EJC Prelim H1 Chemistry Paper 1 Worked Solutions

© EJC [Turn Over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2025 General Certificate of Education Advanced Level Higher 1 CHEMISTRY Paper 1 Multiple Choice 8873/01 19 September 2025 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, civics group and registration number on the Answer Sheet in the spaces provided unless this has been done for you. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this question paper. The use of an approved scientific calculator is expected, where appropriate. This document consists of 12 printed pages.

2 © EJC 8873/01/J2PE/20 1 In which pair of atoms, in their ground states, have the same number of unpaired electrons? A B and Ca B F and Na C Ne and P D Be and Si Answer: B B: 1s22s22p1 (1 unpaired electrons in 2p subshell) Ca: 1s22s22p63s23p63d104s2 (No unpaired electrons) ✓ F: 1s22s22p5 (1 unpaired electrons in 2p subshell) ✓ Na: 1s22s22p63s1 (1 unpaired electrons in 3s subshell) Ne: 1s22s22p6 (No unpaired electrons) P: 1s22s22p63s23p3 (3 unpaired electrons in 3p subshell) Be: 1s22s2 (No unpaired electrons) Si: 1s22s22p63s23p2 (2 unpaired electrons in 3p subshell) 2 Three particles approach an electric field at the same speed. They are deflected as they pass through the electric field. What could be the identities of particles X, Y and Z? X Y Z A 35 17Cl– 37 17Cl– 23 11Na+ B 6 3Li+ 7 3Li+ 1 1H– C 37 17Cl– 35 17Cl– 23 11Na+ D 7 3Li+ 6 3Li+ 1 1H– Answer: A Since particles X and Y are deflected to the positive terminal, they are negatively charged. Since particle Z is deflected to the negative terminal, it is positively charged. + – X Y Z

3 © EJC 9729/01/J2PE/25 [Turn over Hence, X and Y must be Cl− while Z must be Na+ (either option A or C is correct). Between particles X and Y, the larger angle of deflection of particle X indicates that its 𝑒 𝑚 ratio is larger than that of Y. Hence, particle X should have a smaller mass number than particle Y since both X and Y have the same charge. Hence, X is Cl – 17 35 while Y is Cl – 17 37 . 3 Which molecule has the largest bond angle? A BF3 B SiCl4 C H2S D PH3 Answer: A A: BF3 has a trigonal planar shape, 3 bond pairs, 0 lone pair; F‒B‒F angle = 120° (largest) B: SiCl4 has a tetrahedral shape, 4 bond pairs, 0 lone pair; Cl‒Si‒Cl angle = 109.5° C: H2S has a bent shape, 2 bond pairs, 2 lone pairs; H‒S‒H angle = 104.5° D: PH3 has a trigonal pyramidal shape, 3 bond pairs, 1 lone pair; H‒P‒H angle = 107° 4 The boiling points of hydrogen chloride, HCl, and chlorine, Cl2 are shown in the table. boiling point / ºC HCl −85 Cl2 −34 Which pair of statements explain this difference? 1 The forces of attraction between C l2 molecules are instantaneous dipole- induced dipole attractions. 2 The force s of attraction between HC l molecules are permanent dipole - permanent dipole attractions. 3 H−Cl bonds are stronger than Cl−Cl bonds. 4 The forces of attraction between C l2 molecules are stronger because C l2 molecules have more electrons. A 1 and 2 B 1 and 4 C 2 and 3 D 3 and 4 Answer: B Options 1 and 4 are correct.

4 © EJC 8873/01/J2PE/20 Option 2 × : There are also id-id forces of attractions between HCl molecules. In addition, the pd-pd forces of attraction between HCl molecules does not explain its lower boiling point than Cl2. In fact, the difference in their boiling points is explained using option 4. Option 3 × : Covalent bonds are not broken when accounting for boiling points. Hence, option 3 is incorrect in explaining the difference. 5 Which of the following molecule will not form a hydrogen bond with another of its own molecule? A CH3CHO B CH3NH2 C CH3OH D HF Answer: A • For hydrogen bonds to form, a hydrogen atom needs to be bonded to a very electronegative atoms (i.e., F, O or N). There must also be a lone pair of electrons on a F, O or N atom in an adjacent molecule. • Given H atom is not bonded directly to O atom in the aldehyde, CH3CHO (option A), hence there can be no intermolecular hydrogen bonding sformed. 6 Gases can be liquefied under certain conditions. Which statement best describes why a gas liquefies under high pressure? A Increasing the pressure causes the temperature of the gas to decrease to below its boiling point. B Increasing the pressure decreases the kinetic energy of the molecules, creating the ordered liquid state. C Increasing the pressure decreases the size of the molecules , and so the intermolecular forces of attraction become stronger. D Increasing the pressure pushes the molecules closer together, and so the intermolecular forces of attraction become stronger. Answer: D A ×: Increasing the pressure does not cause the temperature to change. B ×: K.E. of a gas is dependent on the temperature, it does not change with pressure. C ×: The size of molecules is not affected by changes in the pressure. D ✓: Increasing the pressure forces the molecules closer together, thereby increasing the strength of the intermolecular forces of attraction.

5 © EJC 9729/01/J2PE/25 [Turn over 7 The table shows the results of experiments in which the halogens X2, Y 2 and Z 2 were added to separate aqueous solutions containing X–, Y– and Z– ions. X– (aq) Y– (aq) Z– (aq) X2 no reaction no reaction no reaction Y2 X2 formed no reaction Z2 formed Z2 X2 formed no reaction no reaction Which set contains the ions X–, Y – and Z – in order of their decreasing strength as a reducing agent? strongest ⎯⎯⎯⎯⎯→ weakest A X– Y– Z– B X– Z– Y– C Y– Z– X– D Z– X– Y– Answer: B The stronger the halide as a reducing agent, the more likely it will be oxidised to a halogen. From the table, X- is able to reduce both Y2 and Z2, itself oxidising to X2. Hence, X- is the strongest reducing agent. Y- is unable to reduce any of the halogen to halides, hence Y- is the weakest reducing agent. 8 Which element forms a chloride in which both covalent bonding and dative bonding are present? A Mg B Al C Si D P Answer: B Chlorides Type of Bonding A MgCl2 Ionic bonding with chlorine B AlCl3 and Al2Cl6 Al is in group 13 and forms AlCl3 via covalent bonding. It dimerises in the gaseous state to form A l2Cl6, consisting covalent and dative bonds:

6 © EJC 8873/01/J2PE/20 Al Cl Cl Cl Cl Al Cl Cl C SiCl4 Si is in group 14 and forms 4 covalent bonds with chlorine. D PCl5 P is in group 15 and forms 5 covalent bonds with chlorine. 9 Use of the Data Booklet is relevant to this question. The sketch below shows the variation of properties of Period 3 elements. Which option shows the correct label for the axes? y x 1 melting point nucleon number 2 electronegativity atomic number 3 atomic radius highest possible oxidation state A 2 only B 2 and 3 C 1 and 3 D 1, 2 and 3 Answer: A Option 1 ×: Not correct, as melting point trend increases for the first 4 elements (i.e., Na, Mg, Al and Si) across Period 3 only. Following which, the elements are simple molecules, hence have low melting points. Option 2 ✓: Correct, as electronegativity increases the period (i.e.,increasing atomic number) Option 3 ×: Not correct, as size of atomic radius decreases across the period. y x

7 © EJC 9729/01/J2PE/25 [Turn over 10 Use of the Data Booklet is relevant to this question. A 5 g sample of a diatomi