SAU-1 C-0 C-1

SAU-1 C-0 C
Mohammad AlHudaithi

IChO
General instructions
Cover sheet

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

IChO
SAU-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

SAU-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
SAU-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
SAU-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

SAU-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
SAU-1 C-0 G-6

Periodic Table

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

1

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

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

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

G0-1

Arabic (SAU) (Saudi Arabia)

IChO

G0-2

SAU-1 C-0 G-2

Arabic (SAU) (Saudi Arabia)

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 for •correct 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 the •nullification 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
Percentage

Total Score

11

24

Hydrogen at a Metal Surface

8

22

Lambert–Beer Law?

60

Mysterious Silicon

36

Playing with Non-benzenoid Aromaticity

23

Likes and Dislikes of Capsules

11
11
12
13
13
11
10

100

Title
1

35

Isotope Time Capsule2

32

The Redox Chemistry of Zinc

45

The Solid-State Chemistry of Transition Metals

26

Dynamic Organic Molecules and Their Chirality

3
4
5
6
7
8
9

Total

IChO
SAU-1 C-0 G-3

G0-3

Arabic (SAU) (Saudi Arabia)

Physical Constants and Equations
Constants𝑐 = 2.99792458 × 108 m s−1ℎ = 6.62607015 × 10

−34

Js

Speed of light in vacuum
Planck constantC

Elementary charge−1Electric constant(permittivity of vacuum)

𝑁A = 6.02214076 × 1023 mol−1

Avogadro constant𝑒 = 1.602176634 × 10𝑚e = 9.10938370 × 10
𝜀0 = 8.85418781 × 10

−12

𝑘B = 1.380649 × 10

−19

−31

kg

Fm

−23

JK

𝐹 = 𝑁A × 𝑒 = 9.64853321233100184 × 10 C mol
4

𝑅 = 𝑁A × 𝑘B = 8.31446261815324 J K= 8.2057366081 × 10

−2

L atm K

−1
−1

−1

Boltzmann constant−1

Faraday constant

mol

−1

mol−1

𝑢 = 1 Da = 1.66053907 × 10−27 kg

𝑝 = 1 bar = 10 Pa
5𝑝atm = 1.01325 × 10 Pa
5

0 C = 273.15 K∘

1 Å = 101 pm = 10−10−12

1 eV = 1.602176634 × 10

m
m

−19

1 ppm = 10J

Gas constantUnified atomic mass unit
Standard pressure
Atmospheric pressure
Zero degree CelsiusÅngstrom
Picometer
Electronvolt−6Part-per-million−9Part-per-billion−12Part-per-trillion

1 ppb = 10

1 ppt = 10

Electron mass𝜋 = 3.141592653589793

pi

𝑒 = 2.718281828459045

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

IChO

G0-4

SAU-1 C-0 G-4

Arabic (SAU) (Saudi Arabia)

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

Coulomb's law

The

first
law
of
thermodynamics

𝐻 = 𝑈 + 𝑃𝑉

Enthalpy 𝐻

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

Entropy based on Boltzmann's principle 𝑆
The change of entropyΔ𝑆

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.

Gibbs free energy 𝐺

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.

Reaction quotient 𝑄

IChO

G0-5

SAU-1 C-0 G-5

Arabic (SAU) (Saudi Arabia)

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) = 𝑒𝑥Heat change Δ𝑞
Nernst equation for redox reaction

Arrhenius 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.

Lambert–Beer 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.

Henderson–Hasselbalch
equation

When 𝑥 ≠ 1,

The sum of a geometricseriesWhen 𝑥 ≪ 1,
1
≃1+𝑥
1−𝑥

Approximation equationthat can be used to solve
problems

1 − 𝑥𝑛+1
𝑛
1 + 𝑥 + 𝑥2 + ⋯ + 𝑥𝑛 = ∑𝑖=0 𝑥𝑖 =1−𝑥Energy of a photon

IChO
SAU-1 C-0 G-6

G0-6

Arabic (SAU) (Saudi Arabia)

Periodic Table

IChO

G0-7

SAU-1 C-0 G-7

Arabic (SAU) (Saudi Arabia)
1

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

SAU-1 C-1 C-1

SAU-1 C-1 C
Mohammad AlHudaithi

IChO
Problem 1
Cover sheet

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

Q1-1

IChO
SAU-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
SAU-1 C-1 Q-2

Q1-2

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

Fig.2

English (Official)

Q1-3

IChO
SAU-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
SAU-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
SAU-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
SAU-1 C-1 Q-1

Arabic (SAU) (Saudi Arabia)‫ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻋﻠﻰ ﺳﻄﺢ ﻣﻌﺪﻧﻲ‬11 % of the total
Total
24

B.4
3

B.3
3

B.2
3

B.1
5

A.2
4

A.1
6

Question
PointsScore

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.
‫ ﺳﻨﻨﻈﺮ‬،‫ ﻓﻲ ﻫﺬﺍ ﺍﻟﺴﺆﺍﻝ‬.‫ﻳﺘﻮﻗﻊ ﺃﻥ ﻳﻜﻮﻥ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻣﺼﺪﺭ ﻃﺎﻗﺔ ﻓﻲ ﺍﻟﻤﺴﺘﻘﺒﻞ ﻻ ﻳﻌﺘﻤﺪ ﻋﻠﻰ ﺍﻟﻮﻗﻮﺩ ﺍﻷﺣﻔﻮﺭﻱ‬.‫ ﻭﻫﺬﺍ ﻳﺘﻌﻠﻖ ﺑﺘﻘﻨﻴﺔ ﻧﻘﻞ ﻭﺗﺨﺰﻳﻦ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‬،‫ﺇﻟﻰ ﻋﻤﻠﻴﺔ ﺗﺨﺰﻳﻦ ﺍﻟﻬﺪﺭﻭﺟﻴﻦ ﻓﻲ ﻣﻌﺪﻥ‬A PartAs 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.
‫ﺃﻭﻻ ﺇﻟﻰ ﻋﻤﻠﻴﺔ ﺍﻣﺘﺼﺎﺹ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻋﻠﻰ ﺳﻄﺢ‬‫ ﻟﻨﻨﻈﺮ‬.‫ُﻳﻤﺘﺺ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺇﻟﻰ ﺩﺍﺧﻞ ﻣﻌﺪﻥ ﻣﺎ ﻣﻦ ﺧﻼﻝ ﺳﻄﺤﻪ‬ً‫ ﻋﻨﺪﻣﺎ‬.‫( ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ‬ad) ‫( ﻭﺍﻟﺤﺎﻟﺔ ﺍﻟﻤﻤﺘﺼﺔ ﻋﻠﻰ ﺍﻟﺴﻄﺢ ﺑـ‬g) ‫ ﺣﻴﺚ ﻳﺮﻣﺰ ﻟﻠﺤﺎﻟﺔ ﺍﻟﻐﺎﺯﻳﺔ ﺑـ‬،H2 (g) → 2H(ad) ،‫ﺍﻟﻤﻌﺪﻥ‬‫( ﺗﺘﻔﻜﻚ ﻋﻠﻰ ﺍﻟﺴﻄﺢ ﻭﺗﻤﺘﺺ ﻋﻠﻰ ﺷﻜﻞ ﺫﺭﺍﺕ ﻫﻴﺪﺭﻭﺟﻴﻦ )ﺷﻜﻞ‬M) ‫( ﺇﻟﻰ ﺳﻄﺢ ﻣﻌﺪﻥ‬H2 ) ‫ﺗﺼﻞ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‬‫‪Q1-2‬‬‫)‪Arabic (SAU) (Saudi Arabia‬‬‫‪IChO‬‬‫‪SAU-1 C-1 Q-2‬‬‫‪ .(1‬ﻫﻨﺎ‪ ،‬ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﺎﻣﻨﺔ ﻟـ ‪ H2‬ﺗﺘﻤﺜﻞ ﺑﻤﺘﻐﻴﺮﻳﻦ‪ :‬ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺍﻟﺬﺭﺍﺕ )‪ (d‬ﻭﺍﻻﺭﺗﻔﺎﻉ ﺑﺎﻟﻨﺴﺒﺔ ﻟﺴﻄﺢ ﺍﻟﻤﻌﺪﻥ )‪ .(z‬ﻳﻤﻜﻦ‬‫ﺍﻓﺘﺮﺍﺽ ﺃﻥ ﺍﻟﻤﺤﻮﺭ ﺍﻟﻤﺎﺭ ﺑﺬﺭﺗﻲ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻣﻮﺍﺯﻱ ﻟﻠﺴﻄﺢ‪ ،‬ﻭﺃﻥ ﻣﺮﻛﺰ ﺍﻟﺜﻘﻞ ﻳﻘﻊ ﺩﺍﺋﻤﺎً ﻋﻠﻰ ﺍﻟﻤﺤﻮﺭ ﺍﻟﻌﻤﻮﺩﻱ ﺍﻟﻤﻨﻘﻂ‬‫ﺍﻟﻤﻮﺿﺢ ﻓﻲ ﺍﻟﺸﻜﻞ ‪ .1‬ﻓﻲ ﺷﻜﻞ ‪ 2‬ﻳﻈﻬﺮ ﻣﺨﻄﻂ )‪ (contour‬ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﺎﻣﻨﺔ ﻟﻌﻤﻠﻴﺔ ﺍﻟﺘﻔﻜﻚ ﻋﻠﻰ ﺍﻟﺴﻄﺢ‪ .‬ﺍﻷﺭﻗﺎﻡ‬‫ﺑﻮﺣﺪﺓ ‪ kJ‬ﻟﻜﻞ ﻣﻮﻝ ﻣﻦ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‪ .‬ﺗﺒﺎﻋﺪ ﺍﻟﺨﻄﻮﻁ ﺍﻟﻤﺘﺼﻠﺔ ﻳﻤﺜﻞ ‪ ،20 kJ mol−1‬ﻭﺗﺒﺎﻋﺪ ﺍﻟﺨﻄﻮﻁ ﺍﻟﻤﻨﻘﻄﺔ ﻳﻤﺜﻞ ‪100‬‬‫‪ ،kJ mol−1‬ﻭﺗﺒﺎﻋﺪ ﺍﻟﺨﻄﻮﻁ ﺍﻟﻤﺘﺼﻠﺔ ﻣﻊ ﺍﻟﻤﻨﻘﻄﺔ ﻳﻤﺜﻞ ‪ .80 kJ mol−1‬ﺗﻢ ﺇﻫﻤﺎﻝ ﺍﻟﻄﺎﻗﺔ ﺍﻻﻫﺘﺰﺍﺯﻳﺔ ﻟﻠﻨﻘﻄﺔ ﺍﻟﺼﻔﺮﻳﺔ‬‫)‪.(zero-point vibration energy‬‬‫ﺷﻜﻞ ‪ .1‬ﺗﻌﺮﻳﻒ ﺍﻟﻤﺘﻐﻴﺮﺍﺕ‪ .‬ﺍﻟﺮﺳﻢ ﻟﻴﺲ ﻟﻠﻘﻴﺎﺱ‪.‬‬IChO
SAU-1 C-1 Q-3

