2023 RI H2 Chem Prelims P3 Answers

© Raffles Institution 2023 9729/03/S/23 2023 Y6 H2 Chemistry Preliminary Exam Paper 3 Suggested Solutions Section A 1(a)(i) Both diamond and silicon have giant covalent structures. During melting, strong Si– Si and C –C covalent bonds between atoms are broken in silicon and diamond respectively. Si–Si covalent bonds are weaker than C –C as the valence orbitals of Si are more diffuse than that of C, thus overlapping of the orbitals of Si is less effective than that of C. Hence less energy is required to break the weaker Si–Si bonds during melting compared to C–C bonds in diamond. 1(a)(ii) Al2O3 is amphoteric/reacts with both acids and bases, while SiO2 is acidic/reacts with bases only. Al2O3(s) + 6H+(aq) → 2Al3+(aq) + 3H2O(l) Al2O3(s) + 2OH–(aq) + 3H2O(l) → 2[Al(OH)4]–(aq) SiO2(s) + 2OH–(conc) → SiO32–(aq) + H2O(l) or Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l) Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2Na+[Al(OH)4]–(aq) SiO2(s) + 2NaOH(conc) → Na2SiO3(aq) + H2O(l) 1(b)(i) For reaction 2, H = 2(−110.5) − (−910.9) = +689.9 kJ mol−1 Assuming that H and S are independent of temperature, G2500K = +689.9 − (2500)(+361 x 10–3) = −213 kJ mol−1 (3s.f.) Reaction 2 has negative G2500K and is spontaneous, whereas reaction 1 has positive G2500K and is non−spontaneous. Hence reaction 2 is the preferred method for the extraction of Si from SiO2. 1(b)(ii) S is positive as there is an increase in the number /moles/amount of gas eous particles (from 0 to 2) in the equation.

© Raffles Institution 2023 9729/03/S/23 1(c) Or 2Ca(s) + Si(s) + 2O2(g) CaSiO4(s) Hatom [Ca] x 2 2Ca(g) + Si(s) + 2O2(g) LE [CaSiO4] 2  (1st IE [Ca] + 2nd IE [Ca]) 2Ca2+(g) + 4e− + Si(s) + 2O2(g) +13872 kJ mol−1 2Ca2+(g) + SiO44−(g) Using Hess’ Law, −2306 = 2(+121) + 2(+590 + 1150) + 13872 + LE [Ca2SiO4] LE [Ca2SiO4] = −19900 kJ mol−1 (3s.f.) 1(d)(i) The standard electrode potential, E, of a half –cell is the electromotive force / potential difference , measured at 298 K, between the half –cell and the standard hydrogen electrode, in which the concentration of any reacting species in solution is 1 mol dm–3 and any gaseous species is at a pressure of 1 bar. 1(d)(ii) Cathode: O2 + 2H2O + 4e− → 4OH− Anode: Fe → Fe2+ + 2e− (x2) Overall: O2 + 2H2O + 2Fe → 4OH− + 2Fe2+ Ecell = +0.40 − (−0.44) = +0.84 V 1(d)(iii) G = –nFEcell = −4(96500)(+0.84) = −324240 J mol−1 = −324 kJ mol−1 Hf [Ca2SiO4]

© Raffles Institution 2023 9729/03/S/23 1(d)(iv) (I) Oiling the bicycle chain minimises contact between H2O / O2 and Fe, thus rusting stops/slows down. (II) Zn2+ + 2e− ⇌ Zn E = −0.76 V Fe2+ + 2e− ⇌ Fe E = −0.44 V Zn is more easily oxidised than Fe, since E(Zn2+/Zn) is more negative than E(Fe2+/Fe). Hence Zn rusts in place of Fe. 1(e) 2(a)(i) High temperature and low pressure. 2(a)(ii) V = (5.2)(8.31)(15+273)/101300 = 0.123 m3 = 123 dm3 2(a)(iii) Hydrogen bonding in ammonia caused the molecules to be closer to one another, thus the gas occupies a volume smaller than expected. 2(b) Step 2: NH3 Step 3: LiAlH4, dry ether A is B is 2(c)(i) 2(c)(ii) In an octahedral environment, lone pairs on the ligands approach the central metal ion along the x, y and z axes. 3dx2-y2 and 3dz2 orbitals have their greatest electron density along the co-ordinate axes on which the ligands are situated. Hence electrons in these orbitals are pointing towards the lone pairs of ligands and will be repelled by them.

