ISL-1 C-0 C-1

ISL-1 C-0 C
Dagur Benediktsson

IChO
General instructions
Cover sheet

Please return this cover sheet together with all the related question sheets.

IChO
ISL-1 C-0 G-1

International Chemistry Olympiad 2021 Japan
53rd IChO2021 Japan
25th July – 2nd August, 2021
https://www.icho2021.org

G0-1
English (Official)

IChO

G0-2

ISL-1 C-0 G-2

English (Official)

General Instruction• You are allowed to use only pen to write the answer.
• Your calculator must be non-programmable.
• This examination has 9 problems.
• You can solve the problems in any order.
• You will have 5 hours to solve all problems.
• You can begin working only after the START command is given.
• All results must be written in the appropriate answer boxes with pen on the answer sheets. Use
the back of the question sheets if you need scratch paper. Remember that answers written outside
the answer boxes will not be graded.
• Write relevant calculations in the appropriate boxes when necessary. Full marks will be given forcorrect answers only when your work is shown.
• The invigilator will announce a 30-minute warning before the STOP command.
• You must stop working when the STOP command is given. Failure to stop writing will lead to thenullification of your examination.
• The official English version of this examination is available on request only for clarification.
• You are not allowed to leave your working place without permission. If you need any assistance(broken calculator, need to visit a restroom, etc), raise your hand and wait until an invigilator arrives.
GOOD LUCK!

Problems and Grading Information
Title

Total Score

Percentage

1

Hydrogen at a Metal Surface

24

11

2

Isotope Time Capsule

35

11

3

Lambert–Beer Law?

22

8

4

The Redox Chemistry of Zinc

32

11

5

Mysterious Silicon

60

12

6

The Solid-State Chemistry of Transition Metals

45

13

7

Playing with Non-benzenoid Aromaticity

36

13

8

Dynamic Organic Molecules and Their Chirality

26

11

9

Likes and Dislikes of Capsules

23

10

Total

100

IChO
ISL-1 C-0 G-3

G0-3
English (Official)

Physical Constants and Equations
Constants
Speed of light in vacuum𝑐 = 2.99792458 × 108 m s−1

Planck constantℎ = 6.62607015 × 10−34 J s

Elementary charge

𝑒 = 1.602176634 × 10−19 C

Electron mass𝑚e = 9.10938370 × 10−31 kg

Electric constant(permittivity of vacuum)

𝜀0 = 8.85418781 × 10−12 F m−1

Avogadro constant

𝑁A = 6.02214076 × 1023 mol−1

Boltzmann constant𝑘B = 1.380649 × 10−23 J K−1

Faraday constant

𝐹 = 𝑁A × 𝑒 = 9.64853321233100184 × 104 C mol−1

Gas constant

𝑅 = 𝑁A × 𝑘B = 8.31446261815324 J K−1 mol−1= 8.2057366081 × 10−2 L atm K−1 mol−1

Unified atomic mass unit𝑢 = 1 Da = 1.66053907 × 10−27 kg

Standard pressure

𝑝 = 1 bar = 105 Pa

Atmospheric pressure𝑝atm = 1.01325 × 105 Pa

Zero degree Celsius

0 ∘ C = 273.15 KÅngstrom

1 Å = 10−10 m

Picometer

1 pm = 10−12 m

Electronvolt

1 eV = 1.602176634 × 10−19 J

Part-per-million

1 ppm = 10−6

Part-per-billion

1 ppb = 10−9Part-per-trillion

1 ppt = 10−12pi

𝜋 = 3.141592653589793

The base of the natural logarithm(Euler's number)𝑒 = 2.718281828459045

IChO
ISL-1 C-0 G-4

G0-4
English (Official)

Equations
The ideal gas law

Coulomb's law

𝑃 𝑉 = 𝑛𝑅𝑇
, where 𝑃 is the pressure, 𝑉 is the volume, 𝑛 is the amount of substance,
𝑇 is the absolute temperature of ideal gas.
𝑞 𝑞
𝐹 = 𝑘e 1 2 2𝑟, where 𝐹 is the electrostatic force, 𝑘e (≃ 9.0 × 109 N m2 C−2 ) is Coulomb's
constant, 𝑞1 and 𝑞2 are the magnitudes of the charges, and 𝑟 is the distance
between the charges.

The first law of thermodynamics

Δ𝑈 = 𝑞 + 𝑤, where Δ𝑈 is the change in the internal energy, 𝑞 is the heat supplied, 𝑤
is the work done.

Enthalpy 𝐻

𝐻 = 𝑈 + 𝑃𝑉

Entropy based on Boltzmann's principle 𝑆

𝑆 = 𝑘B ln 𝑊
, where 𝑊 is the number of microstates.
𝑞
Δ𝑆 = rev𝑇, where 𝑞rev is the heat for the reversible process.

The change of entropy
Δ𝑆
Gibbs free energy 𝐺

𝐺 = 𝐻 − 𝑇𝑆
Δr 𝐺∘ = −𝑅𝑇 ln 𝐾 = −𝑧𝐹 𝐸 ∘, where 𝐾 is the equilibrium constant, 𝑧 is the number of electrons, 𝐸 ∘ isthe standard electrode potential.

Reaction quotient 𝑄

Δr 𝐺 = Δr 𝐺∘ + 𝑅𝑇 ln 𝑄
For a reaction
𝑎A + 𝑏B ⇌ 𝑐C + 𝑑D𝑐𝑑[C] [D]𝑄=𝑎𝑏[A] [B], where [A] is the concentration of A.

IChO

G0-5

ISL-1 C-0 G-5

Heat change Δ𝑞
Nernst equation for redox reaction
Arrhenius equation

English (Official)

Δ𝑞 = 𝑛𝑐m Δ𝑇, where 𝑐m is the temperature-independent molar heat capacity.
𝐶
𝑅𝑇
ln ox
𝐸 = 𝐸∘ +𝑧𝐹
𝐶red, where 𝐶ox is the concentration of oxidized substance, 𝐶red is the concentration of reduced substance.
𝐸
𝑘 = 𝐴 exp (− 𝑎 )𝑅𝑇, where 𝑘 is the rate constant, 𝐴 is the pre-exponential factor, 𝐸𝑎 is theactivation energy.
exp(𝑥) = 𝑒𝑥

Lambert–Beer equation

𝐴 = 𝜀𝑙𝑐, where 𝐴 is the absorbance, 𝜀 is the molar absorption coefficient, 𝑙 is the
optical path length, 𝑐 is the concentration of the solution.

Henderson–Hasselbalch
equation

For an equilibrium
HA ⇌ H+ + A−
, where equilibrium constant is 𝐾a ,[A− ]pH = p𝐾a + log ()[HA]𝑐
𝐸 = ℎ𝜈 = ℎ
𝜆, where 𝜈 is the frequency, 𝜆 is the wavelength of the light.

Energy of a photon
The sum of a geometricseriesWhen 𝑥 ≠ 1,

Approximation equationthat can be used to solve
problems

When 𝑥 ≪ 1,
1
≃1+𝑥
1−𝑥

𝑛

1 + 𝑥 + 𝑥2 + ⋯ + 𝑥𝑛 = ∑𝑖=0 𝑥𝑖 =1 − 𝑥𝑛+1
1−𝑥

IChO
ISL-1 C-0 G-6

Periodic Table

G0-6
English (Official)IChO
ISL-1 C-0 G-7

1

H NMR Chemical ShiftsΔ𝛿 for one alkyl group-substitution: ca. +0.4 ppm

G0-7
English (Official)IChO
ISL-1 C-0 G-1

International Chemistry Olympiad 2021 Japan
53rd IChO2021 Japan
25th July – 2nd August, 2021
https://www.icho2021.org

G0-1
Icelandic (Iceland)

IChO

G0-2

ISL-1 C-0 G-2

Icelandic (Iceland)

General Instructions• You are only allowed to use pen to write the answers.
• Your calculator must be non-programmable.
• This examination has 9 problems.
• You can solve the problems in any order.
• You will have 5 hours to solve all problems.
• You can begin working only after the START command is given.
• All results must be written in the appropriate answer boxes with pen on the answer sheets. Use
the back of the question sheets if you need rough paper. Remember that answers written outside
the answer boxes will not be graded.
• Write relevant calculations in the appropriate boxes when necessary. Full marks will be given forcorrect answers only when your work is shown.
• The invigilator will announce a 30-minute warning before the STOP command.
• You must stop working when the STOP command is given. Failure to stop writing will lead to thenullification of your examination.
• The official English version of this examination is available on request only for clarification.
• You are not allowed to leave your working place without permission. If you need any assistance(broken calculator, need to go to the toilet, etc.), raise your hand and wait until an invigilator arrives.
GOOD LUCK!

Problems and Grading Information
Title

Total Score

Percentage

1

Vetni á yfirborði málms

24

11

2

Samsætutímaglas

35

11

3

Lambert–Beer lögmál?

22

8

4

Redox efnafræði sinks

32

11

5

Hinn dularfulli kísill

60

12

6

Efnafræði fastra efnasambanda hliðarmálma

45

13

7

Playing with Non-benzenoid Aromaticity

36

13

8

Dynamic Organic Molecules and Their Chirality

26

11

9

Likes and Dislikes of Capsules

23

10

Total

100

IChO
ISL-1 C-0 G-3

G0-3
Icelandic (Iceland)

Physical Constants and Equations
Constants
Speed of light in vacuum𝑐 = 2.99792458 × 108 m s−1

Planck constantℎ = 6.62607015 × 10−34 J s

Elementary charge

𝑒 = 1.602176634 × 10−19 C

Electron mass𝑚e = 9.10938370 × 10−31 kg

Electric constant(permittivity of vacuum)

𝜀0 = 8.85418781 × 10−12 F m−1

Avogadro constant

𝑁A = 6.02214076 × 1023 mol−1

Boltzmann constant𝑘B = 1.380649 × 10−23 J K−1

Faraday constant

𝐹 = 𝑁A × 𝑒 = 9.64853321 × 104 C mol−1

Gas constant

𝑅 = 𝑁A × 𝑘B = 8.3144626 J K−1 mol−1= 8.2057366 × 10−2 L atm K−1 mol−1

Unified atomic mass unit𝑢 = 1 Da = 1.66053907 × 10−27 kg

Standard pressure

𝑝 = 1 bar = 105 Pa

Atmospheric pressure𝑝atm = 1.01325 × 105 Pa

Zero degrees Celsius

0 ∘ C = 273.15 KÅngstrom

1 Å = 10−10 m

Picometer

1 pm = 10−12 m

Electronvolt

1 eV = 1.602176634 × 10−19 J

Part-per-million

1 ppm = 10−6

Part-per-billion

1 ppb = 10−9Part-per-trillion

1 ppt = 10−12pi

𝜋 = 3.141592653589793

The base of the natural logarithm(Euler’s number)𝑒 = 2.718281828459045

IChO
ISL-1 C-0 G-4

G0-4
Icelandic (Iceland)

Equations
The ideal gas law

Coulomb’s law

𝑃 𝑉 = 𝑛𝑅𝑇
, where 𝑃 is the pressure, 𝑉 is the volume, 𝑛 is the amount of substance,
𝑇 is the absolute temperature of ideal gas.
𝑞 𝑞
𝐹 = 𝑘e 1 2 2𝑟, where 𝐹 is the electrostatic force, 𝑘e (≃ 9.0 × 109 N m2 C−2 ) is Coulomb’s
constant, 𝑞1 and 𝑞2 are the magnitudes of the charges, and 𝑟 is the distance
between the charges.

The first law of thermodynamics

Δ𝑈 = 𝑞 + 𝑤, where Δ𝑈 is the change in the internal energy, 𝑞 is the heat supplied, 𝑤
is the work done.

Enthalpy 𝐻

𝐻 = 𝑈 + 𝑃𝑉

EntropybasedonBoltzmann’s principle 𝑆

𝑆 = 𝑘B ln 𝑊
, where 𝑊 is the number of microstates.
𝑞
Δ𝑆 = rev𝑇, where 𝑞rev is the heat for the reversible process.

The change of entropy
Δ𝑆
Gibbs free energy 𝐺

𝐺 = 𝐻 − 𝑇𝑆
Δr 𝐺∘ = −𝑅𝑇 ln 𝐾 = −𝑧𝐹 𝐸 ∘, where 𝐾 is the equilibrium constant, 𝑧 is the number of electrons, 𝐸 ∘ isthe standard electrode potential.

Reaction quotient 𝑄

Δr 𝐺 = Δr 𝐺∘ + 𝑅𝑇 ln 𝑄
For a reaction
𝑎A + 𝑏B ⇌ 𝑐C + 𝑑D𝑐𝑑[C] [D]𝑄=𝑎𝑏[A] [B], where [A] is the concentration of A.

IChO

G0-5

ISL-1 C-0 G-5

Heat change Δ𝑞
Nernst equation for aredox reaction
Arrhenius equation

Icelandic (Iceland)Δ𝑞 = 𝑛𝑐m Δ𝑇, where 𝑐m is the temperature-independent molar heat capacity.
𝐶
𝑅𝑇
ln ox
𝐸 = 𝐸∘ +𝑧𝐹
𝐶red, where 𝐶ox is the concentration of the oxidized substance, 𝐶red is theconcentration of the reduced substance.
𝐸
𝑘 = 𝐴 exp (− 𝑎 )𝑅𝑇, where 𝑘 is the rate constant, 𝐴 is the pre-exponential factor, 𝐸𝑎 is theactivation energy.
exp(𝑥) = 𝑒𝑥

Lambert–Beer equation

𝐴 = 𝜀𝑙𝑐, where 𝐴 is the absorbance, 𝜀 is the molar absorption coefficient, 𝑙 is the
optical path length, and 𝑐 is the concentration of the solution.