Q1-3
Arabic (SAU) (Saudi Arabia).2 ‫ﺷﻜﻞ‬6pt

For each of the following items (i)–(iii), select the closest value from A–G.
.A-G ‫( ﺍﺧﺘﺮ ﺃﻗﺮﺏ ﻗﻴﻤﺔ ﻣﻦ ﺍﻟﺨﻴﺎﺭﺍﺕ‬i)-(iii) ‫ﻟﻜﻞ ﻣﻦ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﺘﺎﻟﻴﺔ‬(i) The interatomic distance for a gaseous H2 molecule‫ ﺍﻟﻐﺎﺯﻱ‬H2 ‫( ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺍﻟﺬﺭﺍﺕ ﻓﻲ ﺟﺰﻱﺀ‬i)(ii) The interatomic distance between metal atoms (𝑑M in Fig. 1).(1 ‫𝑑 ﻓﻲ ﺷﻜﻞ‬M ) ‫( ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺫﺭﺍﺕ ﺍﻟﻤﻌﺪﻥ‬ii)(iii) The distance of adsorbed H atoms from the surface (ℎad in Fig. 1).(1 ‫ ﻓﻲ ﺷﻜﻞ‬ℎad ) ‫( ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺫﺭﺓ ﻫﻴﺪﺭﻭﺟﻴﻦ ﻭﺍﻟﺴﻄﺢ‬iii)
D. 0.15 nm

C. 0.11 nm

B. 0.07 nm

A. 0.03 nm

G. 0.27 nm

F. 0.23 nm

E. 0.19 nm

A.1

IChO
SAU-1 C-1 Q-4

4pt

Q1-4
Arabic (SAU) (Saudi Arabia)

For each of the following items (i)–(ii), select the closest value from A–H.
.A-H ‫( ﺍﺧﺘﺮ ﺃﻗﺮﺏ ﻗﻴﻤﺔ ﻣﻦ ﺍﻟﺨﻴﺎﺭﺍﺕ‬i)-(ii) ‫ﻟﻜﻞ ﻣﻦ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﺘﺎﻟﻴﺔ‬(i) the energy required for the dissociation of gaseous H2 to gaseous H (H2 (g) →2H(g))(H2 (g) → 2H(g)) ‫ ﻏﺎﺯﻱ‬H ‫ ﺍﻟﻐﺎﺯﻱ ﺇﻟﻰ‬H2 ‫( ﺍﻟﻄﺎﻗﺔ ﺍﻟﻼﺯﻣﺔ ﻟﺘﻔﻜﻴﻚ‬i)(ii) the energy released during the adsorption of a gaseous H2 (H2 (g) → 2H(ad))(H2 (g) → 2H(ad)) ‫ ﻏﺎﺯﻱ‬H2 ‫( ﺍﻟﻄﺎﻗﺔ ﺍﻟﻤﻨﺒﻌﺜﺔ ﺧﻼﻝ ﻋﻤﻠﻴﺔ ﺍﻣﺘﺼﺎﺹ‬ii)
D. 100 kJ mol−1

H. 400 kJ mol−1

C. 60 kJ mol−1

G. 300 kJ mol−1

B. 40 kJ mol−1

F. 200 kJ mol−1

A. 20 kJ mol−1

E. 150 kJ mol−1

A.2

IChO
SAU-1 C-1 Q-5

Q1-5
Arabic (SAU) (Saudi Arabia)

B Part
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.
‫ ﻛﻤﺎ ﻫﻮ‬،‫ ﺃﻭ ﺗﺘﺤﺪ ﻭﺗﻌﻮﺩ ﺇﻟﻰ ﺍﻟﻄﻮﺭ ﺍﻟﻐﺎﺯﻱ‬،‫ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﻤﻤﺘﺼﺔ ﻋﻠﻰ ﺍﻟﺴﻄﺢ ﺇﻣﺎ ﺃﻥ ﺗﻤﺘﺺ ﺇﻟﻰ ﺩﺍﺧﻞ ﺍﻟﻤﻌﺪﻥ‬.‫ ﻳﺮﻣﺰ ﻟﺬﺭﺓ ﻫﻴﺪﺭﻭﺟﻴﻦ ﻣﻤﺘﺼﺔ ﻓﻲ ﺩﺍﺧﻞ ﺍﻟﻤﻌﺪﻥ‬H(ab) ‫ ﺍﻟﺮﻣﺰ‬.(1b) ‫( ﻭ‬1a) ‫ﻣﻮﺿﺢ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻦ‬The 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:‫ ﻭﺍﻣﺘﺼﺎﺹ‬،(desorption) ‫ ﺍﻧﻔﺼﺎﻝ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻣﻦ ﺍﻟﻤﻌﺪﻥ‬،(adsorption) ‫ﺳﺮﻋﺔ ﺍﻟﺘﻔﺎﻋﻼﺕ ﻟﻜﻞ ﻣﻮﻗﻊ ﺳﻄﺤﻲ ﻟﻼﻣﺘﺼﺎﺹ‬:‫ ﻳﻤﻜﻦ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻨﻬﺎ ﻋﻠﻰ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﻟﻲ‬.‫ ﻋﻠﻰ ﺍﻟﺘﺮﺗﻴﺐ‬،𝑟3 [s−1 ] ‫𝑟 ﻭ‬2 [s−1 ] ،𝑟1 [s−1 ] ‫( ﻫﻲ‬absorption) ‫ﻟﺪﺍﺧﻞ ﺍﻟﻤﻌﺪﻥ‬where 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.
‫ ﻣﻦ‬.‫𝑃 ﻋﺒﺎﺭﺓ ﻋﻦ ﺿﻐﻂ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‬H2 ‫ ﻭ‬،‫𝑘 ﻋﺒﺎﺭﺓ ﻋﻦ ﺛﻮﺍﺑﺖ ﺳﺮﻋﺔ ﺍﻟﺘﻔﺎﻋﻼﺕ‬3 [s−1 ]‫ ﻭ‬،𝑘2 [s−1 ] ،𝑘1 [s−1 Pa−1 ] ‫ﺣﻴﺚ‬‫ ﻳﻤﻜﻦ ﺍﻓﺘﺮﺍﺽ ﺃﻥ‬.‫( ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺸﻐﻮﻟﺔ ﺑﺬﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‬0 ≤ 𝜃 ≤ 1) 𝜃 ‫ ﻳﻤﺜﻞ‬،‫ﺑﻴﻦ ﺍﻟﻤﻮﺍﻗﻊ ﺍﻟﻤﺘﺎﺣﺔ ﻋﻠﻰ ﺍﻟﺴﻄﺢ‬‫( ﻛﺒﻴﺮﺓ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻌﻤﻠﻴﺔ ﺍﻻﻣﺘﺼﺎﺹ ﺇﻟﻰ ﺩﺍﺧﻞ ﺍﻟﻤﺎﺩﺓ‬desorption) ‫ﺳﺮﻋﺔ ﺍﻻﻣﺘﺼﺎﺹ ﻭﺍﻧﻔﺼﺎﻝ ﺍﻟﺠﺴﻴﻤﺎﺕ ﻣﻦ ﺍﻟﻤﻌﺪﻥ‬.‫𝑟( ﻭﺃﻥ 𝜃 ﺗﺒﻘﻰ ﺛﺎﺑﺘﺔ‬1 , 𝑟2 ≫ 𝑟3 )

IChO
SAU-1 C-1 Q-6

5pt

𝑟3 can be expressed as:

Express 𝐶 using 𝑘1 and 𝑘2 .

Q1-6
Arabic (SAU) (Saudi Arabia)

B.1:‫𝑟 ﻋﻠﻰ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﻟﻲ‬3 ‫ﻳﻤﻜﻦ ﺗﻤﺜﻴﻞ‬.𝑘2 ‫𝑘 ﻭ‬1 ‫ ﺑﺎﺳﺘﺨﺪﺍﻡ‬C ‫ﻋﺒﺮ ﻋﻦ‬

IChO
SAU-1 C-1 Q-7

Q1-7
Arabic (SAU) (Saudi Arabia)

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.
‫( ﻣﻊ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‬1.0 × 10−3 m3 ) ‫ ﻟﺘﺮ‬1 ‫ = 𝑆 ﻓﻲ ﻭﻋﺎﺀ ﺑﺤﺠﻢ‬1.0 × 10−3 m2 ‫ﻭﺿﻌﺖ ﻋﻴﻨﺔ ﻣﻌﺪﻧﻴﺔ ﻟﻬﺎ ﻣﺴﺎﺣﺔ ﺳﻄﺢ‬‫ ﺗﻢ ﺍﻟﺤﻔﺎﻅ‬.𝑁 = 1.3 × 1018 m−2 ‫ ﻛﺜﺎﻓﺔ ﻣﻮﺍﻗﻊ ﺍﻣﺘﺼﺎﺹ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻋﻠﻰ ﺍﻟﺴﻄﺢ ﻫﻲ‬.(𝑃H2 = 1.0 × 102 Pa).𝑣 = 4.0 × 10−4 Pa s−1 ‫𝑃 ﺑﻤﻌﺪﻝ ﺛﺎﺑﺖ‬H2 ‫ ﻳﺘﻨﺎﻗﺺ‬،1 ‫ ﻋﻨﺪ ﺍﺳﺘﻤﺮﺍﺭ ﺍﻟﺘﻔﺎﻋﻞ‬.𝑇 = 400 K ‫ﻋﻠﻰ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﺍﻟﺴﻄﺢ ﻋﻨﺪ‬.‫ ﻫﻮ ﻏﺎﺯ ﻣﺜﺎﻟﻲ ﻭﺃﻥ ﺣﺠﻢ ﻋﻴﻨﺔ ﺍﻟﻤﻌﺪﻥ ﺿﺌﻴﻞ‬H2 ‫ﺍﻓﺘﺮﺽ ﺃﻥ‬
3pt

