Questions & Answers

Q1(a).
Short Answer 1 mark

The solubility of the Group 2 hydroxides increases down the group. Explain this trend.

Q1(b).
Short Answer 1 mark

The solubility of Be(OH) in water is 2.40 × 10

Q1(i).
Short Answer 2 marks

Write an expression for the solubility product, K , of Be(OH) and state its units. units =

Q1(ii).
Short Answer 2 marks

Calculate the numerical value of K for Be(OH) =

Q1(c).
Short Answer 1 mark

Be(OH) is soluble in aqueous solutions containing an excess of hydroxide ions and forms the complex ion [Be(OH)

Q1(i).
Short Answer 1 mark

Define the term complex ion.

Q1(ii).
Short Answer 1 mark

Draw a three-dimensional diagram to show the structure of the complex ion [Be(OH) Name the shape of the [Be(OH) complex ion. shape

Q1(d).
Short Answer 1 mark

(i) Explain why transition elements can form complex ions.

Q1(ii).
Short Answer 2 marks

Complete Table 1.1 to show the coordination number of each metal ion, and the shapes and overall polarities of the complex ions listed. Table 1.1 complex ion shape coordination number polar or non-polar cis-[Pt(HNCH ]square planar [Ag(NH non-polar [Fe(C

Q1(e).
Short Answer 1 mark

(i) Define stability constant, K stab

Q1(ii).
Short Answer 2 marks

Nickel can form complexes with the ligands en, HNCH , and tn, HNCH equilibrium 1 [Ni(H + 3en [Ni(en) + 6HO K stab = 6.76 × 10 equilibrium 2 [Ni(H + 3tn [Ni(tn) + 6HO K stab = 1.86 × 10 Construct an expression for the stability constant, K stab , for equilibrium 1. State the units for K stab stab units =

Q1(iii).
Short Answer 1 mark

Describe what the K stab values indicate about the position of equilibrium for equilibrium 1 and 2. Use the K stab values to deduce which complex, [Ni(en) or [Ni(tn) , is more stable. [Total: 16]

Q2(a).
Short Answer 1 mark

Explain why transition elements have variable oxidation states.

Q2(b).
Short Answer 1 mark

Sketch the shape of a 3d orbital.

Q2(c).
Short Answer 1 mark

Explain why transition elements form coloured compounds.

Q2(d).
Short Answer 1 mark

Aqueous solutions of copper( II) salts contain [Cu(H ions. Equilibrium 3 and equilibrium 4 show two reactions of these ions. equilibrium 3 [Cu(H

Q2(aq).
Short Answer 1 mark

+ 2OH

Q2(aq).
Short Answer 1 mark

Cu(OH)

Q2(s).
Short Answer 1 mark

+ 6H O

Q2(s)(l).
Short Answer 1 mark

equilibrium 4 [Cu(H

Q2(aq).
Short Answer 1 mark

+ 4NH

Q2(aq).
Short Answer 1 mark

[Cu(NH

Q2(aq).
Short Answer 1 mark

+ 4H O(l)

Q2(i).
Short Answer 1 mark

State the colour of Cu(OH)

Q2(s).
Short Answer 1 mark

and [Cu(NH

Q2(aq).
Short Answer 1 mark

colour of Cu(OH)

Q2(s).
Short Answer 1 mark

colour of [Cu(NH

Q2(ii).
Short Answer 1 mark

Use Le Chatelier's principle to explain why a precipitate is formed when NaOH(aq) is added dropwise to [Cu(H

Q2(e).
Short Answer 1 mark

There are two possible stereoisomers with the formula [Cu(NH Draw three-dimensional diagrams to show the two stereoisomers. Cu Cu [2] [Total: 9]

Q3(a).
Short Answer 1 mark

(i) Define first electron affinity.

Q3(ii).
Short Answer 1 mark

The first electron affinity of an atom is usually an exothermic process, whereas the second electron affinity is an endothermic process. Suggest why.

Q3(iii).
Short Answer 1 mark

Describe the general trend in first electron affinities for Cl, Br and I. Explain your answer.

