Mock Examination (2008-2009) Chemistry (Paper I)

1.

a. Octahedral [1]

2. 
   
   

   complex I	complex II and complex III
   [1]	[1+1]
   trans-isomer	cis-isomer and its mirror image

3. 
   Complex I: trans-dichlorobis(ethane-1,2-diamine)chromium(III) ion
   

Complex II: cis-dichlorobis(ethane-1,2-diamine)chromium(III) ion

OR cis- bis(ethane-1,2-diamine)dichlorochromium(III) ion

[1+1]

ii. They are optical isomers (or enantiomers) of each other (or optical isomerism) [1]

e. Complex II and complex III [1]. Since they are asymmetric (or The two polar Cr-Cl bonds do not cancel out each other). [1]

2

a. Two amino acids joined together by a peptide/amide/ –CONH– link [1]

2. 
   IUPAC name of aspartic acid: Aminobutanedioic acid [1]
   

c. i. Heat under reflux [1/2 mark]

with HCl (or NaOH) [1/2 mark]

ii. Methanol [1 mark]

iii. Paper chromatography/ thin layer chromatography/ electrophoresis/ Mass spectrometry/… (any one reasonable answer) [1]

4. 
   3D structural formulae of two optical isomers of phenylalanine. [1+1]
   
5. 
   Dipolar ion of phenylalanine, [1]
   
6. 
   Hydrolysis of aspartame takes place in acidic medium. [1]
   

3

a. High resistance voltmeter (or potentiometer) [1/2]

Standard hydrogen electrode (or other reference cell; e.g. Cu²⁺/ Cu) [1/2]

Vanadium ions half-cell (or general ion/ ion half-cell) [1/2]

Labels: Any two [1/2+1/2]

b. i. Stable oxidation state would be expected to be +5 [1]

ii. Both vanadium half-cells are more negative than oxygen half-cell [1].

(or V(+4) to V(+5) is still sufficiently negative to supply electrons to oxygen half-cell)

c. Electrode potential values depend on the ligands present in a complex [1]

d. [Ar] 4s²3d³ (or [Ar] 3d³4s²) [1]

e. V²⁺(g)  V³⁺(g) + e^– [1]

f. After the loss of 2 electrons from calcium, further ionisation requires a large energy input. Since third ionization involves the removal of electron from argon like electronic configuration which carries extra stability. [1]

Vanadium can lose a greater number of electrons before a large energy input is required since electrons are being removed from the 3d-orbitals. [1]

4.

a. i. Energy cycle: [1]

H₃ H₄

H₂ + H₃ = H₄

H₂ + 8 x H_f(CO) = H_f(C₈H₁₈) + 8 x H_f(H₂O)

H₂ + 8 x (-111) = (-169) + 8 x (-242) [1M]

H₂ = -1217 (kJ mol^-1)// [1A]

ii. Temperature: 300-500^oC and pressure: 1-10 atm (or high temperature and low pressure) [1]. Since the forward reaction is endothermic and no. of gaseous products are greater than that of reactant, hence the eqm. pos. will shift to product side at high temp and low pressure [1+1].

b. i. Labelled axes [1/2 + 1/2]

correctly plotted data and straight line [1]

ii. Initial rate of reaction is proportional to starting concentration of H₂O₂ [1], so reaction is first order with respect to hydrogen peroxide [1].

To be continued

5

a. Water ligands around the copper ions are replaced by new ligands (L) [1]

c. K_c = [CuL₂(org)][H⁺(aq)]²/ [Cu²⁺(aq)][HL(org)]² [1M]

= [(0.045)(10^-2)²]/ [(0.002)(0.1)²] (or 1M)

K_c = 0.225 [1A]

d. i. Awareness of Le Chatelier's Principle or reversibility of process

Cu²⁺(aq) + 2 HL(organic)CuL₂(organic) + 2 H⁺(aq) [1]

Correct discussion of effect of [H⁺] on position of equilibrium

e.g. Increase in [H⁺] shifts eqm. pos. to L.H.S. [1]

ii. The process can be reversed by then shaking the organic solution with moderately concentrated acid. This pushes Cu²⁺ ions back into aqueous solution and, again, an increase in concentration can be achieved. [1]

iii. Separating funnel [1].

iv. Good ventilation/ release pressure inside the separating funnel/ restrict the use of naked flame/… (any reasonable answer) [1].
6.

a. i. C₆H₅NH₂ + H₂SO₄  C₆H₅NH₃⁺HSO₄^- (or C₆H₅NH₄SO₄) [1]

ii.

b.

Curly arrow: 0.5 each (max 2 marks)

Intermediate and product: [0.5 + 0.5]

Bonus mark for resonance structures of intermediate: 0.5 each (max 1)

[max 3]

P.T.O.

Form 7

Chemistry (08-09)

P.4

c. Steric effect between the very bulky sulphonic acid group at the 1-position and the hydrogen atom at the adjacent 8-position.

(or The difference in thermodynamic stability of the two isomers can be attributed largely to the destabilising repulsion between the very bulky sulphonic acid group at the 1-position and the hydrogen atom at the adjacent 8-position.) [1]

d.

• 
  Energy barrier (E₁) for the formation of isomer 1 is lower than that of isomer 2 (E₂).
  
• 
  Fewer molecules exceed E₂ at low temperature than that exceeding E₁.
  
