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Date	May 2018	Marks available	2	Reference code	18M.3.sl.TZ2.1
Level	SL	Paper	3	Time zone	TZ2
Command term	Justify	Question number	1	Adapted from	N/A

Question

The table summarizes some properties of graphite and graphene.

M18/4/CHEMI/SP3/ENG/TZ2/01

Graphene is two-dimensional, rather than three-dimensional, material.

Justify this by using the structure of graphene and information from the table.

[2]

a.i.

Show that graphene is over 1600 times stronger than graphite.

[1]

a.ii.

Identify a value from the table which can be used to support the information about graphene given below.

M18/4/CHEMI/SP3/ENG/TZ2/01.a.iii_01

Electrons in a solid are restricted to certain ranges, or bands, of energy (vertical axis). In an insulator or semiconductor, an electron bound to an atom can break free only if it gets enough energy from heat or a passing photon to jump the “band gap”, but in graphene the gap is infinitely small.

M18/4/CHEMI/SP3/ENG/TZ2/01.a.iii_02

[1]

a.iii.

Diamond, graphene, and graphite are all network solids.

Suggest, giving a reason, the electron mobility of diamond compared to graphene.

[2]

The melting point of diamond at 1 × 10⁶ kPa is 4200 K (in the absence of oxygen).

Suggest, based on molecular structure, why graphene has a higher melting point under these conditions.

[2]

Markscheme

consists of single/one sheet/layer «of carbon atoms»

graphene has no density measurement

graphene has no distance between layers data

graphene has large specific surface area «compared to graphite»

Do not accept “sp²” alone without reference to single/one sheet/layer.

Accept “thickness of one atom” OR “consists of a plane” for M1.

[2 marks]

a.i.

Any one of these alternatives:

ALTERNATIVE 1

« $\frac{{1.3 \times {{10}^{11}}}}{{76 \times {{10}^6}}}$ »

1.7 × 10³/1711

ALTERNATIVE 2

1600 × 76 × 10⁶ = 1.2 × 10¹¹ «is less than tensile strength of graphene»

ALTERNATIVE 3

$\frac{{1.3 \times {{10}^{11}}}}{{1600}}$ = 8.1 × 10⁷ «is greater than upper end of tensile strength for graphite»

Accept any value in the range 1700–27 083. Answer may be expressed in scientific notation or otherwise.

Accept any value calculated which is less than the graphene tensile strength based on a value chosen from within the 4.8–76 × 10⁶range.

[1 mark]

a.ii.

«graphene has a high electron mobility of» 15 000–200 000 «cm² V^–1 s^–1»

A specific value or range of values must be given.

Accept any value in the 15 000–200 000 «cm²V^–1s^–1» range.

[1 mark]

a.iii.

smaller/zero

no delocalized electrons/electrons are bound/electrons not free to move/electrons not free to roam

localized electrons «in sigma bonds»

large band gap

Accept “diamond is a dielectric” OR “diamond does not conduct electricity” for M2.

Award [1 max] for just “immobile/less mobile”.

Award [2] for “electrons immobile «in diamond» due to the large band gap” OR "electrons «in diamond» immobile

since electrons are localized «in the sigma bonds»”.

[2 marks]

shorter bonds in graphene

bonds in graphene intermediate between single and double

bond order in graphene is 1.33

delocalization creates stronger bonds

shorter bonds are stronger

stronger/shorter bonds require higher temperature/faster thermal motion to be altered

stronger/shorter bonds require greater energy to be broken

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

Syllabus sections

Core » Topic 4: Chemical bonding and structure » 4.3 Covalent structures

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Deduce the Lewis (electron dot) structures of ozone.
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16N.2.sl.TZ0.2d:
The Lewis (electron dot) structure of the ethanedioate ion is shown below.

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Draw the Lewis (electron dot) structure of the ethanedioate ion, ^–OOCCOO^–.
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Deduce the Lewis (electron dot) structure and molecular geometry of PCl₃.
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18N.2.hl.TZ0.3b.i:
Draw two Lewis (electron dot) structures for BrO₃⁻.
18N.2.sl.TZ0.3b:
Draw the Lewis (electron dot) structure for BrO₃⁻ that obeys the octet rule.
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17M.1.sl.TZ1.12: Which correctly states the strongest intermolecular forces in the compounds below?
17M.1.sl.TZ1.11:
Which combination describes the sulfate(IV) ion, SO₃^2– (also known as sulfite ion)?
17M.2.hl.TZ1.5c:
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Explain the effect of adding ${{\text{H}}^ + }{\text{(aq)}}$ ions on the position of the equilibrium.
17M.1.hl.TZ1.12: Which combination describes the bonding and structure in benzoic acid, C6H5COOH?
18M.3.sl.TZ2.1c:
The melting point of diamond at 1 × 10⁶ kPa is 4200 K (in the absence of oxygen).

Suggest, based on molecular structure, why graphene has a higher melting point under these conditions.
18M.2.sl.TZ2.6b:
Lewis structures show electron domains and are used to predict molecular geometry.

