ChemHL P2 N00 M

ChemHL P2 N00 M - INTERNATIONAL BACCALAURAT BACHILLERATO...

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MARKSCHEME November 2000 CHEMISTRY Higher Level Paper 2 N00/420/H(2)M BACCALAUREATE BACCALAURÉAT INTERNATIONAL BACHILLERATO INTERNACIONAL 17 pages
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SECTION A [1] 1. (a) (i) A, C, D (Must have all three for mark.) [1] (ii) A [1] (iii) van der Waal’s forces (or dispersion or London forces or induced dipole induced dipole (but not dipole–dipole) interaction). [1] (iv) Nitrogen or oxygen or fluorine. [1] (v) F because it has the highest melting/boiling point. (Need explanation for mark.) [1] [1] (vi) E: a metal; F: a metalloid (accept semi-conductor). [1] (b) (i) C CH 3 O OH H O O C CH 3 δ− δ− δ+ δ+ [1] Exists as a dimer in non-polar solvents (because of strong inter-molecular H-bonding) but exists as a monomer in aqueous solution (because of H-bonding with water). (No mark without mention of non-polar solvent.) [1] (ii) C O C OH H C O CO H H δ+ δ− (Award [1] for diagram showing intra-molecular H-bonding) [1] The cis-isomer experiences intra -molecular H-bonding that reduces the chances of H-bonding between molecules; OR the trans-isomer experiences (more) inter -molecular H-bonding that increases the chances of H-bonding between molecules. continued… – 6 – N00/420/H(2)M
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Question 1 (b) continued [1] (iii) O C 2 H 5 H H O C 2 H 5 δ− δ+ [1] Only ethanol experiences H-bonding because H is bonded to O, whereas in the ether, all H atoms are bonded to C OR ether does not exhibit H-bonding as H is not bonded to O. – 7 – N00/420/H(2)M
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[3] 2. (a) (i) Fe 3 + : ↑↓ Fe 2 + : ↑↓ ↑↓ Fe 0 : 3d 4s (Award [1] for each correct electronic configuration.) [1] (ii) +2 [1] (iii) , OR OR OR FeO 4 2 - FeO 3 Fe O 27 2 - 26 Fe O [1] (b) (i) A ligand is an anion or a molecule (having lone electron pairs) that can form a (co-ordinate) bond to a (central) atom or cation. [1] [1] (ii) A Lewis acid–base reaction The ligand is a Lewis base (or an electron pair donor) and the metal ion a Lewis acid (or an electron pair acceptor). OR because donates an pair to form a covalent bond with HO 2 e Fe 2 + (Need both comments for mark.) [1] [1] (iii) Because the d orbitals are split into two sets of different energy levels and electron transitions between them are responsible for their colours. (Need both statements for mark.) Different ligands split to different extent giving different colours. OR They contain different ligands, so the energy difference between the split 3d orbitals is not the same in each case. The transitions between the two absorb different amounts of energy, e corresponding to different wavelength of light in the visible spectrum. – 8 – N00/420/H(2)M
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3. (a) (i) t k 1 2 0 693 = . k = ¥ 0 693 162 10 4 . .s [1] -- 428 10 51 (No mark without units.) (ii) ln ln k = - + E RT A a Therefore E a =- (ln ln ) AR T k [1] - ¥ ¥ - 3219 10 059 8314 1107 1 .( . ) . J K K (Award [1] for correct temperature.) [1] = 122 kJ (No mark without unit. If T taken as 834 , then ! C a E 92.1kJ = [1] (iii) Slow step: OR NO NO NO 22 4 2 N 2 2 2 + [1] Fast step: OR N O 42 2 2 2 Æ+ N O 2 2 [2] (b) Enthalpy OR P.E.
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ChemHL P2 N00 M - INTERNATIONAL BACCALAURAT BACHILLERATO...

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