3pt

3pt

Calculate the amount of H atoms in moles absorbed per unit area of the surface
per unit time, 𝐴 [mol s−1 m−2 ].
‫ﺍﺣﺴﺐ ﻛﻤﻴﺔ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﻤﻤﺘﺼﺔ ﺑﺎﻟﻤﻮﻻﺕ ﻟﻜﻞ ﻭﺣﺪﺓ ﻣﺴﺎﺣﺔ ﻣﻦ ﺍﻟﺴﻄﺢ ﻭﻟﻜﻞ‬.𝐴 [mol s−1 m−2 ] ،‫ﻭﺣﺪﺓ ﺯﻣﻨﻴﺔ‬At 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 .
‫ ﺇﺫﺍ ﻟﻢ‬.400 K ‫𝑘 ﻋﻨﺪ‬3 ‫ ﺍﺣﺴﺐ ﻗﻴﻤﺔ‬.1.0 × 102 Pa−1 ‫ ﺗﻜﻮﻥ 𝐶 ﻣﺴﺎﻭﻳﺔ ﻟـ‬،𝑇 = 400 K ‫ﻋﻨﺪ‬.𝐴 = 3.6 × 10−7 mol s−1 m−2 ‫ ﺍﺳﺘﺨﺪﻡ‬،B.2 ‫ﺗﺤﺼﻞ ﻋﻠﻰ ﺇﺟﺎﺑﺔ ﻓﻲ ﺍﻟﻔﻘﺮﺓ‬At 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).
‫ ﺍﺧﺘﺮ ﺷﻜﻞ‬.𝑘3 = 4.8 × 10−2 s−1 ‫ = 𝐶 ﻭ‬2.5 × 103 Pa−1 ‫ ﺗﻜﻮﻥ‬،‫ﻓﻲ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﻣﺨﺘﻠﻔﺔ‬.‫𝑃 ﻋﻨﺪ ﻫﺬﻩ ﺍﻟﺪﺭﺟﺔ ﻣﻦ ﺍﻟﺤﺮﺍﺭﺓ‬H2 ‫𝑟 ﺑﺎﻟﻨﺴﺒﺔ ﻟﻤﺘﻐﻴﺮ‬3 ‫( ﻟﺪﺍﻟﺔ‬a)–(h) ‫ﺍﻟﻤﻨﺤﻨﻰ ﻣﻦ‬

B.2

B.3

B.4

IChO
SAU-1 C-1 A-1

Hydrogen at a Metal Surface
Part A
A.1 (6 pt)
(i)

(ii)

(iii)

A.2 (4 pt)
(i)

(ii)

A1-1
English (Official)IChO
SAU-1 C-1 A-2

Part B
B.1 (5 pt)

𝐶=
B.2 (3 pt)𝐴=

mol s−1 m−2

A1-2
English (Official)IChO
SAU-1 C-1 A-3

B.3 (3 pt)

𝑘3 =B.4 (3 pt)

s−1

A1-3
English (Official)SAU-1 C-2 C-1

SAU-1 C-2 C
Mohammad AlHudaithi

IChO
Problem 2
Cover sheet

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

Q2-1

IChO
SAU-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
SAU-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
SAU-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
SAU-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
SAU-1 C-2 Q-1

Arabic (SAU) (Saudi Arabia)‫ﺍﻟﻜﺒﺴﻮﻟﺔ ﺍﻟﺰﻣﻨﻴﺔ ﻟﻠﻨﻈﺎﺋﺮ‬11 % of the total
Total
35

A.4
9

A.3
10

A.2
8

A.1
8

Question
PointsScore

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.
‫ ﻳﻌﺘﺒﺮ ﻟﻬﺬﻩ‬.isotopologues ‫ ﺗﺴﻤﻰ‬،CH3 D ‫ ﻭ‬CH4 ‫ ﻣﺜﻞ‬،‫ﺍﻟﻜﻴﺎﻧﺎﺕ ﺍﻟﺠﺰﻳﺌﻴﺔ ﺍﻟﺘﻲ ﺗﺨﺘﻠﻒ ﻓﻘﻂ ﻓﻲ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﻈﻴﺮﻱ‬.‫ ﻳﻮﺟﺪ ﺍﺧﺘﻼﻑ ﻃﻔﻴﻒ‬،‫ ﻓﻲ ﺍﻟﺤﻘﻴﻘﺔ‬.‫ﺍﻟﺠﺰﻳﺌﺎﺕ ﺃﻥ ﻟﻬﺎ ﻧﻔﺲ ﺍﻟﺼﻔﺎﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬Assume that all of the substances shown in this Question are in a gas phase.
.‫ﺍﻓﺘﺮﺽ ﺃﻥ ﺟﻤﻴﻊ ﺍﻟﻤﻮﺍﺩ ﻓﻲ ﻫﺬﺍ ﺍﻟﺴﺆﺍﻝ ﻓﻲ ﺍﻟﻄﻮﺭ ﺍﻟﻐﺎﺯﻱ‬Let us consider the following equilibrium::‫ﻟﻨﻨﻈﺮ ﻓﻲ ﺗﻔﺎﻋﻞ ﺍﻻﺗﺰﺍﻥ ﺍﻟﺘﺎﻟﻲ‬The entropy, 𝑆, increases with increasing the number of possible microscopic states of a system, 𝑊 ::W ،‫ ﺑﺰﻳﺎﺩﺓ ﻋﺪﺩ ﺍﻟﺤﺎﻻﺕ ﺍﻟﻤﺠﻬﺮﻳﺔ ﻟﻠﻨﻈﺎﻡ‬،𝑆 ،‫ﻳﺰﺩﺍﺩ ﺍﻹﻧﺘﺮﻭﺑﻲ‬IChO
SAU-1 C-2 Q-2

Q2-2
Arabic (SAU) (Saudi Arabia)

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.

‫ ﻷﻥ ﺫﺭﺍﺕ ﺍﻷﻛﺴﺠﻴﻦ ﻣﺘﻤﺎﻳﺰﺓ‬12 C16 O18 O ‫ = 𝑊 ﻟﺠﺰﻱﺀ‬2 ،‫ ﻓﻲ ﺍﻟﻤﻘﺎﺑﻞ‬.12 C18 O2 ‫ ﻭ‬12 C16 O2 ‫ = 𝑊 ﺑﺎﻟﻨﺴﺒﺔ ﻟـ‬1 ‫ﺣﻴﺚ‬.𝑊 = 22 = 4 ‫ ﺗﻜﻮﻥ‬،1 ‫ ﻓﻲ ﺍﻟﻄﺮﻑ ﺍﻷﻳﻤﻦ ﻟﻤﻌﺎﺩﻟﺔ ﺍﻻﺗﺰﺍﻥ‬12 C16 O18 O ‫ ﻷﻥ ﻳﻮﺟﺪ ﺟﺰﻳﺌﻴﻦ‬.‫ﻓﻲ ﻫﺬﺍ ﺍﻟﺠﺰﻱﺀ‬8pt

The enthalpy change, Δ𝐻, of eq. 3 is positive regardless of the temperature.
.‫ ﺩﺍﺋﻤﺎً ﻣﻮﺟﺐ ﺑﻐﺾ ﺍﻟﻨﻈﺮ ﻋﻦ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‬3 ‫ ﻟﻠﻤﻌﺎﺩﻟﺔ‬،Δ𝐻 ،‫ﺍﻟﺘﻐﻴﺮ ﻓﻲ ﺍﻹﻧﺜﺎﻟﺒﻲ‬

A.1

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.
‫ → 𝑇( ﻭﻋﻨﺪ‬0 ‫ ﻋﻨﺪ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﻣﻨﺨﻔﻀﺔ ﺟﺪﺍً )ﻓﻜﺮ ﻓﻲ‬3 ‫ ﻟﻠﻤﻌﺎﺩﻟﺔ‬،𝐾 ،‫ﺍﺣﺴﺐ ﺛﺎﺑﺖ ﺍﻻﺗﺰﺍﻥ‬‫ ﺍﻓﺘﺮﺽ ﺃﻥ ﺍﻟﺘﻔﺎﻋﻞ ﻻ ﻳﺘﻐﻴﺮ ﻋﻨﺪ ﺩﺭﺟﺎﺕ‬.(𝑇 → +∞ ‫ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﻣﺮﺗﻔﻌﺔ ﺟﺪﺍً )ﻓﻜﺮ ﻓﻲ‬.‫ ﺗﺼﺒﺢ ﻗﻴﻤﺔ ﺛﺎﺑﺘﺔ ﻋﻨﺪ ﺩﺭﺟﺎﺕ ﺍﻟﺤﺮﺍﺭﺓ ﺍﻟﻤﺮﺗﻔﻌﺔ‬Δ𝐻 ‫ ﻭﺍﻓﺘﺮﺽ ﺃﻥ‬،‫ﺍﻟﺤﺮﺍﺭﺓ ﻫﺬﻩ‬The ΔH of the following process can be explained by molecular vibrations.
.‫ ﻟﻬﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﻣﻦ ﺧﻼﻝ ﺍﻻﻫﺘﺰﺍﺯﺍﺕ ﺍﻟﺠﺰﻳﺌﻴﺔ‬Δ𝐻 ‫ﻳﻤﻜﻦ ﺃﻥ ﻳﺘﻢ ﺗﻔﺴﻴﺮ‬At 𝑇 = 0 K, the vibrational energy of a diatomic molecule whose vibration frequency is 𝜈 [s−1 ] is expressedas::‫[ ﺑﺎﻟﺘﺎﻟﻲ‬s−1 ] 𝜈 ‫ ﻳﻌﺒﺮ ﻋﻦ ﺍﻟﻄﺎﻗﺔ ﺍﻻﻫﺘﺰﺍﺯﻳﺔ ﻟﺠﺰﻱﺀ ﺛﻨﺎﺋﻲ ﺍﻟﺬﺭﺓ ﺫﻭ ﺍﻟﺘﺮﺩﺩ ﺍﻻﻫﺘﺰﺍﺯﻱ‬،0 K ‫ﻋﻨﺪ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‬Wherein 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:IChO
SAU-1 C-2 Q-3

