A – Structural Questions: Question 1. (a)(i) The presence of isotopes 1M (ii) Let the abundance of 63X be a %. The % abundance of 65X. = ( 100 – a ) 1M Relative atomic mass = ( 62. 93 x a) + ( 64. 93 x ( 100 -a) ) 1M 100 63. 55 = 62. 93a + 6493 -64. 3a 100 6355 = -2a + 6500 a = 69. 0% 1M The % abundance of 65X = 100- 69. 0 = 31. 0 % Relative abundance 63X : 65X 1 : 2 1M (iii) Relative Abundance 63 64 65 Relative mass /m/e 2M SpeciesprotonsneutronsElectrons 20 Ne 10 10 10 10 16O2- 8 8 8 10 2 M The species have same number of electrons or isoelectronic. M —————- 10M 2. (a) (i) H2O2 + 2H+ + 2 I- > 2H2O + I21M (ii) Rate = k [H2O2] [I-]1M (iii) 0. 21M 0. 11M (iv) second order1M (b) (i) 121M (ii) 1s2 2s2 2p6 3s2. 1M (iii) +2 , X has two valence electrons2M (iv) X is a better electricity conductor. 1M —————- 10M 3. (a)
Atomic size increases, screening effect increases with more inner shells of electrons 1M effective nuclear charge decreases, ionisation energy lowered, valence electrons are more easily removed. 1M (b) i. Be2+ (aq) + 4H2O (l) > [ Be (H2O)4 ]2+ (aq)1M ii. It is acidic, acting as a Bronsted-Lowry acid1M The Be2+ ion has a high charge density 1M and can strongly polarise large anions due to its smaller size. 1M The ions of other Group 2 elements have larger sizes and charge densities and weaker polarising power (d)i. platinum and rhodium1M ii. 4NH3(g) + 5O2(g) ? 4NO(g) + 6H2O (g)1M iii. low temperature1M low pressure1M ( Note : The reaction is exothermic reaction. According to le Chatelier principle, a low temperature will favour the formation of NO. For gaseous equilibrium, a decrease in pressure will favour the reaction which produces more gaseous molecules. Thus in the above equilibrium a low pressure will avour the formation of NO. ) ________ 10M 4. (a) i.