Chemistry Paper 3 June 2019

1. AGroup II, because the 2nd ionisation energy is large
2. BGroup VII, because the 1st ionisation energies are low
3. CGroup I, because a large jump occurs between the 1st and 2nd ionisation energies
4. DGroup III, because there are three elements listed

1. ABoron's outer electron is in a higher-energy, better-shielded 2p orbital, whereas beryllium's is in the 2s orbital
2. BBoron has a completely filled outer sub-shell
3. CBeryllium has more protons than boron
4. DBoron has a smaller nuclear charge than beryllium

1. A$\left(\dfrac{x}{4}\right)^{1/3}$
2. B$\sqrt{x}$
3. C$\left(\dfrac{x}{2}\right)^{1/3}$
4. D$x^{1/3}$

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1. AThe outer electron is removed from a shell that is progressively closer to the nucleus every time, so it experiences a stronger attraction and is held more tightly
2. BThe nuclear charge decreases down the group, weakening the pull on the outer electron
3. CEach successive element has fewer occupied electron shells, exposing the outer electron more directly to the nucleus
4. DAtomic radius and electron shielding both increase down the group, so the outer electron is less strongly attracted to the nucleus

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1. AAn s orbital is a flattened disc shape around the nucleus, and a p orbital is shaped like a figure-of-eight extending symmetrically outward
2. BAn s orbital is dumb-bell shaped with two lobes, and a p orbital is spherical
3. CAn s orbital is spherical, and a p orbital is dumb-bell shaped with two lobes directed along an axis
4. DBoth the s and p orbitals are spherical, differing only in their radius

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1. ANaOH raises the temperature of the solution, which lowers the solubility of most ionic solids
2. BNaOH provides the common ion OH-, which shifts the Fe(OH)2 solubility equilibrium to the left and lowers its solubility
3. CNaOH lowers the ionic product of water, which directly reduces how much Fe(OH)2 can dissolve
4. DNaOH reacts with the Fe2+ ions already present to form a completely different insoluble complex that has nothing to do with the original hydroxide equilibrium

1. A1s2 2p3
2. B1s2 2s2 2p1
3. C1s1 2s2 2p2
4. D1s2 2s1 2p2

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1. A$+2.46$ V
2. B$-1.23$ V
3. C$+0.62$ V
4. D$+1.23$ V

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1. A$6.18 \times 10^{23}$ mol$^{-1}$
2. B$1.24 \times 10^{24}$ mol$^{-1}$
3. C$6.02 \times 10^{22}$ mol$^{-1}$
4. D$3.09 \times 10^{23}$ mol$^{-1}$

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1. ACopper only dissolves at the cathode
2. BThe anode is heavier and easier to weigh
3. CThe cathode deposit may be loose or contain trapped impurities, whereas the anode loss corresponds exactly to the copper dissolved
4. DThe anode does not change mass, so it is a good control

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1. Arequire no maintenance whatsoever and can run indefinitely without any fuel supply
2. Bproduce large quantities of carbon dioxide but none of the other pollutants that fossil fuels release
3. Cgenerate electricity through a combustion reaction that is generally considered easier to control safely than a fuel cell's electrochemical reaction
4. Dproduce only water as a product and convert chemical energy to electrical energy more efficiently than combustion

1. Aneed a continuous supply of hydrogen gas, which is difficult and costly to store and transport safely
2. Brelease large amounts of carbon monoxide, unburnt hydrocarbons and other harmful pollutants into the surrounding atmosphere
3. Ccannot generate a usable electrical current for more than a few seconds at a time
4. Dproduce a strongly acidic waste product that must be neutralised before disposal

1. ABubbles of oxygen gas are produced vigorously at both the anode and the cathode continuously throughout the entire electrolysis process
2. BThe cathode loses mass steadily while the anode gains an equal mass of copper
3. CThe blue colour of the solution fades to colourless as copper is permanently removed from the circuit
4. DThe blue colour of the solution stays roughly constant while the anode loses mass and the cathode gains a copper coating

1. AIt dissolves away entirely within the first few seconds of electrolysis, well before the cathode shows any visible change
2. BIt gains mass as copper is deposited onto it from solution
3. CIt remains completely unaffected while only the cathode changes
4. DIt loses mass as copper atoms are oxidised and dissolve into solution as Cu2+ ions

