2019 EJC Prelims H2 Chemistry Paper 1 Worked Solution

2019 JC2 Preliminary Examination H2 Chemistry 9729 Paper 1 Worked Solution 1 From Group 1318, the number of unpaired electrons es from 1 (Group 13) to 3 (Group 15), then es to 0 (Group 18). Hence, W, X, Y and Z cannot be main group elements, i.e. they are TM. W must have been 4d8 5s2 and not 4d2 5s2 as the number of unpaired electrons es with more electrons added.  B 2 Angle of deflection, z m For a proton, 1 1H , 1 11 z m    For particle X,  51 115 3 z m     A : 3 1 A ; B : 5 2 B ; C : 6 3 C ; D : 93 4 D  A 3 A : trans-isomer is non -polar (C–Cl bond dipole cancels), while cis-isomer is polar. Hence cis-isomer is more soluble in polar H2O. B : C=O is more polar than C –Cl as O is more electronegative. Also, oxygen of C=O can form H -bond with water via its lone pair.  CH3COCH3 is more soluble. C : CH3CH2CH2CH2CH(OH)CO2H can form intermolecular H -bond with H2O, while CH 3CH2CH2CH2CH(NH3 +)CO2 – is zwitterionic and forms stronger ion - dipole interaction with H 2O, hence more soluble. D : 2-nitrophenol can form intramolecular H-bond, while 4-nitrophenol forms intermolecular H -bond with H 2O, hence more soluble.  B 4 A : Like graphite, each C is sp2 hybridised with one unpaired e– in the remaining p-orbital, which is delocalised, allowing CNT to conduct electricity. B : Like graphite, the C –C bonds within each molecule is stronger than that in diamond as they result from sp 2-sp2 overlap. However, between two molecules, there is only weak intermolecular id -id attraction, hence softer than diamond. C : Like graphite, the high melting point is due to energy required to break the strong intramolecular C –C bond and not to overcome the weak intermolecular id-id attraction. D : Like graphite, the C –C bonds within each molecule is stronger than that in diamond as they result from sp 2-sp2 overlap.  A 5 I : H2SO4 oxidises Br– to Br2, itself reduced to SO2 II : H2SO4 catalyses the condensation reaction between an acid and an alcohol to give an ester (and water), but making the C=O more electrophilic. III : Acid-base reaction where H2SO4 is the acid and NaOH the base, to give salt (Na2SO4) and water.  D 6 1 : A more exothermic lattice energy just implies that it takes more energy to vaporise the ionic lattice into gasous ions. No bearing on the decomposition. 2,4: CO3 2– is more easily polarised as planar CO3 2– has a larger surface area than the near spherical tetrahedral SO4 2– ion. Hence the C–O bond is more readily weakened, causing CaCO3 to decomposing at a lower temperature. 3 : Due to smaller size of the CO3 2– ion compared to SO4 2–, the charge density is higher. However, higher anionic charge density has no bearing on the decomposition.  C 7 Oxidation state of S in SF6 is +6, SCl2 is +2 and S2Br2 is +1, which correlates with the ability of the halogen to oxidise S.  D 8 13 0.13Number of