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 th