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1. AIodine is consumed in the rate-determining step
2. BThe slow, rate-determining step is the keto-enol tautomerism, which does not involve iodine
3. CIodine is not involved in the reaction at all
4. DThe reaction has only one step

1. As$^{-1}$
2. Bmol dm$^{-3}$ s$^{-1}$
3. Cmol$^{-2}$ dm$^{6}$ s$^{-1}$
4. Dmol$^{-1}$ dm$^{3}$ s$^{-1}$

1. AThe particles are too large to touch
2. BThe collision energy is below the activation energy, or the orientation is unfavourable
3. CThe temperature is always too low
4. DThe catalyst has been used up

1. AMeasure the total volume of gas produced at the end of the reaction only
2. BWeigh the sealed reaction flask carefully once at the very start and once again at the very end of the whole timed reaction period
3. CMeasure the melting point of the reaction mixture at regular time intervals
4. DWithdraw samples at time intervals and titrate the remaining iodine against standard sodium thiosulphate solution

1. AThe slow, acid-catalysed conversion of the keto form of butanone into its enol form, before any iodine reacts
2. BThe fast reaction between the enol form of butanone and iodine, which quickly gives the iodo-ketone product and HI
3. CThe dissociation of hydrogen iodide, HI, into hydrogen and iodine gas
4. DThe re-formation of the keto form of butanone from its enol tautomer after the reaction is complete

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1. AA horizontal line, since the rate is independent of the reactant's concentration
2. BA curve that bends upward more steeply as concentration increases
3. CA straight line with a negative gradient, since rate falls as concentration rises
4. DA straight line through the origin, since the rate is directly proportional to concentration

1. AA horizontal line, since the rate does not depend on the reactant's concentration at all under these conditions
2. BA straight line through the origin, since rate is directly proportional to concentration
3. CAn upward-curving line (parabola), since rate is proportional to the square of the concentration
4. DA straight line with a negative gradient as concentration increases

1. AThe reactants change into a completely different set of chemical substances that happen to react together faster than the originals did
2. BA greater proportion of particles have energy exceeding the activation energy, and collisions become more frequent
3. CThe activation energy of the reaction is permanently lowered whenever the temperature rises
4. DParticles become larger at higher temperature, making collisions between them more likely

1. ATrigonal pyramidal; boiling point rises because of hydrogen bonding
2. BTetrahedral; boiling point rises because larger molecules have more electrons and stronger van der Waals forces
3. CTetrahedral; boiling point rises because the covalent bonds get stronger
4. DSquare planar; boiling point rises because the bonds become more ionic

1. ACCl$_4$ is ionic while SiCl$_4$ is covalent
2. BCCl$_4$ is insoluble, so hydrolysis is impossible in principle
3. CCarbon is small and has no available d orbitals, so water cannot coordinate to expand its octet
4. DThe C–Cl bond is weaker than the Si–Cl bond

1. AAcidic only, because it reacts with NaOH
2. BAmphoteric, because germanium lies on the metal/non-metal borderline and the oxide reacts with both acids and alkalis
3. CBasic, because germanium is a metal
4. DNeutral, because germanium is a semiconductor

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1. AStrongly basic -- it reacts readily with acids to form salts and water
2. BAmphoteric -- it reacts with both acids and alkalis to form salts
3. CNeutral -- it does not react with either acids or alkalis to form a salt
4. DStrongly acidic -- it reacts readily with alkalis to form carbonate salts

1. APb(IV) in PbO2 is a strong oxidising agent and oxidises Cl- ions to Cl2 while itself being reduced to Pb(II)
2. BPbO2 acts purely as a base, and the chlorine comes from a side reaction with atmospheric oxygen
3. CThe reaction produces chlorine simply because lead metal is more reactive than chlorine and displaces it directly from the acid
4. DConcentrated HCl decomposes into chlorine gas and hydrogen gas whenever it is gently warmed in the open air

1. AThe chlorides become more ionic across the period
2. BThe relative molecular masses decrease across the period
3. CThe bonding changes from giant ionic (strong electrostatic forces) to simple molecular (weak van der Waals forces)
4. DThe chlorides become more soluble across the period

1. A$\text{PCl}_5 + 3\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_3 + 5\text{HCl}$
2. B$\text{PCl}_5 + 5\text{H}_2\text{O} \rightarrow \text{P(OH)}_5 + 5\text{HCl}$
3. C$\text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}$
4. D$\text{PCl}_5 + \text{H}_2\text{O} \rightarrow \text{POCl}_3 + 2\text{HCl}$

