EJC Prelim P1 Worked Solution

2018 JC2 Preliminary Examination H2 Chemistry 9729 Paper 1 Worked Solution 1 The positive α particles either passes straight through the empty space or are occasionally repelled by the dense positive nucleus. They are not affected by the much smaller electrons and hence does not provide any information about the electrons.  C 2 [O] : 2Am Amnn ne++ −→+ [R] : 32Cu Cunn e n+− ++→ So, 32 2Am Cu Am Cunn nn++ +++→+ 4 Am 25.0 0.0100 2.5 10 mol1000 nn + −=× = × 3 4 Cu 15.0 0.0500 7.5 10 mol1000n + −=× = × CuAm :1 : 1 : 3 3nnn n n+ ==  = Am species formed has O.S. of +6 () 2 2AmO +  D 3 OF 2 : 2 b.p. + 2 l.p.  bent  105º CH4 : 4 b.p. + 0 l.p.  tetrahedral  109.5º SF6 : 6 b.p. + 0 l.p.  octahedral  90º H2O : 2 b.p. + 2 l.p.  bent  105º O3 :  bent  <120º N2O :  linear  180º PF3 : 3 b.p. + 1 l.p.  trig. pyr.  107º BF3 : 3 b.p. + 0 l.p.  trig. planar  120º IF3 :  trig. planar  120º NO2 :  bent  >120º SO2 : 2 b.p. + 1 l.p.  bent  <120º CO2 : 2 b.p. + 0 1.p.  linear  180º  B 4 Bonding in solid NH 4HF2:  A 5 At low pressure, gas particles are far apart. IMF between particles causes force they impinge on wall to be smaller and thus pressure is lower than ideal gas  pVreal < pVideal At higher pressure, gas particles are close together and the volume occupied by particles is significant compared to volume of container. Hence the gas occupies a larger volume than ideal gas  pVreal > pVideal  B 6  Arrhenius acid: Produces H+(aq) in water  Brønsted-Lowry acid: H+ donor; B(OH)3 accepts a OH– and does not donate H+  Lewis acid: Lone pair acceptor; accepts a lone pair fron OH–  Monobasic acid: each mole of B(OH)3 reacts with one mole of OH–  B 7 NaC l : no hydrolysis  pH = 7 MgCl2 : slight hydrolysis  pH ≈ 6.5 22MgC H O Mg(OH)C HC++ll l … AlCl3 : extensive hydrolysis  pH ≈ 3 32 3A C 3H O A (OH) 3HC+→ +ll l l SiCl4 : complete hydrolysis  pH ≈ 1 42 2SiC 2H O SiO 4HC+→ +ll PCl5 : complete hydrolysis  pH ≈ 1 52 3 4PC 4H O H PO 5HC+→ +ll  A 8 A : Graph is for ionic radii B : Melting point of Si should be highest due to extensive strong Si–Si covalent bonds within giant molecular structure C : Nuclear charge increases across period while shielding effect is the same, hence ENC increases across the period resulting in increase in electronegativity D : Silicon being a metalloid should only have conductivity below that of the metals (Na, Mg, Al)  C 9 () () ()() ()() ()() ()() ()() ()() ()() ()() 1f f ff 2 ff 2 f f2 ff 2 products reactants 2N a O H a q H g 2N a s 2H O 2N a O H a q 0 02 H O 2N a O H a q 2H O HH H HH HH H H HH Δ= Δ − Δ =Δ + Δ − Δ− Δ =Δ + − −Δ =Δ −Δ   l l l dd d dd dd d d dd ()() ()()f2 c 2HO H gHHΔ= Δ ldd as both corresponds to the same reaction: H2(g) + ½O2(g) → H2O(l)  B 10 B : At 298 K, C is a solid C : Hydrogen exists as H 2 under std state D : At 298 K, H 2O is a liquid  A 11 Units for rate is concentration per unit time, in this case, mol dm–3 s–1 If k has units of mol–2 dm6 s–1, the reaction must be overall third order: () ( ) 331 2 31 3mol dm s mol dm s mol dm−− − − − =  C 12 Given () ()2Xg 2 X g ,→ [ ] 1 22rate X , 20 minkt== 1 : 12.5% X2 is left after 3 1 2 t (60 min). i.e. 87.5% of X2 had reacted 111 2222 222 22 2 111XX XX 824 50% X 25% X 12.5% X ttt ⎯⎯ → ⎯⎯ → ⎯⎯ → 2 : 2X X reacted 87.522 1 1 . 7 5 m o l100nn == × × = 3 : 2gas X X 12.5 1511 . 