Q2-3
Arabic (SAU) (Saudi Arabia)،‫ﺣﻴﺚ 𝑘 ﻫﻮ ﺛﺎﺑﺖ ﺍﻟﻤﺮﻭﻧﺔ ﻭ 𝜇 ﻫﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﺨﺘﺰﻟﺔ ﻭﻳﻤﻜﻦ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻨﻬﺎ ﺑﻮﺍﺳﻄﺔ ﻛﺘﻠﺔ ﺍﻟﺬﺭﺗﻴﻦ ﻓﻲ ﺍﻟﺠﺰﻱﺀ ﺛﻨﺎﺋﻲ ﺍﻟﺬﺭﺓ‬:‫ ﻛﺎﻟﺘﺎﻟﻲ‬،𝑚2 ‫𝑚 ﻭ‬1

8pt

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 .
‫ ﻟﻠﺘﻔﺎﻋﻞ‬Δ𝐻 ‫ ﺍﺣﺴﺐ‬.‫ ﻋﻨﺪ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻨﻪ ﻛﻌﺪﺩ ﻣﻮﺟﻲ‬4161.0 cm−1 ‫ ﻋﻨﺪ‬H2 ‫ﻳﻜﻮﻥ ﺍﻫﺘﺰﺍﺯ‬.J mol−1 ‫ ﺑﻮﺣﺪﺓ‬0 K ‫ﺍﻟﺘﺎﻟﻲ ﻋﻨﺪ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ‬Assume 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 •:‫ﺍﻓﺘﺮﺽ ﺍﻟﺘﺎﻟﻲ‬.Δ𝐻 ‫• ﻓﻘﻂ ﺍﻟﻄﺎﻗﺔ ﺍﻻﻫﺘﺰﺍﺯﻳﺔ ﺗﺴﺎﻫﻢ ﻓﻲ‬.D2 ‫ ﻭ‬،HD ،H2 ‫• ﻗﻴﻤﺔ 𝑘 ﻣﺘﻄﺎﺑﻘﺔ ﻟـ‬.2 Da ‫ ﻫﻲ‬D ‫ ﻭﻛﺘﻠﺔ‬1 Da ‫ ﻫﻲ‬H ‫• ﻛﺘﻠﺔ‬

A.2

IChO
SAU-1 C-2 Q-4

Q2-4
Arabic (SAU) (Saudi Arabia)

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 .

‫ ﺑﺎﻟﺘﻐﻴﺮ ﻓﻲ ﺍﻟﻨﺴﺒﺔ‬ΔD2 ‫ ﺗﻢ ﺗﻌﺮﻳﻒ‬،‫ ﻫﻨﺎ‬.‫ ﻋﻠﻰ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻟﻠﻨﻈﺎﻡ ﻋﻨﺪ ﺍﻻﺗﺰﺍﻥ‬D2 ‫ ﻭ‬،HD ،H2 ‫ﺗﻌﺘﻤﺪ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻻﺭﻳﺔ ﻟـ‬.D2 ‫ﺍﻟﻤﻮﻻﺭﻳﺔ ﻟـ‬

[D ]
[D2 ]
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 ﻳﺠﺐ ﻣﻼﺣﻈﺔ ﺃﻥ ﺗﻮﺯﻳﻊ ﺍﻟﻨﻈﺎﺋﺮ ﻳﺼﺒﺢ‬.𝑇 → +∞ ‫ﻋﻨﺪ‬

10pt

[D2 ]
[D ]
‫𝑅 ﺍﻟﻨﺴﺒﺔ‬D∗ 2 ‫ ﻓﻲ ﺍﻟﻌﻴﻨﺔ ﻭ ﻳﻤﺜﻞ‬2 ‫𝑅 ﺍﻟﻨﺴﺒﺔ‬D2 ‫ﺣﻴﺚ ﻳﻤﺜﻞ‬
[H2 ]
[H2 ]
.𝑇 → +∞ ‫ﻋﺸﻮﺍﺋﻴﺎً ﻋﻨﺪ‬Calculate 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.
‫ ﻋﻨﺪﻣﺎ ﻳﻜﻮﻥ ﺍﻟﺘﺒﺎﺩﻝ ﺍﻟﻨﻈﻴﺮﻱ ﻓﻲ ﺍﺗﺰﺍﻥ ﻋﻨﺪ ﺩﺭﺟﺔ‬D ‫ ﻣﻦ ﺍﻟﻮﻓﺮﺓ ﺍﻟﻄﺒﻴﻌﻴﺔ ﻟـ‬ΔD2 ‫ﺍﺣﺴﺐ‬‫ ﺍﻓﺘﺮﺽ ﺃﻥ ﻧﺴﺐ ﺍﻟﻮﻓﺮﺓ‬.0.300 ‫ ﺗﺴﺎﻭﻱ‬4 ‫ﺣﺮﺍﺭﺓ ﺗﺠﻌﻞ ﺛﺎﺑﺖ ﺍﻻﺗﺰﺍﻥ 𝐾 ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ‬.‫ ﻋﻠﻰ ﺍﻟﺘﺮﺗﻴﺐ‬،1 − 1.5576 × 10–4 ‫ ﻭ‬1.5576 × 10–4 ‫ ﻫﻲ‬H ‫ ﻭ‬D ‫ﺍﻟﻄﺒﻴﻌﻴﺔ ﻟـ‬

A.3

Q2-5

IChO
SAU-1 C-2 Q-5

Arabic (SAU) (Saudi Arabia)

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:،‫ ﺃﻱ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺫﺭﺗﻴﻦ ﻣﻦ ﺍﻟﻨﻈﺎﺋﺮ ﺍﻟﺜﻘﻴﻠﺔ ﻓﻲ ﺟﺰﻱﺀ ﻭﺍﺣﺪ‬،‫ ﺫﻭ ﺗﺒﺪﻳﻠﻴﻦ‬isotopologue ‫ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻻﺭﻳﺔ ﻟـ‬،‫ﺑﺸﻜﻞ ﻋﺎﻡ‬‫ ﻭﺍﻟﺘﻲ ﻣﻮﺻﻮﻓﺔ‬،47 ‫ ﻭ‬44 ‫ ﺫﺍﺕ ﺍﻟﻜﺘﻞ ﺍﻟﺠﺰﻳﺌﻴﺔ‬CO2 ‫ ﻟﻨﻨﻈﺮ ﻓﻲ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻻﺭﻳﺔ ﻟﺠﺰﻳﺌﺎﺕ‬.‫ﺗﺰﺩﺍﺩ ﻣﻊ ﺍﻧﺨﻔﺎﺽ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‬:‫ ﻣﻌﺮﻓﺔ ﻛﺎﻟﺘﺎﻟﻲ‬Δ47 ‫ ﺍﻟﻘﻴﻤﺔ‬.‫ ﺃﺩﻧﺎﻩ‬CO2 [47] ‫ ﻭ‬CO2 [44] ‫ﺑـ‬

[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 to‫ ﺍﻟﻮﻓﺮﺓ ﺍﻟﻄﺒﻴﻌﻴﺔ ﻟﺬﺭﺍﺕ‬.𝑇 → +∞ ‫ﻋﻨﺪ‬

13[CO2 [47]][CO2 [47]]∗‫𝑅 ﺍﻟﻨﺴﺒﺔ‬47‫ ﻭﻳﻤﺜﻞ‬،‫ﻓﻲ ﺍﻟﻌﻴﻨﺔ‬‫𝑅 ﺍﻟﻨﺴﺒﺔ‬47 ‫ﺣﻴﺚ ﻳﻤﺜﻞ‬[CO2 [44]][CO2 [44]].‫ ﺍﻫﻤﻞ ﺍﻟﻨﻈﺎﺋﺮ ﺍﻟﻐﻴﺮ ﻣﺬﻛﻮﺭﺓ‬.‫ﺍﻟﻜﺮﺑﻮﻥ ﻭﺍﻷﻛﺴﺠﻴﻦ ﻣﻮﺿﺤﺔ ﺑﺎﻷﺳﻔﻞ‬C

12

0.01111218

O

0.0020000

C

0.98888817O

0.0003790

natural abundance

16O

0.997621

natural abundance

The temperature dependence of Δ47 is determined as follows, where 𝑇 is given as the absolute temperature in units of K::K ‫ ﺣﻴﺚ 𝑇 ﻋﺒﺎﺭﺓ ﻋﻦ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﺑﻮﺣﺪﺓ‬،‫ ﺑﺪﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻣﻮﺿﺢ ﺃﺩﻧﺎﻩ‬Δ47 ‫ﺍﺭﺗﺒﺎﻁ‬9pt

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.
.4.50865 × 10–5 ‫𝑅 ﻟﻌﺎﻟﻖ ﺃﺣﻔﻮﺭﻱ ﻣﺄﺧﻮﺫ ﻣﻦ ﻗﺎﻉ ﺍﻟﺒﺤﺮ ﻓﻲ ﺍﻟﻘﻄﺐ ﺍﻟﺠﻨﻮﺑﻲ ﻛﺎﻧﺖ‬47 ‫ﻗﻴﻤﺔ‬‫ ﺗﻔﺴﺮ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻫﺬﻩ ﺑﺄﻧﻬﺎ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﺍﻟﻬﻮﺍﺀ‬.𝑅47 ‫ﻗﺪﺭ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﺑﺎﺳﺘﺨﺪﺍﻡ ﻗﻴﻤﺔ‬‫ ﻓﻲ‬CO2 [47] ‫ ﺍﺳﺘﺨﺪﻡ ﻓﻘﻂ ﺍﻟﻨﻈﻴﺮ ﺍﻷﻛﺜﺮ ﺷﻴﻮﻋﺎً ﻟـ‬.‫ﺧﻼﻝ ﺍﻟﻌﺼﺮ ﺍﻟﺬﻱ ﻋﺎﺵ ﻓﻴﻪ ﺍﻟﻌﺎﻟﻖ‬.‫ﺣﺴﺎﺑﺎﺗﻚ‬

