RI 2022 Y5 Timed Practice Solutions

2022 H2 Chemistry Y5 Term 3 Common Test Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer D C 2&3 D B C A C D C D A B C A Question 1 (D) A Amount of N2 = 2.80 28.0 = 0.100 mol 1 mol of N2 has 2 mol of N atoms Amount of N atoms = 2(0.100) = 0.200 mol No. of atoms = 0.200 x 6.02 x 1023 = 1.204 x 1022 B Amount of Ar = 3.60 39.9 = 0.09023 mol No. of atoms = 0.09023 x 6.02 x 1023 = 5.431 x 1022 C Amount of C2H2 = 950 24000 = 0.03958 mol 1 mol of C2H2 has 4 mol of C and H atoms. Amount of atoms = 4(0.03958) = 0.1583 mol No. of atoms = 0.1583 x 6.02 x 1023 = 9.530 x 1022 D No. of CO2 = 5.10 x 1022 1 mol of CO2 has 3 mol of C and O atoms No. of atoms = 3(5.10 x 1022) = 1.53 x 1023 Question 2 (C) Cl2 + 2e– ⎯→ 2Cl− -- (1) Cl2 ⎯→ Cl–containing product -- (2) The 0.05 mol of Cl− formed in (1) came from ( 0.05 2 ) = 0.025 mol of Cl2. Since 0.03 mol of Cl2 was used, there is (0.03 – 0.025) = 0.005 mol of Cl2 involved in (2). The amount of Cl2 used in equation (1) : equation (2) = 0.025 : 0.005 = 5:1. Therefore, equations (1) and (2) can be written as 5Cl2 + 10e– ⎯→ 10Cl− -- (3) 1Cl2 ⎯→ Cl–containing product -- (4) The 10 mol of e– from equation (3) must be given out in equation (4). So equation (4) becomes 1Cl2 ⎯→ Cl–containing product + 10e– Since 1Cl2 gave 10e–, 1 Cl gives 5 e– to become the chlorine-containing product i.e. the final oxidation state of the Cl in the product is +5. Question 3 (2&3) As 2&3 was not provided as an option, all students were awarded one mark for this question. 1 Incorrect. Electronic configuration of Cr = [Ar] 3d 5 4s1 ⇒ electronic configuration of Cr+ = [Ar] 3d5 2 Correct. All electrons in [Ar] are paired. Cr2+ electronic configuration = [Ar] 3d4 Cr3+ electronic configuration = [Ar] 3d3 For Cr 2+ and Cr 3+, there are no paired electrons. 3 The equation describes the 3 rd IE of Cr which, from the Data Booklet, is +2990 kJ mol–1. Question 4 (D) Since there is a large jump between the 5th and 6th IE for element X, the 6 th electron is removed from an inner shell i.e. X has 5 valence electrons and is from group 15. Since W, X, Y and Z are consecutive elements, W is from group 14, X is from group 15, Y is from group 16 and Z is from group 17. The group 17 element has the highest first IE as it has the highest nuclear charge while having approximately constantly shielding effect as in W, X and Y.

Question 5 (B) The cation of J • is polyatomic (made up of many atoms) • has overall charge of 2+ • contains Pt in a +4 oxidation state • dative bonded to 6 species (NH3 or Cl−) If J is made up of 2 Cl and 4 NH3 dative bonded to Pt, then the overall charge of the cation = O.S. of Pt + 2(charge of Cl−) + 4(charge of NH3) = +4 + 2(−1) + 4(0) = 2+ This satisfies