ADVANCED SUBSIDIARY (AS)General Certificate of Education

2017

ChemistryAssessment Unit AS 1assessingBasic Concepts in Physicaland Inorganic Chemistry

[SCH12]FRIDAY 26 MAY, MORNING

TIME1 hour 30 minutes.

INSTRUCTIONS TO CANDIDATESWrite your Centre Number and Candidate Number in the spaces provided at the top of this page.Answer all fifteen questions.Answer all ten questions in Section A. Record your answers by marking the appropriate letter on the answer sheet provided. Use only the spaces numbered 1 to 10. Keep in sequence when answering.Answer all five questions in Section B. You must answer the questions in the spaces provided.Do not write outside the boxed area on each page or on blank pages.Complete in black ink only. Do not write with a gel pen.

INFORMATION FOR CANDIDATESThe total mark for this paper is 90.Quality of written communication will be assessed in Question 13(c).In Section A all questions carry equal marks, i.e. one mark for each question.In Section B the figures in brackets printed down the right-hand side of pages indicate the marks awarded to each question or part question.A Periodic Table of Elements, containing some data, is included with this question paper.

Centre Number

Candidate Number

10675

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icatio

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Section A – Multiple Choice

Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.

Each multiple choice question is worth 1 mark.

1 Bromine is formed in the reaction below.

Cl2 + 2NaBr → 2NaCl + Br2

Which statement about the reaction is correct?

A Bromide ions lose electrons

B Bromine is reduced by chlorine

C Chloride ions are reduced

D Chlorine is a weaker oxidising agent than bromide

2 Which trend in the Periodic Table is correct?

A Boiling point decreases from fluorine to bromine

B First ionisation energy decreases from lithium to caesium

C First ionisation energy increases from nitrogen to oxygen

D Melting point decreases from sodium to silicon

3 Which of the following is the structure of 55Mn2+ ?

protons neutrons electrons

A 25 30 23

B 25 30 27

C 27 30 25

D 30 25 28

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4 Potassium iodide is formed when potassium is warmed in iodine vapour. Which of the following shows the bonding in the three species?

potassium iodine potassium iodide

A ionic covalent ionic

B metallic ionic covalent

C covalent covalent ionic

D metallic covalent ionic

5 The element astatine lies below iodine in the Periodic Table and is likely to

A be black.

B be a volatile liquid at room temperature and pressure.

C form an astatide ion, At2-.

D oxidise iodide ions to iodine.

6 Which molecule is non-polar?

A H2S

B NH3

C PF3

D SF6

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7 The element boron has a relative atomic mass of 10.8. In this sample, boron exists as two isotopes, 10B and 11B. The percentage abundance of 10B in this sample of boron is

A 10.8%.

B 20.0%.

C 80.0%.

D 89.2%.

8 When burned in oxygen magnesium forms magnesium oxide.

2Mg + O2 → 2MgO

What is the number of molecules of oxygen required for the complete oxidation of 1.2 g of magnesium?

A 1.5 × 1022

B 3.0 × 1022

C 3.0 × 1023

D 6.0 × 1023

9 Which statement describes the trends in electronegativity values in the Periodic Table?

A Decrease across a Period and increase down a Group

B Decrease across a Period and decrease down a Group

C Increase across a Period and increase down a Group

D Increase across a Period and decrease down a Group

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10 Which of the following would exactly neutralise 10.0 cm3 of 1.00 mol dm-3 NaOH(aq)?

A 2.50 cm3 of 1.00 mol dm-3 CH3COOH

B 5.00 cm3 of 1.00 mol dm-3 HCl

C 5.00 cm3 of 1.00 mol dm-3 H2SO4

D 3.00 cm3 of 1.00 mol dm-3 H3PO4

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Section B

Answer all five questions in the spaces provided.

11 Sulfate, hydrogensulfate and thiosulfate ions are formed when sulfuric and thiosulfuric acids ionise.

(a) (i) Write the equation for the complete ionisation of thiosulfuric acid.

[2]

(ii) Write the formula for the hydrogensulfate ion.

[1]

(b) (i) Write the formula for ammonium sulfate.

[1]

(ii) Describe the bonding in ammonium sulfate.

[2]

(c) Describe how you could use chemical tests on an aqueous solution of ammonium sulfate to prove that it contains ammonium ions and sulfate ions.

