NJC 2017 A Level Chem Solution

H2 Chemistry TYS 2017 suggested answers Paper 1 Answer Key 1 C 11 A 21 A 2 B 12 A 22 A 3 D 13 D 23 D 4 C 14 D 24 C 5 D 15 A 25 D 6 B 16 B 26 A 7 B 17 A 27 A 8 B 18 D 28 B 9 C 19 A 29 C 10 B 20 C 30 A 2017 A level Paper 1 suggested answers 1) Let % of 29Si be x, % of 30Si be (100−92.23− x) 92.23×28+𝑥×29+(100−92.23−𝑥)×30 100 = 28.10 2582.44 + 29x + 233.1−30x = 2810 x = 5.54 Ans: C 2) These are the most common isotopes for Li to Mg (Qn stated that the nucleon number is the closest whole number to its Ar) Element Li Be B C N O F Ne Na Mg Nucleon number 7 9 11 12 14 16 19 20 23 24 No. of protons 3 4 5 6 7 8 9 10 11 12 No. of neutrons 4 5 6 6 7 8 10 10 12 12 Only statements 1 and 2 are correct. Ans: B 3) C O HH For methanal, there are 3 bond pair regions and 0 lone pair region around the central C atom. It is trigonal planar with bond angle of 120°. Ans: D

4) Statement 1 is correct. More energy is required to overcome the stronger hydrogen bonding between H2O molecules as compared to the weaker instantaneous dipole-induced dipole attraction between methane molecules. Statement 2 is wrong. During boiling, only the intermolecular forces of attractions are broken. Covalent bonds are not broken. Statement 3 is wrong. Both H2O and CH4 has the same number of electrons in the molecules (10 electrons). Note that size of electron cloud affects strength of id-id. Ans: C 5) Gases liquefies under pressure as the intermolecular forces of attractions becomes more significant when the molecules are brought closer together. Hence the molecules are arranged in a more orderly arrangement in a liquid state as compared to gaseous state. Ans: D 6) Lewis base is an electron pair donor. Bronsted-Lowry acid is a proton(H+) donor. H2O can be BOTH an electron pair donor and H+ donor. Ans: B 7) Using information from the Data Booklet, atomic radius of Mg>Al>Si>P. Electronegativity increases across the period, hence electronegativity of P>Si>Al>Mg. Atomic radius of Si(0.117nm) is only slightly larger than P(0.110nm). Ans: B 8) Fifth I.E. shows the removal of electron for the following process: X4+ → X5+ + e− The sharp change in 5th I.E. from R to S shows that the 5th electron for R is removed from an inner shell as compared to S. Hence R4+ has a configuration of ns2np6. R has configuration of ns2np6(n+1)s2(n+1)p2 and is in Group 14. Q is an element from Group 13. Ans: B 9) Mg2+ has the highest charge density, and hence the strongest polarising power. It is least likely to form a stable peroxide. Ans: C

10) G = (−1.9) − 298(0.0034) = −2.9kJ mol−1. The forward reaction is spontaneous and the reverse reaction is not spontaneous. The reaction does not occur readily at room temperature and pressure even though G is negative, this implies that the rate of reaction must be very slow. Hence the forward reaction must have an extremely small rat