JPJC 2022 JC1 promos P1 Worked Solution

©Jurong Pioneer Junior College 1 2022 JC1 YEE Paper 1 Worked Solutions Suggested Worked Solution for 2022 JC1 YEE Paper 1 (9729/01) 1 C 6 D 11 A 16 C 21 C 26 C 2 B 7 A 12 D 17 B 22 A 27 C 3 D 8 C 13 A 18 D 23 C 28 B 4 B 9 A 14 B 19 D 24 C 29 B 5 D 10 A 15 A 20 A 25 B 30 B 1 Answer: C The shapes of the five d−orbitals are shown below. Of the five, the first 4 have four lobes. 2 Answer: B For W, since there is a sharp increase from the 7 th to 8th IE, it implies that the 8 th electron of W is removed from the next inner quantum shell and hence, W has 7 valence ele ctrons. Thus, W is from Group 17. Since W, X, Y and Z are co nsecutive elements, i t implies th at X is from Group 18 , Y is from Group 1 of the next Period and Z is from Group 2 of the next Period. Hence, for the same nth IE, since the IE generally increases across the Period and decreases down the Group, X has the highest 1st IE. 3 Answer: D Interpret the two information as follow: • X and Y form ionic compounds Na2X and Na2Y respectively.  X and Y form anion of charge 2−. Hence, they are likely from Group 16 and has a valence electronic configuration of ns2 np4. (Reject Option A and B) • Element Y forms YF6 molecules where X is not able to do so.  Y can expand octet but not X. Hence, X is from Period 2 while Y is from Period 3 onwards. (Reject Option C) 4 Answer: B •  bond is formed from head−on overlap between orbitals (atomic orbitals and/or hybrid orbitals). (Reject Option C and D) s s s p p p •  bond is formed from sideway overlap of p−orbitals. (Reject Option A)

©Jurong Pioneer Junior College 2 2022 JC1 YEE Paper 1 Worked Solutions 5 Answer: D Molecular shape Polarity A B is from Grp 13 (3 valence e−). Around central B, there are 3 bp & 0 lp.  Shape is trigonal planar. Since there is no lp and the surrounding atoms are identical (all C l), the dipole moment from each B−Cl bond cancelled out ( i.e. BCl3 has no net dipole moment).  BCl3 is non−polar. B N is from Grp 15 (5 valence e−). Around central B, there are 3 bp & 1 lp.  Shape is trigonal pyramidal. Since there is 1 lp, the dipole moment from each N−Cl bond are not cancelled out even though the surrounding atoms are th e same (i.e. NCl3 has net dipole moment).  NCl3 is polar. C S is from Grp 16 (6 valence e−). Around central S, there are 2 bp & 1 lp.  Shape is bent. Since there is 1 lp, the dipole moment from each S=O bond are not cancelled out even though the surrounding atoms are the same (i.e. SO2 has net dipole moment).  SO2 is polar. ✓D C is from Grp 14 (4 valence e−). Around central C, there are 4 bp & 0 lp.  Shape is tetrahedral. Since there is 1 lp and the surrounding atoms are different, the dipo le moments do not cancelled out ( i.e. CHCl3 has net d ipole moment).  CHCl3 is polar. 6 Answer: D Both Al2Cl6 and AlCl3 have simple molecular/covalent structures. (Reject Option A and B). Al2Cl6 is formed from two AlCl3 monomer through dative bond formation. The Cl of one AlCl3 molecule has a lone pair (i.e. electron−rich) to donate to the empty p−orbital of Al of another AlCl3 molecule (i.e. electron−deficient) through dative bond formation. (Reject Option C) 7 Answer: A Option A is incorrect as compounds with giant covalent structure do not conduct electricity in any physical state due to the absence of mobile charge carrier. The only exception is graphite which conducts electricity in solid state and along the layer. 8 Answer: C Both calcium and sodium have giant covalent structure. More energy is required to overcome the stronger metallic bonds between the mobile valence electrons and Ca 2+ than that between the mobile valence electrons and Na + since Ca has more mobile va lence electrons available for m etallic bonding and the charge density of Ca 2+ is higher (due to higher ionic charge since both Ca2+ and Na+ have similar ionic radius). Hence, Ca has higher melting point than Na. Note: Option D is not the best answer si nce the statement is not specific in mentioning “mobile valence” electron. Thus, Option C is the most relevant.