A.4

IChO
SAU-1 C-2 A-1

Isotope Time Capsule
A.1 (8 pt)

𝑇 →0∶𝐾=, 𝑇 → +∞ ∶ 𝐾 =A2-1
English (Official)IChO
SAU-1 C-2 A-2

A.2 (8 pt)

Δ𝐻 =J mol−1

A2-2
English (Official)IChO
SAU-1 C-2 A-3

A.3 (10 pt)

ΔD2 =

A2-3
English (Official)IChO
SAU-1 C-2 A-4

A.4 (9 pt)

𝑇 =

K

A2-4
English (Official)SAU-1 C-3 C-1

SAU-1 C-3 C
Mohammad AlHudaithi

IChO
Problem 3
Cover sheet

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

Q3-1

IChO
SAU-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
SAU-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
SAU-1 C-3 Q-1

Arabic (SAU) (Saudi Arabia)‫ﻻﻣﺒﺮﺕ؟‬-‫ﻗﺎﻧﻮﻥ ﺑﻴﺮ‬8 % of the total
Total
22

B.2
6

B.1
6

A.1
10

Question
PointsScore

In this problem, ignore the absorption of the cell and the solvent. The temperatures of all solutions andgases are kept constant at 25 °C.
.25 °C ‫ ﺗﻢ ﺍﻟﺤﻔﺎﻅ ﻋﻠﻰ ﺩﺭﺟﺎﺕ ﺍﻟﺤﺮﺍﺭﺓ ﻟﻜﻞ ﺍﻟﻤﺤﺎﻟﻴﻞ ﻭﺍﻟﻐﺎﺯﺍﺕ ﻋﻨﺪ‬.‫ ﺍﻫﻤﻞ ﺍﻣﺘﺼﺎﺹ ﺍﻟﺨﻠﻴﺔ ﻭﺍﻟﻤﺬﻳﺐ‬،‫ﻓﻲ ﻫﺬﺍ ﺍﻟﺴﺆﺍﻝ‬A PartAn 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:،1.00 × 10−2 mol L−1 ‫ ﻫﻲ‬X ‫[ ﻓﻲ ﻣﺤﻠﻮﻝ‬H+ ] ‫ ﻭ‬،[HA] ،[A− ] ‫ ﺗﺮﺍﻛﻴﺰ‬.NaA ‫ ﻭ‬HA ‫ ﺑﺎﺳﺘﺨﺪﺍﻡ‬X ‫ﺗﻢ ﺗﺤﻀﻴﺮ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ‬‫ ﻭﺍﻟﺘﻲ ﻣﺮﺗﺒﻄﺔ ﺑﺒﻌﻀﻬﺎ ﺍﻟﺒﻌﺾ ﻣﻦ ﺧﻼﻝ ﺗﻔﺎﻋﻞ ﺍﻻﺗﺰﺍﻥ‬،‫ ﻋﻠﻰ ﺍﻟﺘﺮﺗﻴﺐ‬،1.00 × 10−4 mol L−1 ‫ ﻭ‬،1.00 × 10−3 mol L−1:‫ﻗﺎﻋﺪﺓ ﺍﻟﺘﺎﻟﻲ‬-‫ﺣﻤﺾ‬

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.
‫ ﺍﻓﺘﺮﺽ ﻋﺪﻡ ﺣﺪﻭﺙ ﺃﻱ‬.‫ ﺍﻫﻤﻞ ﺗﻐﻴﺮ ﺍﻟﻜﺜﺎﻓﺔ ﻋﻨﺪ ﺍﻟﺘﺨﻔﻴﻒ‬.‫ ﻣﻦ ﻫﺬﺍ ﺍﻟﺴﺆﺍﻝ‬A ‫ﻃﻮﻝ ﺍﻟﻤﺴﺎﺭ ﺍﻟﺒﺼﺮﻱ ﻫﻮ 𝑙 ﻓﻲ ﺍﻟﻔﻘﺮﺓ‬.1 ‫ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ ﻋﺪﺍ ﺍﻟﺘﻔﺎﻋﻞ‬10pt

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 .
‫ ﺗﻢ ﺗﺨﻔﻴﻒ ﺍﻟﻤﺤﻠﻮﻝ‬،‫ ﺑﻌﺪ ﺫﻟﻚ‬.𝜆1 ‫𝐴 ﻋﻨﺪ ﺍﻟﻄﻮﻝ ﺍﻟﻤﻮﺟﻲ‬1 ‫ ﻫﻮ‬X ‫ﻛﺎﻥ ﺍﻣﺘﺼﺎﺹ ﺍﻟﻤﺤﻠﻮﻝ‬pH ‫ ﺇﻟﻰ ﺃﻥ ﺃﺻﺒﺢ ﺿﻌﻒ ﺣﺠﻤﻪ ﺍﻻﺑﺘﺪﺍﺋﻲ ﺑﺈﺿﺎﻓﺔ ﺣﻤﺾ ﺍﻟﻜﻠﻮﺭ ﺫﻭ ﺍﻟﺮﻗﻢ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻨﻲ‬X‫ ﺣﻴﺚ‬،𝜀HA /𝜀A− ‫ ﺣﺪﺩ ﺍﻟﻨﺴﺒﺔ‬.𝜆1 ‫𝐴 ﻋﻨﺪ‬1 ‫ ﻻ ﺯﺍﻝ ﺍﻣﺘﺼﺎﺹ ﺍﻟﻤﺤﻠﻮﻝ‬،‫ ﺑﻌﺪ ﺍﻟﺘﺨﻔﻴﻒ‬.= 2.500.𝜆1 ‫ ﻋﻨﺪ‬،‫ ﻋﻠﻰ ﺍﻟﺘﺮﺗﻴﺐ‬،A− ‫ ﻭ‬HA ‫𝜀 ﻣﻌﺎﻣﻞ ﺍﻣﺘﺼﺎﺹ‬A− ‫𝜀 ﻭ‬HA ‫ﻳﻤﺜﻞ‬

A.1

IChO
SAU-1 C-3 Q-2

Q3-2
Arabic (SAU) (Saudi Arabia)

B PartLet us consider the following equilibrium in the gas phase.
:‫ﻟﻨﻔﻜﺮ ﻓﻲ ﺍﻻﺗﺰﺍﻥ ﺍﻟﺘﺎﻟﻲ ﻓﻲ ﺍﻟﻄﻮﺭ ﺍﻟﻐﺎﺯﻱ‬D ⇌ 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.
(‫ ﻓﻲ ﻭﻋﺎﺀ ﻣﺘﻮﺍﺯﻱ ﻣﺴﺘﻄﻴﻼﺕ ﺫﻭ ﺟﺪﺍﺭ ﻣﺘﺤﺮﻙ ﺷﻔﺎﻑ ﺑﻤﻘﻄﻊ ﻋﺮﺿﻲ 𝑆 )ﺍﻧﻈﺮ ﺍﻟﺸﻜﻞ ﻓﻲ ﺍﻷﺳﻔﻞ‬D ‫ﺗﻢ ﻣﻞﺀ ﻏﺎﺯ ﻧﻘﻲ‬‫𝑛(𝜀 = 𝐴 ﺣﻴﺚ ﺗﻤﺜﻞ 𝜀 ﻣﻌﺎﻣﻞ‬/𝑉 )𝑙 ‫ ﺍﻣﺘﺼﺎﺻﻴﺔ ﺍﻟﻐﺎﺯ ﻫﻲ‬.‫ ﻭﻳﺒﻘﻰ ﺍﻟﻀﻌﻂ ﺛﺎﺑﺘﺎً ﻋﻨﺪ 𝑃 ﺣﺘﻰ ﻋﻨﺪ ﺍﻻﺗﺰﺍﻥ‬،𝑃 ‫ﻋﻨﺪ ﺿﻐﻂ‬‫ ﺍﻓﺘﺮﺽ ﺃﻥ ﻛﻞ ﻣﻜﻮﻧﺎﺕ ﺍﻟﻤﺰﻳﺞ ﺍﻟﻐﺎﺯﻱ‬.‫ ﻭ 𝑙 ﻃﻮﻝ ﺍﻟﻤﺴﺎﺭ ﺍﻟﺒﺼﺮﻱ‬،‫ ﻭ 𝑉 ﺣﺠﻢ ﺍﻟﻐﺎﺯ‬،‫ ﻭ 𝑛 ﻛﻤﻴﺔ ﺍﻟﻐﺎﺯ ﺑﺎﻟﻤﻮﻻﺕ‬،‫ﺍﻻﻣﺘﺼﺎﺹ‬.‫ﻋﺒﺎﺭﺓ ﻋﻦ ﻏﺎﺯﺍﺕ ﻣﺜﺎﻟﻴﺔ‬Use the following definitions if necessary.
.‫ﺍﺳﺘﺨﺪﻡ ﺍﻟﺘﻌﺎﺭﻳﻒ ﺍﻟﺘﺎﻟﻴﺔ ﻋﻨﺪ ﺍﻟﺤﺎﺟﺔ‬After equilibrium
D

M

𝑝M

𝑝D

0

𝑃

Partial pressure

𝑛M

𝑛D

0

𝑛0

Amount in moles𝑉
6pt

Initial state

M

D

𝑉0

Volume

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.
‫𝐴 ﻋﻨﺪ ﺍﻟﺤﺎﻟﺔ‬B1 ‫𝜆 ﻣﻦ ﺍﺗﺠﺎﻩ ﺍﻟﻤﺤﻮﺭ 𝑥 )ﺃﻱ 𝑥𝑙 = 𝑙( ﻭﻛﺎﻧﺖ‬B1 ‫ﺗﻢ ﺣﺴﺎﺏ ﺍﻣﺘﺼﺎﺻﻴﺔ ﺍﻟﻐﺎﺯ ﻋﻨﺪ‬‫𝜀 ﻣﻌﺎﻣﻼﺕ‬M ‫𝜀 ﻭ‬D ‫ ﺣﻴﺚ ﺗﻤﺜﻞ‬،𝜆B1 ‫𝜀 ﻋﻨﺪ‬D /𝜀M ‫ ﺣﺪﺩ ﺍﻟﻨﺴﺒﺔ‬.‫ﺍﻻﺑﺘﺪﺍﺋﻴﺔ ﻭﻋﻨﺪ ﺍﻻﺗﺰﺍﻥ ﻛﺬﻟﻚ‬.‫ ﻋﻠﻰ ﺍﻟﺘﺮﺗﻴﺐ‬،M ‫ ﻭ‬D ‫ﺍﻣﺘﺼﺎﺹ‬