Henderson–Hasselbalch
equation

For an equilibrium
HA ⇌ H+ + A−
, where equilibrium constant is 𝐾a ,[A− ]pH = p𝐾a + log ()[HA]𝑐
𝐸 = ℎ𝜈 = ℎ
𝜆, where 𝜈 is the frequency, 𝜆 is the wavelength of the light.

Energy of a photon
The sum of a geometricseriesWhen 𝑥 ≠ 1,

Approximation equationthat can be used to solve
problems

When 𝑥 ≪ 1,
1
≃1+𝑥
1−𝑥

𝑛

1 + 𝑥 + 𝑥2 + ⋯ + 𝑥𝑛 = ∑𝑖=0 𝑥𝑖 =1 − 𝑥𝑛+1
1−𝑥

IChO
ISL-1 C-0 G-6

The Periodic Table

G0-6
Icelandic (Iceland)

IChO
ISL-1 C-0 G-7

1

H NMR Chemical ShiftsΔ𝛿 for one alkyl group-substitution: ca. +0.4 ppm

G0-7
Icelandic (Iceland)ISL-1 C-1 C-1

ISL-1 C-1 C
Dagur Benediktsson

IChO
Problem 1
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q1-1

IChO
ISL-1 C-1 Q-1English (Official)Hydrogen at a Metal Surface
11 % of the total
Question

A.1

A.2

B.1

B.2

B.3

B.4

Total

Points

6

4

5

3

3

3

24

Score

Hydrogen is expected to be a future energy source that does not depend on fossil fuels. Here, we willconsider the hydrogen-storage process in a metal, which is related to hydrogen-transport and -storage
technology.

Part AAs hydrogen is absorbed into the bulk of a metal via its surface, let us first consider the adsorption
process of hydrogen at the metal surface, H2 (g) → 2H(ad), where the gaseous and adsorbed states ofhydrogen are represented as (g) and (ad), respectively. Hydrogen molecules (H2 ) that reach the metal
surface (M) dissociate at the surface and are adsorbed as H atoms (Fig. 1). Here, the potential energyof H2 is represented by two variables: the interatomic distance, 𝑑, and the height relative to the surface
metal atom, 𝑧. It is assumed that the axis along the two H atoms is parallel to the surface and that thecenter of gravity is always on the vertical dotted line in Fig. 1. Fig. 2 shows the potential energy contour
plot for the dissociation at the surface. The numerical values represent the potential energy in units ofkJ per mole of H2 . The solid line spacing is 20 kJ mol−1 , the dashed line spacing is 100 kJ mol−1 , and thespacing between solid and dashed lines is 80 kJ mol−1 . The zero-point vibration energy is ignored.

IChO
ISL-1 C-1 Q-2

Q1-2

Fig.1 Definition of variables. Drawing is not in scale.

Fig.2

English (Official)

Q1-3

IChO
ISL-1 C-1 Q-3

A.1

A.2

English (Official)

For each of the following items (i)–(iii), select the closest value from A–G.
(i) The interatomic distance for a gaseous H2 molecule(ii) The interatomic distance between metal atoms (𝑑M in Fig. 1)(iii) The distance of adsorbed H atoms from the surface (ℎad in Fig. 1)
A. 0.03 nm

B. 0.07 nm

C. 0.11 nm

E. 0.19 nm

F. 0.23 nm

G. 0.27 nm

6pt

D. 0.15 nm

For each of the following items (i)–(ii), select the closest value from A–H.
(i) the energy required for the dissociation of gaseous H2 to gaseous H[H2 (g) → 2H(g)](ii) the energy released during the adsorption of a gaseous H2 [H2 (g) → 2H(ad)]A. 20 kJ mol−1

B. 40 kJ mol−1

C. 60 kJ mol−1

D. 100 kJ mol−1

E. 150 kJ mol−1

F. 200 kJ mol−1

G. 300 kJ mol−1

H. 400 kJ mol−1

4pt

IChO
ISL-1 C-1 Q-4

Q1-4
English (Official)Part B
The adsorbed hydrogen atoms are then either absorbed into the bulk, or recombine and desorb backinto the gas phase, as shown in the reactions (1a) and (1b). H(ab) represents a hydrogen atom absorbed
in the bulk.

The reaction rates per surface site for adsorption, desorption, and absorption are 𝑟1 [s−1 ], 𝑟2 [s−1 ] and𝑟3 [s−1 ] , respectively. They are expressed as:

where 𝑘1 [s−1 Pa−1 ], 𝑘2 [s−1 ] and 𝑘3 [s−1 ] are the reaction rate constants and 𝑃H2 is the pressure of H2 .
Among the sites available on the surface, 𝜃 (0 ≤ 𝜃 ≤ 1) is the fraction occupied by H atoms. It is assumed
that adsorption and desorption are fast compared to absorption (𝑟1 , 𝑟2 ≫ 𝑟3 ) and that 𝜃 remains constant.
B.1

𝑟3 can be expressed as:

Express 𝐶 using 𝑘1 and 𝑘2 .

5pt

IChO
ISL-1 C-1 Q-5

Q1-5
English (Official)

A metal sample with a surface area of 𝑆 = 1.0 × 10−3 m2 was placed in a container (1L = 1.0 × 10−3 m3 )with H2 (𝑃H2 = 1.0 × 102 Pa ). The density of hydrogen-atom adsorption sites on the surface was 𝑁 =1.3 × 1018 m−2 . The surface temperature was kept at 𝑇 = 400 K. As the reaction (1) proceeded, 𝑃H2
decreased at a constant rate of 𝑣 = 4.0 × 10−4 Pa s−1 . Assume that H2 is an ideal gas and that the volumeof the metal sample is negligible.

B.2

Calculate the amount of H atoms in moles absorbed per unit area of the surface
per unit time, 𝐴 [mol s−1 m−2 ].

3pt

B.3

At 𝑇 = 400 K, 𝐶 equals 1.0 × 102 Pa−1 . Calculate the value of 𝑘3 at 400 K. If you
did not obtain the answer to B.2, use 𝐴 = 3.6 × 10−7 mol s−1 m−2 .

3pt

B.4

At a different 𝑇 , 𝐶 = 2.5 × 103 Pa−1 and 𝑘3 = 4.8 × 10−2 s−1 are given. For 𝑟3 as afunction of 𝑃H2 at this temperature, select the correct plot from (a)–(h).

3pt

Q1-1

IChO
ISL-1 C-1 Q-1Icelandic (Iceland)

Vetni á yfirborði málms
11 % of the total
Question

A.1

A.2

B.1

B.2

B.3

B.4

Total

Points

6

4

5

3

3

3

24

Score

Talið er að vetni verði sá orkugjafi framtíðarinnar sem ekki er háður jarðefnaeldsneyti. Í þessu dæmi er
fjallað um geymsluferli vetnis í málmi en það er skylt flutnings- og geymslutækni fyrir vetni.

Hluti AÞegar vetni ísogast (absorbs) inn í málm gegnum yfirborð hans, þá þarf fyrst að huga að ásogsferli(adsorption process) vetnis á málmyfirborðið, H2 (g) → 2H(ad), þar sem gasástand er táknað með (g)og ásogað ástand með (ad). Vetnissameindir (H2 ) sem koma að málmyfirborði (M) klofna þar og ásogast
sem H atóm (mynd 1). Mættisorka (potential energy) H2 er sýnd sem fall af tveimur breytistærðum: fjarlægðinni á milli atómanna, 𝑑, og hæð þeirra frá málmatómi á yfirborði, 𝑧. Gert er ráð fyrir að ásinn milli
H atómanna tveggja sé samsíða yfirborðinu og að massamiðja þeirra sé ávallt á lóðréttu punktalínunni á
mynd 1. Mynd 2 sýnir hæðarlínur mættisorkunnar fyrir klofnunina á yfirborðinu. Tölugildin sýna mættisorkuna í einingunni kJ per mól af H2 . Bil milli heillína er 20 kJ mol−1 , bil milli punktastrikaðra lína er 100
kJ mol−1 og bil milli heillína og punktastrikaðra lína er 80 kJ mol−1 . Ekki þarf að taka tillit til núllstigsorku
titrings ( zero-point vibration energy).

IChO
ISL-1 C-1 Q-2

Q1-2
Icelandic (Iceland)

Mynd 1 Skilgreiningar á breytistærðum. Teikningin er ekki í réttum hlutföllum.

Mynd 2

Q1-3

IChO
ISL-1 C-1 Q-3

A.1

A.2

Icelandic (Iceland)

Veldu það gildi af A-G sem kemst næst eftirfarandi stærðum (i)–(iii).
(i) Fjarlægðin milli atóma í H2 gassameind(ii) Fjarlægðin milli málmatóma (𝑑M á mynd 1)(iii) Fjarlægð ásogaðs H atóms frá yfirborðinu (ℎad á mynd 1)
A. 0,03 nm

B. 0,07 nm

C. 0,11 nm

E. 0,19 nm

F. 0,23 nm

G. 0,27 nm

6pt

D. 0,15 nm

Veldu það gildi A-H sem kemst næst eftirfarandi stærðum (i)–(ii).
(i) orkan sem þarf fyrir klofnun H2 í gasfasa í H atóm í gasfasa[H2 (g) → 2H(g)](ii) orkan sem losnar við ásog H2 gass
[H2 (g) → 2H(ad)]A. 20 kJ mol−1

B. 40 kJ mol−1

C. 60 kJ mol−1

D. 100 kJ mol−1

E. 150 kJ mol−1

F. 200 kJ mol−1

G. 300 kJ mol−1

H. 400 kJ mol−1

4pt

IChO
ISL-1 C-1 Q-4

Q1-4
Icelandic (Iceland)

Hluti BÁsoguð vetnisatóm geta annaðhvort ísogast inn í málminn eða þau tengjast saman aftur og frásogast(desorb) til baka í gasfasa, eins og sýnt er í efnahvörfum (1a) og (1b). H(ab) táknar vetnisatóm sem hefurísogast inn í málminn.

Hvarfhraðar á hvern ásogsstað yfirborðsins fyrir ásog, frásog og ísog (í sömu röð) eru 𝑟1 [s−1 ], 𝑟2 [s−1 ] og
𝑟3 [s−1 ] . Jöfnurnar fyrir hvarfhraðana eru:

þar sem 𝑘1 [s−1 Pa−1 ], 𝑘2 [s−1 ] og 𝑘3 [s−1 ] eru hraðafastarnir og 𝑃H2 er þrýstingur H2 .
Fyrir ásogsstaðina á yfirborðinu þá er 𝜃 (0 ≤ 𝜃 ≤ 1) sá hluti þeirra sem er setinn H atómum. Gert er ráð
fyrir að ásog og frásog gerist hratt miðað við ísog (𝑟1 , 𝑟2 ≫ 𝑟3 ) og að 𝜃 sé fasti.
B.1

𝑟3 er hægt að lýsa með jöfnunni:Sýndu jöfnu fyrir 𝐶 sem fall 𝑘1 og 𝑘2 .

5pt

IChO
ISL-1 C-1 Q-5

Q1-5
Icelandic (Iceland)

Málmsýni með yfirborðsflatarmál𝑆 = 1, 0 × 10−3 m2 var sett í hylki (1L = 1, 0 × 10−3 m3 ) sem innihélt H2(𝑃H2 = 1, 0 × 102 Pa ). Þéttleiki ásogsstaða vetnisatóma á yfirborðinufyrir var 𝑁 = 1, 3 × 1018 m−2 . Hitastiginu á yfirborðinu var haldið við 𝑇 = 400 K. Eftir því sem hvarfi (1) miðaði áfram lækkaði 𝑃H2 með jöfnum
hraða, 𝑣 = 4, 0×10−4 Pa s−1 . Gera má ráð fyrir að H2 sé kjörgas og að rúmmál málmsýnisins sé óverulegt.
B.2

Reiknaðu mólfjölda ásogaðra H atóma á flatarmálseiningu á tímaeiningu,
𝐴 [mol s−1 m−2 ].

3pt

B.3

Við 𝑇 = 400 K, er gildi 𝐶 1, 0 × 102 Pa−1 . Reiknaðu gildi 𝑘3 við 400 K. Ef þú gastekki reiknað svarið í B.2, notaðu þá gildið 𝐴 = 3, 6 × 10−7 mol s−1 m−2 .

3pt

B.4

Við annað hitastig 𝑇 , eru gildi 𝐶 = 2, 5 × 103 Pa−1 og 𝑘3 = 4, 8 × 10−2 s−1 . Veldu
þann feril af (a)–(h) sem sýnir 𝑟3 sem fall af 𝑃H2 við þetta hitastig.

