2019 NJC Prelim P3 ANS

[Turn over Suggested Answer for SH2 H2 Chemistry 2019 Prelim Paper 3 Section A Answer all the questions from this section. 1 Potassium is an element found in Period 4 of the Periodic Table. (a) (i) Write an equation for the second ionisation energy of potassium. [1] K+(g) → K2+(g) + e− Must show state symbol (ii) Fig 1.1 shows the second ionisation energy for the consecutive elements from Period 3 and Period 4. Fig 1.1 Suggest which of the above elements, A to H, is potassium. [2] The second electron of potassium is removed from the 3p subshell (inner quantum shell) as compared to the second electron of calcium which is removed from the 4s subshell (outer quantum shell). Much more energy is required to remove the second electron from K as compared to Ca. Hence element G is potassium. Note: It is insufficient to mention that 2nd I.E. of K is much higher than 1st I.E. as the comparison with 1st I.E. is not reflected in Fig 1.1 (b) When 0.400 g of potassium dichromate(VI), K2Cr2O7, was heated using a strong flame, a yellowish green solid mixture was obtained and a gas was evolved. The gas collected ignited a glowing splint. Upon adding water to the yellowish green solid, part of the solid dissolved to give a yellow solution of K2CrO4, leaving a green solid. (i) State the identity of the gas. [1] Oxygen gas.

NJC/H2 Chem Prelim/03/2019 2 (ii) The green solid that remained was collected and dried. It weighed 0.052 g and was found to be 68.4% chromium by mass and contains only chromium and oxygen. Determine the chemical formula of the green solid. [2] Cr O Mass 68.4 100 × 0.052 = 0.03557 31.6 100 × 0.052 = 0.01643 Ar 52.0 16.0 Moles 0.000684 0.001027 ÷ by smallest no. 1 1.5 Mol ratio 2 3 Hence chemical formula is Cr2O3 (iii) Using your answers to b(i) and b(ii), write the equation for the decomposition of potassium dichromate(VI). [1] 4K2Cr2O7  4K2CrO4 + 2Cr2O3 + 3O2 (c) State two physical properties of chromium that differs from calcium. Explain the reasons for those differences. [2] TM has higher density due to larger atomic mass and smaller radius giving rise to dense close-packed structure (allowing more small size but heavy atoms to be packed within a specific volume.) TM has higher melting point due to more delocalised electrons contributed from 3d and 4s subshells, resulting in stronger metallic bonding. TM has higher electrical/heat conductivities due to more delocalised electrons contributed from 3d and 4s subshells. Note: Variable oxidation states and reactions to give coloured complexes are chemical properties. (d) Vanadate(V) ion, VO3−, changes into pale yellow oxovanadium(V) ions, VO2+, when acid is added. Upon addition of excess zinc, the pale yellow solution changes to blue, to green