2013 MJC H1 Chemistry P2 Mark scheme

1 ©MJC Chemistry 2013 MJC H1 Suggested Answers   Section A: Structured Questions 1(a) (i) Perc entage abundance of Te-130 = 34.48 Isotopic mass x % abundance of Te-128 = 4069 Isotopic mass x % abundance of Te-130 = 4482 Relative atomic mass of Te = 127.7 (ii) There is inter-electron repulsion between the paired electrons in the 5p orbital of Te. (b) (i) (ii) There are 4 bond pairs and 1 lone pair around Te atom. To minimise repulsion, the 5 electron pairs are directed to the corners of a trigonal bipyramid. Since lone pair-bond pair repulsion > bone pair-bond pair repulsion, the shape of TeCl 4 is distorted tetrahedral. (iii) Te has empty and energetically assessible 5d orbitals to accept electrons and expands its octet structure. (c) TeC l4 undergoes complete hydrolysis to form strongly acidic solutions with pH= 2 TeC l4 + 2H2O → TeO2 + 4HCl AlCl3 undergoes hydration and slight hy drolysis to form an acidic solution with pH=3.0 AlCl 3 (s)  OH 2 Al3+ (aq) + 3Cl- (aq) [Al(H2O)6]3+ [Al(H2O)5(OH)]2+ + H+ Te xx xx xx x Cl x xx xx xx Cl x xx xx xx Cl xx xx x xx Cl

2 ©MJC Chemistry 2(a) (i) W. It has the highest q r   value (ii) V (or Cr or W) (b) Adding carbon atoms into th e space between the iron atoms in the lattice will prevent the ir on atoms from sliding ov er each other easily and hence do not go out of shape easily. (c) Cr forms an oxide layer and pr events oxygen from reacting with the iron. (d) Metallic bonds are strong electrostatic forces of attraction between the cations and sea of delocalised electrons in a giant metallic lattice structure. (e) (i) FeO 4 2- -0.4 to -0.6V (ii) 6e + 5H 2O + 2FeO4 2-  Fe2O3 + 10OH (f) (i) No of moles of I-= 1.293 x 10-2 No of mole of Fe 3+ = 1.29x 10-2 (ii) x = 3. (iii) +2 and +3 + e  e e e e   e e  e e metal cations + + + + + + + ++++ + + + + delocalised electrons

3 ©MJC Chemistry (g) Ea is lowered due to an alternative reaction pathway. Number of reactant particles with E ≥ Ea increases. Frequency of effective collisions increases. Rate of reaction is proportional to frequency of effective collisions. 3(a) Reagent Compound A, B or C Structural formula of the organic product Na2CO3 (aq) C CH=CHCH3 COO-Na+

4 ©MJC Chemistry 2,4-DNPH A CH=CHC H N N H NO2 O2N Tollens’ reagent A CH=C H C O O- (b)(i) CH I 3 (ii) COOH COOH

5 ©MJC Chemistry (c)(i) (ii) Step I HCN Conditions: trace amount of NaOH or NaCN, cold Step II Dilute H 2SO4 heat (d) CHO CHOH CHOH CHOH CHOH CH2OH H O H Hydrogen bonding Section B: Free Response Questions C 4(a) Standard enthalpy change of format ion is the energy change when one mole of product is formed from its elements at 298K and 1 atm. (b) Hc (ETBE) = - 3.75 x 103 kJ mol-1 (c) - 610 kJ mol -1 (d) In the internal combustion engine, where the temperature is very high, nitrogen react with oxygen in the ai r to form nitrogen oxide. Nitrogen oxides cause acid rain. (e)(i) K has a molecular formula of C 6H14O K undergoes oxidation with warm al kaline iodine. K contains the structure CH3-CH(OH) - + -