3pt

IChO

A1-1

ISL-1 C-1 A-1

Icelandic (Iceland)

Vetni á yfirborði málms
Hluti A
A.1 (6 pt)
(i)

(ii)

(iii)

A.2 (4 pt)
(i)

(ii)

IChO
ISL-1 C-1 A-2

Hluti B
B.1 (5 pt)

𝐶=
B.2 (3 pt)𝐴=

mol s−1 m−2

A1-2
Icelandic (Iceland)

IChO
ISL-1 C-1 A-3

B.3 (3 pt)

𝑘3 =B.4 (3 pt)

s−1

A1-3
Icelandic (Iceland)ISL-1 C-2 C-1

ISL-1 C-2 C
Dagur Benediktsson

IChO
Problem 2
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q2-1

IChO
ISL-1 C-2 Q-1

English (Official)

Isotope Time Capsule
11 % of the total
Question

A.1

A.2

A.3

A.4

Total

Points

8

8

10

9

35

Score

Molecular entities that differ only in isotopic composition, such as CH4 and CH3 D, are called isotopologues. Isotopologues are considered to have the same chemical characteristics. In nature, however,there exists a slight difference.
Assume that all of the substances shown in this Question are in a gas phase.
Let us consider the following equilibrium:

The entropy, 𝑆, increases with increasing the number of possible microscopic states of a system, 𝑊 :𝑊 = 1 for 12 C16 O2 and 12 C18 O2 . In contrast, 𝑊 = 2 for a 12 C16 O18 O molecule because the oxygen atoms
are distinguishable in this molecule. As the right-hand side of the equilibrium shown in eq. 1 has two
12 16 18
C O O molecules, 𝑊 = 22 = 4.

IChO
ISL-1 C-2 Q-2

A.1

Q2-2
English (Official)

The enthalpy change, Δ𝐻, of eq. 3 is positive regardless of the temperature.

8pt

Calculate the equilibrium constants, 𝐾, for eq. 3 at very low (think of 𝑇 → 0) andvery high (think of 𝑇 → +∞) temperatures. Assume that the reaction remains
unchanged at these temperatures and that Δ𝐻 converges to a constant value
for high temperatures.
The ΔH of the following process can be explained by molecular vibrations.

At 𝑇 = 0 K, the vibrational energy of a diatomic molecule whose vibration frequency is 𝜈 [s−1 ] is expressedas:Wherein 𝑘 is the force constant and 𝜇 the reduced mass, which is expressed in terms of the mass of thetwo atoms in the diatomic molecule, 𝑚1 and 𝑚2 , according to:

A.2

The vibration of H2 is at 4161.0 cm−1 when reported as a wavenumber.
Calculate the Δ𝐻 of the following equation at 𝑇 = 0 K in units of J mol−1 .

Assume that:• only the vibrational energy contributes to the Δ𝐻.
• the 𝑘 values for H2 , HD, and D2 are identical.
• the mass of H to be 1 Da and the mass of D to be 2 Da.

8pt

IChO
ISL-1 C-2 Q-3

Q2-3
English (Official)

The molar ratio of H2 , HD, and D2 depends on the temperature in a system in equilibrium. Here, ΔD2 isdefined as the change of the molar ratio of D2 .

[D2 ]
[D ]
in the sample and 𝑅D∗ 2 to 2 at 𝑇 → +∞. It should be noted here that the
[H2 ]
[H2 ]
distribution of isotopes becomes random at 𝑇 → +∞.
Here, 𝑅D2 refers to
A.3

Calculate ΔD2 with natural D abundance when the isotopic exchange is in equilibrium at the temperature where 𝐾 in eq. 4 is 0.300. Assume that the naturalabundance ratios of D and H are 1.5576 × 10–4 and 1 − 1.5576 × 10–4 , respectively.

10pt

Q2-4

IChO
ISL-1 C-2 Q-4English (Official)

In general, the molar ratio of the doubly substituted isotopologue, which contains two heavy isotope
atoms in one molecule, increases with decreasing temperature. Let us consider the molar ratio of CO2
molecules with molecular weights of 44 and 47, which are described as CO2 [44] and CO2 [47] below. The
quantity Δ47 is defined as:[CO2 [47]][CO2 [47]]∗in the sample and 𝑅47
toat 𝑇 → +∞. The natural abundances of carbon[CO2 [44]][CO2 [44]]and oxygen atoms are shown below; ignore isotopes that are not shown here.

𝑅47 refers to12

natural abundance

16

natural abundanceC

130.988888

0.011112

C

O

17

O

18

0.997621

0.0003790

0.0020000

O

The temperature dependence of Δ47 is determined as follows, where 𝑇 is given as the absolute temperature in units of K:

A.4

The 𝑅47 of fossil plankton obtained from the Antarctic seabed was 4.50865×10–5 .
Estimate the temperature using this 𝑅47 . This temperature is interpreted as the
air temperature during the era in which the plankton lived. Consider only themost common isotopologue of CO2 [47] for the calculation.

9pt

Q2-1

IChO
ISL-1 C-2 Q-1

Icelandic (Iceland)Samsætutímaglas
11 % of the total
Question

A.1

A.2

A.3

A.4

Total

Points

8

8

10

9

35

Score

Sameindir sem eru ólíkar að því eina leyti að hafa mismunandi samsetningar samsæta, eins og CH4 og
CH3 D, eru kallaðar samsætuhverfur (isotopologues). Samsætuhverfur teljast hafa sömu efnaeiginleikaen í raun er á þeim smá munur.
Gerðu ráð fyrir að öll efnin í þessu dæmi séu í gasfasa.
Athugum eftirfarandi efnajafnvægi:Óreiðan, 𝑆, eykst með auknum fjölda mögulegra örástanda kerfis, 𝑊 :𝑊 = 1 fyrir 12 C16 O2 og 12 C18 O2 . Hins vegar er 𝑊 = 2 fyrir 12 C16 O18 O sameind því að hægt er að greina ámilli súrefnisatómanna þeirri sameind.
Hægra megin í jafnvæginu (sjá jöfnu 1) eru tvær 12 C16 O18 O sameindir og þar er því 𝑊 = 22 = 4.

IChO
ISL-1 C-2 Q-2

A.1

Q2-2
Icelandic (Iceland)

Vermibreytingin, Δ𝐻, í efnajöfnu 3 er jákvæð við öll hitastig.

8pt

Reiknaðu jafnvægisfastann, 𝐾, fyrir efnajafnvægi 3 við mjög lág hitastig (þegar
𝑇 → 0) og við mjög há hitastig (þegar 𝑇 → +∞). Gerðu ráð fyrir að hvarfið
breytist ekki við þessi hitastig og að Δ𝐻 nái föstu gildi við há hitastig.
Finna má gildi ΔH fyrir eftirfarandi hvarf út frá sameindatitringum.

Við 𝑇 = 0 K, má lýsa titringsorku tvíatóma sameindar, sem hefur titringstíðnina 𝜈 [s−1 ], með eftirfarandi
jöfnu:

Hér er k kraftstuðullinn (force constant) og 𝜇 er skerti massinn (reduced mass), en hann er samsettur úr
massa atómanna í tvíatóma sameind, 𝑚1 og 𝑚2 , eins og eftirfarandi jafna sýnir:A.2

Titringur H2 er við 4161,0 cm−1 (þegar gildið er gefið á formi bylgjutölu (wavenumber)).
Reiknaðu Δ𝐻 fyrir eftirfarandi efnajöfnu við 𝑇 = 0 K í einingunum J mol−1 .

Gerðu ráð fyrir eftirfarandi:• Δ𝐻 er aðeins háð titringsorku• gildi 𝑘 fyrir H2 , HD, og D2 eru þau sömu.
• massi H er 1 Da og massi D er 2 Da.

8pt

IChO
ISL-1 C-2 Q-3

Q2-3
Icelandic (Iceland)Í kerfi sem er við jafnvægi eru mólhlutföll H2 , HD og D2 háð hitastigi. Hér er ΔD2 skilgreint sem breytingin
á mólhlutfalli D2 .

Hér vísar 𝑅D2 til

[D2 ]
[D ]
í sýninu og 𝑅D∗ 2 til 2 þegar 𝑇 → +∞.
[H2 ]
[H2 ]

Bent skal á að dreifing samsæta er handahófskennd (random) þegar 𝑇 → +∞.
A.3

Reiknaðu ΔD2 þegar samsætuskiptin eru í jafnvægi, við það hitastig þegar K í
jöfnu 4 er 0,300.
Gerðu ráð fyrir að náttúrulegt samsætuhlutfall D sé 1, 5576×10–4 og náttúrulegtsamsætuhlutfall H sé 1 − 1, 5576 × 10–4 .

10pt

Q2-4

IChO
ISL-1 C-2 Q-4Icelandic (Iceland)

Tvísetin samsætuhverfa (doubly substituted isotopologue) inniheldur tvö þung samsætuatóm í einni
sameind. Almennt eykst mólhlutfall tvísetinna samsætuhverfa með lækkandi hitastigi.
Skoðum mólhlutfall CO2 sameinda sem hafa mólmassana 44 og 47, og sem táknuð eru sem CO2 [44] og
CO2 [47] . Stærðin Δ47 er skilgreind á eftirfarandi hátt:

Hér vísar 𝑅47 til[CO2 [47]][CO2 [47]]∗í sýninu og 𝑅47
tilþegar 𝑇 → +∞.
[CO2 [44]][CO2 [44]]Náttúrulegt samsætuhlutfall kolefnis- og súrefnisatóma er sýnt hér að neðan; ekki þarf að taka tillit til
annarra samsæta.
12

náttúrulegt samsætuhlutfall

16

náttúrulegt samsætuhlutfall

C

13

0,988888

0,011112

C

O

17

O

18

0,997621

0,0003790

0,0020000

O

Hvernig Δ47 er háð hitastigi er gefið með jöfnunni hér að neðan, þar sem hitastigið T er gefið með
einingunni K:

A.4

Gildi 𝑅47 í steingervingi svifjurtar frá hafsbotni við Suðurskautslandið er
4, 50865 × 10–5 .
Notaðu þetta gildi á 𝑅47 til að áætla hitastigið þegar svifjurtin var á lífi. Þetta
hitastig er talið vera lofthitinn á þeim tíma þegar svifjurtin var uppi. Í útreikningunum þarf aðeins að nota algengustu samsætuhverfu CO2 [47].

9pt

IChO
ISL-1 C-2 A-1

Samsætutímaglas
A.1 (8 pt)

𝑇 →0∶𝐾=, 𝑇 → +∞ ∶ 𝐾 =A2-1
Icelandic (Iceland)

IChO
ISL-1 C-2 A-2

A.2 (8 pt)

Δ𝐻 =J mol−1

A2-2
Icelandic (Iceland)

IChO
ISL-1 C-2 A-3

A.3 (10 pt)

ΔD2 =

A2-3
Icelandic (Iceland)

IChO
ISL-1 C-2 A-4

A.4 (9 pt)

𝑇 =

K

A2-4
Icelandic (Iceland)ISL-1 C-3 C-1

ISL-1 C-3 C
Dagur Benediktsson

IChO
Problem 3
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q3-1

IChO
ISL-1 C-3 Q-1

English (Official)

Lambert–Beer Law?
8 % of the total
Question

A.1

B.1

B.2

Total

Points

10

6

6

22

Score

In this problem, ignore the absorption of the cell and the solvent. The temperatures of all solutions andgases are kept constant at 25 °C.

Part A
An aqueous solution X was prepared using HA and NaA. The concentrations [A− ], [HA], and [H+ ] in solution X are 1.00 × 10−2 mol L−1 , 1.00 × 10−3 mol L−1 , and 1.00 × 10−4 mol L−1 , respectively, which arecorrelated via the following acid-base equilibrium:
HA ⇌ A− + H+

𝐾=

[A− ][H+ ]
[HA]

(1)

The optical path length is 𝑙 in Part A. Ignore the density change upon dilution. Assume that no chemical
reactions other than eq 1 occur.
A.1

The absorbance of X was 𝐴1 at a wavelength of 𝜆1 . Then, solution X was diluted
to twice its initial volume using hydrochloric acid with pH = 2.500. After the
dilution, the absorbance was still 𝐴1 at 𝜆1 . Determine the ratio 𝜀HA /𝜀A− , where𝜀HA and 𝜀A− represent the absorption coefficients of HA and of A− , respectively,at 𝜆1 .

10pt

Q3-2

IChO
ISL-1 C-3 Q-2English (Official)Part BLet us consider the following equilibrium in the gas phase.
D ⇌ 2M(2)

Pure gas D is filled into a cuboid container that has a transparent movable wall with a cross-section of 𝑆
(see the figure below) at a pressure 𝑃 , and equilibrium is established while the total pressure is kept at 𝑃 .
The absorbance of the gas is 𝐴 = 𝜀(𝑛/𝑉 )𝑙, where 𝜀, 𝑛, 𝑉 , and 𝑙 are the absorption coefficient, amount of
the gas in moles, volume of the gas, and optical path length, respectively. Assume that all componentsof the gas mixture behave as ideal gases.

Use the following definitions if necessary.
Initial state

After equilibrium

D

M

D

M

Partial pressure

𝑃

0

𝑝D

𝑝M

Amount in moles

𝑛0

0

𝑛D

𝑛M

Volume𝑉0

𝑉

B.1

The absorbance of the gas at 𝜆B1 measured from direction 𝑥 (𝑙 = 𝑙𝑥 ) was 𝐴B1
both at the initial state and after the equilibrium. Determine the ratio 𝜀D /𝜀M
at 𝜆B1 , where 𝜀D and 𝜀M represent the absorption coefficients of D and of M,
respectively.

6pt

B.2

The absorbance of the gas at 𝜆B2 measured from direction 𝑦 was 𝐴B2 both at
the initial state (𝑙 = 𝑙𝑦0 ) and after the equilibrium (𝑙 = 𝑙𝑦 ). Determine the ratio
𝜀D /𝜀M at 𝜆B2 .

