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Edexcel Jun 2014 (IAL) Paper 5 Q20
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![SECTION B Answer ALL the questions. Write your answers in the spaces provided. 20 The table below shows some ions of vanadium that are stable in aqueous solution. Ion V(H2O)6 2+ V(H2O)6 3+ VO2+ VO2 Oxidation number of vanadium Colour in aqueous solution violet green blue yellow +5 (a) (i) Complete the table above by adding the missing oxidation numbers. (ii) Complete the electronic configuration of the element vanadium and hence explain the highest oxidation number shown by vanadium. Electronic configuration of the element vanadium: (1) (2) [Ar]Explanation of the highest oxidation number*P42979A01332* 13 Turn over<br />
*(iii) Explain why V(H2O)6 3+ is coloured. (4)(iv) Suggest why the +5 oxidation state of vanadium exists as VO2 solution, rather than V(H2O)6 5+. + in aqueous (2)(v) If V(H2O)6 answer. 5+ did exist in aqueous solution, would it be coloured? Explain your (1)14 *P42979A01432*<br />
(b) (i) Sulfur dioxide dissolves in water to form sulfuric(IV) acid, H2SO3. Determine the feasibility of using sulfur dioxide dissolved in water to reduce VO2+(aq) to V3+(aq). Quote the relevant half equations and standard electrode potentials from page 15 of the Data Booklet and write the overall equation for the reaction. (3)(ii) Write an ionic equation for the disproportionation of vanadium in oxidation state +3 to vanadium in oxidation states +2 and +4. (iii) Calculate the Ecell for this disproportionation and hence determine its feasibility. (1) (2)(Total for Question 20 = 16 marks) *P42979A01532* 15 Turn over<br />
BLANK PAGE 16 *P42979A01632*<br />](https://learnah.org/wp-content/uploads/2018/08/l_smalljpgJune_2014_IAL_QP_-_Unit_5_Edexcel_Chemistry_A-level_20-0.jpg)
![Section B Question Number 20(a)(i) Acceptable Answers Reject Mark Ion Oxidation number of vanadium V(H2O)6 2+ V(H2O)6 3+ VO2+ VO2 +2 +3 +4 (+5) All three correct (1) IGNORE omission of + Question Number 20(a)(ii) Electronic configuration of V: [Ar]3d3 Acceptable Answers Mark Reject Gives electronic structure of Ar Loss of electrons from a (single) d orbital 4s2 ALLOW 1s2 2s2 2p6 3s2 3p6 3d3 4s2 [Ar] 4s23d3 1s2 2s2 2p6 3s2 3p6 4s2 3d3 IGNORE Additional [Ar] (1) 5 electrons in valence shell / available for bonding ALLOW 5 electrons in outer shell (So max ON = +5) OR Uses the 2 4s and 3 3d electrons (1) ALLOW Lose 5 electrons (to form Ar structure) No TE on incorrect electronic configuration except 3d5 (4s0) IGNORE Stability of +5 oxidation state WCH0501 1406<br />
Question Number 20(a)(iii) (3)d orbitals / (3)d subshell split (by Acceptable Answers the attached ligands) (1) Electrons are promoted (from lower to higher energy d orbital(s) / levels) OR Electrons move from lower to higher energy d orbital(s) / levels) ALLOW dd transitions occur (1) Absorbing energy /photons of a certain frequency (in the visible region) ALLOW Absorbing light (1) Reflected / transmitted / remaining light is coloured / in the visible region ALLOW Complementary colour seen Reflected / transmitted / remaining light / frequency is seen (1) Penalise omission of (3)d once only. Ignore reference to electrons relaxing / dropping to the ground state Mark Reject Orbital / shell is split Emitted WCH0501 1406<br />
Mark Reject Just Polarize Ionic bonds Acceptable Answers Question Number 20(a)(iv) V5+ is (small &) highly charged /has a (very) high charge density (1) Would polarize / distort H2O / H2O electron clouds / OH bond ALLOW O-H bond weakening / breaking OR Deprotonation (1) IGNORE References to ionization energy of V / highly electropositive Reject Mark WCH0501 1406 Question Number 20(a)(v) No. Because V5+ has no d electrons / d Acceptable Answers sub-shell is empty / d orbitals are empty. IGNORE Any mention of 4s V5+ has no partially filled d orbitals<br />
Acceptable Answers Question Number 20(b)(i) Either 2 + 2eH2SO3 + H2O Eo = Method 1 (using equations) 4H+ + SO4 +0.17 (V) VO2+ + 2H+ + e V3+ + H2O Eo = +0.34 (V) (1) 2VO2+ + H2SO32V3+ + SO4 2 + H2O (1) Ecell (SO2) = 0.340.17 = (+)0.17 (V) AND So reduces V(IV) to V(III) / reaction is feasible (1) OR Method 2 (using anticlockwise rule) When half reactions are placed in order (more negative first) 4H+ + SO4 = +0.17 V VO2+ + 2H+ + e V3+ + H2O Eo = +0.34 V Required reaction goes in anticlockwise direction Arrows on half equations and explanation (2) 2 + 2eH2SO3 + H2O Eo 2VO2+ + H2SO32V3+ + SO4 2 + H2O (1) Reject Mark Use of thiosulphate half cell =0 Uncancelled electrons Uncancelled electrons WCH0501 1406<br />
Acceptable Answers Question Number 20(b)(ii) 2V3+ + H2OV2+ + VO2+ + 2H+ Reject Question Number 20(b)(iii) ( Relevant electrode potentials are Acceptable Answers Reject Mark Mark ALLOW V(H2O)6 3+ and V(H2O)6 2+ VO2+ + 2H+ + e V3+ + H2O Eo = +0.34 V V3+ + eV2+ Eo = 0.26 V ) Ecell (disproportionation ) =( 0.260.34 ) = 0.6(0) (V) (1) Ecell negative so disproportionation not (thermodynamically) feasible. (1) TE for second mark only if value given Total for Question 20 = 16 marks WCH0501 1406<br />](https://learnah.org/wp-content/uploads/2018/08/June_2014_IAL_MS_-_Unit_5_Edexcel_Chemistry_A-level_20-0.jpg)
