2021 C1 Promo Teaching Solutions Compiled

2021 HCI C1 H2 Chemistry Promotional Exam / Paper 1 HWA CHONG INSTITUTION 2021 C1 H2 CHEMISTRY PROMOTIONAL EXAM SUGGESTED SOLUTIONS Paper 1 1 2 3 4 5 6 7 8 9 10 C C B B D A C A D B 11 12 13 14 15 16 17 18 19 20 D A B C B C B C D A Comments 1 C Since the s and p subshells in inner quantum shells are completely filled, the s and p electrons from the inner quantum shells are all paired. Hence, we only need to consider the valence shell electronic configuration of the particles. Option Explanation 1 Cr: 3d54s1. It contains an unpaired s electron. 2 Ge: 4s24p2. It contains an unpaired p electron. 3 S: 3s23p4. Hence, valence shell electronic configuration of S 2is 3s23p6. It does not contain an unpaired s electron or an unpaired p electron. 4 Sc: 3d14s2. It does not contain an unpaired s electron or an unpai red p electron. General comments: Give answers that are clear, concise, coherent and in legible handwriting. Do not cram your answers or write into the edges of the page. If more space is required, use the additional page at the end of Paper 3, or request for writing paper for Paper 2. Attempt all calculation questions, so that you would be able to move on to subsequent parts. Follow through marks may be awarded (error carried forward) for correct working steps. For Paper 3, answers should always be written on the lines, except when the question specifies answers to be written in the table/figure itself. Avoid the use of ambiguous terms like “it” or “they”, when there are multiple species, groups, and/or interactions in a question. Use terms like C=C carbon / C=O double bond to refer to specific atoms or bonds, especially if there are a few different kinds of carbons / more than one double bond in the molecule. Be careful with the terms atoms, molecules, ions, carbocation, radical, elements, compounds – these differ in meaning and are not interchangeable. For Paper 1, plan your time well so you do not leave any answer unshaded on the OMS.

1 2 C Balancing the mass number and proton number on both sides of the equation, n0 1 + X7 16  C6 14 + H1 3 Total no. of protons present in products = 6 + 1 = 7 = No. protons present in reactants Since a neutron has proton number of zero, X has 7 protons  X is N (identity of an element is based on the no. of protons it possesses)  X contains 16 – 7 = 9 neutrons 3 B Given that Y has at least 19 electrons, the Period number of Y must be 4 or higher. From Period 4 onwards, the d subshell needs to be considered when deducing the electronic configuration of the element from successive ionisation energy data. Options C and D are incorrect. The graph shows a sharp increase from the 15 th to 16th ionisation energy, which indicates that the 16th electron is in an inner quantum shell. If Y is a Period 4 Group 17 element ([Ne]3s 23p63d104s24p5), after