• 
  Isomer 1 is formed more rapidly than isomer 2 at lower temperature.
  
• 
  Sulphonation is reversible.
  
• 
  At high temperature more molecules exceed E₂ (and E₁).
  
• 
  Both isomers are formed.
  
• 
  Fall in energy for isomer 2 is more negative than for isomer 1 (or isomer 2 is more stable). Isomer 1 will convert to isomer 2.
  

1. 
   C₆H₅COOCH₃ + H₂OC₆H₅COOH + CH₃OH
   

b. i. No. of mole of methyl benzoate used = 2.70/ 136 = 0.01985 mole//

ii. % yield of benzoic acid = [(1.5 / 122) / 0.01985] x 100% = 61.94%//

iii. Hydrolysis is not complete/ C₆H₅COOH slightly soluble in water/ Lost of product during manipulation (1 each, any TWO)

c. i. To remove methanol from the reaction mixture.

ii. Since C₆H₅COO^Na⁺ is water soluble/ to precipitate C₆H₅COOH

iii. to remove any impurities/ to purify the product.

d. Benzoic acid: 1680 – 1750 cm^-1 C=O and 2500 – 3300 cm^-1 O-H [0.5+0.5]

Methanol: 3230 – 3550 cm^-1 OH [0.5]

O-H absorption peak of benzoic acid is boarder than that of methanol [0.5].

To be continued

Form 7

Chemistry (08-09)

P.5

8.

SiO₂ + CaO  CaSiO₃ [1]

ii. Otherwise it would be removed as an oxide. [1]

b. Four marks for any four of the following points:

• 
  Select a suitable filter
  
• 
  Use a blank/ reference cell (to set zero absorbance or 100% transmittance)
  
• 
  Make up solutions of known concentration
  
• 
  Measure the absorptions of the solutions
  
• 
  Plot a calibration curve
  
• 
  Measure absorption of the green solution of unknown concentration
  
• 
  Read off its concentration from calibration curve
  

(= 4.84 x 10^–4 moles)

Mass of iron in steel sample = 5 x 4.84 x 10^–4 x 10 x 56.0 = 1.36 g [1A]

Percentage by mass of iron in steel = 1.36/ 1.40 x 100% = 96.8% [1A]

9.

Introduction

A definition of polymer: a very large molecule obtained from monomers.

Types pf polymers (max 4)

This essay requires that some kind of classification be given. There are many ways of doing this. This part also is accompanied by examples of each type.

Definitions and examples of natural polymers (e.g. proteins, polysaccharides and nucleic acids) and synthetic polymers [1+1]

Definition: Formed from intermolecular reactions in which monomers joined through without elimination of small molecules [0.5]

Examples: Polythene, PVC, Polystyrene, …[0.5]

Chemical equation of Addition polymerization (a suitable example) [1]

Definition: Formed from intermolecular reactions in which monomers joined through elimination of small molecules e.g. H₂O, ROH [0.5]

Examples: nylon, PET, urea methanal, …[0.5]

Chemical equation of Condensation polymerization [1]

For example, equations of Nylon (or terylene) P.T.O.

Form 7

Chemistry (08-09)

P.6

Mechanisms of their formation (max 4)

Addition polymers:- Free radical addition polymerization [0.5]

Mechanisms are generally chain reaction involving initiation, propagation and termination. [2]

Illustrate with polythene (or any suitable polymer) as an example [0.5]

Or

Condensation Polymer mechanisms: Nucleophilic addition followed elimination [0.5]

Suitable example [0.5]

Compare ANY ONE of the following pair of polymers

• 
  LDPE and HDPE
  
  • 
    Conditions of polymerization [0.5+0.5]
    
  • 
    Crystalline structures in solid state [0.5+0.5]
    
  • 
    Different in properties [0.5+0.5]
    
• 
  Nylon and Kelvar
  
  • 
    Crystalline structures in solid state [0.5+0.5]
    
  • 
    Extent of intermolecular hydrogen bond [0.5+0.5]
    
• 
  Thermoplastic and thermosetting plastic
  

Examples [0.5 each]

OR

Describe the effect of bonding and structure on properties of …[1+1]

Biopolymer [e.g. poly(lactic acid)]

Photodegradable plastics.

Synthetic biodegradable plastics

10.

1. The oxides of period 3 elements are classified into two types according to the nature of their bonding.

a. Ionic oxides

2. 
   Covalent oxides
   

Na₂O, MgO, Al₂O₃, SiO₂, P₄O₆ (or P₄O₁₀), SO₂ (or SO₃), Cl₂O [0.5]

Correctly matched with …

Bonding. [0.5]

Structures. [0.5]

States (or melting points). [0.5]

Electron diagrams of an ionic oxide and a covalent oxide. [0.5 + 0.5]

[max 3]

3. 
   Expansion of octet for period 3 oxides due to the presence of low lying vacant 3d orbitals. [1]
   

1. The general periodic patterns in the reactions of period 3 oxides with water, dilute acids and alkalis.

1. 
   Ioinc oxides react with water to form hydroxides. [0.5]
   

2. 
   Ionic oxide with covalent character,
   

Examples (or equations) [0.5 + 0.5]

ii. silicon(IV) oxide does not reacts with water but it reacts with hot strong alkalis. [0.5]

Example (or equation) [0.5]

2. 
   Covalent oxides reacts with water to form acids [0.5]