RVHS Prelim H2 Chemistry P1-3 Soln

River Valley High School 9729//PRELIMS/25 2025 Preliminary Examination [Turn over RIVER VALLEY HIGH SCHOOL JC2 H2 CHEMISTRY 9729 PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 1 1 D 6 C 11 C 16 D 21 A 26 C 2 D 7 D 12 D 17 C 22 C 27 B 3 C 8 B 13 C 18 D 23 B 28 D 4 D 9 A 14 D 19 A 24 C 29 B 5 B 10 D 15 D 20 A 25 C 30 C 1 A2+ has the highest third ionisation energy ➔ 3rd electron is removed from the inner principal quantum shell ➔ Electronic configuration of A2+ is 1s22s22p6 ➔ Electronic configuration of A is 1s22s22p63s2 (Period 3 Group 2) ➔ Electronic configuration of I is 1s22s22p63s23p64s2 (Period 4 Group 2) Option A is incorrect. Electronic configuration of B2+ is 1s22s22p63s1 Electronic configuration of C2+ is 1s22s22p63s2 Since B and C has the same number of inner principal quantum shells, shielding effect provided on the valence electrons is the approximately the same. Since C has greater number of protons than B, the ionic size of C2+ is smaller than that of B2+. Option B is incorrect. H is in Group 1, I is in Group 2. Since angle of deflection is directly proportional to charge/ mass, H+ will be deflected to a larger extent. Option C: ➔ Electronic configuration of C2+ is 1s22s22p63s2 ➔ Electronic configuration of D2+ is 1s22s22p63s23p1 Less energy is needed to remove an electron from D2+ as 3p electron is of higher energy than 3s electron. Option D: Correct Answer: D

2 River Valley High School 9729/PRELIMS/25 2025 Preliminary Examination 2 No. of moles of MoOx2− = 50 1000 × 0.300 = 0.0150 mol No. of moles of MnO4− = 45 1000 × 0.200 = 0.00900 mol 0.00900 mol MnO4− is reduced to Mn2+ by gaining 0.0450 mol of electrons. 0.0150 mol Mo3+ is oxidised to MoOx2− by removing 0.0450 mol of electrons. 1 mol Mo3+ is oxidised to MoOx2− by removing 3 mol of electrons Let the oxidation state of Mo in MoOx2− be y. [O]: Mo3+ → MoOx2− + 3e +3 = y− 3 y= +6 +6 −2x = −2, solving x = 4 Answer: D 3 CxHy + (x+ 𝑦 4) O2 → xCO2 + 𝑦 2 H2O 20 cm³ 60 cm³ KOH absorbs CO2 ⇒ VCO₂=60 cm³. Thus, x=60/20=3 VCxHy + Vinitial O2 = 20 + 120cm³ = 140 cm³ After combustion, Vexcess O2 + VCO₂ = 140 – 50 = 90 cm³ Vexcess O2 = 90 – 60 = 30 cm³ VO2 consumed during combustion = 120 – 30 = 90 cm³ x+ 𝑦 4 = 90/ 20 = 4.5 y = 6 Answer: C 4 BF3: 3 bond pair 0 lone pair => 120° SiF4: 4 bond pair 0 lone pair => 109.5° SF2: 2 bond pair 2 long pair => less than 109.5° (actual 98.3°) BrF5: 5 bond pair 1 lone pair => less than 90° (actual 85°-90°) Answer D

3 River Valley High School 9729/01/PRELIMS/25 2025 Preliminary Examination [Turn over 5 Hybridisation of N NO2 : sp2 NO2+: sp NO2−: sp2 NO3−: sp2 Since all species contains a N=O double bond, sp -hybridised N in NO 2+ will have the most effective orbital overlap due to the high s character. Answer: B 6 Answer: C pV = nRT pV  T At a lower T, pV is lower. 7 Amount of CH3CH2OH = 0.86 46 = 0.0187 mol Heat released from combustion of ethanol = 0.0187  1367 = 25.56 kJ Heat absorbed by water = 300  4.18  18 = 22570J = 22.57 kJ Process efficiency = 22.57 25.56  100% = 88.0% Answer: D 8 N2(g) + 4H2(g) + Cl2(g) 2NH4Cl(s) 2NH3(g) + H2(g) + Cl2(g) 2NH3(g) + 2HCl(g) –629 = –92 + 2Hf(HCl(g)) + 2(–176) Hf(HCl(g)) = –92.5 kJ mol−1 Answer: B –629 –92 2Hf(HCl(g)) 2(–176)

