2019 TJC H2 Chem Prelim P2 ANS

2 DO NOT WRITE IN THIS MARGIN [Turn over DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelim / 2019 1 (a) The table below shows the first, second and third ionisation energies of some elements in the Periodic Table. Element First Ionisation energy (kJ mol-1) Second Ionisation energy (kJ mol-1) Third Ionisation energy (kJ mol-1) Ca 590 1150 4940 Mn 716 1510 3250 Fe 762 1560 2960 Co 757 1640 3230 Ni 736 1750 3390 (i) Explain why the first and second ionisation energies of the transition metals are relatively invariant. [2] The first and second I.E of the transition elements involves the removal of 4s electrons. Across the period,  nuclear charge increases due to increasing number of protons. Screening effect increases as electrons are added to the penultimate 3d subshell, providing a shield between nucleus and outer 4s electrons.  Increase in nuclear charge is only slightly more significant than the increase in screening effect. Hence, small increase in both the first and second I.E. (ii) Write the electronic configurations of Ca and Fe atoms at ground state. Explain the differences in the values of the third ionisation energies between iron and calcium. [2]  Electronic Configuration of Fe: 1s2 2s2 2p6 3s2 3p6 3d6 4s2 Electronic Configuration of Ca: 1s2 2s2 2p6 3s2 3p6 4s2  The third ionisation energy of calcium is more than that of iron because an electron is removed from a 3p orbital of Ca2+ whereas an electron is removed from a 3d orbital of the Fe2+.The 3p electron is nearer to the nucleus compared to the 3d electron and hence required more energy to remove. (b) Calcium cyanamide, CaCN2, is used as a fertiliser in agriculture. Through hydrolysis in the presence of carbon dioxide, calcium cyanamide produces cyanamide, NH2CN. Cyanamide can be extracted by organic solvents. (i) Draw the dot–and–cross diagram for the cyanamide molecule, NH2CN. [1]

3 DO NOT WRITE IN THIS MARGIN [Turn over DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelim / 2019 (ii) With reference to structure and bonding, deduce whether CaCN2 has a higher or lower melting point as compared to NH2CN. [2]  NH2CN has simple molecular structure with hydrogen bonds between its molecules. CaCN2 has a giant ionic lattice structure held by strong electrostatic fo rces of attraction/ionic bonds between the oppositely charged ions.  Melting involves the breaking of the stronger electrostatic forces of attraction between the oppositely charges ions (Ca 2+ and CN 22-) as compared to the (or) weaker hydrogen bonds between NH2CN molecules. Hence more energy is required to melt CaCN2, hence higher melting point. [Total: 7] 2 A saturated solution of magnesium methanoate, Mg(HCO 2)2, has a solubility of approximately 143 g dm –3 at room temperature. The exact solubility can be determined by titrating magnesium methanoate solution against a standard potassium manganate(VII) solution. During the titration, the methanoate ion, HCO 2–, is oxidised to carbon dioxide while the manganate(VII) ion, MnO4–, is reduced to Mn2+. (a) (i) Write the overall equation for the reaction between HCO 2– and MnO 4– under acidic conditions. [1] Oxidation :HCO2–  CO2 + H+ + 2e (x5) Reduction : MnO4– + 8H+ + 5e–  Mn2+ + 4H2O (x2)  2MnO4– + 11H+ + 5HCO2–  2Mn2+ + 8H2O + 5CO2 (ii) Calculate the approximate concentration of HCO2– ions present in the saturated solution. [1]  [HCO2–] in saturated solution = 𝟏𝟒𝟑 𝟐𝟒.𝟑+𝟐(𝟏+𝟏𝟐+𝟐×𝟏𝟔) × 𝟐 = 2.50 mol dm3

4 DO NOT WRITE IN THIS MARGIN [Turn over DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelim / 2019 (iii) The titre value from titrating 25.0 cm 3 of saturated magnesium methanoate solution against 0.0500 mol dm–3 potassium manganate(VII) solution is found to be too high. Describe the steps to prepare a suitable solution of magnesium methanoate from the saturated solution. Your plan should include details of quantities measured and apparatus used. [3] Assuming the titre volume of potassium manganate(VII) solution to be 25.00 cm 3 (accept 20.00 to 25.00 cm3), No. of moles of HCO2– in 25.0 cm3 of diluted solution = 𝟐𝟓 𝟏𝟎𝟎𝟎 × 𝟎. 𝟎𝟓 × 𝟓 𝟐 = 3.125 x 103 mol No. of moles of HCO2– in 250 cm3 of diluted solution = 3.125 x 102 mol  Volume of saturated HCO2– solution needed for dilution = 𝟑.𝟏𝟐𝟓 ×𝟏𝟎−𝟐 𝟐.𝟓 × 𝟏𝟎𝟎𝟎 = 12.50 cm3  Using a burette, transfer 12.50 cm3 of saturated magnesium methanoate solution into a 250 cm3 standard flask. Make up to the mark with water and shake well. (b) An industrial chemist introduced 2 atm of carbon dioxide and 2 atm of hydrogen gas into a 1 dm3 container at 300 K. The temperature is then raised to 900 K, at which the Kp is 0.641. CO2(g) + H2(g) ⇌ H2O(g) + CO(g) H < 0 (i) Calculate the partial pressure of H2 present at equilibrium at 900 K. [3] Since P  T,  PCO2 = PH2 = 6 atm at 900K CO2(g) + H2(g) ⇌ H2O(g) + CO(g) Initial partial pressure /atm 6 6 0 0 Change in partial pressure /atm -x -x +x +x Equilibrium partial pressure /atm 6-x 6-x x x Kp = 𝐏𝐇𝟐𝐎×𝐏𝐂𝐎 𝐏𝐂𝐎𝟐×𝐏𝐇𝟐  0.641 = 𝐱𝟐 (𝟔−𝐱)𝟐 x = 2.67 atm  PH2 = 6 – 2.67 = 3.33 atm (ii) Explain if the Kp value at 300 K is higher or lower than 0.641. [2]  The value of Kp at 300 K is higher than 0.641.  By Le Chatelier’s Principle, at the lower temperature of 300 K, the system will favour the exothermic forward reaction to produce more heat . Thus position of equilibrium shifts to the right and Kp increases. [Total: 10]

5 DO NOT WRITE IN THIS MARGIN [Turn over DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 9729 / TJC Prelim / 2019 3 (a) Sulfur tetrachloride decomposes to sulfur dichloride and a gas that bleaches litmus. [3] (i) Using the VSEPR theory, deduce and draw the shape of SCl2. Give a value for the bond angle. There are 2 bond pairs and 2 lone pairs of electrons around S. To minimize repulsion and maximize stability, the 4 electron pairs are directed to the corners of a tetrahedron. The shape is bent.  2 bond pairs and 2 lone pairs around S.  4 electrons pairs arranged tetrahedrally to  minimize repulsion and maximize stability.  bent  diagram  bond angle : any angle between 900 & 109.50. 2  : 1m (ii) State whether the bond angle of SCl2 is expected to be larger or smaller than SF2. Explain your answer. [1] Cl is less electronegative than F. The bond pair of electrons is closer to S for SC l2 resulting in a greater bond pair-bond pair repulsion. Hence the bond angle of SCl2 is larger. (b) (i) There are two possible molecular arrangements for