2024 Y5 H2 Chem TP Solutions

2024 H2 Chemistry Y5 Timed Practice – Suggested Solutions Section A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C C B D B A B A B D D D C D C 1 Answer: C Amt of atoms in 0.2 g O2 = 0.2 / 32.0 x 2 = 0.0125 mol Amt of atoms in 0.225 g H 2O = 0.225 / 18.0 x 3 = 0.0375 mol Amt of ions in 1.275 g of Al2O3 = 1.275 / 102 x 5 = 0.0625 mol Amt of atoms in 0.300 dm3 of Ne at r.t.p. = 0.300 / 24 x 1 = 0.0125 mol Amt of atoms in 0.284 dm3 of N2 at s.t.p. = 0.284 / 22.7 x 2 = 0.0250 mol 2 Answer: C amt of e – transferred = 2 x amt of Sn2+ reacted = 2 x (31.20 / 1000 x 0.0400) = 2. 496 x 10–3 mol amt of IO3− reacted = 25.0 / 1000 x 0.0250 = 6.25 x 10–4 mol mol ratio of e – : IO3− = 2.496 x 10–3 : 6.25 x 10–4 = 3.99 : 1 ≈ 4 : 1 (i.e. 1 mol of IO3− gains 4 moles of e–.) oxidation number of I in product = (+5) – 4 = +1 3 Answer: B There is a large decrease in the 5th IE from R to S ⇒ R4+ has a noble gas configuration after losing 4 electrons. ⇒ R is in Group 14. ⇒ Q is in Group 13. The chloride is QCl3. 4 Answer: D The angle of deflection is proportional to charge to mass ratio. For 7Li+, charge / mass = +1/7 and angle = +xo (towards +ve terminal) Option A: charge / mass = –2/14 = –1/7, so angle = –x o (towards –ve terminal) Option B: charge / mass = –1/16, so angle < –xo (towards –ve terminal) Option C: charge / mass = +3/27 = +1/9, so angle < +xo (towards +ve terminal) Option D: charge / mass = +7/35 = +1/5, so angle > +xo (towards +ve terminal) This document is copyrighted, please do not reproduce it without permission

5 Answer: B BrF2+ is bent with a bond angle of about 105o. NO2 is bent with a bond angle greater than 120o. XeF2 is linear with a bond angle of 180o. Hence bond angle of BrF2+ < NO2 < XeF2. 6 Answer: A Statement 1 is correct as there is an intramolecular hydrogen bond between one of the F atoms and the H atom of the –OH group. Statement 2 is correct as there are 4 bond pairs (3 single bonds and 1 dative bond) around B atom and so the bonds around B are tetrahedrally arranged. Statement 3 is correct as there are 4 bond pairs around N atom but 2 bond pairs and 2 lone pairs around O atom, causing the H–N– O angle to be 109.5° and the H –O–N angle to be 105° . This is because the lone pair exerts greater repulsion than the bond pair. 7 Answer: B Option 1 Si, AlF 3, HCl are giant molecular, ionic, simple molecular respectively. Option 2 SiC l4, Al2O3, HBr are simple molecular, ionic, simple molecular respectively. Option 3 AlC l3, SiO2, BaI2 are simple molecular, giant molecular, ionic respectively. 8 Answer: A • and ○ represent Na+ and Cl− respectively. (Note: Na+ is smaller than Cl− in size.) Only in structure A, all Na+ and Cl− are bonded to oppositely charged ions. 9 Answer: B At constant V and T, p is directly proportional to n