M1. (a) Yellow (solution) 1. Orange solution 1 SO 4. Yellow / purple (solution) Allow orange / brown (solution) 1. Brown precipitate / solid 1 + 3H 2

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Miles Morrison
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1 M. (a) Yellow (solution) range solution Cr + H + Cr 7 + H Allow equation with H S (b) Yellow / purple (solution) Allow orange / brown (solution) Brown precipitate / solid [Fe(H ) 6 ] + + H Fe(H ) (H) + H (c) Blue (solution) Allow pale blue Dark / deep blue solution Ignore any reference to blue ppt [Cu(H ) 6 ] + + NH [Cu(H ) (NH ) ] + + H Can be in two equations (d) Colourless (solution) White precipitate / solid Do not allow grey Bubbles / effervescence / gas evolved / given off Do not allow just C [Al(H ) 6 ] + + C Al(H ) (H) + C + H [] M. (a) A ligand is an electron pair / lone pair donor Allow uses lone / electron pair to form a co-ordinate bond Page of

2 A bidentate ligand donates two electron pairs (to a transition metal ion) from different atoms / two atoms (on the same molecule / ion) QoL (b) CoCl diagram Tetrahedral shape 09 8 Four chlorines attached to Co with net charge correct Charge can be placed anywhere, eg on separate formula Penalise excess charges Allow 09 to 09.5 [Co(NH ) 6 ] + diagram ctahedral shape 90 Six ammonia / NH molecules attached to Co with + charge correct Allow 80 if shown clearly on diagram CE= 0 if wrong complex but mark on if only charge is incorrect (c) In different complexes the d orbitals / d electrons (of the cobalt) will have different energies / d orbital splitting will be different Light / energy is absorbed causing an electron to be excited Different frequency / wavelength / colour of light will be absorbed / transmitted / reflected Page of

3 (d) mol of H oxidises mol of Co + r H + Co + H + Co + M r CoS.7H = 8 If M r wrong, max for M, M, M5 Moles Co + = 9.87 / 8 = 0.05 Moles H = 0.05 / = M is method mark for (M) / (also scores M) Volume H = (moles 000) / concentration = ) / 5.00 =.5 cm / (.5 0 dm ) Units essential for answer M5 is method mark for (M) x 000 / 5 Allow. to.6 cm If no : ratio or ratio incorrect Max for M, M & M5 Note: Answer of 7 cm scores for M, M, M5 (and any other wrong ratio max ) Answer of 6.8 cm scores for M, M, M5 (and any other wrong M r max ) Answer of.5 cm scores for M5 only (so wrong M r AND wrong ratio max ) [6] M. (a) (i) M (yellow precipitate is) silver iodide R AgI (which may be awarded from the equation) M Ag + + I AgI (Also scores M unless contradicted) M sodium chloride R NaCl For M Accept multiples Ignore state symbols Allow crossed out nitrate ions, but penalise if not crossed out Page of

4 (ii) The silver nitrate is acidified to react with / remove ions that would interfere with the test prevent the formation of other silver precipitates / insoluble silver compounds that would interfere with the test remove (other) ions that react with the silver nitrate react with / remove carbonate / hydroxide / sulfite (ions) Ignore reference to false positive (iii) M and M in either order M Fluoride (ion) R F M Silver fluoride / AgF is soluble / dissolves (in water) no precipitate would form / no visible /observable change Do not penalise the spelling fluoride, Penalise fluride once only Mark M and M independently (b) M Ba + + S - BaS (or the ions together) M white precipitate / white solid / white suspension M Barium meal or ( internal ) X-ray or to block X-rays M BaS / barium sulfate is insoluble (and therefore not toxic) For M, ignore state symbols Allow crossed out sodium ions, but penalise if not crossed out For M, ignore milky If BaS R BaS used in M and M, penalise once only For M Ignore radio-tracing For M NT barium ions NT barium NT barium meal NT It unless clearly BaS Page of

5 (c) M (.00000) + (.0079) = M Ethene and C or they have an imprecise M r of 8.0 / 8 R Ethene and C or they have the same M r to one d.p. R These may be shown by two clear, simple sums identifying both compounds M C H + C + H (H C=CH ) M Displayed formula M5 Type of polymer = Addition (polymer) M must show working using 5 d.p.for hydrogen Penalise similar or close to, if this refers to the imprecise value in M, since this does not mean the same For M, accept CH =CH R CH CH For M, all bonds must be drawn out including those on either side of the unit. Penalise sticks Ignore brackets around correct repeating unit but penalise n Penalise additional 5 [5] M. (a) Brown ppt/solid Gas evolved/effervescence [Fe(H ) 6 ] + + C Fe(H ) (H) + C + H Must be stated, Allow C evolved. Do not allow C alone Correct iron product () allow Fe(H) and in equation Balanced equation () Page 5 of

