurate than that calculated from enthalpy of hydrogenation.
19 The methyl group is an electron-releasing (electron pushing) group and so increases the electron density in the ring. This makes the attack by electrophiles have a lower activation energy.
20 Ethylamine and propane have the same number of electrons and hence similar strength London forces. However, ethylamine also forms strong intermolecular hydrogen bonds. Thus more energy is required to separate ethylamine molecules and so it has a higher boiling temperature.
21 The volatile ethylamine reacts with acids to form an involatile ionic salt. Thus the smell disappears. When alkali is then added, the volatile ethylamine is liberated and so the smell returns. The equations are:
C2H5NH2 + HCl → C2H5NH3+Cl−
C2H5NH3+Cl− + OH− → C2H5NH2 + H2O
22 If the temperature drops below 0°C, then the reaction is too slow. If it is allowed to rise above 10°C, the diazonium compound decomposes.
23 By adding aqueous strong acid such as dilute sulfuric acid. The equation is:
RCOO− + H+ → RCOOH 24
25 +NH3CH2COO− + OH− → NH2CH2COO− + H2O
26 NH2CH2CONHCHCH(CH3)2COOH and NH2CHCH(CH3)2CONHCH2COOH
27 CH2OHCH=CHCHO (or CH3C(OH)=CHCHO or CH2=C(CHO)CH2OH). It must have a C=C, an OH and an aldehyde group.
28 The peak at 1740 cm−1 indicates an ester and the absence at between 3300 cm−1 and 2500 cm−1 shows that it is not an acid (or alcohol). The two singlets mean that there are no hydrogen atoms on adjacent carbon atoms, so the compound is CH3COOCH3 or methyl methanoate.
29 Step 1. Add a mixture of concentrated nitric and sulfuric acids at 50°C.
Step 2. Warm the nitrobenzene formed in step 1 with tin and concentrated hydrochloric acid then add alkali and steam distil off the phenylamine formed.
Step 3. Add ethanoyl chloride, CH3COOCl.
30 (a) The distillation apparatus
(b) The heat under reflux apparatus
Index
Note: bold page numbers indicate definitions.
A
acids
acid fuel cells 9 amine reactions 39 hydroxyl group test 48 acyl chlorides 36, 39, 40, 60 addition polymers 43–44 addition reactions 30, 33, 50 alcohols ammonia solution 27, 28, 29 amphoteric hydroxides 29 analysis
functional groups 47–50 molecular formulae 46–47 anticlockwise rule 11
apparatus diagrams 6, 52–53, 55, 56 aqueous sodium hydroxide 28–29, 48, 58 arenes (aromatic compounds) 30–31 61
benzene 31–36
benzene reactions 33–36, 41, 61 bidentate ligands 23
boiling point 50, 55–56
bond enthalpies, benzene 32–33 breathalysers 19
bromine reactions 8, 33–34, 37
C
coloured complex ions 14, 22–23 combinatorial chemistry 56complex ions 22–24
d-block elements 20 see also transition elements delocalisation stabilisation energy 32
delocalised electrons 31, 34, 36 deprotonation 28–29
diazo compounds 38, 40–41 disproportionation 8, 27 distillation 53–54
E
E cell (standard reduction potential) 9 calculation of 10–12
overall redox equations 12–13 predicted and actual value 13–14 EDTA 24, 27, 77–78
electrode potentials 9
electromagnetic radiation 80 electron configurations 20–21
electrophile 31, 34
electrophilic substitution 33–35, 37 elimination reactions 30–31, 42 empirical formulae 47
enthalpy experiments, measurement errors 18 enthalpy of formation 32–33
enthalpy of hydrogenation 32 equations 5, 12–13, 57–61 errors in measurement 17–18 esters 60
ethanoic acid, synthesis of 51 ethanol in breath 19
ethanoyl chloride 36, 40, 41, 58, 60, 61 examination guidance 63–64
experimental techniques 52–54 experiments, risk in 51–52 extent of a reaction 10
F
feasibility of a reaction 10–13 formulae 46–47, 62
free radical 31
Friedel–Crafts reaction 35–36 fuel cells 18–19
fuming sulfuric acid 35, 61 functional-group analysis
synthesis 50, 51 test for 48
heating under reflux 52 hydrogenation enthalpy 32 hydrogen fuel cells 19
hydrolysis, nitriles 41, 51, 60 hydroxyl groups 48
I
infrared spectroscopy 49, 80, 81 benzene structure 33
use in breathalysers 19 iodine titrations 15–16 iodoform reaction 49, 51 ionisation energy 21, 22
K
ketones 58–59
tests identifying 48–49 Kevlar® 42, 69
L
learning skills 5–6
ligand exchange reactions 27–28, 29, 39, 40 ligands 20, 23–24
liquids, purification of 54–55
M
measurement errors 17–18 melting point 50, 55
molecular formulae 46–47 monodentate ligands 23
multiple-choice questions 65–70
N
NH2 group reactions 40–41 ninhydrin 45, 46
nitrogen compounds see organic nitrogen compounds nitrous acid, phenylamine reaction 40–41
NMR spectral analysis 49, 80–83 non-standard conditions, redox 13–14 oxidation number 7, 8, 10
oxidation states, changes in 14, 21–22, 24 oxidising agents 7, 8, 10–11
concentration of 15–16
P
plane-polarised light, amino acids 44 polyamides 42
potassium manganate(VII) 16–17 proteins 45 reducing agents 7, 8, 10–11
concentration of 16–17 vanadium ions 14–15 reduction 7, 10
chromium 25–26
vanadium 14–15
reduction potentials 9, 10, 14–15, 25 reflux distillation 52
repeat units, polymers 42, 43 resonance energy 32, 33 resonance hybrid 31
risk in experiments 51–52
S
shapes of complex ions 24
sodium thiosulfate titrations 15–16 18 solids, purification of 54
solvent extraction 54
standard electrode (reduction) potential 9, 10, 14–15, 25 standard hydrogen electrode 9
steam distillation 53–54 stereospecific reactions 52 stoichiometry 8
structural formulae 62
substitution reactions 30, 33–34, 37, 50 sulfuric acid reactions 34–35
thermodynamically spontaneous (feasible) reactions 10–11, 12–13 thin-layer chromatography 45
ligand exchange 27–28 properties of 21–24 reactions of 25–27 uses of 30
U
UV light/radiation 30, 80, 82
V
vanadium compounds 14–15, 24
W
water reactions
addition polymers 43 amines 38
metal ions 28
weighing errors 17–18
X
X-ray diffraction 33
Z
zwitterions 44