Q3(b).
Short Answer 1 mark

Table 3.1 shows energy changes to be used in this question and in

Q3(b)(c).
Short Answer 1 mark

. Table 3.1 standard enthalpy change of atomisation of zinc +131 first ionisation energy of zinc +906 second ionisation energy of zinc +1733 standard enthalpy change of formation of ZnI

Q3(s).
Short Answer 1 mark

208 lattice energy, , of zinc iodide, ZnI

Q3(s).
Short Answer 1 mark

2605 first ionisation energy of iodine +1008 second ionisation energy of iodine +1846 I-I bond energy +151 enthalpy change of sublimation of iodine, I

Q3(g).
Short Answer 3 marks

+62 Calculate the first electron affinity for iodine. Use relevant data from Table 3.1 in your working. It may be helpful to draw a labelled energy cycle. Show all working.

Q3(c).
Short Answer 1 mark

Predict how of CdI

Q3(s).
Short Answer 1 mark

differs from of ZnI

Q3(s).
Short Answer 1 mark

Table 3.2 of CdI

Q3(s).
Short Answer 1 mark

is less negative than of ZnI

Q3(s).
Short Answer 1 mark

of CdI

Q3(s).
Short Answer 1 mark

is the same as of ZnI

Q3(s).
Short Answer 1 mark

of CdI

Q3(s).
Short Answer 1 mark

is more negative than of ZnI

Q3(s).
Short Answer 1 mark

Explain your answer. [Total: 9]

Q4(a).
Short Answer 1 mark

Calcium carbonate decomposes on heating. CaCO

Q4(g).
Short Answer 2 marks

, for this reaction at various temperatures. Table 4.1 1050 9.9 1085 4.3 1120 -1.3 1148 -5.8 1176 -10.3 , for this reaction remain constant over this temperature range. against T on the grid. 12.5 10.0 7.5 5.0 2.5 0.0 -2.5 -5.0 -7.5 -10.0 -12.5 1040 1060 1080 1100 1120 1140 1160 1180 1200 / kJ mol [2] for this reaction. Show all working.

Q4(b).
Short Answer 1 mark

Group 1 hydrogencarbonates, MHCO , decompose on gentle heating to give the corresponding metal carbonate, carbon dioxide and water vapour.

Q4(i).
Short Answer 1 mark

Write an ionic equation for the decomposition of the hydrogencarbonate ion.

Q4(ii).
Short Answer 1 mark

The thermal stability of Group 1 hydrogencarbonates increases down the group. Suggest an explanation for the trend in thermal stability of the Group 1 hydrogencarbonates.

Q4(c).
Short Answer 1 mark

The buffer system in seawater contains a mixture of HCO - and H2CO equilibrium 5 H + H HCO + H

Q4(i).
Short Answer 1 mark

Define a buffer solution.

Q4(ii).
Short Answer 1 mark

Construct two equations to show how equilibrium 5 acts as a buffer solution.

Q4(iii).
Short Answer 1 mark

The [HCO ] ratio in a sample of seawater is 14.1. Calculate the pH of this sample. [pK: H , 6.35] pH = [3] [Total: 14]

Q5(a).
Short Answer 1 mark

Complete Table 5.1 to predict the substance liberated at each electrode during electrolysis of the indicated electrolyte with inert electrodes. Table 5.1 electrolyte substance liberated at the anode substance liberated at the cathode PbBr

Q5(l).
Short Answer 1 mark

Cu(NO

Q5(b).
Short Answer 1 mark

An electrolytic cell is set up to determine a value for the Avogadro constant, L. The electrolyte is dilute sulfuric acid and both electrodes are copper.

Q5(i).
Short Answer 1 mark

State the relationship between the Faraday constant, F, and the Avogadro constant, L.

