Chemistry Stpm

PAHANG TRIAL 2009 EXAMINATION

CHEMISTRY PAPER 2 MARKING SCHEMES

SECTION 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 -64.93a 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

Species protons neutrons Electrons 20 Ne 10 10 10 10 16 O2- 8 8 8 10 2 M
The species have same number of electrons or isoelectronic. 1M ---------------- 10M

2. (a) (i) H2O2 + 2H+ + 2 I- → 2H2O + I2 1M (ii) Rate = k [H2O2] [I-] 1M (iii) 0.2 1M 0.1 1M (iv) second order 1M

(b) (i) 12 1M (ii) 1s2 2s2 2p6 3s2 . 1M (iii) +2 , X has two valence electrons 2M (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 acid 1M

© 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 rhodium 1M

ii. 4NH3(g) + 5O2(g)  4NO(g) + 6H2O (g) 1M

iii. low temperature 1M low pressure 1M ( 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.)

________ 10M

4.(a) i. A is CH3CH2CH2COOH 1M B is CH3CH2CH2COCl 1M
C is CH3CH2CH2COOCH2CH 1M

ii. butanoyl chloride 1M

iii. Formation of ester:
CH3CH2CH2COCl + CH3CH2OH → CH3CH2CH2COOCH2CH3 + HCl 1M (b) i. H3N+CH2COOH + H2NCH(CH3)COOH →

H2NCH2CONHCH(CH3)COOH + H2O 1M

Glycylalanine 1M

( Note: Alanylglycine can also be formed )

ii. The amino group –NH2 which is basic group reacts with hydrochloric acid to form the ammonium chloride salt of alanine 1M

HOOCCH(CH3)NH2 + HCl → HOOCCH(CH3)NH3+Cl- 1M ___________

10 M

SECTION B - ESSAY

5.(a) (i) Orbitals with the same energy 1M Example : 2p or 3d s orbitals 1M

(ii) Nitrogen atom has 7 electrons 1M Fill 1s orbital with 2 electrons 1M Fill 2s orbital with 2 electrons 1M Fill 2px,2py and 2pz orbitals with 3 electrons 1M / 6

1M

(b) Fe 2+ 1s2 2s2 2p6 3s2 3p6 3d4 1M Fe 3+ 1s2 2s2 2p6 3s2 3p6 3d5 1M

In terms of electronic configuration, Fe 3+ is more stable than Fe 2+ 1M Because it has half-filled 3d orbital which is more stable 1M./ 4

(c ) The valence electronic configuration of the electrons for nitrogen atom is 2s2 2px1 2py1 2pz1 1M Nitrogen atom uses sp3 hybrid orbitals for forming covalent bonds between N and H atoms. Energy

2p

sp3 hybrid porbitals

N(ground state) 1M

In sp3 hybrid orbitals of nitrogen atom,one of the orbitals Is occupied by a lone pair of electrons and three sp3 orbitals are half filled 1M

Each N-H atom is formed by the overlapping of the s orbital of hydrogen atom with one of the half filled sp3 orbitals to give the ammonia molecule 1M

Diagram of the bond formation in NH3 molecule. 1M /.5 ------------------- Total : 15 M
6. (a) Dyanamic equilibrium .... a reversible reaction , in a closed system forward and backward reactions have the same rate of reaction. 2M (b) (i) N2O4 ↔ 2NO2 Kc = [NO2] 2 = [0.12] 2 [N2O4] [0.04]

= 0.36 mol dm-3 5M

(ii) Using PV =nRT where n = 0.12 +0.04 = 0.16 mol

P = 0.16 (8.31) (383) 10 -3

= 509.24 kPa .3M

(c) N2(g) + 3H2(g) 2NH3(g) - at low temperatures, % NH3 is higher - forward reaction is exothermic - equilibrium position shifts to the right at higher temperature

- forward reaction is accompanied by a reduction in volume of gas
- at higher pressures, equilibrium position shifts to the right
- at high pressures, % NH3 is higher 5M
---------------
Total : 15M
No. 7
(a) (i) Aluminium metal is extracted by electrolysis
The electrolyte is molten bauxite in sodium hexafluoroaluminate.
The electroyte has aluminium ion and oxide ions.
Anode : 2O2- --- > O2 + 4e
Cathode : Al3+ + 3e --- > Al 5M
(ii) (Any 2 points) light Resistant to corrosion
Strong alloy 2M

`(b) aluminium : A giant metallic structure, strong metallic bonf. Silicon : giant 3 D covalent structure. Strong covalent bond between silicon atomes. higher melting point
Phosphorus and sulphur - Both are simple molecules. Weak van der waals between molecules Sulphur has a stronger intermolecular forces –
S8 larger than P4 8M

No 8. (a) chlorine – strong oxidation agent Bromide is oxidized to bromine E° of chlorine is more positive than that of bromine. Cl2 + 2Br- ---- > 2Cl- + Br2 4M

(b) iodine forms triodes complex in KI.
I2 + I- ---- > I3-
Iodine does not form any complex ions in water.
I2 + 2H2O --- > I-¬ + HIO + H3O+ 4M

(c) HCl is released in cold acid
NaCl + H2SO4 à NaHSO4 + HCl
If heated more HCl released.
NaHSO4 + NaCl -à Na2SO4 + HCl 4M (d) Iodide is oxidized to iodine Purple Iodine is released Pungent smell of H2S is detected 3M -------------- Total : 15M

9.( a ) ( i ) order : W, Y, X W, Y, X act as Lewis bases.

X is the strongest base because ethyl group is an electron donor by inductive effect. Y is more basic than W because the lone pair electron on the N atom is not delocalised. W is less basic than Y because the lone pair electron on the N atom is delocalised into the benzene ring. 5M

( ii ) pKb value > 9.39 Z is a weaker base than W. Presence of Cl - an electron withdrawing group reduces the donating potential of lone pair electron on the N atom through inductive effect. 4M ( b ) Concentated H2SO4 and HNO3., 55○C Mechanism: HNO3 + H2SO4 NO2+ + HSO4– + H2O NO2+ is an electrophile. H + NO2+ NO2

H NO2 + HSO4– –NO2 + H2SO4

+ HNO3 –NO2 + H2O 6M

Total : 15 marks

10.( a ) ( i ) Terylene/Dacron

~~~~O - CH2 - CH2 - O - C – –C - O - CH2 - CH2 - O - C – –C~~~~ 3M || || || || O O O O ( ii ) Condensation polymerisation To make cloth/sleeping bags, etc 2M

( b ) ( i ) K: functional group : -OH

isomers : CH3CH2CH2OH and CH3CHCH3OH warm isomers separately with alkaline iodine, CH3CHCH3OH gives a yellow precipitate but CH3CH2CH2OH does not. CH3CH2CH2OH + 4I2 + 6OH– CHI3 + 5I– + 5H2O + CH3COO–

5M

(ii ) L : functional group : ―C = O | Isomers : CH3CH2CHO and CH3COCH3

warm isomers separately with Tollen’s reagent. CH3CH2CHO gives a silver mirror but CH3COCH3 does not.

CH3CH2CHO + 2[Ag(NH3)2]2+ + OH– CH3CH2COO– + 2Ag + 2NH4+ + 2NH3 5M

Note: Can also accept other suitable chemical test. Total : 15 marks