EJC Prelim P1_Worked_Solution
Uploaded by admin · 29 August 2025
Preview
2018 JC2 Preliminary Examination H2 Chemistry 9729 Paper 1 Worked Solution 1 The positive α particles either passes straight through the empty space or are occasionally repelled by the dense positive nucleus. They are not affected by the much smaller electrons and hence does not provide any information about the electrons. C 2 [O] : 2Am Amnn ne++ −→+ [R] : 32Cu Cunn e n+− ++→ So, 32 2Am Cu Am Cunn nn++ +++→+ 4 Am 25.0 0.0100 2.5 10 mol1000 nn + −=× = × 3 4 Cu 15.0 0.0500 7.5 10 mol1000n + −=× = × CuAm :1 : 1 : 3 3nnn n n+ == = Am species formed has O.S. of +6 () 2 2AmO + D 3 OF 2 : 2 b.p. + 2 l.p. bent 105º CH4 : 4 b.p. + 0 l.p. tetrahedral 109.5º SF6 : 6 b.p. + 0 l.p. octahedral 90º H2O : 2 b.p. + 2 l.p. bent 105º O3 : bent <120º N2O : linear 180º PF3 : 3 b.p. + 1 l.p. trig. pyr. 107º BF3 : 3 b.p. + 0 l.p. trig. planar 120º IF3 : trig. planar 120º NO2 : bent >120º SO2 : 2 b.p. + 1 l.p. bent <120º CO2 : 2 b.p. + 0 1.p. linear 180º B 4 Bonding in solid NH 4HF2: A 5 At low pressure, gas particles are far apart. IMF between particles causes force they impinge on wall to be smaller and thus pressure is lower than ideal gas pVreal < pVideal At higher pressure, gas particles are close together and the volume occupied by particles is significant compared to volume of container. Hence the gas occupies a larger volume than ideal gas pVreal > pVideal B 6 Arrhenius acid: Produces H+(aq) in water Brønsted-Lowry acid: H+ donor; B(OH)3 accepts a OH– and does not donate H+ Lewis acid: Lone pair acceptor; accepts a lone pair fron OH– Monobasic acid: each mole of B(OH)3 reacts with one mole of OH– B 7 NaC l : no hydrolysis pH = 7 MgCl2 : slight hydrolysis pH ≈ 6.5 22MgC H O Mg(OH)C HC++ll l … AlCl3 : extensive hydrolysis pH ≈ 3 32 3A C 3H O A (OH) 3HC+→ +ll l l SiCl4 : complete hydrolysis pH ≈ 1 42 2SiC 2H O SiO 4HC+→ +ll PCl5 : complete hydrolysis pH ≈ 1 52 3 4PC 4H O H PO 5HC+→ +ll A 8 A : Graph is for ionic radii B : Melting point of Si should be highest due to extensive strong Si–Si covalent bonds within giant molecular structure C : Nuclear charge increases across period while shielding effect is the same, hence ENC increases across the period resulting in increase in electronegativity D : Silicon being a metalloid should only have conductivity below that of the metals (Na, Mg, Al) C 9 () () ()() ()() ()() ()() ()() ()() ()() ()() 1f f ff 2 ff 2 f f2 ff 2 products reactants 2N a O H a q H g 2N a s 2H O 2N a O H a q 0 02 H O 2N a O H a q 2H O HH H HH HH H H HH Δ= Δ − Δ =Δ + Δ − Δ− Δ =Δ + − −Δ =Δ −Δ l l l dd d dd dd d d dd ()() ()()f2 c 2HO H gHHΔ= Δ ldd as both corresponds to the same reaction: H2(g) + ½O2(g) → H2O(l) B 10 B : At 298 K, C is a solid C : Hydrogen exists as H 2 under std state D
Content continues in the PDF.
Related notes
- 2026 H2 Timed Practice Paper 2 Solutions + Examiner Comments (updated 17 July)MYEs/CAs/Other Tests · 2026
- 2026 H2 Timed Practice Paper 2 QP (to upload)MYEs/CAs/Other Tests · 2026
- 2026 H2 Timed Practice Paper 1 MCQ (Question Paper)MYEs/CAs/Other Tests · 2026
- 2026 H2 Timed Practice Paper 1 MCQ Combined + answer (finalised)MYEs/CAs/Other Tests · 2026
- Mock chem paper 2 suggested solutions (corrected)User Mock Papers
- NJC Organic Chem 2026Notes/Practices · 2026