6 (b) (c) White ppt/solid Colourless Solution nly award M if M given or initial ppt mentioned [Al(H ) 6 ] + + H Al(H ) (H) + H Allow [Al(H ) 6 ] + + H Al(H) + 6H Al(H ) (H) + H [Al(H) 6 ] + H Blue ppt/solid Allow formation of [Al(H ) 6 x (H) x ] (x ) where x =,5,6 Allow product without water ligands Allow formation of correct product from [Al(H ) 6 ] + (Dissolves to give a) deep blue solution nly award M if M given or initial ppt mentioned (d) [Cu(H ) 6 ] NH Cu(H ) (H) + NH Allow [Cu(H ) 6 ] + + NH Cu(H) + NH + + H Allow two equations: NH + H NH + + H then [Cu(H ) 6 ] + + H Cu(H) + H etc Cu(H ) (H) + NH [Cu(H ) (NH ) ] + + H + H Allow [Cu(H ) 6 ] + + NH [Cu(H ) (NH ) ] + + H Green/yellow solution [Cu(H ) 6 ] + + Cl [CuCl ] + 6H [] Page 6 of

7 B C M5. (a) A Cr(H ) (H) (or Cr(H) ) [Cr(H ) 6 ] + +C [Cr(H ) (H) ] + C +H (or gives Cr(H) + C + 9H ) (b) (i) NaH (or KH) (ii) +6 (or 6 or +VI or VI) (iii) H (or Na or Ba ) [Cr(H) 6 ] + H Cr +H + e (or [Cr(H) 6 ] Cr +H + H + + e ) (c) (i) At least one H NCH CH NH with correct structure and bonding to Cr via N 6 co ordination with en drawn correctly Correct + charge (Mark independently but must not have 6 monodentate ligands) Page 7 of

8 (d) [Cr(H ) 6 ] + (ii) (iii) Same (or similar) type of bonds broken and made Same number of bonds broken and made (or same co ordination number) Entropy change (or ΔS) is positive (or increase in disorder) Because there are more product particles than reactant particles Reducing agent (mark independently) (e) (i) Ethanal (or CH CH) (not CH CH) (ii) Ethanoic acid (or correct formula) [8] M6. (a) oxidation state of N in Cu(N ) : +5; oxidation state of N in N : +; oxidation product: oxygen; (b) copper-containing species: [Cu(H ) 6 ] + ; shape: octahedral; Page 8 of

9 (c) (i) precipitate B: Cu(H ) (H) or Cu(H) or name; equation: [Cu(H ) 6 ] NH Cu(H ) (H) + NH R NH + H NH + + H and (ii) NH accepts a proton; [Cu(H ) 6 ] + + H Cu(H ) (H) + H ; (d) (i) identity: [Cu(NH ) (H ) ] + ; colour: deep blue; equation: (ii) Cu(H ) (H) + NH [Cu(NH ) (H ) ] + + H + H ; NH is an electron pair donor; (e) identity: [CuCl ] ; colour: shape: yellow-green; tetrahedral; (f) (i) Is s p 6 s p 6 d 0 ; (ii) role of Cu: a reducing agent; [7] M7. (a) d 7 Page 9 of

10 (b) [Co(H ) 6 ] + Pink (c) (i) [Co(NH ) 6 ] + Pale brown or straw (ii) [Co(H ) 6 ] + + 6NH [Co(NH ) 6 ] + + 6H (d) [Co(NH ) 6 ] + An oxidising agent [8] M8. (a) (i) Ammonia If reagent is missing or incorrect cannot score M Starts as a pink (solution) Changes to a yellow/straw (solution) Allow pale brown Do not allow reference to a precipitate (ii) (dark) brown Do not allow pale/straw/yellow-brown (i.e. these and other shades except for dark brown) Page 0 of

11 (b) (i) Ruby/red-blue/purple/violet/green Do not allow red or blue If ppt mentioned contradiction/ce =0 Green If ppt mentioned contradiction/ce =0 [Cr(H ) 6 ] + + 6H [Cr(H) 6 ] + 6H Formula of product Can score this mark in (b) (ii) (ii) H + e H [Cr(H) 6 ] + H Cr + 8H + H Allow mark out of for a balanced half-equation such as Cr(III) Cr(VI) + e or Cr + + H Cr + 8H + + e etc also for Cr(III) + H Cr (unbalanced) Yellow Do not allow orange (c) Mn + 6H + + 5H Mn + + 8H + 5 if no equation and uses given ratio can score M, M, M & M5 Moles Mn = (.5/000) =.55 0 Note value must be quoted to at least sig. figs. M is for.55 0 Moles H = (.55 0 ) 5/ =.8 0 M is for 5/ (or 7/) Mark consequential on molar ratio from candidate's equation Page of

12 Moles H in 5 cm original M is for 0 = (.8 0 ) 0 = 0.08 riginal [H ] = 0.08 (000/5) =.8 mol dm (allow.5-.0) M5 is for consequentially correct answer from (answer to mark ) (000/5) Note an answer of between.5 and.0 is worth marks) If candidate uses given ratio /7 max marks: M: Moles of Mn =.55 0 M: Moles H = (.55 0 ) 7/ = M: Moles H in 5 cm original = ( ) 0 = M: riginal [H ] = (000/5) =. mol dm (allow.0 to.5) [7] Page of

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Part of the practical procedure is given below.

Part of the practical procedure is given below. A peptide is hydrolysed to form a solution containing a mixture of amino acids. This mixture is then analysed by silica gel thin-layer chromatography (TLC) using a toxic solvent. The individual amino acids