4 River Valley High School 9729/PRELIMS/25 2025 Preliminary Examination 9 Option 1 is correct. +1434 2CO2(g) + 3H2O(g) → CH3CH3(g) + 7 2O2(g) 3y –1542 2CO2(g) + 3H2O(l) 3y = –(–1542) – 1434 y = +36 kJ mol−1 Option 2 is correct. For reaction 2, ΔG = 0. 0 = y – (373)ΔS ΔS = 𝑦 373 kJ mol−1 K−1 Option 3 is incorrect. For reaction 1, ΔH > 0, ΔS < 0, –TΔS > 0. ΔG > 0 at all temperatures, and is spontaneous at all temperatures. Answer: A 10 Let half-life of Q be T, and half-life of P be 2T. Let initial number of atoms of P = NP and initial number of atoms of Q = NQ = 4Np. After two half-lives of P, time elapsed = 4T ➔ this is equivalent to four half-lives of Q For P, number of atoms after 4T = NP / 4 For Q, number of atoms after 4T = NQ / 16 = 4NP / 16 = NP / 4 Hence the ratio of number of P to number of Q after two half-lives of P = 1 Answer: D 11 Total number of molecules, at either T1 or T2, is represented by: P + Q Number of molecules with energy equal to or greater than activation energy at T2: Q + R Therefore the fraction will be represented by . Answer: C QR PQ + +

5 River Valley High School 9729/01/PRELIMS/25 2025 Preliminary Examination [Turn over 12 The oxidation state of sulfur increases from 0 to +6 in SF 6, +2 in SC l2 and +1 in S2Br2. Hence this shows that the oxidising ability of F2 is the highest, followed by Cl2 then Br2. Answer: D 13 The reaction gives higher methanol yield at lower T ⇒ forward reaction is exothermic ⇒ H < 0. Decrease in disorder as number of moles of gas molecules decreases from 3 to 1⇒ S < 0 Since G = H − TS, the gradient of G against T graph gives −S. With S < 0, the gradient is positive. (Options B and D eliminated) With the y-intercept as H, and H < 0, the line starts below zero at the y-axis. At low temperature, |−TS| < |H|, G < 0, reaction is sponetaneous. As temperature increases, |−TS| > |H|, G becomes positive. Reaction is non- sponetaneous. Hence, answer is C. 14 Option A is incorrect. N2O4(g) ⇌ 2NO2(g) Initial amount / mol 1 0.2 Change in amount / mol –0.24 +0.48 Eqm amount / mol 0.76 0.68 Option B is incorrect. Kc = [𝑁𝑂2]2 [𝑁2𝑂4] = ( 0.68 𝑉 )2 ÷ ( 0.76 𝑉 ) = 0.608 𝑉 Since the volume of the reaction vessel is not known, it is not possible to calculate the value of Kc. Option C is incorrect. Since V and T are constant, pressure is directly proportional to amount. As the total number of moles of gas increased from 1.2 mol to 1.44 mol, the total pressure in the vessel at equilibrium should also increase. Answer: D

6 River Valley High School 9729/PRELIMS/25 2025 Preliminary Examination 15 Option A is incorrect. O−H bonds are broken. H2O  H+ + OH− Position of equilibrium shifts to absorb heat when T increases. Since Kw increases, forward reaction is favoured. The ionic dissociation process is endothermic. Option B is incorrect. The ionic dissociation of water increases ten-fold. Option C is incorrect as concentration of H+ is the same as the concentration of OH− in water, pH = −lg [H+] [H+] = [OH−] = (Kw)1/2 Option D is correct. When T increases, Kw increases, [H+] (= [OH−]) increases. pH decreases. But water remains neutral because at all temperatures, [OH−] = [H+] Answer: D 16 Ka(C6H5COOH) > Ka(C6H5COOH). C6H5COOH is a stronger acid. Option 1 is incorrect. Since both acids are monoprotic, and as a stronger acid, C6H5COOH dissociates to a larger extent to give H+ ions. The pH of C6H5COOH should be lower. Option 2 is incorrect. 2 a ( C)K (1 )C = − 2 a CK 1 = − Option 3 is correct. Since C 6H5COOH is a stronger acid than CH 3COCO2H, the conjugate base of C6H5COOH is weaker than that of CH3COCO2H. Hence, pKb of C6H5COO− is larger. Answer: D

7 River Valley High School 9729/01/PRELIMS/25 2025 Preliminary Examination [Turn over 17 Option C is incorrect. H3PO4 and HPO42− is not a conjugate acid-base pair, thus they do not form a buffer. −H+ −H+ −H+ H3PO4  H2PO4−  HPO42−  PO43− Option A, B & D are correct. In decreasing order of acidity (thus, magnitude of Ka) H3PO4, H2PO4−, HPO42−, PO43− In increasing order of basicity (thus, magnitude of Kb) PO43−, HPO42−, H2PO4−, H3PO4 Answer: C 18 For precipitation, I.P. = Ksp [Cr3+][OH−]3 = 6.3 × 10