VJC 2022 JC1 Promo H2 Chem P1 MCQ Final Solutions

1 VICTORIA JUNIOR COLLEGE 2022 JC1 PROMOTIONAL EXAM H2 CHEMISTRY PAPER 1 ANSWERS 1 C 6 A 11 B 16 B 21 C 26 C 2 B 7 B 12 B 17 A 22 C 27 D 3 A 8 B 13 A 18 D 23 B 28 D 4 A 9 C 14 C 19 A 24 D 29 A 5 A 10 A 15 D 20 B 25 C 30 D 1 C (1 and 3 only) Option 1: Correct Lv+ has 116 – 1=115 electrons while Fl– has 114 + 1 = 115 electrons Option 2: Wrong Angle of deflection α |charge / mass| Angle of deflection of Lv3+ α |+3/292| = 0.0103 Angle of deflection of Fl2– α |–2/289| = 0.00692 Option 3: Correct From the Data Booklet, Fl having 114 proton no. is in Group 14, Lv having 116 proton no. is in Group 16. Hence, outer electronic configuration of Fl: ns2np2 ; Lv: ns2np4 ; Fl2–: ns2np4 In addition, being in the same period, they have the same number of core electrons , similar to that of the noble gas at the end of the previous period. Hence, one Lv atom has the same electronic configuration as one Fl 2– ion. 2 B Large increase between the 2nd and 3rd IE ⇒ element X is in Group 2. Hence, X forms the cation X2+. To maintain overall charge neutrality, the formula of compound Y should thus be XO2. 3 A The N atom in –CN group of NH 2CN has a lone pair of electrons for donation, while the B atom in BF 3 has a vacant orbital to accept the lone pair of electrons . This will allow the formation of dative bond when the shared pair of electrons is provided by only N. 4 A Option A: Correct POCl3 (4bp) Tetrahedral 109.5o CCl4 (4bp) Tetrahedral 109.5o Option B: Wrong AlCl3 (3bp) Trigonal planar 120o NCl3 (3bp + 1lp) Trigonal pyramidal 107o Option C: Wrong SO2 (2bp + 1lp) Bent shape 118o CO2 (2bp) Linear 180o Option D: Wrong ClF3 (3bp + 2lp) T-shaped 90o SF4 (4bp + 1lp) See-saw 120o & 90o 5 A Option A: Correct Both substances have intermolecular hydrogen bonding but the second substance has larger electron cloud size which is more easily polarised. Hence, there are stronger instantaneous dipole - induced dipole (id-id) interactions between CH3CH2CH2CH2OH molecules. Hence, the first substance has a lower bp. Option B: Wrong I2 is a solid whereas HC l is a gas. This indicates that I2 has a higher bp. This is because I2 has a much larger electron cloud size which is more easily polarised. Hence, larger amount of energy is required to overcome stronger id-id interactions between I2 molecules than weaker permanent dipole-permanent dipole interactions between HCl molecules. Hence, the first substance has a higher bp. Option C: Wrong Larger amount of energy is required to overcome the stronger hydrogen bonding between CH3CH2NHCH3 molecules than the weaker permanent dipole -permanent dipole interactions between CH 3CH2OCH3 molecules. Hence, first substance has a higher boiling point. Option D: Wrong CH3COOH exists as dimers, whereby two CH3COOH molecules bond to each other via intermolecular hydrogen bonds. As such, higher extent of hyd