[4]

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12 Some properties of the metals sodium and aluminium are shown in the table below.

metal charge on metal ion electronic structure of the atom

melting point /°C

sodium 1+ 1s2 2s2 2p6 3s1 98

aluminium 3+ 1s2 2s2 2p6 3s2 3p1 660

(a) Describe, without using a diagram, the bonding in sodium metal.

[2]

(b) Explain why aluminium has a higher melting point than sodium.

[2]

(c) (i) Write the equation, including state symbols, for the first ionisation energy of sodium.

[2]

(ii) The first six ionisation energies, in kJ mol-1, of sodium are 496, 4563, 6913, 9544, 13352 and 16611. Explain which of these values can be used to identify sodium as belonging to Group I of the Periodic Table.

[2]

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(iii) The outer electron in the sodium atom is located in the 3s orbital. Explain what is meant by the term orbital.

[2]

(d) Aluminium forms covalent bonds with chlorine.

(i) Explain what is meant by the term covalent bond.

[2]

(ii) Write the equation for the reaction of aluminium with chlorine to form aluminium chloride, AlCl3.

[1]

(iii) State the octet rule and explain whether the atoms in aluminium chloride obey the rule.

[3]

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13 (a) Zinc reacts with chlorine to form the ionic compound zinc chloride. Draw a dot and cross diagram, using outer electrons only, to show how zinc chloride, ZnCl2, is formed from zinc and chlorine atoms.

[2]

(b) Zinc is an essential trace element. People who have a zinc deficiency can take hydrated zinc sulfate, ZnSO4.xH2O, as a dietary supplement.

The value of x can be determined by heating hydrated zinc sulfate to constant mass.

A student heated 5.65 g of hydrated zinc sulfate and obtained 3.85 g of anhydrous zinc sulfate.

(i) Calculate the number of moles of anhydrous zinc sulfate obtained.

[1]

(ii) Calculate the mass of water present in the hydrated zinc sulfate.

[1]

(iii) Calculate the number of moles of water present in the hydrated zinc sulfate.

[1]

(iv) Calculate the value of x in ZnSO4.xH2O

[1]

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(c) Describe how you would prepare 250.0 cm3 of a 28.7 g dm-3 zinc sulfate solution from the anhydrous solid.

In this question you will be assessed on using your written communication skills including the use of specialist scientific terms.

[6]

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14 Nitrogen and phosphorus are Group V elements. They form the toxic hydrides ammonia and phosphine.

(a) Ammonia is formed by the reversible reaction of nitrogen with hydrogen. Write the equation for this reaction.

[2]

(b) Phosphine is formed by the reaction of phosphorus with aqueous sodium hydroxide.

(i) Balance the equation for the formation of phosphine.

P4 + NaOH + H2O → NaH2PO2 + PH3 [1]

(ii) Deduce the oxidation number of phosphorus in:

P4

NaH2PO2

PH3 [3]

(iii) Explain, using the oxidation numbers of phosphorus, why the reaction is described as disproportionation.

[3]

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(c) The boiling point of ammonia is -33 °C while that of phosphine is -88 °C. Explain why the boiling point of ammonia is higher than that of phosphine.

[3]

(d) Both ammonia and phosphine molecules react with H+ ions.

PH3 + H+ → PH4+

(i) Name the type of bond formed between a phosphine molecule and the H+ ion.

[1]

(ii) Draw and name the shapes of the molecule PH3 and the ion PH4+.

PH3

Shape _________________

PH4+

Shape _________________

[4]

(iii) Explain why the bond angle in PH3 is different from the bond angle in PH4+.

[3]

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(e) Ammonia is very soluble in water. Draw diagrams to show the two ways in which a molecule of ammonia can be attracted to a molecule of water. Include all partial charges and lone pairs in your diagram.

[4]

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15 Ammonia is used to make nitric acid by the Ostwald Process outlined below.

Reaction 1: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Reaction 2: 2NO(g) + O2(g) → 2NO2(g)

Reaction 3: 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)

(a) (i) Calculate the number of moles of oxygen needed to react with 6.8 kg of ammonia.

[3]

(ii) Calculate the number of moles of nitrogen(IV) oxide which can be obtained from 6.8 kg of ammonia.

[2]

(iii) Calculate the concentration of nitric acid, in g dm-3, produced on reacting the nitrogen(IV) oxide obtained in part (ii) with 50 dm3 of water.

