NJC 2015 A Level Chem Solution

1 H2 Chemistry TYS 2015 suggested answers Questions that are no longer in syllabus Paper 1 Paper 2 3(d) Paper 3 2(b), 3(a), 3(c)(i)&(ii) Paper 1 Answer Key 1 B 11 D 21 B 31 D 2 B 12 C 22 D 32 A 3 C 13 D 23 B 33 C 4 C 14 B 24 A 34 A 5 B 15 D 25 D 35 B 6 D 16 D 26 D 36 A 7 B 17 A 27 D 37 C 8 C 18 C 28 C 38 C 9 A 19 B 29 B 39 A 10 B 20 A 30 A 40 B 2015 A level Paper 1 suggested answers 1 order of filling atomic orbitals: 1s, 2s, 2p, 2p, 3s, 3p, 4s, 3d Ground state valence shell e configuration of Group 14 element: ns2 np2; larger n value, higher the energy level. B is the ans. 2 In hydrocarbons: -  bonds in C-C and C-H bonds can be formed from the overlap of either s and/or p orbitals Note: C atoms should be forming  bonds using sp / sp2 / sp3 hybrid orbitals. -  bonds in C=C bonds can only be formed from the side-on overlap of p orbitals, not s orbitals B is the best ans 3 no. of protons in OH− = atomic no. of O + atomic no. of H = 8 + 1 = 9 No. of electrons in OH− = no. of protons in OH− + net -ve charge on OH− = 9 + 1 = 10 No. of neutrons in OH− = (mass no. – atomic no.) of O + (mass no. – atomic no.) of H = (16 – 8) + (1 – 1) = 8 C is the ans

2 4 Both Ca and Na are metallic elements with strong metallic bonding (electrostatic forces of attraction) between delocalized sea of electrons and the cations. Melting point depends on metallic bond strength. Ca (in Group 2) can lose 2 e’s per atom to form Ca2+, Na (in Group 1) can only lose 1 e per atom to form Na+. C is the ans. Note: Option D is wrong as the strength of metallic bond depends on number of delocalized electrons, NOT the total number of electrons. 5 N: O O H 107o N+ O O- 109.5o 120o 120o 105o 109.5o 109.5o B is the ans 6 LE: energy released when 1 mol of an ionic solid is formed from its constituent gaseous ions D is the ans 7 NaOH + HCl → NaCl + H2O (l) Initial / mol 0.000625 (L.R.) 0.0025 0 - Change / mol −0.000625 −0.000625 +0.000625 - Final / mol 0 0.001875 −0.000625 - [HCl ] = 1000 25.012.5 0.001875 + = 0.0500 mol dm−3 B is the ans 8 Based on the production of CO2 and CO in 9:1 ratio 10CH4 + 19½ O2 → 9CO2 + CO + 20 H2O For gases under the same conditions, vol ratio = mol ratio  Vol of O2 needed to burn 1 dm3 CH4 = 1 × 19.5/10 = 1.95 dm3 C is the ans

3 9 CO2(g) + nH2O(l) CO2.nH2O(aq) H bonds are formed  H < 0 Gas turns aq  S < 0 (less disordered arrangement) A is the ans 10 Eo: standard electron potential at 25 oC, 1 atm pressure, 1 mol dm‒3 of the ions/substance involved B is the ans 11 CH3CO2H(g) CH3CO2− + H+ [ ]eqm / mol dm−3 (1 - )C C C Ka = COH][CH ][H ]CO[CH 3 23 -+ = )C (1 C)( 2   - =   1 C - 2 D is the ans 12 CO(g) + 2H2(g) CH3OH(g) + heat H < 0 Increase P, eqm shift right to favour side with less moles of gas to partially offset