OCR A Jan 2013 Paper 5 Q8 (with explained solutions) |

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OCR A Jan 2013 Paper 5 Q8

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8 Vanadium is a transition element that forms compounds and ions in which vanadium has oxidation states +2, +3, +4 and +5. 18 (a) Complete the electron configuration of a vanadium ion in the +3 oxidation state: 1s22s22p6[1] (b) Suggest why vanadium does not form ions in which vanadium has an oxidation state greater than +5.[1] (c) A student carries out an investigation into the oxidation states of vanadium as outlined below. Stage 1 Stage 2 Stage 3 A 0.126 g sample of vanadium metal is completely reacted with acid to form a yellow solution. The solution is made up to 50.0 cm3 in a volumetric flask. This yellow solution contains VO3 The yellow solution is reduced to form a violet solution containing Vn+ ions. This 50.0 cm3 violet solution contains vanadium in the +n oxidation state.ions with vanadium in the +5 oxidation state. 10.0 cm3 of the violet solution is titrated with 2.25102 mol dm3 KMnO4(aq). 13.2 cm3 of KMnO4(aq) are required to reach the end-point. In the titration, Vn+ ions are oxidised back to VO3ions.MnO4ions are reduced: MnO4 (aq) + 8H+(aq) + 5e Mn2+(aq) + 4H2O(l) (i) Why is there no clear colour change at the end-point of this titration?[1]OCR 2013<br />
(ii) Analyse the students results as follows: 19Determine the value of n in the Vn+ ions formed in Stage 2Construct an equation for the reaction that takes place during the titration. Show all your working. n =equation: [7] [Total: 10]OCR 2013 END OF QUESTION PAPER<br />

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Mark Scheme January 2013 Marks Guidance ALLOW 4s0: (1s22s22p6 ) 3s23p63d24s0 ALLOW subscripts for superscripts ALLOW S, P, D (i.e. upper case) ALLOW 3d sub-shell is empty OR no d electrons left ALLOW 6th electron in a 3p sub-shell/orbital ALLOW too much attraction on 3p electrons OR a lot of energy required to remove 3p electrons IGNORE only 5 electrons in outer shell IGNORE full outer shell/noble gas electron configuration IGNORE no 3d sub-shell Note: Key comment about 3d sub-shell being empty OR non-removal/greater attraction of 3p electrons ALLOW KMnO4 AND Vn+/V2+ have similar colours ALLOW KMnO4 is purple and the solution is violet Assumption is that the solution is V2+(aq) ALLOW any reasonable description of purple/mauve/violet colours DO NOT ALLOW just KMnO4 is purple/pink IGNORE reference to Mn2+ being (pale) pink F325 Question 8 (a) (1s22s22p6 ) 3s23p63d2 (cid:57) Answer (b) (Only) 5 electrons in 4s and 3d sub-shells/orbitals (cid:57) (c) (i) KMnO4 is purple/pink AND Vn+/V2+ is violet (cid:57) 21<br />
F325 Mark Scheme January 2013 Question 8 (c) (ii) Marks are for correctly calculated values. Working shows Answer how values have been derived. n(KMnO4) = 2.25 (cid:117)10(cid:16)2 (cid:117)13.2 n(V) = 0.126 50.9 1000 = 2.48 (cid:117) 103 (mol) (cid:57) = 2.97 (cid:117) 104 (mol) (cid:57) 2.48 (cid:117)10(cid:16)3 = 4.96 (cid:117) 104 (mol) OR 1.67 OR 5 3 OR 5 x 2.97 (cid:117) 104 = 1.485 x 103 (mol) (cid:57) 2.97 (cid:117)10(cid:16)4 = 1.67 = 4.96 (cid:117)10(cid:16)4 Factor of 5: ratio n(Vn+) n(MnO4 -) OR 1 mol MnO4 reacts with 1.67 mol Vn+ (cid:57) 5 : 3 ratio seen AND n = 2 (cid:57) Correct equation with all species on both sides cancelled: 5V2+(aq) + 3MnO4 5VO3 5V2+ + 3MnO4 Complete equation correct (cid:57) (aq) + 3H2O(l) (cid:127)(cid:111)+ 3Mn2+ on right (cid:57)on left AND 5VO3 (aq) + 3Mn2+(aq) + 6H+(aq) 22 Marks Guidance FULL ANNOTATIONS MUST BE USED --------------------------------------------------------------------------- ALLOW 2.48 x 103 up to calculator value of 2.475442043 x 103, correctly rounded ALLOW 4.95 (cid:117) 104 (mol) from 2.475442043 x 103 5 3 1.67 1 OR 1.67 OR ALLOW ratio 2.48 (cid:117)10(cid:16)3 1.485 (cid:117)10(cid:16)3 =reacts with 5 mol V2+; V changes oxidation number by 3 OR 3 electrons transferred to V n(Vn+) n(MnO4 -) ALLOW inverse ratio DO NOT ALLOW n = 2 without some justification e.g.: 3 mol MnO4 IGNORE state symbols ------------------------------------------------------- ALLOW any attempted equation using n = 2, 3 OR 4. See correct eqn for n=2 and equations on next page -------------------------------------------------------<br />
F325 Mark Scheme January 2013 (aq) + Mn2+(aq) + 22H+(aq)+ Mn2+ on right (cid:57) Guidance (aq) + 11H2O(l) From V4+ : 5V4+(aq) + MnO4 (cid:111) 5VO3 on left AND 5VO3 5V4+ + MnO4 Complete equation correct (cid:57) ------------------------------------------------------- From V3+ : 5V3+(aq) + 2MnO4 (cid:111) 5VO3 5V3+ + 2MnO4 Complete equation correct (cid:57) on left AND 5VO3 (aq) + 7H2O(l) (aq) + 2Mn2+(aq) + 14H+(aq) (cid:57)(cid:57)+ 2Mn2+ on right (cid:57) Question 8 (c) (ii) Cont. Answer Marks Total 10 23<br />

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