DHS H1 CHEM P2 ANS

2 Section A 1 Antimony (Sb), with atomic number 51 has been known sinc e about 4000BC. Nowadays, its main use is to harden and to strengthen lead alloys. A typical sample of antimony consists of two isotopes and has the following composition by mass: 121Sb, 57.25%, 123Sb, 42.75%. (a) (i) Calculate the relative atomic mass of the antimony sample. Relative atomic mass = (57.25 x 121 + 42.75 x 123) / 100 = 121.9 (II) Define the term relative atomic mass. Relative mass of a particle (egs. atom, molecule, ion, etc.) is defined as the number of times the particle is heavier than the mass of an atom of carbon–12. [3] Antimony is produced in two-stage process from the sulphide ore, Sb2S3. The ore is first roasted in oxygen to form the oxide. 2Sb2S3(s) + 9O2(g)  Sb4O6(s) + 6SO2(g) The oxide is then reduced with carbon. Sb4O6(s) + 3C(s)  4Sb(s) + 3CO2(g) (b) Showing your working clearly, calculat e the volume of carbon dioxide at room temperature and pre ssure that would be produc ed by the processing of 3400 g of Sb2S3. 2 Sb2S3 = S4O6 = 3 CO2 No. of mol of CO2 = 3/2 x 3400/(340.1) = 15.0 mols Volume of CO2 = 15 x 24dm3 = 360 dm3 [3] [Total: 6 marks] [Turn Ove r

3 Section A [Turn Over 2 The figure shows a plot of first ionisation energy against atomic number for the elements of atomic number 10 to 18. (The letters are not the chemical symbols for the elements concerned.) (a) Write an equation to define the first ionization energy of A. A(g)  A+(g) + e [1] (b) (i) Describe the general trend of the ionization energies from B to I as shown in the graph above. The successive I.E. from B to I generally increases. Nuclear charge increases when electrons are removed and there are stronger electrostatic forces of attraction between nucleus and valence electrons. Therefore more energy is required to remove the remaining electrons. [2] A First ionization energy / kJ mol-1 F I H C G E D B Atomic number (ii) Explain briefly why the first ionization energy of D is less than C. The first i onisation energy of D involves the removal of a 3p electron whereas first I.E of C involves the removal of a 3s electron. Smaller amount of energy is r equired to remove the 3p electron in D which is

4 further from the nucleus than 3s electron in C. [2] (b) Draw the shape of the orbital from which electron is lost when (i) Element B forms a singly charged ion, (ii) Element H forms a singly charged ion. [2] (c) Give the full electronic configuration of the element labeled F. 1s 22s22p63s23p3 [1] [Total:8 marks] 3 Each of these elem ents in Period 3 will react with oxygen given suitable conditions. Period 3 Na Mg Al Si P S Cl (a) Which element(s) can exist (i) as diatomic molecules at room temperature, Chlorine (ii) as macromolecular structures? Silicon [2] [Turn Ove r

5 (b) Two elements form chlorides with formulae of the type XC l3. Draw the dot- and-cross diagram for these two chlorides, state the shape and suggest values for the bond angles. [4] Trigonal planar Trigonal pyramidal (c) (i) One element form chloride of the type YC l 2 which reacts with water to give a slightly acidic solution. Name t he element, and account for the pH value of YCl2 in water. Write balanced equation to illustrate your answer. Magnesium. The pH is about 6.5 as Mg 2+ can hydrolyse in water to produce a slightly acidic solution. [Mg(H 2O)6 ]2+ + H2O [Mg(H2O)5(OH)]+ + H3O+ (ii) One element forms a chloride of the type ZC l4, which reacts with water to give a strongly acidic solution. Na me the element and write a balanced equation for the chloride reacting with water. The element is silicon. SiCl4 + 2H2O  SiO2 + 4HCl [5] [Total: 11 marks] [Turn Ove r

5 4 Oseltamivir (Tamiflu) is an antiviral drug that slows the spread of non-resistant strains of the influenz a virus between cells in the body. It blocks the action of a viral enzyme called neuraminidase and has sinc e been indicated for the treatment of H5N1 and H1N1 infection. The standard adult dosage is 75mg twice daily. Compound Z is a derivative of oseltamivir that maybe invest igated for antiviral activities. ON H2 OOC H 3 C 1 CH 3 CH 3 ON H CH 3 Oseltamivir M r : 312.4 OH OOC 3 OC H 3 H Z (a) (i) A male adult patient has been put on a 5-day tamiflu treatment. Calculate the total number of moles of tamiflu tak en by this patient over this period of treatment. No of moles = (75 × 10 -3)/312.4 × 2 × 5 = 0.00240 (3sf) [1] (ii) Describe the hybridisation, geometry and bond angle about C 1 atom. sp 3 tetrahedral 109.5 [3] (iii) Name the functional groups present in Compound Z. Ester, alkene, secondary alcohol, ketone. [2] [Turn Ove r

5 H (iv) Compound X can be synthesised from Z using 2 consecutive reactions. Give the reagents and conditions for both reactions. Name the type of reactions involv ed. OH OOC 3 OHC H 3 OH OH X Step I: Reagent: cold, dilute MnO4 – in OH– (aq) Condition: room temperature Type of reaction: oxidation Step II: Reagent: NaBH4 in methanolic solution Condition: room temperature Type of reaction: reduction [3] (v) Draw the products formed when Compound Z is reacted with dilute HC l under reflex. OH O OC H 3 OH OHC 3 H 1 1 [2] [Turn Ove r

5 (vi) Describe a chemical test to distinguish the products from (v ). State the observations with each compound and write balanced equation(s) fo r reaction(s) involved. Test: Add Br 2/CCl4 in the absence of light at room temperature to both products. Observations: For big molecule: Decolourisation of reddish-brown solution seen. For ethanol: No decolourisation observed. Note: Do not accept ‘No visible change’ for negative observations. Equation: OH O OC H 3 OH + 2Br2 1 OH O OC H 3 OH Br Br + 2HBr OR Test: Add 2,4 DNPH at room temperature to both products. Observations: For big molecule: Orange ppt seen. For ethanol: No orange ppt observed. Note: Do not accept ‘No visible change’ for negative observations. Equation: [Turn Ove r

5 OH O OC H 3 OH + 1 OH O N C H3 OH NO2 NO2 NH NO2 NO2 N H NH2 [4] Total: [15 marks] [Turn Ove r

5 P2 Section B Mark Scheme 1(a)(i) [H+] = 10-3.5 = 3.16 x 10-4 mol dm-3 (ii) No. of mol of NaOH = 25 . 01000 25 . 21 = 5.312 x 10–3 = No. of mol of HA [HA] = 1000 25 10 312 . 53 = 0.2125 = 0.213 mol dm–3 (iii) HA is a weak acid. Since [HA] >>[H+], HA dissociates partially / incompletely to form H+. (iv) phenolphthalein (b) (i) K C = ]][[ ] [ ][ OH CH CHCOOH CH CH O H CHCOOCHCH CH 2 32 3 2 3 22 3 (ii) 5 min (iii) KC = ] . ][ . [ ] . ][ . [ 01 0 03 0 07 0 07 0= 16.3 (iv) When temperature increases, by Le Chatelier’s Principle, the equilibrium shifts left to favour the backward endothermic reaction so as to absorb / remove heat. Hence the yield of the ester decreases. Kc value decreases (c) (i) A solution that maintains a fairly constant pH / resists pH changes when a small amount of acid or base is added to it. (ii) When a small amount of H +