6pt

Q3-1

IChO
ISL-1 C-3 Q-1

Icelandic (Iceland)

Lambert–Beer lögmál?
8 % of the total
Question

A.1

B.1

B.2

Total

Points

10

6

6

22

Score

Í þessu verkefni skaltu hunsa gleypni kúvettunnar og leysisins. Hitastigi allra lausna og gastegunda er
haldið föstu við 25 °C.

Hluti A
Vatnslausn X var útbúin með HA og NaA. Styrkir [A− ], [HA], og [H+ ] í lausninni eru 1, 00 × 10−2 mol L−1 ,1, 00 × 10−3 mol L−1 , og 1, 00 × 10−4 mol L−1 , (talið upp í sömu röð) og um samband styrkjanna gildir
eftirfarandi sýru-basa jafnvægi:HA ⇌ A− + H+

𝐾=

[A− ][H+ ]
[HA]

(1)

Lengd geisla í mælilausn (optical path length) er 𝑙 í hluta A. Hunsaðu eðlismassabreytingu sem kann aðverða við þynningu og gerðu ráð fyrir að jafna 1 sé eina efnahvarfið sem á sér stað.
A.1

Gleypni X var 𝐴1 við bylgjulengd 𝜆1 . Síðan var lausn X þynnt að tvöföldu upphaflegu rúmmáli lausnarinnar með saltsýru (hydrochloric acid) sem hafði pH= 2,500. Eftir þynningu var gleypnin ennþá 𝐴1 við 𝜆1 . Ákvarðaðu hlutfallið
𝜀HA /𝜀A− , þar sem 𝜀HA og 𝜀A− eru gleypnistuðlar (absorption coefficients) HA og
A− , við 𝜆1 .

10pt

Q3-2

IChO
ISL-1 C-3 Q-2Icelandic (Iceland)

Hluti BSkoðum eftirfarandi efnajafnvægi í gasfasa.
D ⇌ 2M(2)

Hreinu gasi D er komið fyrir við þrýsting 𝑃 í ferstrendingslaga íláti sem hefur gegnsæja og færanlega hliðmeð þversnið 𝑆 (sjá mynd að neðan). Jafnvægi er komið á meðan heildarþrýstingnum er haldið föstum
við 𝑃 . Gleypni gassins er 𝐴 = 𝜀(𝑛/𝑉 )𝑙, þar sem 𝜀 er gleypnistuðullinn, 𝑛 er mólfjöldi gass, 𝑉 er rúmmál
gassins og 𝑙 er lengd geisla. Gerðu ráð fyrir að allir hlutar gasblöndunnar hegði sér sem kjörgas.

Notaðu eftirfarandi skilgreiningar eftir þörfum:
UpphafsástandVið jafnvægi

D

M

D

M

Hlutþrýstingur

𝑃

0

𝑝D

𝑝M

Mólfjöldi

𝑛0

0

𝑛D

𝑛M

Rúmmál

𝑉0

𝑉

B.1

Gleypni gassins við 𝜆B1 þegar mæld eftir stefnu 𝑥 (𝑙 = 𝑙𝑥 ) var 𝐴B1 , bæði við upphafsástand og við jafnvægi. Ákvarðaðu hlutfallið 𝜀D /𝜀M við 𝜆B1 , þar sem 𝜀D og
𝜀M eru gleypnistuðlar D og M.

6pt

B.2

Gleypni gassins við 𝜆B2 þegar mæld eftir stefnu 𝑦 var 𝐴B2 , bæði við upphafsástand (𝑙 = 𝑙𝑦0 ) og við jafnvægi (𝑙 = 𝑙𝑦 ). Ákvarðaðu hlutfallið 𝜀D /𝜀M við 𝜆B2 .

6pt

IChO
ISL-1 C-3 A-1

Lambert–Beer Lögmál?
Hluti A
A.1 (10 pt)(Framhald á næstu síðu)A3-1
Icelandic (Iceland)

IChO
ISL-1 C-3 A-2

A.1 (cont.)

𝜀HA /𝜀A− =A3-2
Icelandic (Iceland)

IChO
ISL-1 C-3 A-3

Hluti B
B.1 (6 pt)

𝜀D /𝜀M =A3-3
Icelandic (Iceland)

IChO
ISL-1 C-3 A-4

B.2 (6 pt)

𝜀D /𝜀M =A3-4
Icelandic (Iceland)ISL-1 C-4 C-1

ISL-1 C-4 C
Dagur Benediktsson

IChO
Problem 4
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q4-1

IChO
ISL-1 C-4 Q-1

English (Official)

The Redox Chemistry of Zinc
11 % of the total
Question

A.1

A.2

B.1

B.2

B.3

B.4

Total

Points

6

5

4

3

5

9

32

Score

Zinc has long been used as alloys for brass and steel materials. The zinc contained in industrial wastewater is separated by precipitation to detoxify the water, and the obtained precipitate is reduced to recoverand reuse it as metallic zinc.

Part AThe dissolution equilibrium of zinc hydroxide Zn(OH)2 (s) at 25 ∘ C and the relevant equilibrium constants
are given in eq. 1–4.

IChO
ISL-1 C-4 Q-2

Q4-2
English (Official)

The solubility, 𝑆, of zinc (concentration of zinc in a saturated aqueous solution) is given in eq. 5.

A.1

When the equilibria in eq. 1–4 are established, calculate the pH range
in which [Zn(OH)2 (aq)] is the greatest among [Zn2+ (aq)], [Zn(OH)2 (aq)] and[Zn(OH)2−
4 (aq)].

6pt

A.2

A saturated aqueous solution of Zn(OH)2 (s) with pH = 7.00 was prepared andfiltered. NaOH was added to this filtrate to increase its pH to 12.00. Calculate
the molar percentage of zinc that precipitates when increasing the pH from 7.00
to 12.00. Ignore the volume and temperature changes.

5ptPart B
Next, the recovered zinc hydroxide is heated to obtain zinc oxide according to the reaction below:

The zinc oxide is then reduced to metallic zinc by reaction with hydrogen:

B.1

In order for reaction (7) to proceed at a hydrogen pressure kept at 1 bar, it is necessary to reduce the partial pressure of the generated water vapor. Calculate
the upper limit for the partial pressure of water vapor to allow reaction (7)to proceed at 300 ∘ C. Here, the Gibbs formation energies of zinc oxide andwater vapor at 300 ∘ C and 1 bar for all gaseous species are Δ𝐺ZnO (300∘ C) =−2.90 × 102 kJ mol−1 and Δ𝐺H2 O (300∘ C) = −2.20 × 102 kJ mol−1 , respectively.

4pt

Metallic zinc is used as a negative electrode (anode) material for metal-air batteries. The electrode consists of Zn and ZnO. It uses the following redox reaction to generate electricity with the electromotive
force (e.m.f.) at 25 ∘ C and pressure of 1 bar, 𝐸 ∘ .

B.2

A zinc–air battery was discharged at 20 mA for 24 hours. Calculate the change
in mass of the negative electrode (anode) of the battery.

3pt

IChO
ISL-1 C-4 Q-3

Q4-3
English (Official)

Mt. Fuji
B.3

Consider the change of e.m.f. of a zinc–air battery depending on the environment. Calculate the e.m.f. at the summit of Mt. Fuji, where the temperature
and altitude are −38 ∘ C (February) and 3776 m, respectively. The atmosphericpressure is represented by

5pt

at altitude ℎ [m] and temperature 𝑇 [∘ C]. The molar ratio of oxygen in the atmosphere is 21%. The Gibbs energy change of reaction (8) is Δ𝐺ZnO (−38∘ C) =−3.26 × 102 kJ mol−1 at −38 ∘ C and 1 bar.
B.4

Calculate the Gibbs energy change for reaction (6) at 25 ∘ C. Note that the standard reduction potentials, 𝐸 ∘ (Zn2+ /Zn) and 𝐸 ∘ (O2 /H2 O) at 25 ∘ C and 1 bar are
given as (10) and (11), respectively.

9pt

Q4-1

IChO
ISL-1 C-4 Q-1

Icelandic (Iceland)

Redox efnafræði sinks
11 % of the total
Question

A.1

A.2

B.1

B.2

B.3

B.4

Total

Points

6

5

4

3

5

9

32

Score

Sink hefur löngum verið notað í málmblöndur. Sink sem finna má í úrgangsvatni frá iðnaði er aðskilið
frá vatninu með fellingu (precipitation), til að afeitra vatnið. Fellingin sem fæst er afoxuð svo að hægt sé
að endurheimta sinkið og að lokum endurnýta það sem sinkmálm.

Hluti A
Leysnijafnvægi sinkhýdroxíðs Zn(OH)2 (s) við 25 ∘ C og tilheyrandi jafnvægisfastar er gefið í jöfnum 1-4.

Leysni, 𝑆, sinks (þ.e. styrkur sinks í mettaðri vatnslausn) er gefin með jöfnu 5.

IChO
ISL-1 C-4 Q-2

Q4-2
Icelandic (Iceland)

A.1Þegar jafnvægjum 1-4 er náð, reiknaðu á hvaða pH bili styrkur [Zn(OH)2 (aq)]er hæstur af styrkjum [Zn2+ (aq)], [Zn(OH)2 (aq)] og [Zn(OH)2−
4 (aq)].

6pt

A.2

Mettuð Zn(OH)2 (s) vatnslausn með pH = 7,00 var útbúin og síuð. NaOH varbætt við síuvökvann til að hækka pH hans upp í 12,00.
Reiknaðu mólprósentu sinks sem fellur út þegar pH er hækkað úr 7,00 í 12,00.
Hunsaðu rúmmáls- og hitastigsbreytingar.

5pt

Hluti BNæst var hið endurheimta sinkhýdroxíð hitað og umbreytt í sinkoxíð samkvæmt hvarfinu hér að neðan:Svo var sinkoxíðið afoxað í sinkmálm, með því að hvarfa það við vetni:

B.1

Til þess að hvarf (7) geti gengið við fastan 1 bar þrýsting vetnis, þá er nauðsynlegt að minnka hlutþrýsting vatnsgufunnar sem myndast.
Reiknaðu efri mörk hlutþrýstings vatnsgufu ef hvarf (7) á að geta gengið við 300∘
C. Hér er Gibbs myndunarfríorka sinkoxíðs og vatnsgufu við 300 ∘ C og 1 bar(fyrir öll efni á gasfasa) Δ𝐺ZnO (300∘ C) = −2, 90 × 102 kJ mol−1 og Δ𝐺H2 O (300∘ C) =−2, 20 × 102 kJ mol−1 .

4pt

Sinkmálmur er notaður sem efniviður í neikvætt rafskaut (anóðu) í málm-loft rafhlöðum. Rafskautið
samanstendur af Zn og ZnO. Það notar eftirfarandi oxunar-afoxunarhvarf til að framkalla rafstraum meðíspennu (electromotive force, e.m.f.) 𝐸 ∘ við 25 ∘ C og 1 bar.

B.2

Sink-loft rafhlaða var afhlaðin við 20 mA í 24 klst. Reiknaðu massabreytingu
neikvæða rafskautsins (anóðunnar) í rafhlöðunni.

3pt

IChO
ISL-1 C-4 Q-3

Q4-3
Icelandic (Iceland)

Mt. Fuji
B.3Skoðum nú hvernig breyting íspennu í sink-loft rafhlöðu er háð aðstæðum.
Reiknaðu íspennuna á tindi Mt. Fuji, þar sem hitastig er −38 ∘ C (í febrúar) og
hæð fjallsins er 3776 m. Loftþrýstingur er gefinn með

5ptvið hæð ℎ [m] og hitastig 𝑇 [∘ C]. Mólhlutfall súrefnis í andrúmslofti er 21%. Gibbs
fríorkubreyting hvarfs (8) er Δ𝐺ZnO (−38∘ C) = −3, 26 × 102 kJ mol−1 við −38 ∘ C og
1 bar.
B.4

Reiknaðu Gibbs fríorkubreytingu hvarfs (6) við 25 ∘ C. Staðalafoxunaríspennur,
𝐸 ∘ (Zn2+ /Zn) og 𝐸 ∘ (O2 /H2 O) við 25 ∘ C og 1 bar eru gefnar í jöfnum (10) og (11).

9pt

IChO
ISL-1 C-4 A-1

Redox efnafræði sinks
Hluti A
A.1 (6 pt)< pH <

A4-1
Icelandic (Iceland)

IChO
ISL-1 C-4 A-2

A.2 (5 pt)%

A4-2
Icelandic (Iceland)

IChO
ISL-1 C-4 A-3

Hluti B
B.1 (4 pt)

bar

𝑝H2 O=
B.2 (3 pt)

g

A4-3
Icelandic (Iceland)

IChO
ISL-1 C-4 A-4

B.3 (5 pt)

V

A4-4
Icelandic (Iceland)

IChO
ISL-1 C-4 A-5

B.4 (9 pt)

Δ𝐺∘ =J mol−1

A4-5
Icelandic (Iceland)ISL-1 C-5 C-1

ISL-1 C-5 C
Dagur Benediktsson

IChO
Problem 5
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q5-1

IChO
ISL-1 C-5 Q-1

English (Official)

Mysterious Silicon
12 % of the total
Question

A.1

A.2

A.3

A.4

B.1

B.2

B.3

Total

Points

9

7

6

10

5

15

8

60

Score

Although silicon is also a group 14 element like carbon, their properties differ significantly.

