MI Prelim H1 Chem P1 Ans for exchange

Class Adm No Candidate Name: This question paper consists of 14 printed pages and 2 blank pages 2025 Preliminary Examination Pre-University 2 H1 CHEMISTRY 8873/01 Paper 1 Multiple Choice 17 September 2025 1 hour Additional materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, class and admission number in the spaces provided at the top of this page and on the Multiple Choice Answer Sheet provided. 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 Multiple Choice Answer Sheet provided. Read the instructions on the Multiple Choice 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. FOR EXAMINER’S USE TOTAL (30 marks) H

2 MCQ ANSWERS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C B B A D B C D C C B C A B D 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A B C C B D B A D B A B C C C 1 Which of the following statements about 1 mol of a gas is always correct? A It has the same mass as any gas with 6.02 × 1023 molecules. B It has the same number of atoms as 1 mol of hydrogen gas. C It occupies the same volume as 1 mol of H2S gas under the same conditions. D It contains the same number of particles as 1 12 g of 12C. C - 1 mol of any gas occupies the same volume as each other, under the same conditions. X Option A - 6.02 × 1023 gas molecules  1 mole gas  molar mass of the gas - different gases have different molar masses. X Option B - the gas may exist as monatomic  has 6.02 × 1023 atoms - H2 gas consists of diatomic molecules  1 mol H2 gas  2 x 6.02 × 1023 atoms. X Option D For the 1 mol gas: - if it were a monatomic gas  6.02 × 1023 atoms - if it were a diatomic gas  2 x 6.02 × 1023 atoms - if it were a tri-atomic gas  3 x 6.02 × 1023 atoms Whereas: 1 12 g of 12C  1 12  12 mol of 12C  1 12 g of 12C  0.00694 mol of 12C  0.00694 mol of 12C  0.00694 x 6.02 × 1023 atoms of 12C atoms

3 [Turn over 2 In an experiment, 50 cm3 of a 0.1 mol dm−3 solution of a metallic salt reacts exactly with 25 cm3 of 0. 1 mol dm −3 aqueous sodium sulfite, Na 2SO3. The half -equation for the oxidation of the sulfite ion is shown below: SO32−(aq) + H2O(l) ⟶ SO42−(aq) + 2H+(aq) + 2e− If the original oxidation number of the metal ion for the metallic salt is +3, what is the new oxidation number of the metal ion? A +1 B +2 C +4 D +5 Amount of SO32− = 25/1000 x 0.1 = 0.0025 mol Amount of electrons lost by SO32− = 0.0025 x 2 = 0.005 mol Amount of electrons gained by metal = 0.005 mol Amount of metal present = 50/1000 x 0.1 = 0.005 mol Mole ratio of metal : electrons gained = 1:1 Every one mole of metal gained one mole of electron. Final oxidation number of metal = +3 + (-1) = +2 3 A carbon -containing macromolecule contains a large amount of carbon, mainly of isotopes 12C and 13C. It was found that the relative atomic mass of carbon in the molecule is 12.2. What is the ratio of 12C to 13C in the molecule? A 3:1 B 4:1 C 3:4 D 1:4 Let the % abundance of 12C be x. Therefore, the % abundance of 13C is (100-x). 12.2 = [12x + 13(100-x)] / 100 X = 80 Therefore ratio of 12C to 13C = 80:20 = 4:1 4 Use of the Data Booklet is relevant to this question. What does the ion 31P3– have in common with an 40Ar atom? A They have isoelectronic structures. B Both have more neutrons than electrons. C The electronic configuration of their valence electron shell is 2s2 2p6. D Both particles are deflected to the same extent when passed through an electric field with the same strength. A

4 31P3– ion has a total of (15+3) electrons = 18 electrons 40Ar atom has a total of 18 electrons 31P3– ion and 40Ar atom have isoelectronic structure as each other. X B 31P3– ion has 18 electrons and 16 neutrons (31-15 = 16)  31P3– has less neutrons than electrons  B is incorrect. X C 31P3– ion and 40Ar atom are Period 3 elements, their valence electron shell would be principal quantum shell 3 and not 2, i.e. valence shell electronic configuration is 2s2 2p6. X D 31P3– ion is negatively charged and 40Ar atom is electrically neutral, they will be deflected differently in an electric field. 40Ar atom will not be deflected. 5 Elements X and Y have successive ionisation energies as shown: ionisation energies / kJ mol‒1 1st 2nd 3rd 4th 5th 6th 7th X 787 1577 3232 4356 16091 19805 23780 Y 1310 3400 5300 7500 11000 13300 20300 What could be the formula of the compound of these two elements? A X2Y3 B X3Y2 C XY D XY2 D Element X: Group 14 element (4 valence e); Element Y: Group 16 element (6 valence e)  mole ratio X : Y = 1 : 2  formula is XY2

5 [Turn over 6 What is the electronic configuration of an element with its first ionisation energy higher than each of the elements before and after it in the Periodic Table? A 1s22s22p2 B 1s22s22p3 C 1s22s22p4 D 1s22s22p5 B - All the electronic configurations given in the options, suggest the element is in Period 2 and they are in consecutive order, i.e. A Group 14, B is Group 15, C is Group 16 and D is Group 17. - 1st IE of atoms will increase across Period 2 due to increase nuclear charge and insignificant change in electron shielding effect  effective nuclear charge causes stronger forces of attraction between the nucleus and the valence electron  higher 1st IE across Period 2  1st IE of Option A < 1st IE of Option B. 16 and D is Group - 1st IE of Option C will be less than Option B because in C, there is greater inter - electronic repulsion than B  1st IE of Option C < 1st IE of Option B. -  1st IE Option A < 1st IE Option B > 1st IE Option C  Option B has higher 1st IE than either side of the elements beside it in the PT. 7 An element R has a giant molecular structure. Which statements are true of R? 1 It has strong covalent bonds between the molecules. 2 It has high melting point. 3 It is insoluble in water. 4 It can conduct electricity in the solid state. A 1 and 2 only B 1 and 4 only C 2 and 3 only D 3 and 4 only C X Option 1 - the strong covalent bonds in R, are between atoms and not molecules since R is an element.  Option 2 - melting involves breaking the strong covalent bonds between the atoms in R. - Since R has extensive covalent bonds between the atoms, large amount of energy is required to break the covalent bonds  mp increases.

6 Alkene functional group undergoes addition reaction with H 2 in the presence of platinum catalyst.  Option 3 - all macromolecular substances are insoluble in water. X Option 4 - Element R is unknown, hence unable to determine if it can conduct electricity. 8 In which pair of chemical species, is the bond angle of I smaller than that in II? I II A CO2 SF6 B NO3− SO2 C CO2 NO3− D SF6 NO3− D I II Species e− domains Shape & bond angle Species e− domains Shape & bond angle A CO2 linear 180o SF6 octahedral 90o B NO3− trigonal planar 120o SO2 bent < 120o (~117.5o) C CO2 linear 180o NO3− trigonal planar 120o