1. AStrontium
2. BBarium
3. CMagnesium
4. DCalcium

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1. AA green flame, and formation of a black solid residue
2. BA bright yellow/orange flame, and formation of a white solid
3. CA blue flame, and formation of a strong-smelling gas only
4. DNo visible flame at all, and formation of a colourless liquid

1. AIt lowers the water activity around microbial cells by osmosis, dehydrating and killing the microorganisms
2. BIt directly kills bacteria and other microorganisms by reacting chemically with the proteins in their cell walls
3. CIt raises the pH of food enough to prevent any microorganism from surviving on its surface
4. DIt coats the surface of food in an airtight layer that stops oxygen reaching microorganisms

1. AAluminium chloride simply contains more chlorine atoms per formula unit than sodium chloride
2. BAl3+ is small and highly charged, so it polarises water molecules enough to release H+ ions on hydrolysis, unlike Na+
3. CAlCl3 reacts with atmospheric carbon dioxide in solution, producing carbonic acid
4. DSodium chloride is actually a stronger acid than aluminium chloride, so its aqueous solution must be the more acidic one

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1. AElectrophilic addition; H$^+$ attacks the carbonyl oxygen
2. BNucleophilic substitution; CN$^-$ displaces the oxygen
3. CNucleophilic addition; CN$^-$ attacks the electrophilic carbonyl carbon, breaking the C=O $\pi$ bond
4. DFree-radical addition; a CN radical adds to the carbonyl carbon

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1. Ax: acidified KMnO$_4$/K$_2$Cr$_2$O$_7$, warm; y: NaBH$_4$ (or LiAlH$_4$)
2. Bx: NaBH$_4$; y: acidified K$_2$Cr$_2$O$_7$
3. Cx: Tollens' reagent; y: H$_2$/Ni at 150 $°$C only
4. Dx: concentrated H$_2$SO$_4$; y: aqueous NaOH

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1. AAdd 2,4-DNPH: H gives an orange precipitate, I does not
2. BAdd sodium hydrogencarbonate solution: H effervesces (CO$_2$), I gives no reaction
3. CAdd bromine water: H decolourises it, I does not
4. DAdd Tollens' reagent: I gives a silver mirror, H does not

1. ACN- attacks the electrophilic carbonyl carbon, breaking the C=O pi bond to give an alkoxide, which is then protonated
2. BCN- substitutes directly for a hydrogen atom on the carbon chain next to the carbonyl group
3. CThe carbonyl carbon and CN- form a bond by free-radical addition initiated by ultraviolet light
4. DH+ from HCN attacks the carbonyl oxygen atom of the aldehyde first, and CN- then substitutes directly for that oxygen atom entirely afterwards

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1. AAddition, since the two molecules simply join together directly with no small molecule lost from either one
2. BSubstitution, since a hydrogen atom on H is simply replaced by a fragment of I
3. CCondensation (esterification), since an ester link forms with the elimination of small molecules of water
4. DElimination, since a small molecule is removed from within a single reactant only

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1. AIt forms covalent network bonding in the solid
2. BIt has a very high relative molecular mass
3. CIt contains a ring, which makes the molecule rigid
4. DIt exists as a zwitterion, so strong ionic attractions between the ions must be overcome

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1. AProline cannot act as a buffer because it has only one acidic group
2. B$-\text{COO}^- + \text{H}^+ \rightarrow -\text{COOH}$ and $^+\text{NH}_2 + \text{OH}^- \rightarrow \text{NH} + \text{H}_2\text{O}$
3. C$-\text{COOH} + \text{H}^+ \rightarrow -\text{COOH}_2^+$ and $\text{NH} + \text{OH}^- \rightarrow \text{NHOH}^-$
4. D$-\text{COO}^- + \text{OH}^- \rightarrow -\text{COO}_2^{2-}$ and $^+\text{NH}_2 + \text{H}^+ \rightarrow \text{NH}_3^{2+}$

1. AA cation only, with both the ring nitrogen and the carboxyl group carrying a positive charge
2. BAn anion only, with both the ring nitrogen and the carboxyl group carrying a negative charge
3. CA zwitterion, with the ring nitrogen protonated (+NH2 in the ring) and the carboxyl group as -COO-
4. DA fully neutral molecule overall, with an uncharged ring nitrogen atom and an uncharged, undissociated -COOH group

