Chemistry Paper 3 June 2018

1. A$3.27 \times 10^4$ Pa
2. B$2.37 \times 10^5$ Pa
3. C$8.19 \times 10^5$ Pa
4. D$2.37 \times 10^3$ Pa

1. AMethyl orange
2. BPhenolphthalein
3. CUniversal indicator
4. DStarch, added near the end point

1. A$2.10$ mol dm$^{-3}$
2. B$0.042$ mol dm$^{-3}$
3. C$0.105$ mol dm$^{-3}$
4. D$1.05$ mol dm$^{-3}$

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1. A$0.525 \times 10^{-3}$ mol
2. B$1.05 \times 10^{-3}$ mol
3. C$2.10 \times 10^{-3}$ mol
4. D$4.20 \times 10^{-3}$ mol

1. AOCl$^-$ + I$^-$ + 2H$^+$ $\rightarrow$ Cl$^-$ + HOI + H$_2$O
2. BOCl$^-$ + 2I$^-$ + H$_2$O $\rightarrow$ Cl$^-$ + I$_2$ + 2OH$^-$
3. COCl$^-$ + 2I$^-$ + 2H$^+$ $\rightarrow$ Cl$^-$ + I$_2$ + H$_2$O
4. D2OCl$^-$ + 2I$^-$ + 4H$^+$ $\rightarrow$ 2Cl$^-$ + 2I$_2$ + 2H$_2$O

1. AThree delocalised, equivalent N–O bonds with the charge spread over the oxygens
2. BThree dative bonds from the oxygens to the nitrogen
3. COne double bond, one single bond and one dative bond, all of different length
4. DThree identical single bonds with the charge on one oxygen

1. AAny attraction involving a hydrogen atom
2. BAn intermolecular attraction between a hydrogen bonded to N, O or F and a lone pair on N, O or F of a neighbouring molecule
3. CA covalent bond between hydrogen and a metal
4. DThe attraction between a hydrogen ion and a delocalised electron cloud

1. AHI < MgCl$_2$ < MgO
2. BMgCl$_2$ < MgO < HI
3. CHI < MgO < MgCl$_2$
4. DMgO < MgCl$_2$ < HI

1. Aattraction between the nucleus of one atom and the bonding electron pair of the other; repulsion between the two nuclei and between the electrons
2. Battraction between the two overlapping electron clouds only; repulsion between each atom's protons and neutrons in the nucleus
3. Cattraction between each atom's protons and neutrons; repulsion acting only between the two positively charged nuclei
4. Dattraction between the two positively charged nuclei only; repulsion between the two clouds of bonding electrons around them

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1. AMgO is least soluble because the oxide ion simply repels water molecules electrostatically; MgCl2 and HI are both essentially covalent in character, so the two dissolve to almost exactly the same extent in water
2. BSolubility here depends only on the molar mass of the formula unit, so the heaviest compound, HI, is always the least soluble of the three and MgO is always the most soluble one
3. CMgO, MgCl2 and HI are all equally soluble in principle, and the stated order only reflects differences in particle size after grinding, not any difference in the type of bonding present
4. DMgO has the smallest, most highly charged ions and so the highest lattice energy, making it least soluble; MgCl2's larger anion lowers its lattice energy so it is more soluble; HI is a covalent molecule that ionises readily in water, making it the most soluble

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1. AAll three
2. BNone, all three are different
3. COne
4. DTwo

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1. AMgN
2. BMg$_3$N$_2$
3. CMg$_2$N$_3$
4. DMg$_3$N

1. A$3\text{Mg} + \text{N}_2 \rightarrow \text{Mg}_3\text{N}_2$
2. B$2\text{Mg} + 3\text{N}_2 \rightarrow \text{Mg}_2\text{N}_6$
3. C$3\text{Mg} + 2\text{N}_2 \rightarrow \text{Mg}_3\text{N}_4$
4. DMg + N $\rightarrow$ MgN

1. ANitrogen is more reactive than oxygen at all temperatures
2. BMagnesium nitride is more stable than magnesium oxide
3. CThe very high burning temperature supplies enough energy to break the strong N$\equiv$N triple bond
4. DAir contains more nitrogen than oxygen, so nitrogen always reacts first

1. AMg + O $\rightarrow$ MgO
2. B4Mg + O$_2$ $\rightarrow$ 2Mg$_2$O
3. CMg + O$_2$ $\rightarrow$ MgO$_2$
4. D2Mg + O$_2$ $\rightarrow$ 2MgO

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1. AIt conducts only as a solid, because the rigid lattice channels charge along fixed, permanently ordered ion positions
2. BIt conducts electricity both as a solid and when molten, because ionic compounds conduct in the solid state in much the same way that metals do
3. CIt does not conduct electricity as a solid, because the ions are held in fixed positions in the lattice, but conducts when molten, because the ions become free to move
4. DIt does not conduct electricity in either state, because it contains no delocalised electrons anywhere in its ionic lattice structure