Deduce the electron domain geometry and the molecular geometry for the NH₂⁻ ion.
18M.3.sl.TZ2.1a.ii:
Show that graphene is over 1600 times stronger than graphite.
18M.2.hl.TZ1.1b:
The structural formula of urea is shown.

Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
19M.2.hl.TZ1.5c(iv):
State the type of bond formed when chloramine is protonated.
19M.2.sl.TZ1.5c(i):
Draw a Lewis (electron dot) structure of chloramine.
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Predict with a reason, whether the molecule PF₃ is polar or non-polar.
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Estimate the H−N−H bond angle in methanamine using VSEPR theory.
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Outline why all the C–O bond lengths in the ethanedioate ion are the same length and suggest a value for them. Use section 10 of the data booklet.
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Draw a Lewis (electron dot) structure for ozone.
19M.2.sl.TZ2.1b(i):
Deduce the Lewis (electron dot) structure of ethyne.
17M.2.sl.TZ1.4a:
Estimate the H−N−H bond angle in methanamine using VSEPR theory.
17N.2.sl.TZ0.3a:
Draw the Lewis (electron dot) structures of PF₃ and PF₄⁺ and use the VSEPR theory to deduce the molecular geometry of each species.
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State the electron domain geometry around the nitrogen atom and its hybridization in methanamine.
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What is the structure and bonding in SiO₂(s)?
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18M.2.hl.TZ2.7e:
Carbon and silicon are elements in group 14.

Explain why CO₂ is a gas but SiO₂ is a solid at room temperature.
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Draw a Lewis (electron dot) structure of chloramine.
16N.3.sl.TZ0.6c:
(i) Suggest why incomplete combustion of plastic, such as polyvinyl chloride, is common in industrial and house fires.

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With reference to bonding, suggest a reason why many adults have measurable levels of phthalates in their bodies.
17N.2.hl.TZ0.4a:
Draw the Lewis (electron dot) structures of PF₃ and PF₅ and use the VSEPR theory to deduce the molecular geometry of each species including bond angles.
17N.2.hl.TZ0.4b:
Predict whether the molecules PF₃ and PF₅ are polar or non-polar.
18M.3.sl.TZ2.1a.iii:
Identify a value from the table which can be used to support the information about graphene given below.

Electrons in a solid are restricted to certain ranges, or bands, of energy (vertical axis). In an insulator or semiconductor, an electron bound to an atom can break free only if it gets enough energy from heat or a passing photon to jump the “band gap”, but in graphene the gap is infinitely small.
18M.3.sl.TZ2.1b:
Diamond, graphene, and graphite are all network solids.

Suggest, giving a reason, the electron mobility of diamond compared to graphene.
18M.2.sl.TZ1.1b:
The structural formula of urea is shown.

Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
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Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.

Molecular geometry:

H–N–H bond angle:
21N.2.sl.TZ0.3a(i):
Sketch the Lewis (electron dot) structure of the P₄ molecule, containing only single bonds.
21N.2.sl.TZ0.3b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the...
21M.2.sl.TZ1.2a(i):
Draw the Lewis (electron dot) structure of hydrogen sulfide.
21M.2.sl.TZ1.2a(ii): Predict the shape of the hydrogen sulfide molecule.
21M.2.sl.TZ2.2c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur dichloride, SCl₂.
22M.2.hl.TZ2.6c(ii):
Explain the electron domain geometry of SO₃.
20N.1.sl.TZ0.11:
Which combination correctly describes the geometry of the carbonate ion, ${CO}_{3}^{2 -}$ ?
19M.2.hl.TZ2.1b(i):
Deduce the Lewis (electron dot) structure of ethyne.
21M.2.hl.TZ2.2c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur tetrafluoride, SF₄, and sulfur dichloride, SCl₂.
19M.2.hl.TZ2.3d(ii):
State, giving a reason, the shape of the dinitrogen monoxide molecule.
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20N.2.hl.TZ0.2a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A...
22M.2.hl.TZ2.8a(i):
Outline two differences between the bonding of carbon atoms in C₆₀ and diamond.
22M.2.hl.TZ2.8a(ii):
Explain why C₆₀ and diamond sublime at different temperatures and pressures.
22M.2.sl.TZ2.3d(ii): Explain the electron domain geometry of NO3−.
22M.2.sl.TZ2.4a(i): Outline one difference between the bonding of carbon atoms in C60 and diamond.
19M.2.hl.TZ1.5c(iii):
Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.

Molecular geometry:

H–N–H bond angle:
20N.1.sl.TZ0.10:
Which molecule is most polar?

A.   ${CHF}_{3}$

B.   ${CF}_{4}$

C.   ${CClF}_{3}$

D.   ${CCl}_{4}$
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Deduce a Lewis (electron dot) structure of the nitric acid molecule, HNO₃, that obeys the octet rule, showing any non-zero formal charges on the atoms.
22M.1.sl.TZ2.10: What is the type of bonding in a compound that has high boiling and melting points, poor...
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Core » Topic 4: Chemical bonding and structure

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