B.1

IChO
SAU-1 C-3 Q-3

6pt

Q3-3
Arabic (SAU) (Saudi Arabia)

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 .
‫𝐴 ﻋﻨﺪ ﺍﻟﺤﺎﻟﺔ ﺍﻻﺑﺘﺪﺍﺋﻴﺔ‬B2 ‫𝜆 ﻣﻦ ﺍﺗﺠﺎﻩ ﺍﻟﻤﺤﻮﺭ 𝑦 ﻭﻛﺎﻧﺖ‬B2 ‫ﺗﻢ ﺣﺴﺎﺏ ﺍﻣﺘﺼﺎﺻﻴﺔ ﺍﻟﻐﺎﺯ ﻋﻨﺪ‬.𝜆B2 ‫𝜀 ﻋﻨﺪ‬D /𝜀M ‫ ﺣﺪﺩ ﺍﻟﻨﺴﺒﺔ‬.(𝑙 = 𝑙𝑦 ) ‫𝑦𝑙 = 𝑙( ﻭﻛﺬﻟﻚ ﺑﻌﺪ ﺍﻻﺗﺰﺍﻥ‬0 )

B.2

IChO
SAU-1 C-3 A-1

Lambert–Beer Law?
Part A
A.1 (10 pt)(Continued on the next page)A3-1
English (Official)IChO
SAU-1 C-3 A-2

A.1 (cont.)

𝜀HA /𝜀A− =A3-2
English (Official)IChO
SAU-1 C-3 A-3Part B
B.1 (6 pt)

𝜀D /𝜀M =A3-3
English (Official)IChO
SAU-1 C-3 A-4

B.2 (6 pt)

𝜀D /𝜀M =A3-4
English (Official)SAU-1 C-4 C-1

SAU-1 C-4 C
Mohammad AlHudaithi

IChO
Problem 4
Cover sheet

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

Q4-1

IChO
SAU-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
SAU-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
SAU-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
SAU-1 C-4 Q-1

Arabic (SAU) (Saudi Arabia)‫ﺍﺧﺘﺰﺍﻝ ﺍﻟﺰﻧﻚ‬-‫ﻛﻴﻤﻴﺎﺀ ﺍﻛﺴﺪﺓ‬11 % of the total
Total
32

B.4
9

B.3
5

B.2
3

B.1
4

A.2
5

A.1
6

Question
Points
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.
‫ ﻳﺘﻢ ﻓﺼﻞ ﺍﻟﺰﻧﻚ ﺍﻟﻤﻮﺟﻮﺩ ﻓﻲ ﻣﻴﺎﻩ ﺍﻟﺼﺮﻑ ﺍﻟﺼﻨﺎﻋﻲ‬.‫ﻳﺴﺘﺨﺪﻡ ﺍﻟﺰﻧﻚ ﻣﻨﺬ ﻓﺘﺮﺓ ﻃﻮﻳﻠﺔ ﻛﺴﺒﺎﺋﻚ ﻟﻤﻮﺍﺩ ﺍﻟﻨﺤﺎﺱ ﻭﺍﻟﻔﻮﻻﺫ‬‫ ﻭﻳﺘﻢ ﺍﺧﺘﺰﺍﻝ ﺍﻟﺮﺍﺳﺐ ﺍﻟﻤﺘﺤﺼﻞ ﻋﻠﻴﻪ ﻻﺳﺘﺮﺩﺍﺩﻩ ﻭﺇﻋﺎﺩﺓ ﺍﺳﺘﺨﺪﺍﻣﻪ ﻛﺰﻧﻚ‬،‫ﻋﻦ ﻃﺮﻳﻖ ﺍﻟﺘﺮﺳﻴﺐ ﻻﻧﺘﺰﺍﻉ ﺍﻟﺴﻤﻮﻡ ﻣﻦ ﺍﻟﻤﺎﺀ‬.‫ﻓﻠﺰﻱ‬A Part

The dissolution equilibrium of zinc hydroxide Zn(OH)2 (s) at 25 ∘ C and the relevant equilibrium constants
are given in eq. 1–4.
.4-1 ‫ ﻭﺛﻮﺍﺑﺖ ﺍﻻﺗﺰﺍﻥ ﺍﻟﻤﻮﺍﻓﻘﺔ ﻣﻌﻄﺎﺓ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻻﺕ‬25 ∘ C ‫ ﻋﻨﺪ‬Zn(OH)2 (s) ‫ﺍﺗﺰﺍﻥ ﺍﻟﺬﻭﺑﺎﻧﻴﺔ ﻟﻬﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺰﻧﻚ‬IChO
SAU-1 C-4 Q-2

Q4-2
Arabic (SAU) (Saudi Arabia)

The solubility, 𝑆, of zinc (concentration of zinc in a saturated aqueous solution) is given in eq. 5.
.5 ‫ )ﺗﺮﻛﻴﺰ ﺍﻟﺰﻧﻚ ﻓﻲ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ﻣﺸﺒﻊ( ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ‬,𝑆، ‫ﻛﻤﺎ ﺗﻌﻄﻰ ﺫﻭﺑﺎﻧﻴﺔ ﺍﻟﺰﻧﻚ‬

6pt

5pt

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)].
‫ ﺑﺤﻴﺚ ﻳﻜﻮﻥ‬pH ‫ ﺍﺣﺴﺐ ﺍﻟﻤﺠﺎﻝ ﻓﻲ ﻗﻴﻢ‬، 4-1 ‫ﻋﻨﺪ ﺍﻟﻮﺻﻮﻝ ﺍﻟﻰ ﺣﺎﻟﺔ ﺍﻻﺗﺰﺍﻥ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻻﺕ‬2+.[Zn(OH)2−
4 (aq)] ‫[ ﻭ‬Zn(OH)2 (aq)] ‫[ ﻭ‬Zn (aq)] ‫[ ﻫﻮ ﺍﻷﻋﻠﻰ ﺑﻴﻦ‬Zn(OH)2 (aq)] ‫ﺗﺮﻛﻴﺰ‬A 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.
‫ ﺛﻢ ﺃﺿﻴﻒ‬.‫ ﻭﺗﻢ ﺗﺮﺷﻴﺤﻪ‬pH=7.00 ‫ ﺫﻭ ﻗﻴﻤﺔ‬Zn(OH)2 (s) ‫ﺗﻢ ﺗﺤﻀﻴﺮ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ﻣﺸﺒﻊ ﻣﻦ‬‫ ﺍﺣﺴﺐ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺍﻟﻤﻮﻻﺭﻳﺔ‬.12.00 ‫ ﺇﻟﻰ‬pH ‫ ﺇﻟﻰ ﻫﺬﻩ ﺍﻟﺮﺷﺎﺣﺔ ﻟﺰﻳﺎﺩﺓ ﻗﻴﻤﺔ‬NaOH‫ ﺗﺠﺎﻫﻞ ﺍﻟﺘﻐﻴﺮﺍﺕ ﺍﻟﺤﺎﺻﻠﺔ ﻓﻲ ﺍﻟﺤﺠﻢ‬.12.00 ‫ ﺍﻟﻰ‬7.00 ‫ ﻣﻦ‬pH ‫ﻟﻠﺰﻧﻚ ﺍﻟﻤﺘﺮﺳﺐ ﻋﻨﺪ ﺍﺯﺩﻳﺎﺩ‬.‫ﻭﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‬

A.1

A.2

B PartNext, the recovered zinc hydroxide is heated to obtain zinc oxide according to the reaction below::‫ﻭﻓﻘﺎ ﻟﻠﺘﻔﺎﻋﻞ ﺃﺩﻧﺎﻩ‬‫ ﻳﺘﻢ ﺗﺴﺨﻴﻦ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺰﻧﻚ ﺍﻟﺬﻱ ﺗﻢ ﺍﺳﺘﺮﺩﺍﺩﻩ ﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ ﺃﻛﺴﻴﺪ ﺍﻟﺰﻧﻚ‬،‫ﺑﻌﺪ ﺫﻟﻚ‬ًThe zinc oxide is then reduced to metallic zinc by reaction with hydrogen::‫ﻳﺘﻢ ﺑﻌﺪ ﺫﻟﻚ ﺍﺧﺘﺰﺍﻝ ﺃﻛﺴﻴﺪ ﺍﻟﺰﻧﻚ ﺇﻟﻰ ﺍﻟﺰﻧﻚ ﺍﻟﻔﻠﺰﻱ ﺑﺎﻟﺘﻔﺎﻋﻞ ﻣﻊ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‬IChO
SAU-1 C-4 Q-3

4pt

Q4-3
Arabic (SAU) (Saudi Arabia)