©Jurong Pioneer Junior College 3 2022 JC1 YEE Paper 1 Worked Solutions 9 Answer: A (1 and 2 only) ✓1 Statement is true. At high pressure, volume of gas is smaller. Hence, the volume of gas particle becomes significant compared to the volume of gas. Also, at low temperature, the gas particles has less kinetic energy which is insufficient for them to o vercome the intermolecular forces of attraction. Hence, gases deviate more from ideal gas behaviour at high pressur e and low temperature. ✓2 Using pV = nRT = r mRT M , density () = m V = rp RT M . Since p, Mr and R are constant,  = 1k T   (i.e. 1 T  ). 3 Using pV = nRT, when n, R and p are constant, V = kT (i.e. V  T (in K)) 2 21 1 TVV T=  21 50 273VV 25 273 += + = 1.08V1 (not double!) 10 Answer: A No. of particle = n(particles)  Avogadro’s constant (i.e. no. of particles  n(particles)) n(HCO2CH2CH3) = 2.00  74.0 = 0.0270 mol n(Br2) = 4.00  2(79.9) = 0.0250 mol n(H2) at rtp = 550  24000 = 0.0229 mol n(H2O) = (1.55  1022)  (6.02  1023) = 0.0257 mol 11 Answer: A [R]: BrO3−(aq) + 6H+(aq) + 6e− → Br−(aq) + 3H2O(l) n(BrO3−) used = 0.02  20.0 1000 = 0.0004 mol n(NH2OH) used = 0.01  80.0 1000 = 0.0008 mol mole ratio BrO3− : e− : NH2OH since 1BrO3−  6e− 0.0004 : 6(0.0004) since n(e−) lost = n(e−) gained 6(0.0004) : 0.0008 3 : 1  Each N of NH2OH lost 3 electrons. initial OS of N in NH2OH = −1 final OS of N = (−1) − 3(−1) = +2 OS of N in NO = +2 OS of N in NO2 = + 4 OS of N in N2O = +1 OS of N in NO3− = +5

©Jurong Pioneer Junior College 4 2022 JC1 YEE Paper 1 Worked Solutions 12 Answer: D A The equation represents sum of 1st and 2nd EA of oxygen atom. The equation for 2nd EA of O should be O−(g) + e− → O2−(g). B The equation represents 2  Hneutralisation since 2 moles of H2O are formed. C The equation represents 8  Hformation of SO2(g) since 8 moles of SO2 are formed. 13 Answer: A 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g) 6(−355) 6(+90) (−710) 6(0) Using formula, Hr = mHf(products) − nHf(reactants) = [(−710) + 0] − [6(−355) + 6(+90)] = +880 kJ mol−1 > 0 (Option C and D rejected.) S < 0 as there is a decrease in disorderliness of the system since the number of liquid particles decreases from 6 to 0 mol (or number of solid particles increases from 0 to 6 mol). Note: The number of gaseous particles remains unchanged at 6 mol after the reaction . Hence, the impact of gaseous particle on the entropy of the system is the less significant factor. 14 Answer: B (2 only) 1 ✓2 3  =  −  + − + T ve ve ve G H S Since G > 0 at all T, the reaction is not spontaneous at all T.  =  −  − + − T ve ve ve G H S Since G < 0 at all T, the reaction is spontaneous at all T.  =  −  − − + T ve ve ve G H S At low enough T, G < 0 since |H| > |−TS|. Hence, the reaction is spontaneous at low enough T. 15 Answer: A (1 and 2 only) [As2O3] = 0.05 mol dm−3 [As2O3] = 0.10 mol dm−3 [VO2+] / mol dm−3 time / min t1/2,1 = 32 min t1/2,1 = 30 min 32 62 44 88

©Jurong Pioneer Junior College 5 2022 JC1 YEE Paper 1 Worked Solutions ✓1 Using the graph, where [As2O3] = 0.10 mol dm−3, (see construction lines in blue) average t1/2 = 32 30 2 + = 31 min Since the two half−lives are approximately constant, the order of reaction wrt [VO2+] is one. ✓2 initial rate = |gradient of tangent at t = 0 min| initial rate when [As2O3] is 0.05 = 0 0.20 88 0 − − = 0.00227 mol dm−3 min−1 initial rate when [As2O3] is 0.10 = 0 0.20 44 0 − − = 0.00455 mol dm−3 min−1 Hence, when [As 2O3] is doubled, the initial rate is also doubled . Thus, the order of reaction wrt [As2O3] is one. 3 H2SO4 is not a catalyst since H+ is reacted and