Q5(ii).
Short Answer 1 mark

Use the experimental information in

Q5(ii)(b).
Short Answer 1 mark

and data from the table on page 23 to calculate a value for the Avogadro constant, L. Show all working. Avogadro constant, L = [4] [Total: 8]

Q6(a).
Short Answer 1 mark

The reagent and conditions required for the nitration of benzene, benzoic acid and phenol are shown in Table 6.1. Table 6.1 compound reagents and conditions for nitration benzene concentrated HNO catalyst benzoic acidconcentrated HNO catalyst phenol dilute HNO Concentrated HNO reacts with concentrated H to generate the electrophile NO

Q6(i).
Short Answer 3 marks

Complete Fig. 6.1 to show the mechanism of the reaction between benzene and NO Include all relevant curly arrows and charges.

Q6(ii).
Short Answer 1 mark

Write an equation to show how H is regenerated.

Q6(b).
Short Answer 2 marks

Draw the major products from the mononitration of benzoic acid and of phenol.major product from benzoic acid major product from phenol

Q6(c).
Structured 4 marks

Compare the relative ease of nitration of benzene, benzoic acid and phenol. Explain your reasoning; include reference to the structures of the three compounds in your answer. easiest least easy

Q6(d).
Short Answer 1 mark

The azo compound Congo Red is used as an acid-base indicator and can be made by the route shown in Fig. 6.2. In step 3 of this synthesis, compound Y reacts with compound Z. Compound Z is made from compound X. Assume that the -SO groups do not react. step 1 step 2 step 3 Congo Red NaOH(aq)X CY CSNa

Q6(i).
Short Answer 3 marks

Suggest structures for compounds X, Y and Z and draw them in the boxes in Fig. 6.2.

Q6(ii).
Short Answer 1 mark

Give the reagents and conditions for step 1 and step 2. step 1 . step 2 . [3] [Total: 16]

Q7(a).
Short Answer 1 mark

State the uses of TMS and D O in NMR spectroscopy. TMS O

Q7(b).
Short Answer 1 mark

The three isomeric ketones with molecular formula C O are:

Q7(i).
Short Answer 2 marks

Complete Table 7.1 to show the number of peaks observed in the proton ( H) NMR spectrum and in the carbon-13 NMR spectrum for each compound listed. Table 7.1 ketone number of peaks observed in the proton ( H) NMR spectrum number of peaks observed in the carbon-13 NMR spectrum 3-methylbutanone

Q7(ii).
Short Answer 2 marks

State all the ketones with molecular formula C O that have: a doublet in their proton ( H) NMR spectrum a singlet in their proton ( H) NMR spectrum.

Q7(c).
Short Answer 1 mark

Cortisone, C , is a naturally occurring chemical that contains chiral carbon atoms. cortisone

Q7(i).
Short Answer 1 mark

Deduce the number of chiral carbon atoms in one molecule of cortisone.

Q7(ii).
Short Answer 1 mark

Cortisone is reacted with an excess of NaBH State the molecular formula of the organic compound formed.

Q7(iii).
Short Answer 1 mark

Cortisone is an optically active molecule. Explain what is meant by optically active. [Total: 8]

Q8(a).
Structured 4 marks

Compare the relative acidities of ethanol, ethanoic acid, chloroethanoic acid and phenol. Explain your reasoning. most acidic least acidic

Q8(b).
Short Answer 1 mark

An excess of ethanedioic acid, HOOCCOOH(aq), is reacted with warm acidified KMnO

Q8(aq).
Short Answer 2 marks

State the type of reaction undergone by ethanedioic acid. Describe what you would observe. Write an equation for this reaction. Your equation can use [O] or [H] as necessary. type of reaction observations equation

Q8(c).
Short Answer 2 marks

A section of a polyester is shown.O C OCCH polyester OCCH Draw the structures of the two monomers that form this polyester.

Q8(d).
Short Answer 2 marks

Serine can polymerise to form two different types of condensation polymer; a polyester and a polypeptide. serine Draw the structure of the polypeptide showing two repeat units. The peptide linkage should be shown displayed.