Part A
Unlike the carbon–carbon triple bond, the silicon–silicon triple bond in a compound formulated as
R1 –Si ≡ Si–R1 (R: organic substituent) is extremely reactive. For example, it reacts with ethylene to forma cyclic product that contains a four-membered ring.

When R1 –Si ≡ Si–R1 is treated with an alkyne (R2 –C ≡ C–R2 ), the four-membered-ring compound A is
formed as an initial intermediate. Further reaction of another molecule of R2 –C ≡ C–R2 with A affords
isomers B and C, both of which have benzene-like cyclic conjugated structures, so-called ‘disilabenzenes’
that contain a six-membered ring and can be formulated as (R1 –Si)2 (R2 –C)4 .

IChO
ISL-1 C-5 Q-2

Q5-2
English (Official)

The 13 C NMR analysis of the corresponding six-membered ring skeletons Si2 C4 shows two signals for B
and one signal for C.
A.1

Draw the structural formulae of A, B, and C using R1 , R2 , Si, and C, with one ofthe possible resonance structures.

9pt

A.2

Calculate the aromatic stabilization energy (ASE) for benzene and C (in the caseof R1 = R2 = H) as positive values, considering the enthalpy change in somehydrogenation reactions of unsaturated systems shown below (Fig. 1).

7pt

Fig. 1

Q5-3

IChO
ISL-1 C-5 Q-3English (Official)

When a xylene solution of C is heated, it undergoes isomerization to give an equilibrium mixture ofcompounds D and E. The molar ratio is D : E = 1 : 40.0 at 50.0 ˚C and D : E = 1 : 20.0 at 120.0 ˚C.
Calculate Δ𝐻 for the transformation of D to E. Assume that Δ𝐻 does not depend on temperature.

A.3

6pt

The isomerization from C to D and to E proceeds via transformations of 𝜋-bonds into 𝜎-bonds without
breaking any 𝜎-bonds. A 13 C NMR analysis revealed one signal for the Si2 C4 skeleton of D and two signals
for that of E. The skeleton of D does not contain any three-membered rings, while E has two threemembered rings that share an edge.

Draw the structural formulae of D and E using R1 , R2 , Si, and C.

A.4

10pt

Part B
Silicon is able to form highly coordinated compounds (> four substituents) with electronegative elements
such as fluorine. As metal fluorides are often used as fluorination reagents, highly coordinated silicon
fluorides also act as fluorination reagents.
The fluorination reaction of CCl4 using Na2 SiF6 was carried out as follows.
• Standardization of Na2 SiF6 solution :· Preparation
Aqueous solution F: 0.855 g of Na2 SiF6 (188.053 g mol–1 ) dissolved in water (total volume: 200 mL).
Aqueous solution G: 6.86 g of Ce2 (SO4 )3 (568.424 g mol–1 ) dissolved in water (total volume: 200 mL).
· Procedure
Precipitation titration of a solution F (50.0 mL) by dropwise adding solution G in the presence of xylenol
orange, which coordinates to Ce3+ , as an indicator. After adding 18.8 mL of solution G, the color of thesolution changes from yellow to magenta. The generated precipitate is a binary compound that contains
Ce3+ , and the only resulting silicon compound is Si(OH)4 .
B.1

Write the balanced equation for the reaction of Na2 SiF6 with Ce2 (SO4 )3 .

5pt

• Reaction of CCl4 with Na2 SiF6 :(Substance losses by e.g. evaporation are negligible during the following operations.)
Na2 SiF6 (𝑥 [g]) was added to CCl4 (500.0 g) and heated to 300˚C in a sealed pressure-resistant reaction
vessel. The unreacted Na2 SiF6 and generated NaCl were removed by filtration. The filtrate was diluted
to a total volume of 1.00 L with CCl4 (solution H). The 29 Si and 19 F NMR spectra of solution H showed SiF4
as the only silicon compound. In the 19 F NMR spectrum, in addition to SiF4 , signals corresponding to
CFCl3 , CF2 Cl2 , CF3 Cl, and CF4 were observed (cf. Table 1). The integration ratios in the 19 F NMR spectrum
are proportional to the number of fluorine nuclei.
Table 119

F NMR data

CFCl3

CF2 Cl2

CF3 Cl

CF4Integration ratio

45.0

65.0

18.0

2.0

IChO
ISL-1 C-5 Q-4

Q5-4
English (Official)

SiF4 is hydrolyzed to form H2 SiF6 according to the following eq. 8:

Solution H (10 mL) was added to an excess amount of water, which resulted in the complete hydrolysis of
SiF4 . After separation, the H2 SiF6 generated from the hydrolysis in the aqueous solution was neutralized
and completely converted to Na2 SiF6 (aqueous solution J).
The precipitate of unreacted Na2 SiF6 and NaCl, which was removed by filtration in the initial step (underlined), was completely dissolved in water to give an aqueous solution (solution K; 10.0 L).
Then, additional precipitation titrations using solution G were carried out, and the endpoints of the titrations with G were as follows:·For solution J (entire amount): 61.6 mL.
·For 100 mL of solution K: 44.4 mL.
It should be noted here that the coexistence of NaCl or SiO2 has no effect on the precipitation titration.
B.2

Calculate the mass of the NaCl produced in the reaction vessel (information
underlined), and calculate the mass (𝑥 [g]) of the Na2 SiF6 used as a starting
material.

15pt

B.3

77.8% of the CCl4 used as a starting material was unreacted. Calculate the massof CF3 Cl generated.

8pt

Q5-1

IChO
ISL-1 C-5 Q-1

Icelandic (Iceland)

Hinn dularfulli kísill
12 % of the total
Question

A.1

A.2

A.3

A.4

B.1

B.2

B.3

Total

Points

9

7

6

10

5

15

8

60

Score

Þótt kísill sé frumefni í flokki 14, eins og kolefni, þá er verulegur munur á eiginleikum þessara efna.

Hluti AÓlíkt kolefnis-kolefnis þrítengi, þá er kísil-kísil þrítengi í efnasambandi (táknað með R1 –Si ≡ Si–R1 (R: lífrænn sethópur)), gríðarlega hvarfgjarnt. Til dæmis hvarfast það við etýlen svo það myndast hringlaga
myndefni sem inniheldur fjórhring.

Þegar R1 –Si ≡ Si–R1 er meðhöndlað með alkýn (R2 –C ≡ C–R2 ), myndast fyrst milliefnið A, sem er fjórhringur. A hvarfast áfram við aðra R2 –C ≡ C–R2 sameind og það leiðir til myndunar á ísómerum B og C,
en þær hafa báðar bensen-líka (arómatíska) byggingu. B og C eru svokölluð dísílabensen sem innihalda
sexhring og þau má tákna með (R1 –Si)2 (R2 –C)4 .

IChO
ISL-1 C-5 Q-2

13

Q5-2
Icelandic (Iceland)

C NMR greining á byggingum sexhringjanna Si2 C4 sýnir tvö merki (signals) fyrir B og eitt merki fyrir C.
A.1

Teiknaðu byggingar A, B og C, með því að nota R1 , R2 , Si, og C. Það nægir aðteikna eina vokmynd hvers efnis.

9pt

A.2

Reiknaðu arómatíska stöðgunarorku (aromatic stabilization energy (ASE)) fyrir
bensen og fyrir C (fyrir tilfellið R1 = R2 = H) sem jákvætt gildi. Notaðu vermibreytingu sumra vetnunarhvarfanna á ómettuðum efnum, sem sýnd eru aðneðan á mynd 1.

7pt

Mynd 1

Q5-3

IChO
ISL-1 C-5 Q-3Icelandic (Iceland)Þegar xýlen (xylene) lausn efnis C er hituð, ísómerast efnið og fæst þá jafnvægisblanda af efnasamböndum D og E. Mólhlutfallið er D : E = 1 : 40,0 við 50,0 ˚C og D : E = 1 : 20,0 við 120,0 ˚C.
A.3

Reiknaðu Δ𝐻 fyrir umbreytinguna úr D í E. Gerðu ráð fyrir að Δ𝐻 sé ekki háð
hitastigi.

6ptÍsómerunin úr C í D og E verður vegna umbreytinga 𝜋-tengja yfir í 𝜎-tengi án þess að nokkur 𝜎-tengi
rofni. 13 C NMR greining sýndi eitt merki fyrir Si2 C4 hringinn í D og tvö merki í E. Byggingargrind hrings
D inniheldur enga þríhringi, en byggingargrind hrings E inniheldur tvo þríhringi sem hafa sameiginlegt
tengi.

Teiknaðu byggingu efna D og E með því að nota R1 , R2 , Si, og C.

A.4

10pt

Hluti BÍ efnasamböndum hefur kísill mikla tengigetu með rafneikvæðum (electronegative) frumefnum eins og
flúori (getur tengst fleiri en fjórum atómum). Nú eru málmflúoríð gjarnan notuð sem flúorunar-hvarfefni.
Kísilflúoríð má einnig nota sem flúorunarefni þegar kísillinn tengist mörgum tengihópum.
Na2 SiF6 var notað í flúorunarhvarf CCl4 sem framkvæmt var á eftirfarandi hátt:• Stöðlun Na2 SiF6 lausnar :· Undirbúningar
Vatnslausn F: 0,855 g af Na2 SiF6 (188,053 g mol–1 ) leyst upp í vatni (heildarrúmmál 200 mL).
Vatnslausn G: 6,86 g af Ce2 (SO4 )3 (568,424 g mol–1 ) leyst upp í vatni (heildarrúmmál 200 mL).
· Framkvæmd
Fellitítrun (precipitation titration) á lausn F (50,0 mL) með því að bæta lausn G út í í dropatali, í viðurvist
xylenol orange efnavita, sem bindst Ce3+ . Eftir viðbót 18,8 mL af lausn G, breyttist litur lausnarinnar úr
gulum yfir í fjólubleikan. Botnfallið sem myndaðist er tvíefni (binary compound) sem inniheldur Ce3+ , ogeina kísil-efnasambandið sem myndaðist er Si(OH)4 .
B.1

5pt

Skrifaðu stillta efnajöfnu fyrir hvarf Na2 SiF6 með Ce2 (SO4 )3 .

• Hvarf CCl4 með Na2 SiF6 :(Magn þess efnis sem kann að tapast t.d. vegna uppgufunar er hverfandi í eftirfarandi framkvæmd.)Na2 SiF6 (𝑥 [g]) var bætt út í CCl4 (500,0 g) og hitað að 300˚C í lokuðu þrýstiþolnu hvarfhylki. Óhvarfað
Na2 SiF6 og NaCl sem myndaðist var síað frá. Síuvökvinn var þynntur að heildarrúmmáli 1,00 L með CCl4
(lausn H). 29 Si og 19 F NMR róf af lausn H sýndu að SiF4 var eina kísil-efnasambandið. Í 19 F NMR rófinu sást,að auk SiF4 voru merki sem svara til CFCl3 , CF2 Cl2 , CF3 Cl, og CF4 (Tafla 1). Heildunarhlutföll (integration
ratio) í 19 F NMR rófinu eru í réttu hlutfalli við fjölda flúorkjarna.
Tafla 119

F NMR

Heildunarhlutfall

CFCl3

CF2 Cl2

CF3 Cl

CF4

45,0

65,0

18,0

2,0

IChO
ISL-1 C-5 Q-4

Q5-4
Icelandic (Iceland)

SiF4 er vatnað svo það myndast H2 SiF6 samkvæmt efnajöfnu 8:Þegar lausn H (10 mL) var bætt út í yfirmagn vatns, vatnaðist SiF4 að fullu. Það H2 SiF6 sem myndaðist við
hvarfið var aðskilið frá lausninni. Það var hlutleyst og umbreytt að fullu í Na2 SiF6 (vatnslausn J).
Botnfallið sem innihélt hið óhvarfaða Na2 SiF6 og NaCl, sem var síað frá í upphafsskrefinu (undirstrikað í
texta) var leyst upp að fullu í vatni, og fékkst þá vatnslausn K (10,0 L)Næst voru framkvæmdar fellitítranir með lausn G. Endapunktar þessara títrana voru eftirfarandi:· Fyrir lausn J (öll lausnin) þurfti 61,6 mL af lausn G· Fyrir 100 mL af lausn K þurfti 44,4 mL af lausn G
Vert er að taka fram að NaCl og SiO2 hafa engin áhrif á fellitítrunina.
B.2

Reiknaðu massa NaCl sem myndaðist í hvarfhylkinu (undirstrikað í texta), og
reiknaðu massa (𝑥 [g]) af Na2 SiF6 sem var notað sem upphafsefni.

15pt

B.3

77,8% af því CCl4 sem var notað sem upphafsefni, var óhvarfað. Reiknaðu
massa CF3 Cl sem myndaðist.

8pt

IChO

A5-1

ISL-1 C-5 A-1

Icelandic (Iceland)

Hinn dularfulli kísill
Hluti A
A.1 (9 pt)
A (3 pt)

B (3 pt)

C (3 pt)

A.2 (7 pt)

C6 H6 ∶

kJ mol–1 , C ∶kJ mol–1

IChO
ISL-1 C-5 A-2

A.3 (6 pt)

kJ mol–1

Δ𝐻 =A.4 (10 pt)
D (5 pt)

E (5 pt)

A5-2
Icelandic (Iceland)

IChO
ISL-1 C-5 A-3

Hluti B
B.1 (5 pt)

B.2 (15 pt)(Framhald á næstu síðu)A5-3
Icelandic (Iceland)

IChO
ISL-1 C-5 A-4

B.2 (cont.)

