NJC 2018 A Level Chem Solution

H2 Chemistry TYS 2018 suggested answers Paper 1 Answer Key 1 D 11 C 21 C 2 D 12 D 22 A 3 D 13 A 23 B 4 B 14 A 24 D 5 C 15 A 25 A 6 D 16 A 26 B 7 C 17 B 27 B 8 C 18 A 28 C 9 C 19 C 29 D 10 B 20 D 30 D 2018 A level Paper 1 suggested answers 1) 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: D 2) AlCl3 is a covalent compound with simple molecular structure. [Note: Al3+ has a high charge density with strong polarising power that can polarise the large electron cloud of Cl− significantly, leading to overlapping of electron cloud between Al and Cl. This causes the sharing of electron cloud, thus covalent bond is formed between Al and Cl] AlCl3 dimerises to Al2Cl6 as the empty 3p orbital of Al can accept a pair of electrons from Cl. Dative bond is from Cl to Al. Ans: D 3) CSeSe Se ClCl  C Cl ClClCl Around C: 2 b.p. 0 l.p. (linear) Around Se: 2 b.p. 2 l.p. (bent) Around C: 4 b.p. 0 l.p. (tetrahedral) non-polar polar non-polar Ans: D 4) Cs+, I− and Xe are isoelectronic with 54 electrons. Nuclear charge increases from I− < Xe < Cs+ while shielding effect remains constant. Nuclear attraction towards the valence electrons increases from I− < Xe < Cs+. Largest amount of energy is required to remove the most loosely held electron from Cs+. Ans: B

5) Option A is false as 1 mole of compound like Br2, contains 2 mol of atoms. Option B is false as we should only consider mass of lithium−7 isotope instead of all isotopes. Option D is false as molecular mass is the sum of average mass of the atoms in compound E. Ans: C 6) Atomic radius increases down a Group and decreases across the Period. Rb and In are both in Period 5. Atomic radius of Rb > In Ans: D 7) PO2 from air at sea level = 0.2 bar Percentage of O2 in tank = PO2 Ptotal × 100% = 0.2 4 × 100% = 5% Ans: C 8) Option A is wrong. It represents L.E. of Al2O3(s) [L.E. is when one mole of ionic solid is formed from its constituent gaseous ions] Option B is wrong. It represents 2×ΔHn [ΔHn is when one mole of H2O is formed from acid-base reaction] Option D is wrong. It represents 8×ΔHf SO2(g) [ΔHf is when one mole of compound is formed from its constituent elements in their standard state] Ans: C 9) 2×ΔHvap H2O = 890 + 410×4 + 496×2 − 805×2 − 460×4 = +72 ΔHvap H2O = +36 kJ mol−1 Ans: C

10) 0.001 mol of iodine oxide reacts with 0.01 mol of I − to gives 0.006 mol of I 2. Balancing number of I atoms on both sides, 0.006 mol of I 2 contains 0.012 mol of I atoms. Hence, 0.001 mol of iodine oxide contains 0.002 mol of iodine atoms. 0.002 mol Ix+ reacts with 0.010 mol I− to produce 0.006 mol I2. [O] 2I− ⎯→ I2 + 2e− 0.010 mol I− gives 0.010 mol of e− [R] Ix+ + ne− ⎯→ I2 0.002 mol of Ix+ gains 0.010 mol of e− 1 mol of Ix+ gains 5 mol of e− to give I2 During reduction, oxidation state of Ix+ decreases by 5 units from +5 to 0. Ans: B 11) It is given that order of reaction wrt NO2 is 2, the slow step should involve two molecules of NO2 and overall equation of step 1 and 2 must be the same as the given equation. Ans: C 12) For enzyme catalysed reaction, the rate of reaction increases with the concentration of reactants and reach a maximum rate when the active sites of the enzyme are saturated. After which, any increase in concentration of reactants would not increase the rate any further. Ans: D 13) High charge density cation polarises the electron cloud of H2O ligands and weaken the O−H covalent bond to produce H+. Al3+ has the highest charge density, the O−H bond is weakened most significantly and produce the largest concentration of H+. [Al(H2O)6]3+ [Al(H2O)5(OH)]2+ + H+ Ans: A 14) H2O H+ + OH− When temperature increases from 10 to 40 C, the position of equilibrium shifts to the right. [H+] and [OH−] increases by the same extent. pH will become lower as there are more [H+]. pH = −lg[H+] Ans: A Note: pOH will also become lower. pKw is a smaller value at higher temperature. pH + pOH = pKw, pKw = 14 at 25 C. pKw = 14.5 at 10 C pKw = 13.5 at 40 C 15) We will need a buffer solution containing a weak base and its conjugate acid to maintain pH at about 10. Ans: A

