ASRJC 2021 J2Prelims H2Chem P1 soln

ASRJC JC2 PRELIMS 2021 9729/01/H2 [Turn over Anderson Serangoon Junior College 2021 Preliminary Examination H2 Chemistry Paper 1 Suggested Worked Solution 1 Two particles, A+ and B3+ are fired with equal velocity into an electric field. The information on two particles are given in the table below. particle number of electrons number of neutrons angle of deflection in an electric field A+ 10 12 2.0° B3+ ? 14 5.1° What is the number of electrons for B3+? A 10 B 11 C 12 D 13 Answer: A e p n (given) Mass (n+ p) Charge, q q (given) A+ 10 10+1 = 11 12 23 +1 2.0° B3+ ? ? 14 m +3 5.1° By proportion, for B3+ = ÷2 x 5.1 = 0.1109 \ = 0.1109 m = 27 Number of protons for B3+ = 27 -14 =13 Number of electrons for B3+ = 13-3 = 10 q m 1 23 3 m q m 1 23 3 m

2 ASRJC JC2 PRELIMS 2021 9729/01/H2 2 A to H are consecutive elements with atomic numbers less than 20. The graph below shows their second ionisation energies (2nd I.E.). Which of the following statements is correct? A The 2nd IE of G is lower than that of F due to the inter-electronic repulsion between its paired s electrons. B C exists as diatomic molecules at room temperature. C The compound formed between A and E has a low melting point. D Element B is from Group 17. Answer: D Since there is sharp decrease in 2nd IE between elements D & E, the electron removed from E+(g) is in an electron shell that is further from the nucleus than that from D+(g). D+ has noble gas configuration. D has 1 valence electron, hence it is from Group 1. Counting backwards, B is from Group 17. 0500100015002000250030003500400045005000 ABCDEFGH 2nd I.E. element

3 ASRJC JC2 PRELIMS 2021 9729/01/H2 [Turn over 3 The graph shows the logarithm of the first twelve ionisation energies (I.E.) for element J. What can be deduced about element J from the graph? 1 It can form a compound with oxygen with the formula J2O3. 2 J is likely to have a lower first I.E. than the element preceding it in the same period. 3 It is in the second period (Li to Ne) of the Periodic Table. 4 J has a half-filled p-subshell. A 1, 2 and 3 B 1, 2 and 4 C 1 and 2 only D 3 and 4 only Answer: C 1 Correct Element J belongs to Group 13. Largest energy difference between the 3rd and 4th I.E. This implies that the removal of the 4th electron is from an inner quantum shell which requires more energy. Hence, element J has 3 valence electrons and can form a compound with oxygen with the formula J2O3. 2 Correct Valence electronic configuration for J: ns2 np1 Valence electronic configuration for element preceding J: ns2 The first ionisation energy for element J requires removing 1 electron from p subshell which is further from the nucleus than s subshell and experience weaker electrostatic attraction. Hence, the first I.E of J is lower than the element preceding it in the same period. 3 Incorrect Period