DHS H2 CHEM P1 SOLUTIONS Prelim

2011 DHS H2 Chemistry Paper 1 (Solutions) © DHS 2011 1 9647/01(Solutions) 1 Answer: D Topic: mole concept Solution: CnH2n+2 + ((3n+1)/2)O 2 nCO2 + (n+1)H 2O Residual gas, V = CO2 and unreacted O2 at rtp. For 10 cm3 CH4  CO2 = 10 cm3 O2 unreacted = 70 – 20 = 50 cm3   V = 60 cm3 Only option D starts with 60 cm3. 2 Answer: A Topic: Mole Concept Solution: Ammonium cyanate  CO(NH2)2 Ammonium, NH 4 + To balance the atoms and charge: Cyanate should have 1 C, 1N, 1O; overall –1 charge. Hence, CNO–. 3 Answer: D Topic: Energetics Solution: A : O2 is a covalent molecule. B : Atomising O2 requires breaking strong O=O bond. C: Irrelevant to explai n existence of oxides. D: Refer to Born–Haber cycle. Lattice enthalpy is highly exothermic and it counterbalances the endothermic process including 1 st electron affinity. 4 Answer: D Topic: Energetics Solution: C(s) + O 2(g)  CO2(g) (1) Hc  (C(graphite)) CO 2(g)  CO(g) + ½O2(g) (2) –Hc  (CO) (1) + (2) (3) C(s) + ½O 2(g)  CO(g) (3) Hf  (CO) 5 Answer: B Topic: Electrochem Solution: Analysis: Ni oxidised, cation reduced. From data booklet, E ox Ni 2+ /Ni = – 0.25 V E Co 2+ /Co = – 0.28 V E Fe 3+ /Fe 2+ = + 0.77 V E Mn 2+ /Mn = – 1.18 V E V 2+ /V = – 1.20 V E cell is >0 in reaction with Fe3+. 6 Answer: A Topic: Kinetics, Ionic Eqm Solution: pH 3 [H+] = 10–3 mol dm–3 pH 1 [H+] = 10–1 mol dm–3 Given, reaction is 1st order wrt [H+], ratio of rate = 0.001/0.1 = 0.1 7 Answer: C Topic: Ionic Eqm Solution: When iodide is added to Hg 2+, mass of ppt (HgI2) increases as volume of iodide increases. As more iodide is added, mass of precipitate decreases due to formation of soluble complex, HgI4 2–. Decrease in [iodide] due to formation of complex in (2) causes eqm in (1) to shift left to produce more iodide. Hence ppt dissolves. [Concept is similar to PbI 2 dissolving in excess KI.] Hg2+ + 2I– HgI2 (1) HgI2 + 2I– HgI4 2– (2) 8 Answer: C Topic: Group II Solution: Decomposition of group II nitrate forms an oxide. Hence only A and C are possible answers: BaO is formed. However, question states that an unreactive gas is produced with heat evolved, N2 (high bond energy released N N) is the gas and not NO. Also, NO is rapidly oxidised by air to produce NO 2, hence unlikely to be the product in this reaction. 9 Answer: D Topic: Periodic Table Solution: A : Atomic size decreases across period  N > O > F B, C : Ionic radius across isoelectronic series decreases across period  N3– > O2– >F– Na+ > Mg 2+ > Al 3+ (ionic radius in Data Booklet) D: Na+, Ne and F – are isoelectronic (10 electrons) but nuclear charge is highest for Na+, followed by Ne than F– (refer to proton no.). Hence, electrostatic attraction for the valence electrons decreases from Na+ to Ne to F–.  size: Na+ > Ne > F– 10 Answer: C Topic: Periodic Table Solution: Y is phosphorus, X is silicon, Z is sulfur. Sulfur has lower 1 st IE than phosphorus due to inter–electron repulsion in 3p4 configuration. In terms of melting point: P 4 < S8 < Si Si is giant molecular compound. P 4 and S 8 has weak intermolecular vdw forces of attraction and S 8 has larger no. of electrons hence stronger intermolecul ar vdw forces than P4. 11 Answer: C Topic: Periodic Table Solution: Property 1: SiO 2 is insoluble in water while SO2/SO3 forms H2SO3/H2SO4. Property 2: Si, [Ne] 3s 2 3p2; S [Ne] 3s2 3p4 12 Answer: D Topic: Periodic Table Solution: Oxides and chlorides of phosphorus form acidic solutions. P4O6 (s) + 6H2O (l)  4H3PO3 (aq) (pH = 1–2) P 4O10 (s) + 6H2O(l)  4H3PO4 (aq) PCl 3 (l) + 3H2O (l)  H3PO3 (aq) + 3HCI (aq) PCl 5 (l) + 4H2O (l)  H3PO4 (aq) + 5HCI (aq) 13 Answer: A Topic: Periodic Table/Chem. Bonding Solution: Al in AlCl 3 has only 6 electrons, hence can accept lone pair from Cl – ion via dative bond to form AlCl4 –, achieving octet configuration. 