ASRJC 2026 Prelim P1 Ans

ASRJC JC2 PRELIM 2026 9476/01/H2 [Turn over ANDERSON SERANGOON JUNIOR COLLEGE 2026 JC2 PRELIMINARY EXAMINATION NAME:______________________________ ( ) CLASS: 26 /____ CHEMISTRY 9476/01 Paper 1 Multiple Choice SUGGESTED SOLUTIONS 1 The following statements are made by a student about an atom with atomic number 31. Which statements are correct? 1 There are a total of 8 electrons in the s orbitals. 2 There is 1 occupied dumbbell-shaped orbital in the outermost shell. 3 The d electrons have the same principal quantum number as the outermost p electrons. A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 1 only Ans: B Topic: Atomic Structure LO 6: Able to describe and draw the shapes, energy levels of s, p and d orbitals on the x, y and z axis. LO 7: Able to write the electronic configuration of atoms and ions Electronic configuration: Atomic no 31: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1 Statement 1 is correct as there are 2 electrons each in 1s, 2s, 3s and 4s orbitals. p orbital has dumbbell shape. Statement 2 is correct as only one 4p orbital is occupied in the outermost shell. Statement 3 is incorrect as the outermost p orbital is 4p which has a different principal quantum number as the 3d orbital.

2 ASRJC JC2 PRELIM 2026 9476/01/H2 2 Nuclear Magnetic Resonance (NMR) techniques can be used to detect atomic nuclei that have an odd number of protons or an odd number of neutrons. For which element may both isotopes be detected by NMR techniques? A 28Si and 29Si B 16O and 18O C 12C and 13C D 7Be and 9Be Ans: D Topic: Atomic Structure LO 2: Able to determine the numbers of protons, neutrons and electrons present in both atoms (including isotopes) and ions from the proton and nucleon (mass) numbers (and charge). Isotope No of protons No of neutrons Detected by NMR? 28Si 14 14 No 29Si 14 15 Yes 16O 8 8 No 18O 8 10 No 12C 6 6 No 13C 6 7 Yes 7Be 4 3 Yes 9Be 4 5 Yes

3 ASRJC JC2 PRELIM 2026 9476/01/H2 [Turn over 3 The element tin exists in two forms, grey tin and white tin. Some properties of grey tin and white tin are shown. grey tin white tin boiling point / K 2543 2533 electrical conductivity none in solid or liquid good in solid and liquid malleability brittle malleable Which structural change might take place when grey tin changes to white tin? A giant covalent to giant ionic B giant covalent to giant metallic C giant ionic to giant covalent D giant ionic to giant metallic Ans: B Topic: Chemical Bonding LO 24: Able to suggest the type of structure and bonding present in a substance from given information. • The very high boiling points of both grey tin and white tin indicate the two allotropes have giant structures. • In grey tin , no conductivity in solid and liquid states implies that no free mobile charged particles and electrons are available to conduct electricity. In the solid state, the Sn atoms are held by an extensive network of covalent bonds, and in the liquid state, the atoms are bonded together by electrons through covalent bonding. • In white tin, good conductivity in solid and liquid indicates that white tin has a giant metallic structure. The valence electrons and free cations are mobile in the giant metaillic solid structure and molten state repectively, and they act as charge carrier to conduct electricity.