1. ABr2 adds directly across a carbon-carbon double bond somewhere in the ring system, by a typical electrophilic addition mechanism
2. BThe imide rings of thalidomide are completely hydrolysed by the bromine water
3. CNo reaction occurs, since thalidomide's rings contain no sites reactive towards bromine
4. DElectrophilic substitution occurs on the benzene ring of the phthalimide portion, giving a bromo-substituted product and HBr

1. AThe benzene ring of the phthalimide group becomes completely saturated by the direct addition of hydrogen atoms right across it
2. BNo observable reaction occurs, since amide linkages are unaffected by aqueous acid
3. CThe molecule is reduced, converting its ring nitrogen atoms into ring oxygen atoms
4. DThe amide/imide linkages in its rings are hydrolysed, breaking the rings open into carboxylic acid and amine fragments

1. AIt has a chiral carbon atom bonded to four different groups, so it exists as a pair of enantiomers
2. BIt can lose or gain a water molecule reversibly, giving two structurally different compounds
3. CIt has two separate rings that can rotate freely relative to each other, giving conformational isomers only
4. DIt contains a carbon-carbon double bond that can exist in cis and trans geometric forms

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1. ADissolve the sample in water only and measure how long it takes to fully dissolve at room temperature
2. BAdd the sample to concentrated sulphuric acid and observe carefully whether or not it chars on contact, since this simple test can sometimes help narrow down which functional groups a compound contains
3. CMeasure the melting point of the sample only, since solubility itself cannot help identify an unknown compound
4. DTest its solubility in a range of solvents (e.g. water, dilute acid, dilute alkali, an organic solvent) and compare the pattern with reference data for thalidomide

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1. APhenylalanine and ethanoic acid
2. BGlycine, alanine and ethanol
3. CAspartic acid, phenylalanine and methanol
4. DAspartic acid and methanol only

1. AAll three have the same basic strength
2. BC$_6$H$_5$NH$_2$ > NH$_3$ > CH$_3$CH$_2$NH$_2$
3. CNH$_3$ > CH$_3$CH$_2$NH$_2$ > C$_6$H$_5$NH$_2$
4. DCH$_3$CH$_2$NH$_2$ > NH$_3$ > C$_6$H$_5$NH$_2$

1. ACondensation polymers are lighter, so they take up less space
2. BCondensation polymers dissolve completely in rainwater
3. CAddition polymers release toxic gases in landfill
4. DCondensation polymers contain hydrolysable ester/amide links so they are biodegradable, whereas addition polymers have inert C–C backbones that persist

1. AAn ester group and an amide group
2. BA ketone group and a nitrile group
3. CAn alkene group and a halogenoalkane group
4. DA nitro group and a sulphonic acid group

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1. AEthylamine has a larger relative molecular mass than ammonia, and this alone is sometimes mistakenly thought to explain basic strength
2. BThe ethyl group pushes electron density onto the nitrogen atom, making its lone pair more available to accept a proton
3. CEthylamine's nitrogen lone pair is delocalised into a pi system, making it more available to accept a proton
4. DAmmonia's nitrogen atom carries a small partial positive charge that is sometimes said to repel any incoming proton in solution

1. APhenylamine has no lone pair of electrons on its nitrogen atom at all
2. BThe benzene ring withdraws electron density from nitrogen through an inductive effect only, with no delocalisation involved
3. CPhenylamine's nitrogen atom already carries a full positive charge before any proton is added
4. DThe nitrogen lone pair is delocalised into the benzene ring's pi system, making it less available to accept a proton

1. AA condensation polymer has no obvious repeat unit; an addition polymer has a clearly defined repeat unit in every case
2. BA condensation polymer forms with elimination of a small molecule and contains hydrolysable links; an addition polymer forms from C=C monomers with no small molecule lost
3. CA condensation polymer generally contains additional elements such as nitrogen or oxygen beyond carbon and hydrogen in its backbone; an addition polymer is typically made of carbon and hydrogen only, with no such extra heteroatoms present as a rule
4. DA condensation polymer is typically considered a natural material; an addition polymer is typically considered synthetic and man-made

1. AMagnesium is less reactive than iron, so it protects it from the soil
2. BMagnesium is more reactive than iron, so it acts as a sacrificial anode and corrodes in preference to the pipe
3. CMagnesium forms an impermeable coating over the iron surface
4. DMagnesium reacts with water to release hydrogen that displaces oxygen

