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 the increase in P. Decrease T, eqm shift right to favour forward exothermic reaction to partially offset decrease in T by producing heat. C is the ans 13 H2O(g) + CO(g) H2(g) + CO2(g) Eqm amount / mol 1 – y 1 – y y y At eqm, total amt = 2(1 – y) + 2y = 2 mol proportion of H2(g) in eqm mixture = 2 y x 100 = 33.3%  y= 3 2 Kc = 2 22 2 22 [] [ ] (0.666)3.997 4.0[ ][O ] (0.334) H CO HO ==  D is the ans 14 Given: rate = k[H2] [NO]2 also 6.0 = k (0.5)(2.0)2  k = 3 Thus, x = 3 (0.5) (1.0)2 = 1.5 y = 3 (1.0) (1.0)2 = 3 and 0.75 = 3 (1.0) (z)2  z = 0.5 B is the ans

4 15 Elements in the same group have the same valence shell e configuration. Ionic radius increases down a group. Down Group 2, charge-density of M2+ and hence its polarizing power decreases, there is less distortion of electron cloud of NO3− anion; less weakening of N-O bond in NO3− anion. Hence the nitrates become more difficult to decompose. D is the ans 16 A is not the ans: S forms SO2, SO3 B is not the ans: Ar, has the highest 1st IE (Refer to Data Booklet values) C is not the ans: SiCl4 + 2H2O → SiO2 + 4HCl D is the ans P4, 17) Option A is correct, [Cu(NH3)4(H2O)2]2+ can also be represented as [Cu(NH3)4]2+, a dark blue solution. Option B is wrong, NH3 complex of Cu2+ is [Cu(NH3)4]2+, a dark blue solution. Option C is wrong, Cu(OH)2 is a pale blue precipitate. Option D is wrong, [Cu(H2O)6]+Cl− is CuCl, a white precipitate as Cu+ has an electronic configuration of 3d10. Ans: A 18 atomic radius decreases across the period: Mg > Al > Si > P (Refer to Data Booklet) P is a simple covalent substance → lowest mp Mg and Al have metallic lattice with strong metallic bonds between cations and sea of delocalised electrons, Al has higher mp than Mg Si has a covalent lattice with strong covalent bonds between Si atoms and has the highest mp in Period 3 mp: P < Mg < Al < Si C is the ans 19 Given: Cr2O72− is reduced to [Cr(H2O)6]3+  Cr2O72− is oxidising only in acidic medium  Y is either 1o alcohol or 2o alcohol or aldehyde B is the ans (X is sulfuric acid & Y is ethanol) 20 PCl5 can react with –OH group in both alcohol and carboxylic acid. M is CH3CHBrCHClCOCl A is the ans

5 21 isomers of non-cyclic C5H8 with 1 CC bond: B is the ans 22 D is the ans; all 7 sp2 hybridised C are circled O OCCH3 O CH3 O O CH3 CH3 23 NaBH4 can reduce aldehydes to 1o alcohols, but not alkenes (C=C). CH2=CHCH2CHO is reduced to CH2=CHCH2CH2OH B is the ans 24 A is the ans CO2H CO2H HO CO2H C6H8O7 C9H12O8 + C2H5CO2H CO2H CO2H C2H5CO2 CO2H(+ H2O) B is not the ans: CO2H CO2H HO CO2H C6H8O7 C10H14O8 + C3H7CO2H CO2H CO2H C3H7CO2 CO2H(+ H2O) C is not the ans: (only 1 ROH to react with a –CO2H group)

6 CO2H CO2H HO CO2H C6H8O7 C10H14O8 + C3H7OH CO2H CO2H HO CO2C3H7(+ H2O) Or (3 ROH’s should have reacted with all 3 –CO2H groups) (+ 3H2O) CO2H CO2H HO CO2H C6H8O7 C16H26O7 + 3C3H7OH CO2C3H7 CO2C3H7 HO CO2C3H7 D is not the ans: C6H8O7 C9H14O7 25 Both CCl4 and C6H5Cl are inert towards hydrolysis; no ppt will form when each is shaken with AgNO3(aq). Carbon in CCl4 does not have energetically accessible orbitals to accept a lone pair from water. C-Cl bond in C6H5Cl cannot be broken due to partially double bond character. p-orbital of Cl can overlap with the unhybridised p orbitals of carbon atoms in the benzene ring, hence lone pair in C l can delocalised into the benzene ring pi electron cloud, strengthening the C-Cl bond. D is the ans CH3COCl undergoes hydrolysis much more readily than CH3CH2CHClCH3 due to the presence of more highly e-deficient carbon centre (C with electronegative O and Cl attached); thus prone to attack by water (nucleophile). Acyl chlorides can react with water readily at room temperature. CH3COCl + H2O → CH3COOH + HCl 26 In solution at pH 11, deprotonation of acidic protons occurs readily. Predominant structure of aspartic acid is: H2NCO2- CO2- D is the ans

7 27 Upon oxidation, there is an increase of 14 units in Mr  1 O is added & 2 H removed. A: aldehyde, RCHO → carboxylic acid, RCOOH change in mass = +16  A is not the ans B: aldehyde cannot be oxidized to ketone  B is not the ans C: 1o alcohol, RCH2OH → aldehyde, RCHO change in mass = -2  C is not the ans D: 1o alcohol, RCH2OH → carboxylic acid, RCOOH change in mass = +16 – 2 = 14  D is the ans. 28 C is the ans Phenol is acidic, can be neutralised by cold NaOH(aq); alcohol is neutral and does not react with cold NaOH(aq). OH +Na -O 29 Electrophilic substitution of phenol using Br2(aq) B is the ans 30 Reaction involved is electrophilic addition of alkene. Br+ from Br2 acts as the electrophile. A is the ans Br 31 valence shell e configuration of Mn2+: 3d5 ; no paired 3d e’s valence shell e