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1. AAtomisation enthalpies of the elements, plus ionisation energies, plus electron affinities, plus lattice energy
2. BOnly the lattice energy together with the ionisation energies, since the atomisation and electron affinity terms cancel each other out exactly
3. CAtomisation enthalpies alone, since ionisation energy and electron affinity are not really part of a Born-Haber cycle at all
4. DIonisation energies minus electron affinities only, ignoring both atomisation steps and the lattice energy term entirely

1. AThe ability of an atom to attract the bonding pair of electrons in a covalent bond towards itself
2. BThe energy needed to remove an electron from a gaseous atom
3. CThe tendency of an atom to lose electrons and form cations
4. DThe energy released when an atom gains an electron

1. AAtomic radius and shielding both increase down the group, so the nucleus attracts bonding electrons less strongly
2. BNuclear charge decreases down the group
3. CThe halogens become more metallic and gain electrons more easily
4. DBond energies increase down the group

1. A$+6$ in SO$_2$ and $+4$ in H$_2$S
2. B$+4$ in SO$_2$ and $-2$ in H$_2$S
3. C$+4$ in SO$_2$ and $+2$ in H$_2$S
4. D$-2$ in SO$_2$ and $+4$ in H$_2$S

1. Ahalf the distance between the nuclei of two different, unlike atoms joined together by a double bond
2. Bthe straight-line distance from the nucleus out to the outermost occupied electron orbital of the atom
3. Chalf the distance between the nuclei of two atoms of the same element joined by a single covalent bond
4. Dthe radius of the ion formed when the neutral atom loses all of its outer valence-shell electrons

1. AA white precipitate forms while the mixture itself becomes completely colourless as the reaction proceeds to completion
2. BNo visible change occurs at all, since electron transfer between iodide and sulphuric acid produces no observable physical effect
3. CA colour change to brown/black (or purple vapour) as I2 forms, along with steamy fumes and a rotten-egg smell
4. DThe mixture turns bright blue as the iodide ions are oxidised and the iodine formed dissolves into the surrounding solution

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1. AI$^-$ + H$_2$SO$_4$ $\rightarrow$ IO$^-$ + SO$_2$ + H$_2$O
2. B2I$^-$ + 2H$_2$SO$_4$ $\rightarrow$ I$_2$ + SO$_2$ + 2H$_2$O + SO$_4^{2-}$
3. C8I$^-$ + 5H$_2$SO$_4$ $\rightarrow$ 4I$_2$ + H$_2$S + 4H$_2$O + 4SO$_4^{2-}$
4. D2I$^-$ + H$_2$SO$_4$ $\rightarrow$ I$_2$ + SO$_3$ + H$_2$O

1. A2I$^-$ + 5H$_2$SO$_4$ $\rightarrow$ I$_2$ + H$_2$S + 4H$_2$O + 4SO$_4^{2-}$
2. B2I$^-$ + 2H$_2$SO$_4$ $\rightarrow$ I$_2$ + SO$_2$ + 2H$_2$O + SO$_4^{2-}$
3. C8I$^-$ + 5H$_2$SO$_4$ $\rightarrow$ 4I$_2$ + H$_2$S + 4H$_2$O + 4SO$_4^{2-}$
4. D8I$^-$ + 3H$_2$SO$_4$ $\rightarrow$ 4I$_2$ + S + 3H$_2$O + 3SO$_4^{2-}$

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1. A$\text{CO} + \text{O}_2 \rightleftharpoons \text{CO}_2$
2. B$\text{CO}_2 + \text{C} \rightleftharpoons 2\text{CO}$ only
3. C$3\text{CO} \rightleftharpoons \text{CO}_2 + 2\text{C}$
4. D$2\text{CO} \rightleftharpoons \text{CO}_2 + \text{C}$

1. AThe sulphate ion becomes smaller down the group
2. BBarium sulphate is more soluble, so it loses its water
3. CCation charge increases down the group, repelling water molecules
4. DCation size increases down the group, so charge density falls and fewer water molecules are attracted

1. A$11.3$ dm$^3$
2. B$22.7$ dm$^3$
3. C$28.3$ dm$^3$
4. D$56.6$ dm$^3$

1. A+2 and +4 states are equally stable, since the reaction is reversible
2. B+4 oxidation state is more thermodynamically stable than the +2 state
3. C+2 oxidation state is more thermodynamically stable than the +4 state
4. D0 oxidation state is less stable than the +2 state

1. ACO$_2$ + NaOH $\rightarrow$ NaHCO$_3$, forming the hydrogencarbonate salt
2. BCO$_2$ + 2NaOH $\rightarrow$ Na$_2$CO$_3$ + H$_2$O
3. CCO$_2$ + 2Na $\rightarrow$ Na$_2$CO$_3$, reacting directly with the metal
4. D2CO$_2$ + NaOH $\rightarrow$ NaCO$_3$ + CO, releasing carbon monoxide gas

