HCI H2 Chem 2012 Prelim P3 Soln

Suggested Answers Page 1 of 9 © Hwa Chong Institution 2012 9647 / 03 / JC2 Prelim 2012 Hwa Chong Institution 2012 C2 H2 Chemistry Prelim Paper 3 Answers 1(a)(i) (ii) P 4 is highly reactive due to strong repulsion between bond pairs of electrons / highly strained bond angle of 60o (iii) P 4 is a simple non-polar covalent molecule and is unable to form favorable interactions with polar water molecules but able to form favorable interactions with non-polar carbon disulfide through dispersion forces. (iv) White phosphorus burns with a bright white flame / light. (v) P 4(s) + 3O2(g)  P4O6(s) or P 4(s) + 5O2(g)  P4O10(s) P 4O6(s) + 6H2O(g)  4H3PO3(s) or P 4O10(s) + 6H2O(g)  4H3PO4(s) (b)(i) - negligible intermolecular forces of attraction or negligible forces of attraction between the gas particles - volume of gas particle/molecules is negligible compared to the volume of its container (ii) Each gaseous P 4 molecule must have dissociated / decomposed / disintegrated into two gaseous molecules (c)(i) They consist of metal atoms with few valence electrons which can be readily lost to form the delocalised sea of electrons to ac t as mobile charges (resulting in metallic bonding) & have high electrical conductivity. (ii) s–block elements can only lose 1-2 e’s per atom while TM can lose more e’s from both 3d and 4s subshells per atom to form metallic bonding due to close proximity of the 3d and 4s subshells. Thus TM’s have stronger metallic bonding and more energy is needed to melt the metal and thus they have higher mp. (iii) electronic configurations of Fe: [Ar] 3d 6 4s2; of Mn: [Ar] 3d5 4s2 or electronic configurations of Fe 2+: [Ar] 3d6; of Mn2+: [Ar] 3d5 Due to inter-electronic repulsion in 3d subshell of Fe2+, 3d electrons are higher in energy, thus the 3rd e’s are easier to be removed (d)(i) formed MnOof  molof  No formed KMnOof  molof  No 2 4 = 16.0x254.9 0.174 1.00 x 1000 20.0 x 5 1  = 1 2 Let O.N of Mn in A be n+ [R] Mn n+ + 2H2O + (n  4)e  MnO2 + 4H+ (i) [O] Mn n+ + 4H2O  MnO4  + 8H+ + (7  n)e (ii)

Suggested Answers Page 2 of 9 © Hwa Chong Institution 2012 9647 / 03 / JC2 Prelim 2012 From equations (i) & (ii): 2 ‐ 4 MnOof  molof  no MnOof  molof  no = n ‐ 7 4 ‐ n = 1 2  n = 6 Oxidation number of Mn = +6 A is K 2MnO4 (ii) Equation: 3MnO 4 2 + 4H+  MnO2 + 2MnO4  + 2H2O or 3K 2MnO4 + 2H2SO4  MnO2 + 2KMnO4 + 2K2SO4 + 2H2O or 3MnO 3 + H2O  MnO2 + 2MnO4  + 2H+ Disproportionation (MnO 4 2 is simultaneously reduced to MnO2 and oxidised to MnO4 ) (iii) Mn is a TM and can form compounds of variable oxidation numbers due to close proximity of energy levels of 4s and 3d s ubshells and electrons can be easily lost or gained. K and Ca can only form compounds of +1 and +2 respectively as a lot