PAHANG TRIAL 2009 EXAMINATION
CHEMISTRY PAPER 2 MARKING SCHEMES
SECTION A - Structural Questions:
(i) The presence of isotopes
(ii) Let the abundance of 63X be a %.
The % abundance of 65X. = ( 100 – a )
Relative atomic mass
= ( 62.93 x a) + ( 64.93 x ( 100 -a) )
63.55 = 62.93a + 6493 -64.93a 100
6355 = -2a + 6500
a = 69.0%
The % abundance of 65X = 100- 69.0
= 31.0 %
Relative abundance 63X : 65X
1 : 2
63 64 65 Relative mass /m/e 2M
2 M The species have same number of electrons or isoelectronic.
2. (a) (i) H2O2 + 2H+ + 2 I- → 2H2O + I2
(ii) Rate = k [H2O2] [I-]
(iv) second order
(b) (i) 12
(ii) 1s2 2s2 2p6 3s2
(iii) +2 , X has two valence electrons
2M (iv) X is a better electricity conductor.
3.(a) Atomic size increases, screening effect increases with more inner shells of
effective nuclear charge decreases, ionisation energy lowered, valence
electrons are more easily removed.
Be2+ (aq) + 4H2O (l) → [ Be (H2O)4 ]2+ (aq)
It is acidic, acting as a Bronsted-Lowry acid
The Be2+ ion has a high charge density
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
platinum and rhodium
4NH3(g) + 5O2(g) 4NO(g) + 6H2O (g)
( 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
favour the formation of NO.)
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