SRJC_H1_CHEM_P2_ANSWER

2011 H1 MYE Section A Marker’s Remarks 1 Elements D and E are period 3 elements. (a) D is a white solid with melting point of 317 K. When exposed to air, it reacts spontaneously, producing a white powder of empirical formula D2O5. D2O5 readily undergoes neutralisation with aqueous NaOH to form salt and water. (i) From the observations , deduce an identity of D. Include equations, with state symbols, for any reactions that has occurred. Low melting point hence D must be of simple molecular structure: sulfur or phosphorus. The oxide of D is a white powder with empirical formula D2O5 hence must be phosphorus as sulf ur is yellow and its oxide does not have empirical formula of D2O5. P4 (s) + 5 O2 (g) P4O10 (s) P4O10(s) +12 NaOH(aq) 4 Na3PO4(aq) + 6 H2O(l) (ii) Suggest and account for a pH value for the resultant solution when the oxide of D dissolves in water. P4O10(s) + 6H2O(l) 4 H3PO4(aq) Resulting pH: 2 or 3 [6] (b) The successive ionisation energies of element E is shown below. [2] 1st 2nd 3rd 4th 5th 6th 7th 8th I.E. kJmol-1 999 2265 3331 5071 7008 8487 27107 31719 (i) Deduce the group number in which E belongs. Group VI (ii) Write down the electronic configuration of E in the +4 oxidation state. 1s2 2s2 2p6 3s2 [2] (c) Compare the 1st ionisation energy of E with the element just before it. Explain. E will have a lower 1st I.E. compared to the element just before it. Due to inter-electronic repulsion between the pair of electrons sharing the same orbital, thus lesser amount of energy is required to remove an electron from Element E. [2] Total: [10]

2 Manganese dioxide, MnO2 is a strong oxidising agent and it is also commonly used as a catalyst. (a) On the grid below, sketch and label the energy distribution of gas molecules in the presence of a catalyst. Your sketch should clearly illustrate the effect of catalyst on rate of reaction. [2] (b) The lattice energy of MnO 2 can be calculated based on the energy cycle shown below. Mn2+ (g) + 2 O2− (g) Mn2+ (g) + G + 2e Mn2+ (g) + O2 (g) + 2e Mn+ (g) + O2 (g) + e Mn(g) + O2 (g) Mn(s) + O2 (g) MnO2 (i) Identify the species G, with appropriate state symbol G: O(g) Energy Fraction of molecules BE (O2) 1 (Mn) 2 3 1st + 2nd E.A (O) 4 0 Energy

(ii) Identify the enthalpy changes labeled by the numbers 1 to 4. 1 H 2 1st ionisation of Mn 3 2nd ionisation of Mn 4 lattice energy of MnO2 [3] (c) Potassium manganate (VII), KMnO 4, can be manufactured from MnO 2. During the process, MnO 2 is reacted with concentrated KOH to produce potassium manganate ( VI), K 2MnO4. When a solution of K 2MnO4 is acidified, a disproportionation reaction occurs. This produces a dark brown MnO2 precipitate and a