Chemistry Paper 2 November 2005

1. Aa curve which rises steeply at low pressure
2. Ba horizontal straight line
3. Ca straight line through the origin
4. Da curve which passes through a minimum

1. Aneon
2. Bnitrogen
3. Call deviate equally
4. Dammonia

1. Aare monatomic and are highly polar
2. Bare small and attract one another only very weakly
3. Care held together by hydrogen bonds
4. Dhave a full outer shell of eight electrons

Tap to enlarge

1. AA = neon, B = ammonia, C = nitrogen
2. BA = nitrogen, B = ammonia, C = neon
3. CA = ammonia, B = neon, C = nitrogen
4. DA = neon, B = nitrogen, C = ammonia

1. Avan der Waals forces decrease with molecular size
2. Bthe number of $N-H$ bonds available for hydrogen bonding falls
3. Cthe relative molecular mass falls along the series
4. Dthe tertiary amine is the most basic

1. A$CH_3NH_2 \rightleftharpoons CH_3NH^- + H^+$
2. B$CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-$
3. C$CH_3NH_2 + H^+ \rightarrow CH_3NH_3^+$
4. D$CH_3NH_2 + H_2O \rightarrow CH_3OH + NH_3$

1. Aa hydrogen bond
2. Ba non-polar covalent bond
3. Cionic
4. Da dative covalent bond

1. A$AlO_3$, one aluminium ion to three oxide ions
2. B$AlO$, one aluminium ion to one oxide ion
3. C$Al_2O_3$, two aluminium ions to three oxide ions
4. D$Al_3O_2$, three aluminium ions to two oxide ions

1. Aa dative covalent bond; 8 electrons
2. Ban ionic bond; 8 electrons
3. Ca dative covalent bond; 6 electrons
4. Da normal covalent bond; 8 electrons

1. A$-1287$
2. B$-597$
3. C$-261$
4. D$-75$

1. A$Rb^+$ carries a higher charge than $Fr^+$
2. Brubidium has a lower first ionisation energy
3. Cfrancium is radioactive
4. D$Rb^+$ is smaller, so the ions are closer together

1. Aone mole of gaseous chlorine atoms each gain an electron
2. Bone mole of $Cl_2$ molecules each gain two electrons
3. Cone mole of gaseous chloride ions each lose an electron
4. Done mole of chlorine atoms is formed from $Cl_2$

1. Athe sum of the atomisation, ionisation, electron affinity and lattice enthalpy steps combined
2. Bthe lattice enthalpy alone, since it is the largest single term in the cycle
3. Cthe first ionisation energy minus the electron affinity, with the atomisation steps left out
4. Dthe average of the two atomisation energies only, ignoring the ionisation and electron affinity steps

1. Athe molecules become larger, so van der Waals forces increase
2. Bthe electronegativity of the atoms increases
3. Chydrogen bonding becomes stronger
4. Dthe bond energy of the $X-X$ bond increases

1. A$Cl_2 + 2Cl^- \rightarrow 2Cl^- + Cl_2$
2. B$Br_2 + 2Cl^- \rightarrow 2Br^- + Cl_2$
3. C$Cl_2 + 2Br^- \rightarrow 2Cl^- + Br_2$
4. D$Cl_2 + 2Br^- \rightarrow 2BrCl$

1. Apurple, because iodine is displaced
2. Bcolourless, because the ether does not dissolve halogens
3. Corange-brown, because bromine dissolves in the non-polar solvent
4. Dpale green, because chlorine dissolves in the ether

1. AThe solution fades from orange back to colourless as the liberated bromine is immediately reduced again, and only a single layer forms once the mixture is shaken with ether
2. BThe solution turns orange-yellow as bromine is liberated by the more reactive chlorine, and two layers form because the non-polar ether is immiscible with water and dissolves the bromine
3. CThe solution stays colourless throughout because no reaction occurs, and the two layers form because ether reacts with the dissolved salts to form an insoluble solid barrier between them
4. DThe solution turns deep purple as iodine is liberated by the chlorine, and the two layers form because the denser iodine sinks below both liquids and separates them mechanically

1. Asodium hydroxide, widely used to manufacture soap and paper pulp
2. Bethanol, widely used as an industrial solvent and as a fuel additive
3. Cpoly(vinyl chloride), PVC, used to manufacture plastic piping
4. Dammonia, widely used to manufacture nitrogen-based fertiliser