6 ©MJC Chemistry K undergoes acid-metal displacement with sodium. K is an alcohol K undergoes elimination to give alkenes L and M. L undergoes oxidation with acidified KMnO 4 to produce N. N is a carbonyl compound. Since N undergoes condensation wit h 2,4-DNPH, but does not undergo oxidation with Tollen’s reagent,N is not an aldehyde or is a ketone K CH3 CC CH3H CH2CH3 L M N (ii) Functional group isomerism. (f) MgO is a basic oxide. M g O + 2 H + Mg2+ + H2O P 4O10 is an acidic oxide. P 4O10 + 12OH- 4PO4 3- + 6H2O A l2O3 is an amphoteric oxide. A l2O3 + 6H+ 2 Al3+ + 3H2O A l2O3 + 2OH- + 3H2O 2[Al(OH)4]- The in-between behavior of A l2O3 is due to the high charge density of Al3+ ion polorising the O 2- anion hence resulting in partial covalent nature of Al-O interaction. 5(a) 322 32 2 [ C HC HC HC H O ] [CH CH CH ][CO][H ]  cK Unit: mol -2 dm6 CO CH 3 CH 3 CH 2 CH3 CC H CH3 CH2CH3 OH H H3CH2CCC H CH3 H CH2

7 ©MJC Chemistry (b)(i) CH 3CH=CH2 + CO + H 2  CH3CH2CH2CHO Initial conc / mol dm-3 6.25 6.25 6.25 0 Change in conc / mol dm-3 -3 -3 -3 +3 Final conc / mol dm-3 3.25 3.25 3.25 3.00 (ii) 0.0874 mol-2 dm6 (iii) (iv) -137 kJ mol -1 (v) By Le Chatelier’s Prin ciple, an increase in temperature would cause the equilibrium position to shift to th e left towards the endothermic reaction to absorb heat. The formation of butanal is not favoured. (c) (i) Reaction II: KMnO4, H2SO4 (aq), heat Or K 2Cr2O7, H2SO4 (aq), heat Reaction III: LiAlH4 in dry ether, r.t.p. Or NaBH 4 in ethanol, r.t.p. (ii) Reagents & conditions: PCl 5, r.t.p. (absence of water) Reagents & conditions for step 1: Conc H2SO4 at 180 oC Reagents & conditions for step 2: HCl gas, r.t.p. t time 6.25 3.25 3 concentration reactant product

8 ©MJC Chemistry (d) (i) Bond strength: C-C l bond > C-I bond .Bond energy: C-I < C-Cl (ii) Heat each mixture with aqueous NaOH followed by the addition of dilute HNO 3 and AgNO 3(aq).If a white ppt of AgC l is formed, the compound is 2-chlorobutane CH 3CH(Cl)CH2CH3 + NaOH CH 3CH(OH)CH2CH3+ Na+ + Cl- Ag+ + Cl- AgBr (e) 3 : 2. 6(a) (i) 0.0960 mol dm 3 (ii) 1.72 x 10 4 mol dm3 (iii) No. Since this is a strong base – weak acid titration, the pH at equivalence point is basic. The pH transition range of chlorophenol red does not lie within the range of r apid pH change over the equivalence point. (iv) When a small amount of acid, H+ is added: CH3CH(OH)COO- + H+  CH3CH(OH)COOH The added H + is removed as CH 3 CH(OH)COOH. Hence pH remains fairly constant When a small amount of base, OH- is added: CH3CH(OH)COOH + OH -  CH3CH(OH)COO- + H2O The added OH - is removed as CH 3CH(OH)COO- and H 2O. Hence pH remains fairly constant

9 ©MJC Chemistry (b) D does not undergo acid-metal displacement with Na hence does not contain an alcohol or carboxylic acid group. Lactic acid undergoes self- esterification to form D with no of C atoms doubled hence D is a diester. (c) (i) E undergoes oxidation with Fehli ng’s solution. E is an aliphatic aldehyde. E Ξ H 2 There are two alcohol groups. One mole of E undergoes substi tution with two moles of PC l5. There are two alcohol groups. E is (ii) In the carboxylate anion, the negative charge is delocalised over the 2 O atoms. Thereby stabilising the carboxylate ion relative to lactic acid (d) (i) Cis-trans isomerism arises when ro tation of a double bond is restricted due to the presence of  bond It also arises since different substituent groups are bonded to each C atom of the C=C bond. (ii) Due to the prox imity of the two –COOH groups, cis-somer is capable of intramolecular hydrogen bonding. Hence the cis-isomer possesses less extensive intermolecular hydrogen bonding. Since more energy is required to overcome the more extensive hydrogen bonds be