16) AgCl is more soluble than AgBr, meaning ΔG1 is more negative than ΔG3. Option 4 is true ΔG2 = ΔG4 as formation of complex ion is due to the following equation and unaffected by the identity of the halide ions. Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+(aq) Option 2 is true and option 3 is false. Since ΔG2 = ΔG4, Option 1 is true. Ans: A 17) [O] 2I− → I2 + 2e− [R] H2O2 + 2H+ + 2e− → 2H2O Eꝋcell = +1.77 – (+0.54) = +1.23V (reaction is spontaneous) Brown I2(aq) is produced. No effervescence as the products are I2(aq) and H2O Ans: B 18) 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, NOT ppt. Option C is wrong, Cu(OH)2 is a pale blue precipitate, NOT solution Option D is wrong, [Cu(H2O)6]+Cl− is CuCl, a white precipitate as Cu+ has an electronic configuration of 3d10, hence it does not allow d-d transition and hence it would appear colourless solution/white solid. Ans: A 19) Reaction A is redox reaction and ligand exchange reaction. Cu + Cu2+ + 4Cl− → 2[CuCl2]− Oxidation state of Cu changes from 0 (in Cu) and +2 (in Cu2+) to +1 in [CuCl2]−. Reaction B is redox reaction. Cu2+ + Zn → Cu + Zn2+ Reaction C is ligand exchange reaction. [Cr(H2O)6]3+ + SO42− [Cr(H2O)5SO4]+ + H2O Reaction D is a redox reaction. MnO4− + 5Fe2+ + 8H+ → Mn2+ + 4H2O + 5Fe3+ Ans: C 20) Transition metals have high density due to close packing of heavy atoms of small size. Atomic and ionic radii in the Data Booklet shows the V is smaller in size than Ca. Options 1&2 suggest there are stronger repulsion for V, this would have implied a larger radius for V than C. Only option 3 explains the smaller size of Vanadium as compared to Calcium. Ans: D

21) C O CH3 + O H CH3H + CCH3 CH3 CH3 + NCH3 CH3 CH3 H3C+ Ans: C 22) Tollens’ reagent can oxidize aldehydes to R−COO−. Ans: A 23) Note: The initial way of drawing DOES NOT allow us to see the internal plane of symmetry. Compounds 3 & 4 can also be drawn as the followings. BrBr BrBr Compound 2 does not have an internal plane of symmetry. BrBr All four compounds contain chiral carbon. However, compounds 3 & 4 are meso compound with an internal plane of symmetry, rotation of plane−polarised light by each chiral carbon gets cancelled out completely. Ans: B 24) 4 phenol groups and 1 carboxylic acid group would react with 5 mol of NaOH(aq). 1 ester group would react with 1 mol of NaOH(aq) with heating. Hence, 1 mol of rosmarinic acid reacts with 6 mol of NaOH(aq) with heating. Ans: D 25) Option 1 is correct. In SN2 mechanism, the nucleophile can only attack from the opposite of the leaving group. This causes the inversion of stereo configuration of the chiral carbon. Option 2 is correct. In SN1 mechanism, a trigonal planar carbocation is produced after the first step. In step 2, the nucleophile can attack from either side of the trigonal planar carbocation with equal probability, giving rise to a racemic mixture containing equal amount of the two enantiomers. Option 3 is correct. Tertiary chloroalkanes undergo SN1 mechanism and involve a carbocation that is stabilized by three electron donating alkyl groups. Ans: A

26) HA H+ + A− The conjugate base of 2−hydroxybenzoic acid is stabilized by intramolecular H−bonding due to the close proximity of the –COO− and –OH group. This cause the dissociation equilibrium to favour the forward reacti