1. A$SnCl_2$ and $PbCl_2$
2. B$SnCl_2$ and $PbCl_4$
3. C$SnCl_4$ and $PbCl_4$
4. D$SnCl_4$ and $PbCl_2$

1. Areacts with chlorine to give $SnCl_2$
2. Bis hydrolysed by water to $SnO_2$ and HCl
3. Cis insoluble in water
4. Dis oxidised by water to $SnO_3$

1. Athe chlorides become increasingly hydrolysed
2. Bthe chlorides become increasingly ionic
3. Cthe bonds become more polar
4. Dthe molecules become larger, giving stronger van der Waals forces

1. A$SnCl_4 + 2H_2O \rightarrow SnO_2 + 2HCl + Cl_2$
2. B$SnCl_4 + 2H_2O \rightarrow SnO_2 + 4HCl$
3. C$SnCl_4 + H_2O \rightarrow SnO + 4HCl$
4. D$SnCl_4 + 4H_2O \rightarrow Sn(OH)_4 + 4HCl$

1. A$Sn + 2Cl_2 \rightarrow SnCl_4$
2. B$Sn + Cl_2 \rightarrow SnCl_2$
3. C$2Sn + 3Cl_2 \rightarrow 2SnCl_3$
4. D$Sn + 4Cl_2 \rightarrow SnCl_4 + 3Cl_2$

Answer this when you sit the paper.

1. Aalways carries a negative charge
2. Bcontains an unpaired electron
3. Cdonates a pair of electrons
4. Daccepts a pair of electrons

1. A$NO_3^-$
2. B$NO_2^+$
3. C$NO_2^-$
4. D$HNO_3$

1. Arefluxing with acidified $KMnO_4$, then with ethanol and concentrated $H_2SO_4$
2. Bwarming with alkaline iodine, then with ethanol
3. Cnitrating, then reducing with tin and hydrochloric acid
4. Drefluxing with ethanol and concentrated $H_2SO_4$ only

1. Aconcentrated sulfuric acid reacting with aqueous sodium hydroxide instead of nitric acid
2. Bconcentrated nitric acid reacting directly with methylbenzene, with no sulfuric acid present
3. Cconcentrated nitric acid reacting with concentrated sulfuric acid
4. Ddilute nitric acid mixed only with cold water, without any acid catalyst

1. A0 overall, since the positive and negative charges exactly cancel each other out
2. B-1, localised entirely on the single carbon atom that was attacked by the electrophile
3. C+2, shared equally between exactly two adjacent ring carbon atoms only
4. D+1, delocalised over the remaining part of the ring by resonance

1. Aan aldehyde group; a reducing agent
2. Ba ketone group; a dehydrating agent only, with no catalytic role
3. Ca carboxylic acid group; a catalyst and dehydrating agent
4. Dan alcohol group; an oxidising agent

1. Aionic attractions
2. Bcovalent bonds
3. Cvan der Waals forces
4. Dhydrogen bonds

1. Asideways overlap of two parallel p orbitals
2. Bhead-on overlap of two $sp^2$ hybrid orbitals
3. Coverlap of an s orbital with a p orbital
4. Ddonation of a lone pair into an empty orbital

1. A$C_3H_6 + 5O_2 \rightarrow 3CO_2 + 3H_2O$
2. B$2C_3H_6 + 7O_2 \rightarrow 6CO + 6H_2O$
3. C$2C_3H_6 + 9O_2 \rightarrow 6CO_2 + 6H_2O$
4. D$C_3H_6 + 3O_2 \rightarrow 3CO_2 + 3H_2O$

1. Asigma: head-on overlap of two $sp^2$ orbitals; pi: sideways overlap of two parallel p orbitals
2. Bsigma: donation of a lone pair into an empty orbital; pi: head-on overlap of two p orbitals
3. Csigma: sideways overlap of two parallel p orbitals; pi: head-on overlap of two $sp^2$ orbitals
4. Dsigma and pi are both formed by head-on overlap of $sp^2$ orbitals, differing only in bond strength

1. A120 degrees and 109.5 degrees respectively
2. B109.5 degrees and 120 degrees respectively
3. C180 degrees and 109.5 degrees respectively
4. D120 degrees and 120 degrees respectively