NaCl ∶

g, Na2 SiF6 ∶g

A5-4
Icelandic (Iceland)

IChO
ISL-1 C-5 A-5

B.3 (8 pt)

CF3 Cl ∶g

A5-5
Icelandic (Iceland)ISL-1 C-6 C-1

ISL-1 C-6 C
Dagur Benediktsson

IChO
Problem 6
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q6-1

IChO
ISL-1 C-6 Q-1

English (Official)

The Solid-State Chemistry of Transition Metals
13 % of the total
Question

A.1

A.2

A.3

B.1

B.2

B.3

B.4

C.1

C.2

C.3

Total

Points

6

3

3

6

4

4

4

5

5

5

45

Score

Volcano at Sakurajima island

Part A
Japan is one of the countries with the highest numbers of volcanos worldwide. When silicate minerals
crystallize from magma, a part of the transition-metal ions (Mn+ ) in the magma is incorporated into thesilicate minerals. The Mn+ studied in the problem are coordinated by oxide ions (O2− ) and adopt a fourcoordinate tetrahedral (𝑇d ) geometry in the magma and six-coordinate octahedral (𝑂h ) geometry in thesilicate minerals, both of which exhibit a high-spin electron configuration. The distribution coefficient of
Mn+ between the silicate minerals and magma, 𝐷, can be expressed by:where [M]s and [M]l are the concentrations of Mn+ in the silicate minerals and the magma, respectively.
The table below shows the 𝐷 values of Cr2+ and Mn2+ as examples.

IChO
ISL-1 C-6 Q-2

Q6-2
English (Official)

Let ΔO and CFSEO be the energy separation of the d-orbitals of Mn+ and the crystal-field stabilizationenergy in a 𝑂h field, respectively. Let ΔT and CFSET be those in a 𝑇d field.
A.1

Calculate |CFSEO −CFSET | = ΔCFSE in terms of ΔO for Cr2+ , Mn2+ , and Co2+ ;assume ΔT = 4/9ΔO .

6pt

A.2

A linear relationship is observed by plotting ln𝐷 against ΔCFSE / ΔO in the Cartesian coordinate system shown below.
Estimate 𝐷 for Co2+ .

3pt

Metal oxides MO (M: Ca, Ti, V, Mn, or Co) crystallize in a rock-salt structure wherein the Mn+ adopts an 𝑂hgeometry with a high-spin electron configuration. The lattice enthalpy of these oxides is mainly governedby the Coulomb interactions based on the radius and charge of the ions and some contributions fromthe CFSE of Mn+ in the 𝑂h field.
A.3

Choose the appropriate set of lattice enthalpies [kJ mol−1 ] from one of the options (a) to (f).

3pt

IChO
ISL-1 C-6 Q-3

Q6-3
English (Official)Part B
A mixed oxide A, which contains La3+ and Cu2+ , crystallizes in a tetragonal unit cell shown in Fig.1. In the[CuO6 ] octahedron, the Cu–O length along the z-axis (𝑙𝑧 ) is longer than that of the x-axis (𝑙𝑥 ), and [CuO6 ]is distorted from the regular 𝑂h geometry. This distortion removes the degeneracy of the eg orbitals(d𝑥2 −𝑦2 and d𝑧2 ).

Fig. 1
A can be synthesized by thermal decomposition (pyrolysis) of complex B, which is formed by mixing metal
chlorides in dilute aqueous ammonia solution containing squaric acid C4 H2 O4 , i.e., a diacid. The pyrolysis
behavior of B in dry air shows a weight loss of 29.1% up to 200 °C due to the loss of crystallization water,
followed by another weight loss up to 700 °C due to the release of CO2 . The total weight loss during theformation of A from B is 63.6%. It should be noted that only water and CO2 are released in the pyrolysis
reaction.
B.1

Write the chemical formulae for A and B.

6pt

B.2

Calculate 𝑙𝑥 and 𝑙𝑧 using Fig. 1.

4pt

B.3

For Cu2+ in the distorted [CuO6 ] octahedron in A of Fig. 1, write the names of the
split eg orbitals (d𝑥2 −𝑦2 and d𝑧2 ) in (i) and (ii), and draw the electron configurationin the dotted box in your answer sheet.

4pt

Q6-4

IChO
ISL-1 C-6 Q-4English (Official)

A is an insulator. When one La3+ is substituted with one Sr2+ , one hole is generated in the crystal latticethat can conduct electricity. As a result, the Sr2+ -doped A shows superconductivity below 38 K. When asubstitution reaction took place for A, 2.05 × 1027 holes m−3 were generated.
B.4

Calculate the percentage of Sr2+ substituted for La3+ based on the mole ratio
in the substitution reaction. Note that the valences of the constituent ions andthe crystal structure are not altered by the substitution reaction.

4pt

Part C−2+

Cu2 (CH3 CO2 )4 is composed of four CH3 CO2 coordinated to two Cu (Fig. 2A). Cu2 (CH3 CO2 )4 exhibits−high levels of structural symmetry, with two axes passing through the carbon atoms of the four CH3 CO2
2+
and an axis passing through the two Cu , all of which are oriented orthogonal relative to each other.
−
When a dicarboxylate ligand is used instead of CH3 CO2 , a “cage complex” is formed. The cage com2+
L1
plex Cu4 (L1L1)4 is composed of planar dicarboxylate L1 (Fig. 2B) and Cu (Fig. 2C). The angle θ between
the coordination directions of the two carboxylates, indicated by the arrows in Fig. 2B, determines the
structure of the cage complex. The θ is 0° for L1. Note that hydrogen atoms are not shown in Fig. 2.

Fig. 2IChO
ISL-1 C-6 Q-5

C.1

Q6-5
English (Official)

The θ of the planar dicarboxylate L2 below is fixed to 90˚. If the composition
2+
L2of the cage complex formed from L2 and Cu is Cu𝑛 (L2L2)𝑚 , give the smallest−
integer combination of n and m. Assume that only the CO2 groups of L2 form2+
a coordination bond to Cu ions.

5pt

Q6-6

IChO
ISL-1 C-6 Q-6English (Official)

2+−

A zinc complex, Zn4 O(CH3 CO2 )6 , contains four tetrahedral Zn , six CH3 CO2 , and one O2− (Fig. 3A). In
Zn4 O(CH3 CO2 )6 , the O2− is located at the origin, and the three axes passing through the carbon atoms of−
CH3 CO2 are oriented orthogonal relative to each other. When p-benzenedicarboxylate (Fig. 3B, L3, θ =−2+180°) is used instead of CH3 CO2 , the Zn clusters are linked to each other to form a crystalline solid (X)L3that is called a “porous coordination polymer” (Fig. 3C). The composition of X is [Zn4 O(L3L3)3 ]𝑛 , and it hasa cubic crystal structure with nano-sized pores. One pore is represented as a sphere in Fig. 3D, and each2+
tetrahedral Zn cluster is represented as a dark gray polyhedron in Fig. 3C and 3D. Note that hydrogenatoms are not shown in Fig. 3.

Fig. 3
C.2

X has a cubic unit cell with a side length of 𝑎 (Fig. 3C) and a density of 0.592
g cm−3 . Calculate 𝑎 in [cm].

5pt

C.3

X contains a considerable number of pores, and 1 g of X can accommodate
3.0 × 102 mL of CO2 gas in the pores at 1 bar and 25 °C. Calculate the average
number of CO2 molecules per pore.

5pt

Q6-1

IChO
ISL-1 C-6 Q-1

Icelandic (Iceland)

Efnafræði fastra efnasambanda hliðarmálma
13 % of the total
Question

A.1

A.2

A.3

B.1

B.2

B.3

B.4

C.1

C.2

C.3

Total

Points

6

3

3

6

4

4

4

5

5

5

45

Score

Eldfjall á Sakurajima eyju

Part A
Japan er eitt þeirra landa í heiminum sem hefur flest eldfjöll. Þegar silíkatasteindir kristallast frá
hraunkviku þá fer hluti af hliðarmálmjónum (Mn+ ) úr hraunkvikunni í silikatsteindirnar. Í þessu dæmieru Mn+ jónir girtar (coordinated) oxíðjónum (O2− ) með fjórtengdri tetrahedral (𝑇d ) byggingu í hraunkvikunni og sextengdri oktahedral (𝑂h ) byggingu í silíkatsteindum. Í bæði 𝑇d og 𝑂h byggingum hafa Mn+
jónirnar háspuna rafeindaskipan. Dreifingarstuðullinn fyrir Mn+ milli silíkatsteinda og kvikubráðar, 𝐷,
er gefinn með eftirfarandi jöfnu:þar sem [M]s styrkur Mn+ í silíkatsteindum og [M]l er styrkur Mn+ í hraunkvikunni. Í töflunni hér að neðan
eru gefin dæmi um gildi D fyrir Cr2+ og Mn2+ .

IChO
ISL-1 C-6 Q-2

Q6-2
Icelandic (Iceland)

Látum ΔO vera orkumun d-svigrúma Mn+ og CFSEO vera kristalsviðsorkuna (the crystal-field stabilizationenergy) í 𝑂h sviði. Látum ΔT og CFSET vera samsvarandi stærðir fyrir 𝑇d svið.
A.1

A.2

Reiknaðu |CFSEO −CFSET | = ΔCFSE miðað við ΔO fyrir Cr2+ , Mn2+ og Co2+ ;gerðu ráð fyrir að ΔT = 4/9ΔO .

Línulegt samband kemur fram á grafi þegar ln𝐷 er teiknað á móti ΔCFSE / ΔO í
hnitakerfinu hér fyrir neðan.
Reiknaðu gildi 𝐷 fyrir Co2+ .

6pt

3pt

Málmoxíð MO (M: Ca, Ti, V, Mn eða Co) hafa samskonar kristalbyggingu og matarsalt þar sem Mn+ er í 𝑂humhverfi og með háspuna rafeindaskipan. Grindarorka þessara oxíða er aðalega vegna Coulombkrafta
á milli jóna, en þeir ráðast af radíus og hleðslu jónanna, auk dálítils framlags frá CFSE fyrir Mn+ í 𝑂h sviði.
A.3

Veldu viðeigandi röð (a) til (f) fyrir grindarorku [kJ mol−1 ] málmoxíðanna.

3pt

IChO
ISL-1 C-6 Q-3

Q6-3
Icelandic (Iceland)Part B
Blandað oxíð A, sem inniheldur La3+ og Cu2+ , kirstallast með tetragonal grindareiningu eins og sýnt er á
mynd 1. Í [CuO6 ] oktahedroninum er Cu–O lengdin eftir z-ásnum (𝑙𝑧 ) lengri en sú sem er á x-ásnum (𝑙𝑥 ).
[CuO6 ] er því bjagað frá reglulegri 𝑂h byggingu og hverfur þá margfeldni (degeneracy) eg svigrúmanna(d𝑥2 −𝑦2 og d𝑧2 ).

Mynd 1A er búið til með því að hitarjúfa (pyrolysis) komplex B. B er búið til með því að blanda saman málmklóríðum í þynntri ammoníks vantslausn sem inniheldur squariksýru C4 H2 O4 , en hún er tvísýra. Þegar
B er hitað upp í 200 °C í þurru lofti tapar það 29,1% af massa sínum vegna uppgufunar kristalvatns(crystallization water), og ef það er hitað áfram upp í 700 °C þá rofnar CO2 frá efninu. Heildarmassatapið
við myndun A frá B er 63,6%. Bent skal á að aðeins vatn og CO2 rjúka burt við hitarofshvarfið.
B.1

B.2

B.3

Ritaðu efnaformúlur A og B.

Reiknaðu 𝑙𝑥 og 𝑙𝑧 frá mynd 1.
Ritaðu nöfn klofnu eg svigrúmanna (d𝑥2 −𝑦2 og d𝑧2 ) í (i) og (ii) fyrir Cu2+ bjagaða
[CuO6 ] oktahedroninn í A á mynd 1 og teiknaðu rafeindaskipanina í punktsrikuðu reitina á svarblaðinu þínu.

6pt

4pt

4pt

Q6-4

IChO
ISL-1 C-6 Q-4Icelandic (Iceland)

A er rafeinangrandi efni. Þegar einu La3+ er skipt út fyrir eitt Sr2+ , þá myndast ein hola í kristagrindinni
sem getur leitt rafmagn. A sem íbætt (doped) er með Sr2+ verður þá ofurleiðandi fyrir neðan 38 K. Þegar
slíkt skiptihvarf var framkvæmt fyrir A þá mynduðust 2, 05 × 1027 holur á m3 .
B.4

Reiknaðu prósentuna af Sr2+ sem skipt var út fyrir La3+ frá mólhlutfallinu ískiptihvarfinu. Athugaðu að hleðsla jónanna og kristalbyggingin breytast ekkivið skiptihvarfið.

