SRJC H2 CHEM P2 ANS

SRJC 2009 9746/02/PRELIM/2009 [Turn Over SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2 CHEMISTRY 9746/02 Preliminary Examination 26 th August 2009 Paper 2 Structured 1 hr 30min M A R K S C H E M E CONFIDENTIAL DOCUMENT Teachers to keep this document under lock and key until 26 August 2009, 12pms RECOMMENDED F ORMAT OF PRINTING 2 PER PAGE This document consists of 15 printed pages and 1 blank page

SRJC 2009 9746/02/PRELIM/2009 [Turn Over BLANK PAGE

3 SRJC 2009 9746/02/PRELIM/2009 [Turn Over 1 The atomic and cationic radii of the Period 3 elements, Na to Cl, are plotted in the graph below, in order of increasing atomic number. Na Mg Al Si P S ClNa+ Mg2+ Al3+ Si4+ P3- S2- Cl- 0.00 0.05 0.10 0.15 0.20 0.25 10 11 12 13 14 15 16 17 18 Radius / nm A tomic & Ionic Radii vs. Proton no. for Elements in Period 3 Proton no. Atomic Radii Cationic Radii Anionic Radii (a) Explain each of the following: (i) The atomic radius decreases across the period from Na to Cl. Across the period from Na to Cl,  nuclear charge increases [1/2M];  but increase in shielding effect is negligible as electron is added to the same quantum shell; [1/2M]  Hence, the effective nuclear charge increases OR the electrostatic forces of attraction between the nucleus and the outermost/ valence electrons increases across the period. [1M]  Therefore, the atomic radius decreases across the period from Na to Cl. [2] (ii) The cationic radius of Na+ is smaller than its atomic radius. Na+ has one electron less than Na or Na + has one quantum shell less than Na. Nuclear charge remains the same . [1M] The electrostatic forces of attraction between the nucleus and the remaining electrons are greater for Na+. [1M] [2]

4 (b) (i) Indicate, on the graph, the relative positions of the ions, P3–, S2– and Cl–. Refer to Graph [1M] [1] (ii) Explain your answer in (b)(i). The anionic radii of P 3–, S 2– and C l– are larger than their respective atoms. This is due to electrons are being added to the same valence shell , resulting in greater inter-electronic repulsion s between the valence electrons. [1M] Across the period from P 3–, S 2– and C l–, the anionic radii decrease. The nuclear charge increases across the period with negligible increase in shielding effect resulting in the corresponding increase in the effective nuclear charge OR electrostatic forces of attraction between the nucleus and the valence electrons. [1M] [2] (c) (i) Write down the electronic configuration in the ground state for an atom of sulphur. 1s 22s22p63s23p4 or [Ne] 3s23p4 [1M] [1] (ii) Explain the following phenomenon: sul phur can react with fluorine to form SF2, SF4 and SF6 but oxygen can only form OF2. Sulphur is in Period 3 and hence has the availability of 3d orbitals to expand beyond its octet structure while oxygen cannot expand its octet due to the absence of d-orbitals. [1M] [1] (d) The successive ionisation energies in kJ mol –1 of an element A are as follows: 740, 1500, 7700, 10500, 13600, 18000, 21700 State the Group in which A belongs to and explain your answer. Suggest the formula of the chloride of A. 740, 1500, 7700, 10500, 13600, 18000, 21700 760 6200 2800 3100 4400 3700 (difference in IE) - Element A belongs to Group II [1M] - Largest energy difference between the 2nd and 3rd IE. [1/2M] - Removal of the 3 rd electron is from an inner quantum shell which requires more energy. [1/2M] Formula: ACl 2 [1M] [Total: 12] SRJC 2009 9746/02/PRELIM/2009 [Turn Over

5 2 A pheromone is a chemical s ubstance that, when secreted by an individual of a species, for example the insects, can elic it a certain type of behaviour in other individuals. Compound B below is an alarm pheromone secreted by several species of ants, to send out warning signals to other ants. CH3CH2CH2CH=CHCHO Compound B (a) Suggest the type of isomerism that exists in compound B [1] Geometric Isomerism or Cis-Trans Isomerism [1M] (b) Compound B can undergo the following conversion: CH3CH2CH2CH CHCHO Reaction I CHCCH3CH2CH2CH OH CN H Compound B Compound C State the r eagents and conditions required for the above reaction. Outline the mechanism of the reaction, in cluding curly arrows, showing the movement of electrons, and all charges. [4] Reagents: HCN [1/2 M] Conditions: NaOH (or trace of NaCN), 10-20oC or cold [1/2 M] CO H CN CH3CH2CH2CH=CH HC N Fast Step CO H H CN CH3CH2CH2CH=CH +   CN-   CN- CO CH3CH2CH2CH=CH H Slow Step C O- H CN CH3CH2CH2CH=CH M] for curly arrows termediate drawn [1 [1M] all charges [1M] for correct in SRJC 2009 9746/02/PRELIM/2009 [Turn Over

6 (c) (i) Compound C has a chiral carbon. Illust rate the isomerism exhibited with the aid of structural formulae. [1] C H CN OHCH3CH2CH2CH=CH C H CN HO CH3CH2CH2CH=CH [1M] [Comments: invert the molecule] (ii) Compound C, however, does not rotate plane polarised light. Explain this phenomenon with reference to the mechanism mentioned in (b). - Carbonyl carbon of compound B CO CH3CH2CH2CH=CH H is trigonal planar [1/2 M] or Ca 2rbonyl carbon atom in compound A is sp hybridised. Hence, the C atom and the three atoms attached to it lie on the same plane. - ucleophilic attackN on the carbonyl carbon can occur from either above or below the plane [1/2 M] and the chance is 50%-50% [1/2 M] (equally likely) - acemic mixture is formedR [1/2 M] which is optically inactive. h optical forms of compound B will be formed in equimolar or Bot which is optically inactive. (d) The reaction of hydroxide ion with chloromethane to yield methanol and chloride ion is an example of nucleophilic substitution reaction: OH─ + CH3Cl ⇌ CH3OH +Cl─ The value of ∆H for the reaction is –75 kJ mol -1, and the value of ∆ S is +54 J mol-1. (i) What i s the value of ∆G at 298 K? ∆G = ∆H – T∆S = (– 75 x 1000) – (298)(54) = – 91.1 kJ mol–1 [1M] (ii) Predict whether reaction is feasible at 298 K? ∆G = – 91.1 kJ mol–1 < 0, the reaction is feasible. [1M] (iii) Will temperature affect the feasibility of the reaction? Explain. No. [1M] o This is bec ause ∆G < 0 at all temperatures. Hence, the reaction is feasible at all temperatures. [1M ] [Total: 12] SRJC 2009 9746/02/PRELIM/2009 [Turn Over

7 3 Ammonia is manufactured in the Haber process. N2 (g) + 3H 2NH3 (g) ∆H = -92 kJ mol-1 2 (g) Le Chatelier’s principle predicts that the highest yield of ammonia is obtained at high pressure and low temperature. Ho wever, in practice, these conditions are not used. (a) Using Le Chatelier’s principle, explain why the above conditions are used. [3]  reduce the pressure.At high pressure, the system will try to lier’s Principle, position of the equilibrium [1/2 M]  By Le Chate shifts to the right/forward [1/2 M] towards a reduction in the number of moles of gas [1/2 M] to decrease the pressure.  t low temperature, the system will try to produce more heat A . [1/2 M]  By Le Chatelier’s Pr