[3]

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(b) Ammonia reacts with nitric acid according to the equation below.

NH3 + HNO3 → NH4NO3

The following results were obtained by diluting 25.0 cm3 of a concentrated ammonia solution to 250.0 cm3 in a volumetric flask and then titrating 25.0 cm3 portions of the diluted ammonia solution using 0.100 mol dm-3 nitric acid.

titration initial burette reading /cm3

final burette reading /cm3 titre /cm3

rough 0.00 22.00 22.00

first accurate 0.10 21.40 21.30

second accurate 0.20 21.60 21.40

(i) Name a suitable indicator for the titration and state the colour change at the end point.

[3]

(ii) Calculate the mean titre.

[1]

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(iii) A burette has an uncertainty of ±0.05 cm3. Calculate the uncertainty when two burette readings are used to calculate a titre value.

[1]

(iv) Calculate the concentration of the concentrated ammonia solution in mol dm-3.

[5]

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THIS IS THE END OF THE QUESTION PAPER

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Permission to reproduce all copyright material has been applied for.In some cases, efforts to contact copyright holders may have been unsuccessful and CCEAwill be happy to rectify any omissions of acknowledgement in future if notified.

221023

For Examiner’suse only

QuestionNumber Marks

Section A1–10

Section B1112131415

TotalMarks

DO NOT WRITE ON THIS PAGE

CHEMISTRY DATA SHEETGCE A/AS EXAMINATIONS CHEMISTRY

Including the Periodic Table of the Elements

For the use of candidates takingAdvanced Subsidiary and Advanced LevelChemistry Examinations

Copies must be free from notes or additions of any kind.

No other type of data booklet or information sheet is

authorised for use in the examinations.

For first teaching from September 2016For first award of AS Level in Summer 2017For first award of A Level in Summer 2018Subject Code: 1110

GCE

NewSpe

cifi ca

tionGeneral Information

1 tonne = 106 g1 metre = 109 nmOne mole of any gas at 293 K and a pressure of 1 atmosphere (105 Pa) occupies a volume of 24 dm3

Avogadro Constant = 6.02 1023 mol–1

Planck Constant = 6.63 10–34 J sSpecifi c Heat Capacity of water = 4.2 J g–1 K–1

Speed of Light = 3 108 m s–1

Characteristic absorptions in IR spectroscopy

Wavenumber/cm–1 Bond Compound550–850 C–X (X = Cl, Br, I) Haloalkanes750–1100 C–C Alkanes, alkyl groups1000–1300 C–O Alcohols, esters, carboxylic acids 1450–1650 C ̿ C Arenes1600–1700 C ̿ C Alkenes1650–1800 C ̿ O Carboxylic acids, esters, aldehydes, ketones, amides, acyl chlorides2200–2300 C N Nitriles2500–3200 O–H Carboxylic acids2750–2850 C–H Aldehydes2850–3000 C–H Alkanes, alkyl groups, alkenes, arenes3200–3600 O–H Alcohols3300–3500 N–H Amines, amides

Proton Chemical Shifts in Nuclear Magnetic Resonance Spectroscopy (relative to TMS)

Chemical Shi Structure

0.5–2.0 –CH Saturated alkanes0.5–5.5 –OH Alcohols1.0–3.0 –NH Amines2.0–3.0 –CO–CH Ketones –N–CH Amines C6H5–CH Arene (alipha c on ring)2.0–4.0 X–CH X = Cl or Br (3.0–4.0) X = I (2.0–3.0)4.5–6.0 –C ̿ CH Alkenes5.5–8.5 RCONH Amides6.0–8.0 –C6H5 Arenes (on ring)9.0–10.0 –CHO Aldehydes10.0–12.0 –COOH Carboxylic acids

These chemical shi s are concentra on and temperature dependent and may be outside the ranges indicated above.