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.
‫ ﻟﺬﻟﻚ ﻣﻦ ﺍﻟﻀﺮﻭﺭﻱ ﺧﻔﺾ ﺍﻟﻀﻐﻂ‬،1 bar ‫( ﻋﻨﺪ ﺿﻐﻂ ﻫﻴﺪﺭﻭﺟﻴﻦ‬7) ‫ﺣﺘﻰ ﻳﺴﺘﻤﺮ ﺍﻟﺘﻔﺎﻋﻞ‬‫ ﺍﺣﺴﺐ ﺍﻟﺤﺪ ﺍﻷﻋﻠﻰ ﻟﻠﻀﻐﻂ ﺍﻟﺠﺰﺋﻲ ﻟﺒﺨﺎﺭ ﺍﻟﻤﺎﺀ ﺍﻟﺬﻱ ﻳﺴﻤﺢ‬.‫ﺍﻟﺠﺰﺋﻲ ﻟﺒﺨﺎﺭ ﺍﻟﻤﺎﺀ ﺍﻟﻤﺘﻮﻟﺪ‬‫ ﺣﻴﺚ ﺗﻜﻮﻥ ﻃﺎﻗﺎﺕ ﺗﻜﻮﻳﻦ ﺟﻴﺒﺲ ﻷﻛﺴﻴﺪ ﺍﻟﺰﻧﻚ ﻭﺑﺨﺎﺭ‬. 300 ∘ C ‫( ﻋﻨﺪ‬7) ‫ﺑﺎﺳﺘﻤﺮﺍﺭ ﺍﻟﺘﻔﺎﻋﻞ‬‫ ﻭ‬Δ𝐺ZnO (300∘ C) = −2.90 × 102 kJ mol−1 ،‫ ﻫﻤﺎ ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‬1 bar ‫ ﻭ‬300 ∘ C ‫ﺍﻟﻤﺎﺀ ﻋﻨﺪ‬. Δ𝐺H2 O (300∘ C) = −2.20 × 102 kJ mol−1

B.1

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 ∘ .
‫ ﻳﺴﺘﺨﺪﻡ ﻓﻲ‬. ZnO ‫ ﻭ‬Zn ‫ ﻳﺘﻜﻮﻥ ﺍﻟﻘﻄﺐ ﻣﻦ‬.‫ﻳﺴﺘﺨﺪﻡ ﺍﻟﺰﻧﻚ ﺍﻟﻤﻌﺪﻧﻲ ﻛﻤﺎﺩﺓ ﻗﻄﺐ ﺳﺎﻟﺐ )ﺍﻧﻮﺩ( ﻟﺒﻄﺎﺭﻳﺎﺕ ﻫﻮﺍﺀ–ﻓﻠﺰ‬.𝐸 ∘ ،1 bar ‫ ﻭﺿﻐﻂ‬25 ∘ C ‫( ﻋﻨﺪ‬e.m.f) ‫ﺍﻻﺧﺘﺰﺍﻝ ﺍﻟﺘﺎﻟﻲ ﻟﺘﻮﻟﻴﺪ ﻛﻬﺮﺑﺎﺀ ﺑﻘﻮﺓ ﺩﺍﻓﻌﺔ ﻛﻬﺮﺑﺎﺋﻴﺔ‬-‫ﺗﻔﺎﻋﻞ ﺍﻻﻛﺴﺪﺓ‬3pt

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.
‫ ﺍﺣﺴﺐ ﺍﻟﺘﻐﻴﺮ ﻓﻲ ﻛﺘﻠﺔ ﺍﻟﻘﻄﺐ‬.‫ ﺳﺎﻋﺔ‬24 ‫ ﻟﻤﺪﺓ‬20 mA ‫ﻫﻮﺍﺀ ﻋﻨﺪ‬-‫ﺗﻢ ﺗﻔﺮﻳﻎ ﺑﻄﺎﺭﻳﺔ ﺯﻧﻚ‬.‫ﺍﻟﺴﺎﻟﺐ )ﺍﻻﻧﻮﺩ( ﻟﻠﺒﻄﺎﺭﻳﺔ‬Mt. Fuji

B.2

IChO
SAU-1 C-4 Q-4

5pt

Q4-4
Arabic (SAU) (Saudi Arabia)

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 ﺍﺣﺴﺐ‬.‫ﻫﻮﺍﺀ ﻳﻌﺘﻤﺪ ﻋﻠﻰ ﺍﻟﻮﺳﻂ ﺍﻟﻤﺤﻴﻂ‬-‫ ﻓﻲ ﺑﻄﺎﺭﻳﺔ ﺯﻧﻚ‬e.m.f ‫ﺑﻔﺮﺽ ﺍﻥ ﺍﻟﺘﻐﻴﺮ ﻓﻲ‬
،3776 m ‫ )ﻓﺒﺮﺍﻳﺮ( ﻭ‬−38 ∘ C ‫ ﺣﻴﺚ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻭﺍﻻﺭﺗﻔﺎﻉ‬،‫ ﻋﻨﺪ ﻗﻤﺔ ﺟﺒﻞ ﻓﻮﺟﻲ‬e.m.f ‫ﻗﻴﻤﺔ‬‫ ﺿﻐﻂ ﺍﻟﻐﻼﻑ ﺍﻟﺠﻮﻱ ﻳﻤﺜﻞ ﺑﺎﻟﻌﻼﻗﺔ‬.‫ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‬B.3

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.
‫ ﺗﻜﻮﻥ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻸﻛﺴﺠﻴﻦ ﻓﻲ ﺍﻟﻐﻼﻑ ﺍﻟﺠﻮﻱ‬. 𝑇 [∘ C] ‫ ﻭﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ‬ℎ [m] ‫ﻋﻨﺪ ﺍﺭﺗﻔﺎﻉ‬Δ𝐺ZnO (−38∘ C) = −3.26 × 102 kJ mol−1 ‫( ﻫﻮ‬8) ‫ ﻭﺍﻟﺘﻐﻴﺮ ﻓﻲ ﻃﺎﻗﺔ ﺟﻴﺒﺲ ﻟﻠﺘﻔﺎﻋﻞ‬. 21%.1 bar ‫ ﻭ‬−38 ∘ C ‫ﻋﻨﺪ‬
9pt

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.
‫ ﻻﺣﻆ ﺃﻥ ﻗﻴﻢ ﺟﻬﻮﺩ ﺍﻻﺧﺘﺰﺍﻝ‬. 25 ∘ C ‫( ﻋﻨﺪ‬6) ‫ﺍﺣﺴﺐ ﺍﻟﺘﻐﻴﺮ ﻓﻲ ﻃﺎﻗﺔ ﺟﻴﺒﺲ ﻟﻠﺘﻔﺎﻋﻞ‬،(11) ‫( ﻭ‬10) ‫ ﻣﻌﻄﺎﺓ ﻓﻲ‬1 bar ‫ ﻭ‬25 ∘ C ‫( ∘ 𝐸 ﻋﻨﺪ‬O2 /H2 O) ‫( ∘ 𝐸 ﻭ‬Zn2+ /Zn) ، ‫ﺍﻟﻘﻴﺎﺳﻴﺔ‬.‫ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‬

B.4

IChO
SAU-1 C-4 A-1

The Redox Chemistry of Zinc
Part A
A.1 (6 pt)< pH <

A4-1
English (Official)IChO
SAU-1 C-4 A-2

A.2 (5 pt)%

A4-2
English (Official)IChO
SAU-1 C-4 A-3

Part B
B.1 (4 pt)

bar

𝑝H2 O=
B.2 (3 pt)

g

A4-3
English (Official)IChO
SAU-1 C-4 A-4

B.3 (5 pt)

V

A4-4
English (Official)IChO
SAU-1 C-4 A-5

B.4 (9 pt)

Δ𝐺∘ =J mol−1

A4-5
English (Official)SAU-1 C-5 C-1

SAU-1 C-5 C
Mohammad AlHudaithi

IChO
Problem 5
Cover sheet

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

Q5-1

IChO
SAU-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
SAU-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
SAU-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
SAU-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
SAU-1 C-5 Q-1

Arabic (SAU) (Saudi Arabia)‫ﺍﻟﺴﻴﻠﻜﻮﻥ ﺍﻟﻐﺎﻣﺾ‬
Total
60

B.3
8

B.2
15

total the of % 12

B.1
5

A.4
10

A.3
6

A.2
7

A.1
9

Question
Points
Score

Although silicon is also a group 14 element like carbon, their properties differ significantly.
.‫ ﺇﻻ ﺃﻥ ﺧﺼﺎﺋﺼﻪ ﺗﺨﺘﻠﻒ ﺑﺸﻜﻞ ﻛﺒﻴﺮ‬، ‫ ﻣﺜﻞ ﺍﻟﻜﺮﺑﻮﻥ‬14 ‫ﺃﻳﻀﺎ ﻋﻨﺼﺮ ﻣﻦ ﺍﻟﻤﺠﻤﻮﻋﺔ‬ً ‫ﻋﻠﻰ ﺍﻟﺮﻏﻢ ﻣﻦ ﺃﻥ ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﻫﻮ‬

A PartUnlike 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.
‫ ﻓﺈﻥ ﺍﻟﺮﺍﺑﻄﺔ ﺍﻟﺜﻼﺛﻴﺔ ﻟﻠﺴﻴﻠﻴﻜﻮﻥ ﻭﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﻓﻲ ﻣﺮﻛﺐ ﺗﻤﺖ‬، ‫ﻋﻠﻰ ﻋﻜﺲ ﺍﻟﺮﺍﺑﻄﺔ ﺍﻟﺜﻼﺛﻴﺔ ﺑﻴﻦ ﺍﻟﻜﺮﺑﻮﻥ ﻭﺍﻟﻜﺮﺑﻮﻥ‬R1 –Si ≡ Si–R1 ‫ﺻﻴﺎﻏﺘﻪ ﻋﻠﻰ ﺃﻧﻪ‬ring ‫ ﻳﺘﻔﺎﻋﻞ ﻣﻊ ﺍﻹﻳﺜﻴﻠﻴﻦ ﻟﺘﺸﻜﻴﻞ ﻣﻨﺘﺞ ﺣﻠﻘﻲ ﻳﺤﺘﻮﻱ ﻋﻠﻰ‬، ‫ ﻋﻠﻰ ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ‬.‫( ﺷﺪﻳﺪ ﺍﻟﺘﻔﺎﻋﻞ‬R:‫) ﺑﺪﻳﻞ ﻋﻀﻮﻱ‬membered-four.

IChO
SAU-1 C-5 Q-2

Q5-2
Arabic (SAU) (Saudi Arabia)

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 .