4 ASRJC JC2 PRELIM 2026 9476/01/H2 4 Which changes occur when ethoxyethane, (CH 3CH2)2O, reacts with boron trichloride, BC l3, to form (CH3CH2)2OBCl3? 1 The Cl−B−Cl bond angle in boron trichloride decreases. 2 The boron atom accepts a pair of electrons from oxygen. 3 The C−O−C bond angle in ethoxyethane decreases. A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 1 only Ans: B Topic: Chemical Bonding LO 4: Able to apply VSEPR theory to predict the shapes of, and bond angles in molecules with respect to each central atom by referring to the number and type of electron pairs (i.e. bonding and non–bonding electron pairs). LO 6: Able to apply the concept of dative covalent (co –ordinate) bonding and explain how and why it arises between electron deficient molecules like A lCl3 or BeCl2 and molecules with lone pair of electrons e.g. NH3. + B Cl Cl Cl O O B Cl Cl Cl 1 Correct B in BC l3 has three bond pairs. The shape is trigonal planar with bond angle 120 o. On forming (CH3CH2)2OBCl3, number of bond pairs increase to four. The shape is changed to tetrahedral with bond angle decrease to 109.5o. 2 Correct The oxygen in (CH3CH2)2O donates one of its lone pairs to electron deficient boron in BCl3. 3 Incorrect O in (CH3CH2)2O has two bond pairs and two lone pairs. The shape is bent with bond angle 105o. On forming (CH3CH2)2OBCl3, number of bond pairs increase to three. The shape is changed to trigonal pyramidal with bond angle increase to 107o.

5 ASRJC JC2 PRELIM 2026 9476/01/H2 [Turn over 5 Which pair of molecules have a greater permanent dipole in molecule I than in molecule II? I II A CCl4 CH2Cl2 B CO Cl Cl CO CH3 CH3 C C C Cl H3C CH3 Cl C C Cl H3C Cl CH3 D C C F F CH3 CH3 C C F H3C CH3 F Ans: D Topic: Chemical Bonding LO 8: Able to predict whether a molecule is polar or non –polar based on bond polarity and its molecular shape. I II A C Cl Cl Cl Cl There is no net dipole as the dipoles will be cancelled. It is a non-polar molecule. C Cl Cl H H + − − Since there is a net dipole, it is a polar molecule, and it is more polar than I. B CO Cl Cl − + − − The resulting dipole moments due to the 2 Cl atoms partially cancels out the dipole moment due to O atom. CO CH3 CH3 +− Since there is a net dipole, it is a polar molecule, and it is more polar than I since the dipole moment due to O atom is not partially cancelled out like in I.

6 ASRJC JC2 PRELIM 2026 9476/01/H2 C C C Cl H3C CH3 Cl+ + − − There is no net dipole as both dipoles will cancel out. It is a non-polar molecule. C C Cl H3C Cl CH3 + + − − Since there is a net dipole, it is a polar molecule, and it is more polar than I. D C C Cl F CH3 CH3− + − Since there is a net dipole, it is a polar molecule, and it is more polar than I. C C F H3C CH3 F − + − There is no net dipole as both dipoles will cancel out. It is a non -polar molecule.

7 ASRJC JC2 PRELIM 2026 9476/01/H2 [Turn over 6 A given mass of ideal gas occupies a volume V and exerts a pressure p at 37 °C. At which temperature will the same mass of the ideal gas occupy the volume 0.5V and exert a pressure 3p? A 55.5 °C B 55.5 K C 465 °C D 465 K Ans: D Topic: The Gaseous State LO 3: Able to apply p1V1 n1T1 = p2V2 n2T2 for gases under 2 sets of conditions (pressure and volume can be of any units but they must be consistent since it is a ratio. Temperature must be in K!. Method 1 piVi Ti = pfVf Tf pV (37+273) = 3p×0.5V Tf Tf = 1.5pV×310 pv = 465 K Method 2 Since mass of gas remains unchanged, from first set of data we have pV = nRT where n= pV RT = pV R(273+37) -----(1) From second set of information, we have 3p and 0.5V Using pV=nRT, substitute n= pV R(273+37) , 3p and 0.5V to solve for the required temperature. 3p(0.5) = pV R(273+37) RT Solving T = 465 K

8 ASRJC JC2 PRELIM 2026 9476/01/H2 7 Use of the Data Booklet is relevant to this question. Which graph correctly describes the behaviour of equal masses of N 2 and Ne? The gases behave as ideal gases under the conditions chosen. A B C D Ans: C Topic: The Gaseous State LO 6: Able to interpret and sketch graphs derived from the general gas equation pV = nRT