227

89

139

57

256

101

223

87

226

88

261

104

262

105

266

106

264

107

277

108

268

109

271

110

272

111

140

58

141

59

144

60

145

61

150

62

152

63

157

64

159

65

162

66

165

67

167

68

169

69

173

70

175

71232

90

231

91

238

92

237

93

242

94

243

95

247

96

245

97

251

98

254

99

253

100

254

102

257

103

133

55

137

56

178

72

181

73

184

74

186

75

190

76

192

77

195

78

197

79

201

80

89

39

91

40

103

45

85

37

88

38

93

41

96

42

98

43

101

44

106

46

108

47

112

48

131

54222

86

210

85

210

84

209

83

207

82

204

81

84

36

79

34

73

32

40

20

39

19

45

21

48

22

51

23

52

24

55

25

56

26

59

27

59

28

64

29

65

30

11

5

12

6

14

7

16

8

19

9

20

10

4

2

40

18

35.5

17

32

16

31

15

28

14

27

1370

31

75

33

80

35115

49

119

50

122

51

128

52

127

53

23

11

24

12

7

3

9

4

* 58–71 Lanthanum series† 90–103 Actinium series

a = relative atomic mass (approx)x = atomic symbolb = atomic number

THE PERIODIC TABLE OF ELEMENTSGroup

ab

x

*

†

1

1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

III III IV VI VII 0V

285

112Copernicium

MARKSCHEME

10675.01 F

ADVANCED SUBSIDIARY (AS)General Certificate of Education

2017

ChemistryAssessment Unit AS 1

assessing

Basic Concepts in Physicaland Inorganic Chemistry

[SCH12]

FRIDAY 26 MAY, MORNING

New

Specifi

catio

n

10675.01 F 2 [Turn over

General Marking Instructions

IntroductionMark schemes are published to assist teachers and students in their preparation for examinations. Through the mark schemes teachers and students will be able to see what the examiners are looking for in response to questions and exactly where the marks have been awarded. The publishing of the mark schemes may help to show that examiners are not concerned about fi nding out what a student does not know but rather with rewarding students for what they do know.

The purpose of mark schemesExamination papers are set and revised by teams of examiners and revisers appointed by the Council. The teams of examiners and revisers include experienced teachers who are familiar with the level and standards expected of students in schools and colleges.

The job of the examiners is to set the questions and the mark schemes; and the job of the revisers is to review the questions and mark schemes commenting on a large range of issues about which they must be satisfi ed before the question papers and mark schemes are fi nalised.

The questions and the mark schemes are developed in association with each other so that the issues of differentiation and positive achievement can be addressed right from the start. Mark schemes, therefore, are regarded as part of an integral process which begins with the setting of questions and ends with the marking of the examination.

The main purpose of the mark scheme is to provide a uniform basis for the marking process so that all the markers are following exactly the same instructions and making the same judgements in so far as this is possible. Before marking begins a standardising meeting is held where all the markers are briefed using the mark scheme and samples of the students’ work in the form of scripts. Consideration is also given at this stage to any comments on the operational papers received from teachers and their organisations. During this meeting, and up to and including the end of the marking, there is provision for amendments to be made to the mark scheme. What is published represents the fi nal form of the mark scheme.

It is important to recognise that in some cases there may well be other correct responses which are equally acceptable to those published: the mark scheme can only cover those responses which emerged in the examination. There may also be instances where certain judgements may have to be left to the experience of the examiner, for example where there is no absolute correct response – all teachers will be familiar with making such judgements.

10675.01 F 3

AVAILABLEMARKS

Section A

1 A

2 B

3 A

4 D

5 A

6 D

7 B

8 A

9 D

10 C [1] for each correct answer [10] 10

Section A 10

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AVAILABLEMARKS

Section B

11 (a) (i) H2S2O3 + 2H2O → 2H3O+ + S2O32–

or

H2S2O3 → 2H+ + S2O32– [2]

(ii) HSO4

– [1] (b) (i) (NH4)2SO4 [1] (ii) oppositely charged ions are held together by electrostatic forces of attraction error [–1] [2]

(c) NH4+: warm with (dilute) sodium hydroxide solution [1]

(the gas produced) gives white fumes with (a glass rod dipped in) concentrated HCl [1] SO4

2–: add a solution of barium chloride [1] a white precipitate forms [1] [4] 10

12 (a) (attraction) between positive sodium ions [1] and delocalised electrons [1] [2]

(b) increase in charge of metal ion/aluminium ion has 3+ compared to 1+ in sodium [1] increase in the number of electrons in delocalised cloud/sea [1] [2]

(c) (i) Na(g) → Na+(g) + e– [2]

(ii) 1st and 2nd values [1] large gap between [1] [2] (iii) A region within an atom that can hold up to two electrons [1] with opposite spin [1]. [2]