4pt

Hluti C−2+

Cu2 (CH3 CO2 )4 er samsett úr CH3 CO2 sem tengjast tveimur Cu (mynd 2A). Bygging Cu2 (CH3 CO2 )4 hef−
ur mikla samhverfu (symmetry) þar sem tveir ásar fara í gegnum kolefnisatóm fjögurra CH3 CO2 auk
2+áss sem fer í gegnum tvö Cu , en ásarnir þrír eru hornréttir hver á annan. Þegar díkarboxýlat tengill−
L1(ligand) er notaður í stað CH3 CO2 , myndast ”búrkomplex” ( “cage complex”) . Búrkomplexinn Cu4 (L1L1)42+
er samsettur úr flötu díkarboxýlati L1 (mynd 2B) og Cu (mynd 2C). Bygging búrkomplexins ákvaðast af
horninu θ milli girðistefnu (coordination directions) karboxýlathópanna tveggja, en θ-hornið sem er sýntmeð örvunum á mynd 2B. Gildi θ er 0° fyrir L1. Athugaðu að vetnisatómin eru ekki sýnd á mynd 2.

Mynd 2IChO
ISL-1 C-6 Q-5

C.1

Q6-5
Icelandic (Iceland)

Hornið θ fyrir flata díkarboxýlatið L2 hér að neðan helst fast við 90˚. Ákvaðaðu
minnstu gildin fyrir n og m ef samsetning búrkomplexins sem myndast frá L2 og
−
2+
2+
L2
Cu er Cu𝑛 (L2L2)𝑚 . Gerðu ráð fyrir að aðeins CO2 hópar L2 tengist Cu jónum.

5pt

IChO
ISL-1 C-6 Q-6

Q6-6
Icelandic (Iceland)

2+

Zinkkomplex, Zn4 O(CH3 CO2 )6 , inniheldur fjögur Zn sem mynda tetrahedral-laga klasa, ásamt sex−
CH3 CO2 og einu O2− (mynd 3A). Í Zn4 O(CH3 CO2 )6 er O2− í miðjunni og þrír ásar fara í gegnum kolefn−
isatóm CH3 CO2 og eru þeir hornréttir hver á annan. Þegar p-benzenedíkarboxýlat (mynd 3B, L3, θ =−2+180°) er notað í stað CH3 CO2 , tengjast Zn klasarnir sama í kristallað fast efni (X) sem kallast ”gljúp
L3girðifjölliða” ( “porous coordination polymer”, mynd 3C). Samsetning X er [Zn4 O(L3L3)3 ]𝑛 , og hefur efnið
teningslaga kristalbyggingu með holur af nanóstærð. Ein slík nanóhola er sýnd sem kúla á mynd 3D og
2+er hver tetrahedral-laga Zn klasi táknaður sem dökkgrár fjölflötungur á myndum 3C og 3D. Athugaðu
að vetnisatómin eru ekki sýnd á mynd 3.

Mynd 3
C.2

C.3

X hefur teningslaga grindareiningu með kantlengdina a (mynd 3C) og eðlismassann 0,592 g cm−3 . Reiknaðu gildi a í [cm].

X inniheldur aragrúa fjölda nanóhola og 1 g af X getur hýst 3, 0 × 102 mL af CO2
sem er við 1 bar og 25 °C. Reiknaðu meðalfjölda CO2 sameinda sem eru í hverrinanóholu.

5pt

5pt

IChO
ISL-1 C-6 A-1

A6-1
Icelandic (Iceland)

Efnafræði fastra efnasambanda hliðarmálma
Hluti A
A.1 (6 pt)

Cr2+ ∶ΔO , Mn2+ ∶ΔO , Co2+ ∶

ΔO

IChO
ISL-1 C-6 A-2

A.2 (3 pt)

𝐷∶

A.3 (3 pt)

A6-2
Icelandic (Iceland)

IChO

A6-3

ISL-1 C-6 A-3

Icelandic (Iceland)Part B
B.1 (6 pt)

A∶

,B∶

B.2 (4 pt)

𝑙𝑥 =nm, 𝑙𝑧 =nm

IChO
ISL-1 C-6 A-4

B.3 (4 pt), (ii) ∶(i) ∶B.4 (4 pt)%

A6-4
Icelandic (Iceland)

IChO
ISL-1 C-6 A-5

Hluti C
C.1 (5 pt)𝑛=,𝑚=

C.2 (5 pt)

𝑎=

cm

A6-5
Icelandic (Iceland)

IChO
ISL-1 C-6 A-6

C.3 (5 pt)

A6-6
Icelandic (Iceland)ISL-1 C-7 C-1

ISL-1 C-7 C
Dagur Benediktsson

IChO
Problem 7
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q7-1

IChO
ISL-1 C-7 Q-1

English (Official)

Playing with Non-benzenoid Aromaticity
13 % of the total
Question

A.1

A.2

A.3

B.1

Total

Points

5

2

19

10

36

Score

Prof. Nozoe (1902–1996) opened the research field of non-benzenoid aromatic compounds, which are
now ubiquitous in organic chemistry.

Photo courtesy: Tohoku Univ.

Part A
Lineariifolianone is a natural product with a unique structure, which was isolated from Inula linariifolia.
From valencene (1), a one-step conversion yields 2, before a three-step conversion via 3 yields ketone 4.
Eremophilene (5) is converted into 6 by performing the same four-step conversion.

IChO
ISL-1 C-7 Q-2

Q7-2
English (Official)

Inula linariifolia

A.1

Draw the structures of 2 and 6 and clearly identify the stereochemistry where
necessary.

5pt

Then, ketone 4 is converted into ester 15. Compound 8 (molecular weight: 188) retains all the stereocenters in 7. Compounds 9 and 10 have five stereocenters and no carbon-carbon double bonds. Assume

IChO
ISL-1 C-7 Q-3

Q7-3
English (Official)that H2 18 O is used instead of H2 16 O for the synthesis of 18 O-labelled-lineariifolianones 13 and 14 from 11
and 12, respectively. Compounds 13 and 14 are 18 O-labelled isotopomers. Ignoring isotopic labelling,both 13 and 14 provide the same product 15 with identical stereochemistry.

IChO
ISL-1 C-7 Q-4

Q7-4
English (Official)

A.2

Choose the appropriate structure for A.

2pt

A.3

Draw the structures of 8–14 and clearly identify the stereochemistry where necessary. Also, indicate the introduced 18 O atoms for 13 and 14 as shown in theexample below.

19pt

IChO
ISL-1 C-7 Q-5

Q7-5
English (Official)Part B
Compound 19 is synthesized as shown below. In relation to non-benzenoid aromaticity, 19 can be used
as an activator for alcohols, and 20 was converted to 22 via ion-pair intermediate 21. Although theformation of 21 was observed by NMR, 21 gradually decomposes to give 18 and 22.

B.1

Draw the structures of 17–19 and 21. Identifying the stereochemistry is notnecessary.

10pt

Q7-1

IChO
ISL-1 C-7 Q-1

Icelandic (Iceland)

Playing with Non-benzenoid Aromaticity
13 % of the total
Question

A.1

A.2

A.3

B.1

Total

Points

5

2

19

10

36

Score

Prof. Nozoe (1902–1996) founded the research field of non-benzenoid aromatic compounds.

Photo credit: Tohoku Univ.

Part A
Lineariifolianone is a natural product which was isolated from Inula linariifolia. From valencene (1), a
one-step conversion yields 2, before a three-step conversion via 3 yields ketone 4. Eremophilene (5) isconverted into 6 by performing the same four-step conversion.

IChO
ISL-1 C-7 Q-2

Q7-2
Icelandic (Iceland)

Inula linariifolia

A.1

Draw the structures of 2 and 6 and clearly identify the stereochemistry where
necessary.

5pt

IChO
ISL-1 C-7 Q-3

Q7-3
Icelandic (Iceland)

Then, ketone 4 is converted into ester 15. Compound 8 (molecular weight: 188) retains all thestereocentres in 7. Compounds 9 and 10 have five stereocentres and no carbon-carbon double bonds.
Assume that H2 18 O is used instead of H2 16 O for the synthesis of 18 O-labelled-lineariifolianones 13 and 14
from 11 and 12, respectively.
Compounds 13 and 14 are 18 O-labelled isotopomers. Both 13 and 14 provide the same product 15 withidentical stereochemistry ignoring isotopic labelling.

IChO
ISL-1 C-7 Q-4

Q7-4
Icelandic (Iceland)

IChO
ISL-1 C-7 Q-5

Q7-5
Icelandic (Iceland)

A.2

Choose the appropriate structure for A.

2pt

A.3

Draw the structures of 8–14 and clearly identify the stereochemistry where
necessary. Also, indicate the introduced 18 O atoms for 13 and 14 as shownin the example below.

19pt

IChO
ISL-1 C-7 Q-6

Q7-6
Icelandic (Iceland)Part B
Compound 19 is synthesised as shown below. Linking to non-benzenoid aromaticity, 19 can be used as anactivator for alcohols, and 20 was converted to 22 via ion-pair intermediate 21. Although the formationof 21 was observed by NMR, 21 gradually decomposes to give 18 and 22.

B.1

Draw the structures of 17–19 and 21. Stereochemistry is not required.

10pt

IChO
ISL-1 C-7 A-1

Playing with Non-benzenoid Aromaticity
Part A
A.1 (5 pt)
2 (2 pt)

A.2 (2 pt)

6 (3 pt)

A7-1
Icelandic (Iceland)

IChO
ISL-1 C-7 A-2

A.3 (19 pt)

A7-2
Icelandic (Iceland)

IChO
ISL-1 C-7 A-3

Part B
B.1 (10 pt)
17 (2 pt)

18 (2 pt)

19 (3 pt)

21 (3 pt)

A7-3
Icelandic (Iceland)ISL-1 C-8 C-1

ISL-1 C-8 C
Dagur Benediktsson

IChO
Problem 8
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q8-1

IChO
ISL-1 C-8 Q-1

English (Official)

Dynamic Organic Molecules and Their Chirality
11 % of the total
Question

A.1

A.2

A.3

B.1

B.2

Total

Points

9

3

7

3

4

26

Score

Part A
Polycyclic aromatic hydrocarbons with successive ortho-connections are called [n]carbohelicenes (here,n represents the number of six-membered rings) (see below). [4]Carbohelicene ([4]C) is efficiently prepared by a route using a photoreaction as shown below, via an intermediate (Int.) that is readily oxidized
by iodine.

The photoreaction proceeds in a manner similar to the following example.

IChO
ISL-1 C-8 Q-2

Q8-2
English (Official)

Note: For all of Question 8, please draw alternating single and double bonds in your answers to
the problems as depicted in the examples of carbohelicene. Do not use circles for conjugated 𝜋systems.
A.1

Draw the structures of A–C. Stereoisomers should be distinguished.

9pt

A.2

Attempts to synthesize [5]carbohelicene from the same phosphonium salt andan appropriate starting compound resulted in the formation of only a trace
amount of [5]carbohelicene, instead affording product D whose molecularweight was 2 Da lower than that of [5]carbohelicene. The 1 H NMR chemical
shifts of D are listed below. Draw the structure of D.
[D (𝛿, ppm in CS2 , r.t.), 8.85 (2H), 8.23 (2H), 8.07 (2H), 8.01 (2H), 7.97 (2H), 7.91
(2H)]

3pt

[5]- and larger [n]carbohelicenes have helical chirality and interconversion between enantiomers of thesehelicenes is significantly slow at room temperature. The chirality of [n]carbohelicenes is defined as (M)or (P) as shown below.

[n]Carbohelicenes with n larger than 4 can be enantiomerically separated by a chiral column chromatography, which was developed by Prof. Yoshio Okamoto.

Photo courtesy: The Japan Prize Foundation

IChO
ISL-1 C-8 Q-3

Q8-3
English (Official)Multiple helicenes are molecules that contain two or more helicene-like structures. If its helical chirality
is considered, several stereoisomers exist in a multiple helicene. For example, compound E contains
three [5]carbohelicene-like moieties in one molecule. One of the stereoisomers is described as (P, P, P)
as shown below.

A.3

The nickel-mediated trimerization of 1,2-dibromobenzene generates triphenylene. When the same reaction is applied to an enantiomer of F, (P)-F, multiple
helicene G (C66 H36 ) is obtained. Given that interconversion between stereoisomers does not occur during the reaction, identify all the possible stereoisomers of G formed in this process, without duplication. As a reference, one isomer should be drawn completely with the chirality defined as in the example
above, with numerical labels; the other stereoisomers should be listed with location numbers and M and P labels according to the same numbering. For instance, the other stereoisomers of E should be listed as (1, 2, 3) = (P, M, P), (P,M, M), (P, P, M), (M, M, M), (M, M, P), (M, P, P), and (M, P, M).

7pt

IChO
ISL-1 C-8 Q-4

Q8-4
English (Official)Part B
Sumanene is a bowl-shaped hydrocarbon that was first reported in Japan in 2003. The name "sumanene"derives from a Sanskrit-Hindi word "suman" that means sunflower. The synthesis of sumanene wasachieved by a reaction sequence that consists of a ring-opening and a ring-closing metathesis.

Representative metathesis reactions catalyzed by a ruthenium catalyst (Ru*) are shown below.

B.1

Draw the structure of intermediate I (its stereochemistry is not required).

3pt

IChO
ISL-1 C-8 Q-5

B.2

Q8-5
English (Official)

Starting from the optically active precursor J, the same reaction sequence gives
the optically active sumanene derivative K. The stereocenters in J suffer no inversion during the metathesis reaction. Draw the structure of K with the appropriate stereochemistry.

4pt

Q8-1

IChO
ISL-1 C-8 Q-1

Icelandic (Iceland)

Dynamic Organic Molecules and Their Chirality
11 % of the total
Question

A.1

A.2

A.3

B.1

B.2

Total

Points

9

3

7

3

4

26

Score

Part A
Polycyclic aromatic hydrocarbons with successive ortho-connections are called [n]carbohelicenes (here,n represents the number of six-membered rings) (see below).
[4]Carbohelicene ([4]C) is efficiently prepared by a route using a photoreaction as shown below, via anintermediate (Int.) that is readily oxidized by iodine.

