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CIE Jun 2014 v3 Paper 4 Q2
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![The ions of transition elements form complexes by reacting with ligands. PMT (a) (i) State what is meant by the terms: complex,ligand.(ii) Two of the complexes formed by copper are [Cu(H2O)6]2+ and CuCl 4 Draw three-dimensional diagrams of their structures in the boxes and name their shapes. 2. [Cu(H2O)6]2+ CuCl 4 2 shape:shape:(iii) Platinum forms square-planar complexes, in which all four ligands lie in the same plane as the Pt atom. There are two isomeric complexes with the formula Pt(NH3)2Cl 2. Suggest the structures of the two isomers, and, by comparison with a similar type of isomerism in organic chemistry, suggest the type of isomerism shown here. Structures of isomers: isomer 1 isomer 2 Type of isomerism:[7]UCLES 2014 9701/43/M/J/14<br />
PMT (b) Copper forms two series of compounds, one containing copper(II) ions and the other containing copper(I) ions. (i) Complete the electronic structures of these ions. Cu(II) Cu(I) [Ar][Ar](ii) Use these electronic structures to explain why copper(II) salts are usually coloured,copper(I) salts are usually white or colourless.[5]UCLES 2014 9701/43/M/J/14 [Turn over<br />
(c) Copper(I) oxide and copper(II) oxide can both be used in the ceramic industry to give blue, green or red tints to glasses, glazes and enamels. The table lists the values for some compounds. PMT compound Cu2O(s) CuO(s) Cu(NO3)2(s) NO2(g) / kJ mol1 168.6 157.3 302.9 +33.2 (i) Copper(II) oxide can be produced in a pure form by heating copper(II) nitrate. Use suitable values from the table to calculate the H o for this reaction. Cu(NO3)2(s)CuO(s) + 2NO2(g) + 2 1 O2(g) H o =kJ mol1 (ii) Copper(I) oxide can be produced from copper(II) oxide. Use suitable values from the table to calculate H o for the reaction. 2CuO(s) Cu2O(s) + 2 1 O2(g) H o =kJ mol1 Hence suggest whether a low or a high temperature of oxidation would favour the production of copper(I) oxide. Explain your reasoning.[4] [Total: 16]UCLES 2014 9701/43/M/J/14<br />](https://learnah.org/wp-content/uploads/2018/08/June_2014_v3_QP_-_Paper_4_CIE_Chemistry_A-level_2-0.jpg)
![Page 2 Mark Scheme Syllabus Paper GCE A LEVELMay/June 2014 9701 43 Section A (a) (i) m. pt. is high(er) / large(r) / greater (for iron) density is high(er) / large(r) / greater (for iron) (ii) (higher m. pt. due to) strong attraction between cations and electrons or more delocalised electrons (higher density due to) greater Ar and smaller radius [Total: 16] (a) (i) A complex is a compound / molecule / species / ion formed by a central metal atom / ion surrounded by / bonded to one or more ligands / groups/ molecules / anions A ligand is a species that contains a lone pair of electrons that forms a dative bond to a metal atom / ion / or a lone pair donor to metal atom / ionCambridge International Examinations 2014 PMT [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] (b) (i) components to be added: voltmeter or V salt bridge [must be labelled] (ii) M1: A and B copper (metal) or Cu and iron (metal) or Fe M2: either C or D as 1 mol dm 3 / 1 M M3 C and D Cu or CuSO4 or CuCl 2 or Cu (NO3)2 etc. and 2+ 2+ Fe or FeSO4 etc. (iii) E cell = 0.34 + 0.44 = 0.78 (V) 2+ (iv) if C is Fe ; (as [C] increases), the E of the Fe 2+ / Fe increases / becomes more positive / less negative [1] so the overall cell potential / Ecell would decrease / become less positive / more negative or 2+ if C is Cu ; (as [C] increases), the E of the Cu positive / less negative 2+ /Cu increases / becomes more so the overall cell potential / Ecell would increase / become more positive / less negative (c) (i) (colour change is) colourless to pink/pale purple or (end point is the first) permanent (pale) pink/pale purple colour (ii) {n(MnO4 ) = 0.0218.1/1000 = 3.6210 mol} n(Fe ) = 5n(MnO4 ) = 1.8110 2+ 3 -4 mol mass of Fe = 55.8 x 1.8110 3 = 0.101 g (M255.8) ecf Mr = mass / moles = 0.500/1.8110 3 = 276.2 ecf<br />
Page 3 Mark Scheme Syllabus Paper GCE A LEVELMay/June 2014 9701 43 PMT Cl Cu Cl Cl 2- Cl Cl NH [1] + [1] [1] NH Pt or Cl Pt Cl Cl NH [1] [1] [1] [1] [2] [1] [2] [1] [1] (ii) 2+ OH Cu and OH correct 3D structures: octahedral and tetrahedral (iii) Cl Cl Pt or NH both structures geometric or cis-trans (b) (i) Cu(II) is [Ar] 3d Cl Pt Cl NH Cu(I) is [Ar] 3d 10 (ii) Cu(II): d orbitals / subshell are split (in ligand field) and Cu(I): in doing so it absorbs a photon / light no gap in upper orbital / all orbitals are full electron moves from lower to upper orbital or an electron is promoted / excited (c) (i) H = +233.2157.3 + 302.9 = (+) 212 kJ mol 1 ecf (ii) H = 168.6 + 2157.3 = (+)146 kJ mol 1 allow ecf from (c)(i) high T / temperature since H is positive / endothermic [Total: 16] (a) heat in dilute HCl(aq) (or H2SO4(aq)) (b) (i) four isomers [1] [1]Cambridge International Examinations 2014<br />](https://learnah.org/wp-content/uploads/2018/08/June_2014_v3_MS_-_Paper_4_CIE_Chemistry_A-level_2-0.jpg)