(d) (i) (Electrostatic) attraction between a shared pair of electrons and the nuclei of bonded atoms. [2]

(ii) 2Al + 3Cl2 → 2AlCl3 [1] (iii) When forming a compound, an atom tends to gain, lose or share electrons [1] to achieve eight in its outer shell. [1] Aluminium has six electrons in the outer shell [1]

[3] 16 Chlorine has eight

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13 (a)

Zn +

Cl

Cl

→ Zn2+

Cl

–

Cl

–

Outer shell of zinc could show ten electrons [2]

(b) (i) 3.85 ÷ 161 = 0.024/0.0239 [1]

(ii) 5.65 – 3.85 = 1.80g [1]

(iii) 1.80 ÷ 18 = 0.1/0.10 [1]

(iv) 0.0239 : 0.10 1 : 4.18/4.2/4.0/4 X = 4.18/4.2/4.0/4 [1]

(c) Indicative content • (Weigh out) 7.18g of zinc sulfate * • in a beaker/suitable container • Dissolve the solid in a small amount (50–100cm3) of distilled/deionised water • Transfer the solution/with washings • to the 250.0 cm3 volumetric flask * • and make up to the mark • Stopper and invert the flask

* essential for [6] in Band A If either * missing max [5]

Band Response Mark

A

Candidates must use appropriate specialist terms to fully explain the preparation of the standard solution using 6 points of indicative content. They must use good spelling, punctuation and grammar and the form and style are of an excellent standard.

[5]–[6]

B

Candidates must use appropriate specialist terms to explain the preparation of the standard solution using a minimum of 4 points of indicative content. They must use satisfactory spelling, punctuation and grammar and the form and style are of a good standard.

[3]–[4]

C

Candidates must partially explain the preparation of the standard solution using a minimum of 2 points of indicative content. They use limited correct spelling, punctuation and grammar and the form and style are of a basic standard.

[1]–[2]

D Response not worthy of credit. [0]

[6] 12

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AVAILABLEMARKS

14 (a) N2 + 3H2 2NH3 [2] ([–1] each error)

(b) (i) P4 + 3NaOH + 3H2O → 3NaH2PO2 + PH3 [1] (ii) P4 = 0 [1] NaH2PO2 = +1 [1] PH3 = –3 [1] [3]

(iii) 0 to +1 is oxidation; 0 to –3 is reduction [1] oxidation and reduction of the same element [1] in the same reaction [1] [3]

(c) Between molecules of NH3 there are van der Waals’ forces and H-bonds [1] Between molecules of PH3 there are only van der Waals’ forces [1] Hydrogen bonds are stronger and require more energy to break. [1] [3]

(d) (i) dative covalent/co-ordinate bond [1]

(ii) H

P

HH H

+

P

HH H

••

[2]

pyramidal tetrahedral [2]

(iii) PH3 has one lone pair/3 bond pairs [1] greater repulsion between lone pair–bond pair [1] pushes bond pairs closer together/reduces bond angle [1] [3] (e)

O O

N••H+

H+

H+

–H+

H+

•• ••• •

••

–

N••–

H+

H+

H+

H+

H+

–and

[2] [2] [4] 24

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15 (a) (i) 6.8 × 1000 = 6800g 6800 ÷ 17 = 400 NH3 : O2 4 : 5 Moles of O2 = 500 ([–1] each error) [3]

(ii) NO : NO2 1 : 1 Moles of NO2 = 400 [2]

(iii) 400 × (32 ) = 266.6667

Concentration = 266.667 ÷ 50 = 5.33 mol dm–3

5.33 × 63 = 336 g dm–3

([–1] each error) [3] (b) (i) Methyl orange [1] Yellow [1] to red [1] [3]

(ii) 2

21.30 21.40+] g = 21.35 cm3 [1] (iii) 0.05 + 0.05 = 0.1 cm3 [1]

(iv) 1000

21.35 0.100#^ h = 2.135 × 10–3

NH3 : HNO3 1 : 1 moles of diluted ammonia in 25 cm3 = 0.002135 moles of diluted ammonia in 250 cm3 = 0.02135 concentration of diluted ammonia = 0.02135 ÷ 0.25 = 0.0854 concentration of undiluted ammonia = 0.0854 × 10 = 0.854 mol dm–3

([–1] each error) [5] 18

Section B 80

Total 90