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1. A2Sr(NO$_3$)$_2$ $\rightarrow$ 2SrNO$_2$ + 3O$_2$, keeping strontium bonded to nitrogen
2. BSr(NO$_3$)$_2$ $\rightarrow$ SrO + N$_2$O$_5$, releasing dinitrogen pentoxide gas
3. C2Sr(NO$_3$)$_2$ $\rightarrow$ 2SrO + 4NO$_2$ + O$_2$
4. DSr(NO$_3$)$_2$ $\rightarrow$ SrO + 2NO$_2$, releasing no separate oxygen gas at all

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1. AElectrophilic addition; (1,2-dibromoethyl)benzene
2. BNucleophilic substitution; bromobenzene
3. CElectrophilic substitution; 2-bromophenylethene
4. DFree-radical addition; 1,1-dibromoethylbenzene

1. ANitrobenzene and methanol
2. B2-nitrotoluene and 4-nitrotoluene
3. CBenzoic acid and nitrobenzene
4. D3-nitrotoluene and 3,5-dinitrotoluene

1. ABr2 is polarised as it approaches the C=C; the pi electrons attack one Br atom, forming a bromonium ion (or carbocation) intermediate and a Br- ion, which then attacks the adjacent carbon to give the dibromide
2. BA Br radical abstracts a hydrogen atom from the ring first, generating a carbon radical that then reacts with a second molecule of Br2 in a chain process
3. CBr- acts directly as a nucleophile and simply displaces a hydride ion from the double bond in a single, one-step substitution
4. DThe alkene's pi bond is protonated first by trace acid present in the water, and only then does Br- add on to the resulting carbocation

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1. Apurely acidic
2. Bpurely basic
3. Cneutral, since the two groups cancel
4. Damphoteric

1. Aa free-radical intermediate
2. Ba benzyne-type intermediate
3. Ca bromonium ion intermediate
4. Da carbanion intermediate

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1. AAlcohol and carboxylic acid
2. BNitrile and ether
3. CAmide and ester
4. DAmine and ketone

1. APhenylamine, methanol and carbon monoxide
2. BBenzene, ammonia and methanoic acid
3. CBenzoic acid and methanol
4. DPhenol, methylamine and carbon dioxide

1. AConc. H$_2$SO$_4$; then NaOH; then NaNO$_2$ at $70\,°$C
2. BCH$_3$Cl/AlCl$_3$; then KMnO$_4$; then NaNO$_2$/HCl at $0$–$5\,°$C
3. CBr$_2$/FeBr$_3$; then NH$_3$; then NaNO$_2$/HCl at $0$–$5\,°$C
4. DConc. HNO$_3$/H$_2$SO$_4$; then Sn/conc. HCl then NaOH; then NaNO$_2$/HCl at $0$–$5\,°$C

1. AAqueous bromine solution, used at room temperature
2. BDilute HCl (or NaOH), heated under reflux
3. CCold, dilute HCl, with no heating required at all
4. DConcentrated H2SO4, used cold at room temperature

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1. AAbove about 5-10 degrees C, the benzenediazonium salt decomposes, so the low temperature is needed to keep it stable
2. BThe reaction between NaNO2 and HCl does not occur at all above 5 degrees C
3. CLow temperature is needed only to slow the rate so the exotherm can be controlled, with no effect on the product itself
4. DPhenylamine is insoluble in water above 5 degrees C, so the reaction cannot proceed

1. ALiAlH4 in dry ether
2. BZn and dilute H2SO4 followed by distillation
3. CSn and concentrated HCl, then NaOH
4. DH2/Pd catalyst only, no acid needed

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1. AC$_{10}$H$_{12}$
2. BC$_{11}$H$_{14}$
3. CC$_{10}$H$_{14}$
4. DC$_9$H$_{12}$

1. AStep II: conc. H$_2$SO$_4$; Step III: acidified KMnO$_4$
2. BStep II: warm aqueous NaOH; Step III: acidified K$_2$Cr$_2$O$_7$, heat
3. CStep II: ethanolic KOH, reflux; Step III: LiAlH$_4$
4. DStep II: warm aqueous NaOH; Step III: NaBH$_4$

1. ACondensation polymerisation; it hydrogen-bonds with water, raising the boiling point and lowering the freezing point
2. BCondensation polymerisation; it is immiscible with water and forms a protective layer
3. CAddition polymerisation; it conducts heat better than any metal
4. DAddition polymerisation; it is volatile and evaporates readily

1. Astructural (positional) isomerism
2. Boptical isomerism
3. Cgeometric (cis-trans) isomerism
4. Dno isomerism, since they are the same compound

1. AFree rotation about a single C-C bond located immediately next to the aromatic ring system
2. BThe presence of a chiral carbon atom bonded to four completely different substituent groups
3. CDelocalisation of the pi electrons spread evenly around the whole benzene ring system
4. DRestricted rotation about the C=C double bond, with two different groups attached to each of the double-bond carbons

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1. AElimination
2. BNucleophilic substitution
3. CElectrophilic addition
4. DFree-radical substitution

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