NJC 2026 Prelim P2 Ans

NJC SH2 Prelim 9476/2/26 [Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME ANSWERS SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 2 Structured Questions Candidates answer on Question Paper. Additional Materials: Data Booklet 9476/02 24 Aug 2026 2 hours READ THE INSTRUCTIONS FIRST Write your subject class, registration number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. Marks Weighting Paper 1 /30 15% Paper 2 /75 30% Paper 3 /75 35% Paper 4 /50 20% For Examiner’s Use 1 /8 2 /10 3 /16 4 /17 5 /16 6 /8 Paper 2 Total /75 Overall Percentage Grade This document consists of xx printed pages.

2 NJC SH2 Prelim 9476/2/26 1 Elements in Period 3 of the Periodic Table show trends in their properties. (a) Fig. 1.1 shows the ionic radii of the anions and cations of Period 3. Atomic number Fig. 1.1 Explain the shape of the graph in Fig. 1.1. [2] • Ionic radius decreases across isoelectronic series. Nuclear charge across isoelectronic series increases while number of electrons remains the same. Nuclear attraction for the valence electrons increases. • Anions have bigger ionic radius than the cations. Anions have one additional filled electron shell than cations in the same period. the valence electrons are further away from nucleus and have greater shielding effect which outweighs the higher nuclear charge in anions. (b) The third ionisation energies of consecutive Period 3 elements represented by letters A to F are shown in Fig. 1.3. third ionisation energy A B C D E F Fig. 1.3 (i) Write an equation, including state symbols, to represent the third ionisation energy of element, B. B2+(g) → B3+(g) + e–

3 NJC SH2 Prelim 9476/2/26 [Turn over (ii) Identify which element is phosphorus. Explain your answer. There is a significant decrease for 3rd IE for C compared to B, implying that the 3rd electron removed for B is from an electron shell that is closer to the nucleus/ 2 nd electron shell. OR 3 rd electron removed for C is from an electron shell that is further from the nucleus/ 3rd electron shell. Hence B is Mg (Group 2)/ C is element Al (Group 13) and phosphorus will be E./ E will be group 15, phosphorus. (c) Carbon tetrach loride, CCl4, a nd phosphorus pentachloride , PC l5, a re both non -metal chlorides. (i) When added to water, PCl5 undergoes hydrolysis while CCl4 does not. Explain the difference in behaviour of both chlorides. P atom (from Period 3) in PCl5 possesses vacant orbitals in the valence shell/ vacant 3d orbital, but not for C atom (from Period 2) in CCl4, to accept a lone pair of electrons from a water molecule for hydrolysis. (ii) Write a balanced equation for the hydrolysis of PCl5 with water. PCl5 + 4H₂O → H₃PO4 + 5HCl Phosphorus pentachloride exhibits distinct structural behaviors depending on its physical state. In the gas and liquid phases, PCl5 exists as discrete covalent molecules, while in solid phase, it exists as giant ionic lattice structure. (iii) When PCl5 solidifies, it undergoes auto-ionisation to form two distinct ions, both of which contain phosphorus and chlorine atoms only. Suggest the formulae of the two oppositely charged ions. cation : ………[PCl4]+………….. anion : ………[PCl6]–………….. [ Total:8]

4 NJC SH2 Prelim 9476/2/26 2 Hydrogen halides are important industrial chemicals used in processes such as etching, synthesis, and chemical manufacture. (a) The boiling points of hydrogen iodide, H I, and hydrogen chloride, HC l are as shown in Table 2.1. Table 2.1 Species Boiling Point / oC HI –35.4 HCl –85.1 (i) Explain the difference in the boiling points observed for HI and HCl. HI has a larger electron cloud size and is more easily distorted compared to HCl, resulting in a stronger instantaneous dipole–induced dipole interaction (id–id) between HI molecules. More energy is required to break the stronger id -id and permanent dipole – permanent dipole (pd –pd) between H I molecules compared to the id -id and permanent dipole–permanent dipole (pd–pd) interactions between HCl molecules, resulting in higher boiling point of HI. (ii) A 25.0 cm3 sample of 0.2 mol dm–3 HI (aq) has a pH of 0.70 at 298 K. State t he pH of a 25.0 cm 3 sample of HC l (aq) of the same concentration and temperature. Explain your reason. pH of HCl = 0.70 Both HI and HCl are strong acids and are taken to dissociate completely in water. Since the two solutions have the same concentration, they produce the same concentration of H⁺ ions and hence have the same pH. Hydrogen iodide can react with alcohols to form alkanes under suitable conditions. CH3OH + 2HI CH4 + I2 + H2O (iii) Suggest the role of HI in this reaction. Reducing Agent. Oxidation state of I is increased from –1 in HI to 0 in I2. or Oxidation state of C has changed from –2 to –4 [1]

5 NJC SH2 Prelim 9476/2/26 [Turn over (b) Hydrogen fluoride, HF, has a very low acid strength. The dissociation of HF in aqueous solution involves a series of steps. HF(aq) H+(aq) + F–(aq) Hdissociation The Hdissociation can be calculated using the enthalpy change of hydration for H+ and F– ions and other data found in Table 2.2. Table 2.2 H / kJ mol–1 first electron affinity for F(g) –328 enthalpy change of solution of HF(g) –48 enthalpy change of hydration for H+ ion –1091 enthalpy change of hydration for F– ion –515 (i) Explain what is meant by enthalpy change of hydration for F– ion. Enthalpy change of hydration is the heat released when 1 mol of gaseous F– is dissolved in water to form an infinitely diluted solution. (ii) Complete the energy level diagram in Fig. 2.3 and calculate the Hdissociation for HF, using relevant data from Table 2.2 and the Data Booklet. [3] Hdissociation = +48 + 562 + 1310 – 328 – 1606 = –14 kJ mol–1

6 NJC SH2 Prelim 9476/2/26 (iii) Calculate the pH for 25.0 cm3 sample of 0.20 mol dm–3 HF (aq), given that pKa of HF = 3.14. [1] Ka = 10–3.14 = 7.24 × 10–4 Ka = [H+] [F−] [HF] = x2 0.20 - x = 7.24 × 10–4 Since HF is a weak acid, x << 0.20, 0.20- x  0.20, [H+] = √𝐾𝑎 × [𝑤𝑒𝑎𝑘 𝑎𝑐𝑖𝑑] [H+] = √7.24 × 10−4 × 0.2 = 0.012 mol dm−3 pH = − lg [H+] = 1.92 (2d.p.) (iv) Neutron diffraction studies of solid HF show that there are two different distances between hydrogen and fluorine nuclei, 0.092 nm and 0.163 nm. Draw a diagram containing two HF molecules and suggest a reason for the two different distances. Your diagram should show and label both distances. [2] In solid HF, the molecules are held in a lattice state by hydrogen bonding, which is weaker than the c