Q8(e).
Short Answer 1 mark

Explain why condensation polymers normally biodegrade more readily than addition polymers. [Total: 11]

Q9.
Short Answer 1 mark

The structure of cyclohexylamine is shown in Fig. 9.1. cyclohexylamine

Q9(a).
Short Answer 3 marks

Compare the relative basicities of ammonia, cyclohexylamine and phenylamine. Explain your reasoning. most basic least basic

Q9(b).
Short Answer 1 mark

Cyclohexylamine reacts with ethanoyl chloride to form the corresponding amide, L. COCl

Q9(i).
Short Answer 1 mark

Name the mechanism for the reaction shown in Fig. 9.2.

Q9(ii).
Structured 4 marks

Complete the mechanism of the reaction between cyclohexylamine and CH COCl. R-NH is used to represent cyclohexylamine. Include all relevant lone pairs of electrons, curly arrows, charges and partial charges. CC productsRNH

Q9(iii).
Short Answer 1 mark

The reaction between cyclohexylamine and an excess of CH COCl forms compound M. Compound M has the molecular formula C Suggest and draw the structure of M. [1] [Total: 9] Important values, constants and standards Faraday constant F = 9.65 u 10 Avogadro constant L = 6.022 u 10 electronic charge e = ?1.60 u 10?19 molar volume of gas V at room conditions ionic product of water K = 1.00 u 10?14 To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Group The Periodic Table of Elements hydrogen 1.0 He helium 4.0 12131415161718 34567891011123Li lithium 6.9 beryllium 9.0 atomic number atomic symbolKey name relative atomic mass sodium 23.0 magnesium 24.3 potassium 39.1 calcium 40.1 rubidium 85.5 strontiu m 87.6 caesium 132.9 barium 137.3 francium Ra radium boron10.8 13 Al aluminium 27.0 31 Ga gallium 69.7 49 In indium 114.8 thallium 204.4 carbon 12.0 14 Si silicon 28.1 32 Ge germanium 72.6 50 Sntin 118.7 82 Pb lead 207. 2 titanium 47.9 40 Zr zirconium 91.2 72 Hf hafnium178.5 rutherfordium vanadium 50.9 41 Nb niobium 92.9 73 Ta tantalum 180.9 105 Db dubnium chromium 52.0 42 Mo molybdenum 95.9 74 W tungste n 183. 8 106 Sg seaborgium manganese 54.9 43 Tc technetium rhenium 186.2 107 Bh bohrium iron 55.8 44 Ru ruthenium 101.1 76 Os osmium 190.2 108 Hs hassium Co cobalt 58.9 45 Rh rhodium 102. 9 iridium192. 2 109 Mt meitnerium nicke l58.7 46 Pd palladium 106. 4 platinum 195. 1 110 Ds darmstadtium copper 63.5 47 Ag silver 107. 9 79 Au gold 197. 0 111 Rg roentgenium zinc 65.4 48 Cd cadmium 112.4 80 Hg mercury 200. 6 112 Cn copernicium flerovium livermorium nitrogen 14.0 phosphorus 31.0 33 As arsenic 74.9 51 Sb antimony 121.8 bismuth 209.0 oxygen 16.0 sulfur 32.1 34 Se selenium 79.0 52 Te tellurium 127.6 84 Po polonium fluorine 19.0 17 Cl chlorine 35.5 35 Br bromine 79.9 iodine 126.9 Ne neon 20.2 18 Ar argo n39.9 krypto n 83.8 54 Xe xenon131. 3 86 Rn radon nihonium moscovium tennessine oganesson scandium 45.0 yttriu m 88.9 57?71 lanthanoids 89?103 actinoids 57La lanthanum 138.9 89 Ac lanthanoids actinoids actinium Ce cerium 140.1 90 Th thoriu m232.0 praseodymium 140.9 91 Pa protactinium 231.0 neodymium 144.4 uranium 238.0 promethium neptunium samarium 150.4 94 Pu plutonium europium 152.0 95 Am americium gadolinium 157.3 96 Cm curium terbiu m158. 9 97 Bk berkelium dysprosium 162. 5 californium holmium 164. 9 99 Es einsteinium erbium 167. 3 100 Fm fermium Tm thulium168. 9 101 Md mendelevium ytterbiu m 173.1 102 No nobelium lutetium 175.0 lawrencium