NJC 2019 A Level Chem Solution

H2 Chemistry TYS 2019 suggested answers Paper 1 Answer Key 1 B 11 A 21 D 2 C 12 B 22 C 3 B 13 D 23 A 4 C 14 D 24 C 5 C 15 D 25 B 6 A 16 A 26 B 7 D 17 B 27 C 8 B 18 A 28 C 9 D 19 A 29 B 10 D 20 D 30 B/C 2019 A level Paper 1 suggested answers 1) Sr2+ W X─ Y2─ No of electrons *36 *36 *36 *36 No of protons 38 36 35 34 No of neutrons *46 *46 *46 *46 Nucleon Number (no of protons + neutrons) *84 82 81 80 * told in the question Ans: B 2) Angle of deflection  𝐶ℎ𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 Mass of electron is approximately 1 2000 the mass of a proton, hence electron has a larger angle of deflection. The charged particles are attracted to the oppositely charged plate. Ans: C 3) For Group 15 and Group 16 element Electronic configuration where electrons are lost are as such: Group 15 element Group 16 element 1st IE ns2 np3 ns2 np4 2nd IE ns2 np2 ns2 np3 3rd IE ns2 np1 ns2 np2 4th IE ns2 ns2 np1 4th IE in group 15 element will experience a slightly large spike as electrons are removed from an inner subshell. Comparing the values given in the table, 4th IE of option A and C experience a slightly larger spike than that of B and D => A and C belong to Group 15, B and D belong to Group 16. Between Se and Te, Te is in a period further down the periodic table. This means that it has lower IE (since increase in no. of electron shell and shielding effect outweighs increase in nuclear charge) => Te must be B Ans: B

4) BaO2: ionic compound : Ba2+ O22─ (peroxide ion) For O22─ dot and cross: Each O─ is bonded to another O─ by a single covalent bond (told in question). Ans: C 5) Molecule No of bonds 1 Cl Al ClCl Cl Al Cl Cl 6 Sigma bonds 2 COO 2 Sigma bonds 2 Pi bonds 3 C H H H C H O 6 Sigma bonds 1 Pi bond 4 CC H H H CH HH 8 Sigma bonds 1 Pi bond Ans: C 6) Assuming the compounds behave as ideal gases, apply pV = nRT, rearranging, p = 𝑛𝑅𝑇 𝑉 = 𝑚𝑅𝑇 𝑀𝑟𝑉 Since T, V and m is constant, the smaller the Mr, the higher the pressure. Molecule CH4 HCHO CH3Cl HCO2H Mr 16 30 50.5 46 Ans: A 7) Follow the question closely for the reaction that happens: NaN3 (s) ⎯→ Na (s) + 3 2 N2 (g) Amount of NaN3 = 5.00 65.0 = 0.07692 mol Amount of N2 = 1.5 × 0.07692 = 0.1154 mol pV = nRT V = 𝑛𝑅𝑇 𝑝 = 0.1154 × 8.31 × (30+273) 9.85 × 104 = 0.00295 m3 = 2.95 dm3 Ans: D 8) Context of question: Period 3 and Period 4 elements behave similarly. As such, Ge behaves similarly to Silicon, which has high melting point due to its giant covalent lattice structure with strong covalent bonds and is also a semiconductor. Ans: B

9) Clue 1: X dissolves in HCl to give Solution Y, which gives a white ppt insoluble in excess NaOH (aq) Possible Metal Solution Y it forms White ppt it forms Al Al3+ Al(OH)3 White ppt dissolves in excess NaOH to give [Al(OH)4]− Mg Mg2+ Mg(OH)2 White ppt insoluble in excess NaOH Clue 2: Give off gas Z, which reacts with O2(g) t