14 Answer: A Topic: Periodic Table Solution: A: mass number increases from Na to Ar. (Ar – atomic no.  mass no.) B : C  1 st IE D: 1 st IE 15 Answer: A Topic: Group II Solution: A: Li 2CO3  Li2O + CO2. B : Oxide not nitrate is formed. C : Li reacts vigorously with O2. D : Like Mg, Li reacts slowly with cold water. 16 Answer: B Topic: Group II Solution: 2 nd IE decreases down the group due to decreasing ENC. A, C, D: Incorrect. Reactivity increases down the group. Solubility of group II hydroxides increases and thermal stability increases. 17 Answer: A Topic: Group II Solution: Formation of soluble complex with thiosulfate enables AgCl to dissolve (concept similar to AgCl soluble in NH3). B, C, D: Incorrect. This is not a redox reaction. Oxidation no. of Ag + (+1) remains the same before and after. 18 Answer: B Topic: Transition Elements Solution: NH 3 is produced to form alkaline solution. Grey–green ppt is Fe(OH) 2. 19 Answer: A Topic: Organic Chem Solution: Ethanoic acid is soluble in water but not ethyl ethanoate. It reacts with aqueous NaOH but not ethanol. 20 Answer: A Topic: Organic Chem. Solution: A: Alcohols: C nH2n+2O B: Aldehydes: C nH2nO C: Carboxylic acids: C nH2nO2 D: Halogenoalkanes: C nH2n+1X 21 Answer: D Topic: Arenes Solution: Length of C–C bond in benzene is in between that of C–C and C=C bonds in aliphatic compounds due to delocalisation of electrons. B: Incorrect. Benzene does not conduct electricity unlike graphite with delocalised electrons. In graphite, the –electrons are delocalised over the entire plane of C–atoms in a giant molecular structure, while in a benzene molecule, they are only delocalised within the molecule (ring). The delocalised electrons in one benzene molecule cannot ‘move’ to another molecule to conduct electricity. 22 Answer: C Topic: Alkane Solution: Three different forms of radical: CH3CH2 CC H 3 CH3 CH2 CH3CH C CH 3 CH3 CH3 CH2CH2 CC H 3 CH3 CH3 23 Answer: B Topic: Amine, benzene, alcohol Solution: CC N C H 3 H OH H CH3 H ephedrine A : CH 3COOH is neutralised by amine group. CH3COCl reacts with alcohol to form ester. Na Mg Al Si P S Cl Ar

2011 DHS H2 Chemistry Paper 1 (Solutions) © DHS 2011 2 9647/01(Solutions) B: amine group in ephedrine can act as nucleophile to substitute –Br in CH 3Br. Alcohol and amine do not react with NaOH. C : HCl reacts with amine group. PCl 5 reacts with alcohol group. D : Tollens’ reagent does not react since no aldehyde present. Br 2 does not react since catalyst is necessary for ES of benzene. 24 Answer: D Topic: Carbonyl compounds Solution: NHNH2CH3COCH2CO2CH2CH3 + S T Ketone group undergoes condensation with phenylhydrazine to form NH N C CH3 CH2CO2CH2CH3 25 Answer: B Topic: Amides Solution: CH2CH3 N CH2CH3 RC O Protonated amine, (CH 3CH2)2NH2 + is least likely to be obtained after basic hydrolysis. 26 Answer: D Topic: Alcohol, acids and derivatives Solution: P is an alcohol. Q is CH 3CH2COBr and R is carboxylic acid which reacts less readily than Q to form an ester. 27 Answer: B Topic: Alcohols Solution: A: True. (CH3)3COH2 (CH3)3COH + H++ acid conjugate base B: Untrue. H +–Cl–. The alcohol group in (CH3)3COH is protonated by HCl in step 1. HCl is acting as an (acid) proton donor. The nucleophile in this reaction (step 3) is Cl–. C: True. Add all three equations up to get the overall eqn. D: True. From overall eqn, it can be seen that