The photoreaction proceeds in a manner similar to the following example.

IChO
ISL-1 C-8 Q-2

Q8-2
Icelandic (Iceland)Note: For all of Question 8, please draw alternating single and double bonds in your answers to
the problems as depicted in the examples of carbohelicene. Do not use circles for conjugated 𝜋systems.
A.1

Draw the structures of A–C. Stereoisomers should be distinguished.

9pt

A.2

Attempts to synthesize [5]carbohelicene from the same phosphonium salt andan appropriate starting compound resulted in the formation of only a trace
amount of [5]carbohelicene, instead affording product D whose molecularweight was 2 Da lower than that of [5]carbohelicene.

3pt

Draw the structure of D.
The 1 H NMR chemical shifts of D are listed below.
[D (𝛿, ppm in CS2 , r.t.), 8.85 (2H), 8.23 (2H), 8.07 (2H), 8.01 (2H), 7.97 (2H), 7.91
(2H)]

IChO
ISL-1 C-8 Q-3

Q8-3
Icelandic (Iceland)[5]- and larger [n]carbohelicenes have helical chirality and interconversion between enantiomers of thesehelicenes is very slow at room temperature. The chirality of [n]carbohelicenes is defined as (M) or (P) as
shown below.

[n]Carbohelicenes with n larger than 4 can be enantiomerically separated by chiral column chromatography, which was developed by Prof. Yoshio Okamoto.

Photo credit: The Japan Prize Foundation

IChO
ISL-1 C-8 Q-4

Q8-4
Icelandic (Iceland)

A molecule that contains two or more helicene-like structures is called a multiple-helicene. If helicalchirality is considered there can be several stereoisomers of a multiple-helicene.
For example, compound E contains three [5]carbohelicene-like moieties in one molecule. One of thestereoisomers is described as (P, P, P) as shown below.

IChO
ISL-1 C-8 Q-5

A.3

Q8-5
Icelandic (Iceland)

The nickel-mediated trimerisation of 1,2-dibromobenzene generates triphenylene. When the same reaction is applied to an enantiomer of F, (P)-F, multiplehelicene G (C66 H36 ) is obtained.
Given that interconversion between stereoisomers does not occur during thereaction, identify all the possible stereoisomers of G formed in this process,without duplicating stereoisomers.
As a reference, one isomer should be drawn completely with the chirality
defined as in the example above, with numerical labels.
The other stereoisomers should be listed with location numbers and M and P
labels according to the same numbering.
For instance, the other stereoisomers of E should be listed as (1, 2, 3) = (P, M, P),(P, M, M), (P, P, M), (M, M, M), (M, M, P), (M, P, P), and (M, P, M).

7pt

IChO
ISL-1 C-8 Q-6

Q8-6
Icelandic (Iceland)Part B
Sumanene is a bowl-shaped hydrocarbon that was first reported in Japan in 2003. The name ”sumanene”
derives from a Sanskrit-Hindi word ”suman” that means sunflower. The synthesis of sumanene wasachieved by a reaction sequence that consists of a ring-opening and a ring-closing metathesis.

Representative metathesis reactions catalysed by a ruthenium catalyst (Ru*) are shown below.

B.1

Draw the structure of intermediate I (its stereochemistry is not required).

3pt

IChO
ISL-1 C-8 Q-7

B.2

Q8-7
Icelandic (Iceland)Starting from the optically active precursor J, the same reaction sequence gives the optically active sumanene derivative K. The stereocentres in J suffer no
inversion during the metathesis reaction. Draw the structure of K with the appropriate stereochemistry.

4pt

IChO

A8-1

ISL-1 C-8 A-1

Icelandic (Iceland)

Dynamic Organic Molecules and Their Chirality
Part A
A.1 (9 pt)
A (3 pt)

A.2 (3 pt)

B (3 pt)

C (3 pt)

IChO
ISL-1 C-8 A-2

A.3 (7 pt)

A8-2
Icelandic (Iceland)

IChO
ISL-1 C-8 A-3

Part B
B.1 (3 pt)

B.2 (4 pt)

A8-3
Icelandic (Iceland)ISL-1 C-9 C-1

ISL-1 C-9 C
Dagur Benediktsson

IChO
Problem 9
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q9-1

IChO
ISL-1 C-9 Q-1

English (Official)

Likes and Dislikes of Capsule
10 % of the total
Question

A.1

A.2

A.3

A.4

A.5

Total

Points

13

2

2

3

3

23

Score

Good kids don’t do this, but if you unseam a tennis ball, you can disassemble it into two U-shaped pieces.

Based on this idea, compounds 1 and 2 were synthesized as U-shaped molecules with different sizes.
Compound 3 was prepared as a comparison of 1 and the encapsulation behavior of these compounds
was investigated.

IChO
ISL-1 C-9 Q-2

Q9-2
English (Official)

The synthetic route to 2 is shown below. The elemental composition of compound 9: C; 40.49%, H; 1.70%,
and O; 17.98% by mass.

IChO
ISL-1 C-9 Q-3

A.1

Q9-3
English (Official)

Draw the structures of 4–9; the stereochemistry can be neglected. Use "PMB"
as a substituent instead of drawing the whole structure of p-methoxybenzyl
group shown in the scheme above.

13pt

In the mass spectrum of 1, the ion peak corresponding to its dimer (112 ) was clearly observed, whereas
an ion peak for 3 2 was not observed in the spectrum of 3. In the 1 H NMR spectra of a solution of 1 2 ,all the NH protons derived from 1 were observed to be chemically equivalent, and their chemical shift
was significantly different from that of the NH protons of 3. These data indicate that hydrogen bonds
are formed between the NH moieties of 1 and atoms X of another molecule of 1 to form the dimericcapsule.
A.2

Circle all the appropriate atom(s) X in 1.

2pt

A.3

Give the number of the hydrogen bonds in the dimeric capsule (112 ).

2pt

IChO
ISL-1 C-9 Q-4

Q9-4
English (Official)

The dimeric capsule of 1 (112 ) has an internal space wherein an appropriate small molecule Z can be
encapsulated. This phenomenon is expressed by the following equation:

The equilibrium constant of the encapsulation of Z into 12 is given as below:

Encapsulation of a molecule into a capsule could be monitored by NMR spectroscopy. For example, 1 2
in C6 D6 gave different signals in the 1 H NMR spectra before and after addition of CH4 .
Compound 2 also forms a rigid and larger dimeric capsule (222 ). The 1 H NMR spectrum of 22 was measured
in C6 D6 , C6 D5 F, and a C6 D6 /C6 D5 F solvent mixture, with all other conditions being kept constant. The
chemical shifts for the Ha proton of 2 in the above solvents are summarized below, and no other signals
from the Ha in 2, except for the listed, were observed. Assume that the interior of the capsule is always
filled with the largest possible number of solvent molecules and that each signal corresponds to one
species of the filled capsule.

A.4

solvent

𝛿 (ppm) of Ha

C6 D6

4.60

C6 D5 F

4.71

C6 D6 / C6 D5 F

4.60, 4.71, 4.82

Determine the number of C6 D6 and C6 D5 F molecules encapsulated in 2 2 giving
each Ha signal.

3pt

Q9-5

IChO
ISL-1 C-9 Q-5English (Official)

H NMR measurements in C6 D6 revealed that 2 2 can incorporate one molecule of 1-adamantanecarboxylic
acid (AdA), and the association constants (𝐾a ) which are expressed below were determined for various
temperatures. [solvent@222 ] denotes a species containing one or more solvent molecules.

1

Similarly, the 𝐾a values of CH4 and 1 2 given as eq (2) at various temperatures in C6 D6 were also determined by 1 H NMR measurements. The plots of the two association constants (as ln 𝐾a vs 1/T) are shown
below.

No C6 D6 molecule is encapsulated in 1 2 . In line II, the entropy change (Δ𝑆) is ( 1 ) and enthalpy change(Δ𝐻) is ( 2 ), indicating that the driving force for the encapsulation in line II is ( 3 ). Therefore, line I
corresponds to ( 4 ), and line II corresponds to ( 5 ).
A.5

Choose the correct options in gaps (1)–(5) in the following paragraph from A
and B.
AB(1)

positive

negative(2)

positive

negative

(3)

Δ𝑆

Δ𝐻

(4)

1 2 and CH4

2 2 and AdA(5)

1 2 and CH4

2 2 and AdA

3pt

Q9-1

IChO
ISL-1 C-9 Q-1

Icelandic (Iceland)

Likes and Dislikes of Capsule
10 % of the total
Question

A.1

A.2

A.3

A.4

A.5

Total

Points

13

2

2

3

3

23

Score

If you cut a tennis ball along the seam, you can disassemble it into two U-shaped pieces.

Compounds 1 and 2 are U-shaped molecules with different sizes, inspired by this idea. Compound 3 was
prepared for comparison with 1 and the encapsulation behaviour of these compounds was investigated.

IChO
ISL-1 C-9 Q-2

Q9-2
Icelandic (Iceland)

The synthetic route to 2 is shown below. Elemental composition of compound 9: C 40.49%, H 1.70%, andO 17.98% by mass.

IChO
ISL-1 C-9 Q-3

A.1

Q9-3
Icelandic (Iceland)

Draw the structures of 4–9; stereochemistry can be neglected. Use the ”PMB”
abbreviation instead of drawing the whole structure of p-methoxybenzyl groupshown in the scheme above.

13pt

In the mass spectrum of 1, the ion peak corresponding to its dimer (112 ) was clearly observed, whereas
an ion peak for 32 was not observed in the spectrum of 3.
In the 1 H NMR spectrum of a solution of 1 2 , all the NH protons derived from 1 were observed to bechemically equivalent, and their chemical shift was significantly different from that of the NH protons of
3. These data indicate that hydrogen bonds are formed between the NH moieties of 1 and atoms X ofanother molecule of 1 to form the dimeric capsule.
A.2

Circle all the appropriate atom(s) X in 1.

2pt

A.3

Give the number of the hydrogen bonds in the dimeric capsule (112 ).

2pt

IChO
ISL-1 C-9 Q-4

Q9-4
Icelandic (Iceland)

The dimeric capsule of 1 (112 ) has an internal space wherein an appropriate small molecule Z can be
encapsulated. This phenomenon is expressed by the following equation:

The equilibrium constant of the encapsulation of Z into 12 is shown below:

Encapsulation of a molecule into a capsule can be monitored by NMR spectroscopy. For example, 1 2 inC6 D6 gave different signals in the 1 H NMR spectra before and after addition of CH4 .
Compound 2 also forms a larger, rigid dimeric capsule (222 ). The 1 H NMR spectrum of 22 was measured
in C6 D6 , C6 D5 F, and a C6 D6 /C6 D5 F solvent mixture, with all other conditions being kept constant. The
chemical shifts for the Ha proton of 2 in the above solvents are summarized below, and no other signals
from Ha in 2, except for those listed, were observed. Assume that the interior of the capsule is always
filled with the largest possible number of solvent molecules and that each signal corresponds to one
species of filled capsule.

A.4

solvent

𝛿 (ppm) of Ha

C6 D6

4.60

C6 D5 F

4.71

C6 D6 / C6 D5 F

4.60, 4.71, 4.82

Determine the number of C6 D6 and C6 D5 F molecules encapsulated in 2 2 giving
each Ha signal.

3pt

Q9-5

IChO
ISL-1 C-9 Q-5

Icelandic (Iceland)

H NMR measurements in C6 D6 revealed that 2 2 can incorporate one molecule of 1-adamantanecarboxylic
acid (AdA).

1

The association constants (𝐾a ) which are expressed below were determined for various temperatures.
[solvent@222 ] denotes a species containing one or more solvent molecules.

Similarly, the 𝐾a values of CH4 and 1 2 shown in eq (2) were also determined by 1 H NMR measurements
at various temperatures in C6 D6 . The plots of the two association constants (as ln 𝐾a vs 1/T) are shown
below.

No C6 D6 is encapsulated in 1 2 .
A.5

Choose the correct options in gaps (1)–(5) in the following paragraph from A
and B.
In line II, the entropy change (Δ𝑆) is ( 1 ) and enthalpy change (Δ𝐻) is ( 2 ),indicating that the driving force for the encapsulation in line II is ( 3 ). Therefore,line I corresponds to ( 4 ), and line II corresponds to ( 5 ).
A

B

(1)

positive

negative(2)

positive

negative

(3)

Δ𝑆

Δ𝐻

(4)

1 2 and CH4

2 2 and AdA(5)

1 2 and CH4

2 2 and AdA

3pt

IChO
ISL-1 C-9 A-1

Likes and Dislikes of Capsule
A.1 (13 pt)
4 (2 pt)

5 (3 pt)

6 (2 pt)

7 (2 pt)

8 (2 pt)

9 (2 pt)

A9-1
Icelandic (Iceland)

IChO
ISL-1 C-9 A-2

A9-2
Icelandic (Iceland)

A.2 (2 pt)

A.3 (2 pt)
A.4 (3 pt)
𝛿 (ppm) of Ha

numbers of C6 D6numbers of C6 D5 F

4.60 ppm

4.71 ppm

4.82 ppm

A.5 (3 pt)
(1) ∶

(2) ∶

(4) ∶

(5) ∶

(3) ∶