EJC 2020 GCE A Level 8873 H1 Chemistry Paper 1 Suggested Solution

Eunoia Junior College 8873 H1 Chemistry 2020 Paper 1 Suggested Solution 1 Isotopes have the same number of protons, but different number of neutrons  A or C Since the two particles have different charges, the difference between the number of protons and electrons must be different.  C 2 Examine the number of p, n and e 58 3 28Ni  60 2 28Ni  62 2 28Ni  64 3 28Ni  p 28 28 28 28 n 30 32 34 36 e 25 26 26 25 m 9.712×1026 1.005×1025 1.038×1025 1.072×1025  C 3 A : The 3d subshell can hold a maximum of 10 electrons. The shell with principal quantum number 3 can hold a maximum of 2 (3s) + 6 (3p) + 10 (3d) = 18 electrons. B : According to Aufbau principle, electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. C : The order of filling the orbitals is 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s … … D : In atoms of transition elements, the electronic configuration is [Ar] 3dn 4s2 or [Ar] 3d5 4s1 (Cr), with n = 1–9  B 4 Within the same principal quantum shell n, the energy of the orbitals is in the order: ns < np < nd < nf and there are 1 s orbital, 3 p orbitals, 5 d orbitals and 7 f orbitals in each principal quantum shell.  A 5 Gaseous HC l has a simple molecular structure with a polar covalent bond between H and the electronegative atom Cl, i.e. uneven sharing of a pair of electrons between the two nuclei, with a resultant partial plus charge on the H and a partial minus charge on the Cl.  C 6  C 7  A 8 By definition, bond energy is the energy needed to break one mole of the gaseous bond. Covalent bond length:  A 9 Arrhenius acid produces H+(aq), while Arrhenius base produces OH–(aq).  D 10 For an indicator to show the end-point of a titration, the working pH range of the indicator must lie within the region of rapid pH change during the titration.  A 11 Blood contains the H2CO3/HCO3– buffer. To maintain the pH of blood, the conjugate base HCO 3– must remove H 3O+ from the lactic acid produced during exercise: HCO3– + H3O+  H2CO3 + H2O  C 12 A : Since both I and Xe does not conduct electricity and have rather low boiling points, both I and Xe exists as simple molecules. B : Te is only a semi-conductor (like Si), so it is a metalloid with a giant covalent structure. C : Period 3 only contains 3 metal lic elements, Na, Mg and Al, while there are at least 5 metallic elements (high electrical conductivity and high boiling point) in Period 5. D : Rubidium in Group 1 is a metal despite having a melting point below 1000 K.  C 13 1 : Electronegativity of the halogen decreases down the group, all are more electronegative than H. Hence the difference in electronegativity decreases down the group. 2 : Down the group, the atomic (covalent) radius increases, hence the bond length is longer, with the bonding pair of electrons getting further from the halogen nucleus going down the group. 3 : Down the group, as H–X bond length increases, the H –X bond strength decreases, leading to a decrease in thermal stability down the group.  D 14 A : More electrons  stronger id-id  higher boiling point  less volatile B : Volatility is related to strength of intermolecular forces of attraction and not the X–X bond strength. C : More electrons  stronger id -id  higher boiling point  less volatile D : Volatility is related to strength of intermolecular forces of attraction and not the X–X bond strength.  A 15 Atomic radius decreases across the period due to increase in ENC (nuclear charge increases while shielding is ~constant)  Since E is larger  E is in Group 15 Going from Group 15 (ns2 np3) to Group 16 (ns2 np4), there is a drop in first I.E. since there is inter -electronic repulsion between the pair np electrons in Group 16  E should have a higher first I.E.  B 16 A : 2C 71.0 1.0 mol35.5 2n   l  6.02×1023 Cl2 molecules B : 1 mol of Mg(NO 3)2 contains 3 mol of ions = 3 × 6.02 × 10 23 = 1.81 × 10 24 ions C : At s.t.p., Vm = 22.7 dm mol–1 2O 22.7 1.0 mol22.7n  1 mol of O 2 contains 2 mol of O atoms = 2 × 6.02 × 1023 = 1.20 × 1023 O atoms D : 2Be 4.50 0.50 mol9.0n   Each Be2+ contains 4–2 = 2 electrons. 0.50 mol of Be2+ contains 0.50 × 2 × 6.02 × 1023 = 6.02 × 1023 electrons  C 17 2 2 2C H O CO H O42 xy yyxx    2O 0.720 0.030 mol24.0n  2HO 0.36 0.020 mol1.0 2 16.0n   22Z O H O: : 0.0050 : 0.030 : 0.020 1: : 1: 6 : 442 n n n yyx    y = 8 and x = 4  C 18 A : Energy needed to overcome the stronger intermolecular forces of attraction in the liquid. B : Energy is released when stronger intermolecular forces of attraction is formed in the solid. C : Enthalpy change of neutralisation. Energy is released when a O –H is formed between H+ and –OH. D : Opposite of bond dissociation . Energy is released when 2 O –H bonds are formed between the H and O atoms.  A 19 lattice energy qq rr     2Mg Na 2qq  and 2Mg Narr   lattice energy of MgX2 is more negative than that of NaX Br Crr  l  lattice energy of MgCl2 is more negative than that of MgBr2  B

20   31 3 mol dm sunit of mol dm nk    , where n is the overall order of reaction  D 21 Given          0 1 2 2 rate X Y Z Y Zkk       4 2 2 rate 4.68 10 0.0500 0.0400YZ 5.85 k    B 22 Since the decomposition of N 2O5 involves only N2O5 molecules, and is first order w.r.t. N2O5, the half-life of the decomposition must be a constant ln2 k  at the same temperature.  B 23 The presence of a catalyst does not affect the shape of the Boltzmann distribution, hence Emp will not change. However, presence of a catalyst lowers Ea, hence increasing the frequency of effective collisions between the molecules.  D 24 The Kc of a reaction is only affect by changes in temperature (provided H  0)  A 25 Considering the equilibrium:       2C 2NO 2NOC initial amt 1 2 0 change 1 2 1 2 amt eqm 1 22am 2t x x x xx x         ll The change in amt of Cl2 is  1 x . By the stoichiometry of the reaction, changes in the amt of NO and NOCl are       2C 2NO 2NOC initial amt 1 2 0 change amt 1 2 1 2 1 eq 22 2m amt x x x x xx         ll Hence the eqm amts are:       2C 2NO 2NOC initial amt 1 2 0 change amt 1 2 1 2 1 eqm amt 2 2 2 x x x x x x         ll  C 26 The possible constitutional isomers are:  B 27 The two straight-chain C4H8 are  B 28  D 29 Due to the presence of –OH and –CO2H groups in polymer 2 and 3, which can form hydrogen bonds with water, they are likely to be water soluble.  C 30 1 : Surface area is important as catalytic converters relies on heterogeneous catalysis where the reactant molecules are adsorbed and react on the surface of the catalyst. 2 : The ability of gecko to climb a wall depends on sum of instantaneous dipole-induced dipole attractions between the millions of microscopic hairs on the feet of the gecko and the wall. The hairs increases the surface area allowing more extensive id -id attractions to form. 3 : The high tensile strength of graphene is due to the strong C –C bond within the graphene sheet and is not related to the surface area.  B Answer Key Qn Ans Qn Ans Qn Ans 1 C 11 C 21 B 2 C 12 C 22 B 3 B 13 D 23 D 4 A 14 A 24 A 5 C 15 B 25 C 6 C 16 C 26 B 7 A 17 C 27 B 8 A 18 A 28 D 9 D 19 B 29 C 10 A 20 D 30 B