© Raffles Institution 2023 9729/03/S/23 3dxy, 3dyz, 3dxz orbitals have their greatest electron density in between the co-ordinate axes. Hence the repulsion between electrons in these orbitals and those of the approaching ligands will be less compared to electrons in 3dx2-y2 or 3dz2 orbitals. Hence, 3dx2-y2 and 3dz2 orbitals are at a higher energy level and 3dxy, 3dyz, 3dxz orbitals are at a lower energy level. 2(d)(i) Cr(OH)3 2(d)(ii) Oxidation 2(d)(iii) 2CrO42− + 2H+ → Cr2O72− + H2O This reaction involves the combination of two ions with the elimination of a small molecule, H2O. 2(d)(iv) OH− and H 2O are different ligands of different strength . The d orbitals of Cr 3+ are hence split into two sets of slightly different energy levels to different extents, creating different energy gaps for different complexes, which in turn absorb energies of different wavelengths from the visible light spectrum for d -d transitions , thus displaying different colours. 2(d)(v) The ability to display variable oxidation states in their compounds. This is due to the close similarity in energy of the 3d and 4s electrons, which thus allows for different number of these electrons to participate in chemical bonding. 2(e)(i) AgCl(s) + 2NH3(aq) → [Ag(NH3)2]+Cl−(aq) 2(e)(ii) Cation: [Cr(NH3)4Cl2]+ Anion: Cl− 2(e)(iii) 3(a)(i) The carbon atoms in C≡C are sp hybridised. The  bond is formed from head-on overlap of sp hybrid orbitals of each atom. 2  bonds are formed. Each  bond is formed from the side-on overlap of one unhybridised p-orbital from each atom.

© Raffles Institution 2023 9729/03/S/23 3(a)(ii) 3(a)(iii) But-2-yne and hydrogen adsorb onto the surface of the Lindlar’s catalyst through the formation of weak bonds / interactions with the surface of the catalyst. This increases the surface concentration of the reactants and brings the reactants in closer proximity and proper orientation for reaction to take place. This also weakens the covalent bonds within reactant molecules, lowering the activation energy for reaction. Once formed, the product molecules can easily desorb from the surface of the catalyst. 3(a)(iv) Upon adsorption of but -2-yne and H2 on the catalyst surface , H2 collides and forms bonds with one side of the alkyne molecule at the same time , adding two H atoms to the same side. OR

© Raffles Institution 2023 9729/03/S/23 Upon adsorption of but-2-yne and H2 on the catalyst surface, the bulky methyl groups on but-2-yne experience electronic repulsion from the catalyst surface and point away from the catalyst surface. The H atoms then add to the side of the alkyne that was previously temporily bonded to the catalyst surface. 3(b)(i) 3(b)(ii) excess CH3I in ethanol, heat in a sealed tube 3(b)(iii)

© Raffles Institution 2023 9729/03/S/23 2,4-DNPH alkaline I2 3(b)(iv) 3(c) Evidence Deduction R (C10H18O) ⎯⎯⎯→ product Nucleophilic substitution - R contains an alcohol. 1 R + 1 Br2 ⎯→ product Electrophilic addition - R contains 1 C=C. R (C10H18O) ⎯⎯⎯⎯→ S (C10H16) only product Elimination of 1 H2O / dehydration - Formed a C=C in S or R contains an alcohol S ⎯⎯⎯⎯→ T + U (C5H8O4) (C5H8O3). Strong oxidation / oxidative cleavage - S contains C=C - T and U do not contain any primary/secondary alcohol or aldehydes 1 T + 2NaOH ⎯→ product 1 U + 1NaOH ⎯→ product Acid-base reaction - T contains 2 –COOH groups - U contains 1 –COOH group - T has 5 carbons, 2 –COOH and 1 chiral centre. U ⎯⎯⎯⎯→ orange ppt Condensation reaction - U contains carbonyl / ketone group U ⎯⎯⎯⎯→ no yellow ppt No oxidation reaction - does not contain COCH3 U ⎯⎯→ 2 products. Free radical substitution - U contains two types of hydro