1. AThe scratch lets air into a sealed can
2. BTin is more reactive than iron and protects it sacrificially
3. CTin dissolves in the food acids and exposes the iron
4. DTin is less reactive than iron, so the exposed iron becomes the anode and corrodes faster

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1. AN $=$ Cu(OH)$_2$; M $=$ [Cu(NH$_3$)$_4$(H$_2$O)$_2$]$^{2+}$
2. BN $=$ CuO; M $=$ [Cu(NH$_3$)$_6$]$^{2+}$
3. CN $=$ CuCl$_2$; M $=$ [CuCl$_4$]$^{2-}$
4. DN $=$ Cu(NH$_3$)$_2$; M $=$ [Cu(H$_2$O)$_6$]$^{2+}$

1. AZinc forms an unreactive alloy with iron at the scratch that cannot be attacked by oxygen or water
2. BZinc is more reactive than iron, so it continues to corrode preferentially and protects the exposed iron (sacrificial protection)
3. CThe scratch reseals itself almost immediately as a fresh, protective layer of zinc oxide reforms and fully covers the exposed area of iron
4. DIron is more reactive than zinc, so the iron corrodes instead and protects the remaining zinc coating

1. ASteam is bubbled through the cold mixture only, with no heating applied at any stage of the process
2. BThe mixture is frozen and then sublimed directly, with phenylamine collected as a solid on a cold surface
3. CThe mixture is heated strongly and continuously under reflux until phenylamine alone eventually boils off directly at its own true boiling point, unaided by steam
4. DSteam is passed into the heated mixture; phenylamine co-distils with the steam below its normal boiling point, and the vapours are condensed and collected

1. APhenylamine has a high boiling point and would decompose or oxidise if heated strongly enough to distil it directly
2. BPhenylamine reacts explosively with glass apparatus when heated above room temperature
3. CPhenylamine is actually completely insoluble in the reaction mixture at any temperature and simply cannot be vaporised under any laboratory conditions
4. DSimple distillation can only ever separate a solid from a liquid, not two liquids with reasonably close boiling points from each other

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1. AA bright orange precipitate forms and effervescence is observed
2. BA deep blue solution forms directly with no precipitate stage
3. CA pale blue precipitate forms as Cu(OH)2 is produced
4. DThe solution turns immediately colourless with no precipitate at all

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1. AA $=$ graphite, B $=$ diamond; A is a nanomaterial because it conducts electricity
2. BA $=$ fullerene, B $=$ diamond; A is a nanomaterial because it is a molecule
3. CA $=$ carbon nanotube, B $=$ graphite; A is a nanomaterial because its diameter lies in the $1$–$100$ nm range
4. DA $=$ graphene, B $=$ charcoal; A is a nanomaterial because it is only one atom thick

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1. AThey are radioactive and emit ionising radiation
2. BThey dissolve in water, contaminating drinking supplies
3. CThey are chemically inert, so they cannot react with anything
4. DTheir very small size lets them penetrate skin, lungs and cells, and their large surface area makes them reactive, so toxicity and long-term effects are uncertain

1. AIt generally makes waste disposal considerably cheaper overall for most households in the long run
2. BRecyclable materials can be recovered and reused, reducing the demand for new raw materials
3. CIt substantially reduces how much waste eventually needs to be sent to landfill sites over time
4. DIt reduces the need for some further waste treatment processes to be carried out afterwards

1. AHazardous or biodegradable waste can be separated out and treated safely, reducing pollution
2. BIt automatically converts all organic waste into clean drinking water
3. CIt prevents any landfill site from ever needing to be monitored again
4. DIt removes most of the pressure on households to reduce how much waste they actually produce day to day

1. ALandfill sites are sometimes claimed to increase the fertility of the surrounding soil, though this is not generally true
2. BLandfill sites produce no gases of any kind as waste decomposes
3. CLeachate produced within the landfill can contaminate groundwater supplies
4. DLandfill is the only waste disposal method that produces no odour at all

1. ALandfill waste rarely attracts significant numbers of pests or vermin under normal, well-managed conditions
2. BLandfill sites remove unpleasant odours from the surrounding area entirely
3. CDecomposing waste produces methane, a flammable greenhouse gas
4. DLandfill sites take up no land area, since waste is compacted to almost nothing

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