7 5 m o l100 8nn n=+ = × + = At constant temperature and volume, 12 1 22 12 1 15 15 18 8 pp p pp npnn n=  =×= × =  D 13 From the graph, As pressure increases, % products decreases. Hence, there must be more gaseous particles on the product side, since backward reaction is favoured to decreases the pressure As temperature increases, % products decreases. Hence the forward reaction must be exothermic, since backward reaction is favoured to absorb heat  C 14 A : ; 1.0nn n n n== = −QR LM R Since nR at eqm can be read from the graph, the partial pressures of L, M, Q and R can be obtained. Hence Kp can be obtained. B : Since the amount of product R decreases when temperature is increased, the forward reaction must be exothermic since the backward reaction is favoured to absorb heat. C : The position of eqm does not provide information about the kinetics of the reaction D : From the graph, it can seen that it takes a shorter time to plateau at higher temperature, hence eqm is achieved at a faster rate  C 15 pH = 4  pH 4 3H1 0 1 0 m o l d m+− − − ==  X is fully dissociated, i.e. strong acid pH = 3  pH 3 3H1 0 1 0 m o l d m+− − − ==  Y is not fully dissociated, i.e. weak acid pH = 14  ()14 pHpOH 0 3 OH 10 10 10 1 m ol dm −−−− −  == = =  Z is a fully ionised, i.e. strong base  D 16 Let the solubility of Ag3PO4 be s mol dm–3 3 3 sp 4 Ag POK +−  =    A : () ( ) 3 4 sp sp 32 7Ks s s K=  = sp 54 4.26 1027 Ks −== × B : () ( ) 3 3 sp 3 0.10 0.10Ks s s=+ ≈ sp 14 3 8.89 100.10 Ks −== × C : ()33 2Ag 2NH Ag NH ++  +→   Formation of complex increases solubility of Ag3PO4 D : () ( ) () 33 sp 30 . 1 0 3 0 . 1 0Ks s s=+ ≈ sp 63 3.21 1027 0.10 Ks −== × ×  B 17 B : No chiral centre C : =CH 2  no cis-trans isomerism D : No chiral centre  A 18 Structure of 2,2-dimethylpentane:  C

19 1 : 2 : 3 :  C 20 Friedel-Crafts acylation, similar to Friedel- Crafts alkyation, using acyl chloride instead of alkyl chloride:  D 21 A : Second mechanism is single-step SN2 involving only a transition state B : Second mechanism is a nucleophilic substitution reaction C : Since both mechanisms involves two reacting species in the rate determining step, both reactions are second order D : The first mechanism involves addition of the CN– to a trigonal planar C=O which can take place on both faces with equal chance, hence a racemic product will be obtained  C 22 The reaction involves  C 23 Tollens’ reagent, () 3 2Ag NH OH + − , only oxidises the –CHO into –COO– (the salt is obtained and not –CO2H since Tollens’ reagent is alkaline), while itself is reduced to silver metal. Alcohols are not oxidised.  B 24 A : NH 3 donates a pair of electrons to the electron-deficient acyl carbon, hence a nucleophile B : The arrow pushing is correct, leading to the regeneration of C=O and expulsion of the Cl– leaving group C : The acyl carbon is electron-deficient as it is bonded to two electronegative atoms, O and Cl, hence attracting the NH3 nucleophile D : The arrow pushing is wrong as the pair of N–H electrons end up on the H, which will lead to formation of H– and a doubly positive N instead  D 25  B 26 Amides are made from the reaction between an amine and an acyl chloride: Amine and carboxylic acid gives an ammonium salt:  A 27 Basicity (availability of lone pair of electrons on N for donation to H+) : Generally R2NH > RNH2 > NH3 > ArNH2 as R groups exert electron-donating effect, rendering the lone pair more available for donation, while the lone pair is delocalised into the benzene ring of ArNH 2 rendering the lone pair less available for donation. ↓ing basicity  ↑ing pKb So pKb : R2NH < RNH2 < NH3 < ArNH2  B 28 W : Nitrogen in 3NO− is already in the maximum oxidation of +5 and cannot be further oxidised. Oxidation of H2O 222H O O 4H 4 e+−→+ + Y : Due to the high concentration of chloride, oxidation of Cl– occurs instead of H2O ()222C C 2 4C 2C 4ee−− − −→+ → +ll l l Z : Oxidation of Cu instead of H 2O due to the less positive E d . No gas evolved 2Cu Cu 2 e+−→+ 0.34 VE =+d 222H O O 4H 4 e+−→+ + 1.23 VE =+d Since the same amount of charge passes through all three cells, 22OC :1 : 2VV =l  D 29 1 : [ ] 2R: C 2 2 Ce−−+ll ƒ