‫ ﺗﻔﺎﻋﻞ‬.‫ ﻛﻮﺳﻴﻂ ﺃﻭﻟﻲ‬A four-membered-ring ‫ ﻳﺘﺸﻜﻞ ﻣﺮﻛﺐ‬R2 –C ≡ C–R2 ‫ﺑﺄﻟﻜﻴﻦ‬R1 –Si ≡ Si–R1 ‫ﻋﻨﺪﻣﺎ ﻳﺘﻢ ﻣﻌﺎﻟﺠﺔ‬‫ ﻭﻛﻼﻫﻤﺎ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﻫﻴﺎﻛﻞ ﻣﺘﺮﺍﻓﻘﺔ ﺩﻭﺭﻳﺔ ﺗﺸﺒﻪ‬، C ‫ ﻭ‬B ‫ ﻳﻮﻓﺮ ﺃﻳﺰﻭﻣﺮﻳﻦ‬A ‫ ﻣﻊ‬R2 –C ≡ C–R2 ‫ﺇﺿﺎﻓﻲ ﻟﺠﺰﻱﺀ ﺁﺧﺮ ﻣﻦ‬(R1 –Si)2 (R2 –C)4 ‫ ﻭﻳﻤﻜﻦ ﺻﻴﺎﻏﺘﻬﺎ‬six-membered-ring ‫ ” ﺍﻟﺘﻲ ﺗﺤﺘﻮﻱ ﻋﻠﻰ‬disilabenzenes “ ‫ ﻭﺗﺴﻤﻰ‬، ‫ﺍﻟﺒﻨﺰﻳﻦ‬The 13 C NMR analysis of the corresponding six-membered ring skeletons Si2 C4 shows two signals for B
and one signal for C.
.C ‫ ﻭﺇﺷﺎﺭﺓ ﻭﺍﺣﺪﺓ ﻟـ‬B ‫ ﺇﺷﺎﺭﺗﻴﻦ ﻟـ‬Si2 C4 ‫ ﺍﻟﻤﺘﻘﺎﺑﻠﺔ‬six-membered-ring ‫ ﻟﻠﻬﻴﺎﻛﻞ ﺍﻟﻌﻈﻤﻴﺔ‬13 C NMR ‫ُﻳﻈﻬﺮ ﺗﺤﻠﻴﻞ‬
9pt

7pt

Draw the structural formulae of A, B, and C using R1 , R2 , Si, and C, with one ofthe possible resonance structures.
‫ ﻣﻊ ﺃﺣﺪ ﻫﻴﺎﻛﻞ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻤﻤﻜﻨﺔ‬, C ‫ ﻭ‬Si ,R2 ,R1 ‫ ﺑﺎﺳﺘﺨﺪﺍﻡ‬C ‫ ﻭ‬B, A ‫ﺍﺭﺳﻢ ﺍﻟﺼﻴﻎ ﺍﻟﺒﻨﺎﺋﻴﺔ ﻟـ‬

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).
‫( ﻛﻘﻴﻢ ﻣﻮﺟﺒﺔ‬R1 = R2 = H ‫ )ﻓﻲ ﺣﺎﻟﺔ‬C ‫( ﻟﻠﺒﻨﺰﻳﻦ ﻭ‬ASE) ‫ﺍﺣﺴﺐ ﻃﺎﻗﺔ ﺍﻟﺘﺜﺒﻴﺖ ﺍﻟﻌﻄﺮﻱ‬‫ ﻣﻊ ﺍﻷﺧﺬ ﻓﻲ ﺍﻻﻋﺘﺒﺎﺭ ﺗﻐﻴﺮ ﺍﻟﻤﺤﺘﻮﻯ ﺍﻟﺤﺮﺍﺭﻱ ﻓﻲ ﺑﻌﺾ ﺗﻔﺎﻋﻼﺕ ﺍﻟﻬﺪﺭﺟﺔ ﻟﻸﻧﻈﻤﺔ ﻏﻴﺮ‬،.(1 ‫ﺍﻟﻤﺸﺒﻌﺔ ﺍﻟﻤﻮﺿﺤﺔ ﺃﺩﻧﺎﻩ )ﺍﻟﺸﻜﻞ‬

A.1

A.2

IChO
SAU-1 C-5 Q-3

1 Fig.

Q5-3
Arabic (SAU) (Saudi Arabia)

Q5-4

IChO
SAU-1 C-5 Q-4

Arabic (SAU) (Saudi Arabia)

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.
‫ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ‬.E ‫ ﻭ‬D ‫ ﻓﺈﻧﻪ ﻳﺨﻀﻊ ﻷﺯﻣﺮﺓ ﻹﻋﻄﺎﺀ ﺧﻠﻴﻂ ﺗﻮﺍﺯﻥ ﻣﻦ ﺍﻟﻤﺮﻛﺒﺎﺕ‬، C ‫ ﻣﻦ‬xylene ‫ﻋﻨﺪﻣﺎ ﻳﺘﻢ ﺗﺴﺨﻴﻦ ﻣﺤﻠﻮﻝ‬‫ﻫﻲ‬D : E = 1 : 40.0 at 50.0 ˚C and D : E = 1 : 20.0 at 120.0 ˚C.
6pt

Calculate Δ𝐻 for the transformation of D to E. Assume that Δ𝐻 does not depend on temperature.
‫ ﻻ ﺗﻌﺘﻤﺪ ﻋﻠﻰ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‬Δ𝐻 ‫ ﺍﻓﺘﺮﺽ ﺃﻥ‬. E ‫ ﺇﻟﻰ‬D ‫ ﻟﻠﺘﺤﻮﻳﻞ ﻣﻦ‬Δ𝐻 ‫ﺍﺣﺴﺐ‬A.3

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.
NMR ‫ ﻛﺸﻒ ﺗﺤﻠﻴﻞ‬. 𝜎-bonds ‫𝜎ﺩﻭﻥ ﻛﺴﺮ ﺃﻱ‬-bonds ‫𝜋 ﺇﻟﻰ‬-bonds ‫ ﻋﺒﺮ ﺗﺤﻮﻳﻼﺕ ﻣﻦ‬E ‫ ﻭ‬D ‫ ﺇﻟﻰ‬C ‫ﺗﺴﺘﻤﺮ ﺍﻷﺯﻣﺮﺓ ﻣﻦ‬‫ ﻻ ﻳﺤﺘﻮﻱ ﻋﻠﻰ‬D ‫ ﺍﻟﻬﻴﻜﻞ ﺍﻟﻌﻈﻤﻲ ﻟـ‬.E ‫ ﻭ ﺇﺷﺎﺭﺗﻴﻦ ﻟﺘﻠﻚ ﺍﻟﺨﺎﺻﺔ ﺑـ‬D ‫ ﻟـ‬Si2 C4 ‫ ﻋﻦ ﺇﺷﺎﺭﺓ ﻭﺍﺣﺪﺓ ﻟﻠﻬﻴﻜﻞ ﺍﻟﻌﻈﻤﻲ‬13 C.‫ ﺗﺸﺘﺮﻙ ﻓﻲ ﺍﻟﺤﺎﻓﺔ‬three-membered-rings ‫ ﻋﻠﻰ ﺍﺛﻨﻴﻦ‬E ‫ ﺑﻴﻨﻤﺎ ﻳﺤﺘﻮﻱ‬, three-membered-ring ‫ﺃﻱ‬
10pt

Draw the structural formulae of D and E using R1 , R2 , Si, and C.
C ‫ ﻭ‬, Si ,R2 ,R1 ‫ ﺑﺎﺳﺘﺨﺪﺍﻡ‬E ‫ ﻭ‬D ‫ﺍﺭﺳﻢ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺒﻨﺎﺋﻴﺔ ﻝ‬

A.4

B PartSilicon 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.
‫ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﻗﺎﺩﺭ ﻋﻠﻰ ﺗﻜﻮﻳﻦ ﻣﺮﻛﺒﺎﺕ ﻋﺎﻟﻴﺔ ﺍﻟﺘﻨﺴﻴﻖ‬(> four substituents)‫ ﻓﺈﻥ ﻓﻠﻮﺭﻳﺪﺍﺕ‬، ‫ﻏﺎﻟﺒﺎ ﻣﺎ ﺗﺴﺘﺨﺪﻡ ﻛﻮﺍﺷﻒ ﻟﻠﻔﻠﻮﺭﺓ‬‫ﻧﻈﺮﺍ ﻷﻥ ﺍﻟﻔﻠﻮﺭﻳﺪﺍﺕ ﺍﻟﻤﻌﺪﻧﻴﺔ‬.‫ﻣﻊ ﻋﻨﺎﺻﺮ ﻛﻬﺮﺑﻴﺔ ﻣﺜﻞ ﺍﻟﻔﻠﻮﺭ‬ًً.‫ﺃﻳﻀﺎ ﻛﻜﻮﺍﺷﻒ ﻓﻠﻮﺭﺓ‬ً ‫ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﻋﺎﻟﻴﺔ ﺍﻟﺘﻨﺴﻴﻖ ﺗﻌﻤﻞ‬The fluorination reaction of CCl4 using Na2 SiF6 was carried out as follows.

‫ ﺗﺘﻢ ﻋﻠﻰ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﻟﻲ‬Na2 SiF6 ‫ ﺑﺎﺳﺘﺨﺪﺍﻡ‬CCl4 ‫ﺗﻔﺎﻋﻼﺕ ﺍﻟﻔﻠﻮﺭﺓ ﻝ‬

standardization 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).

:‫ﺍﻟﺘﺤﻀﻴﺮ‬.( 200mL :‫( ﻣﺬﺍﺏ ﻓﻲ ﺍﻟﻤﺎﺀ )ﺍﻟﺤﺠﻢ ﺍﻟﻜﻠﻲ‬188.053 g mol–1 ) Na2 SiF6 ‫ ﻣﻦ‬0.855 g : F ‫ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ‬( 200mL :‫( ﻣﺬﺍﺏ ﻓﻲ ﺍﻟﻤﺎﺀ )ﺍﻟﺤﺠﻢ ﺍﻟﻜﻠﻲ‬568.424 g mol–1 ) Ce2 (SO4 )3 ‫ ﻣﻦ‬6.86 g : G ‫ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ‬

Q5-5

IChO
SAU-1 C-5 Q-5Arabic (SAU) (Saudi Arabia)· 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 .

:‫ﺍﻟﻄﺮﻳﻘﺔ‬‫ ﻭﺍﻟﺬﻱ ﻳﻨﺴﻖ‬، ‫ ﺍﻟﺒﺮﺗﻘﺎﻟﻲ‬xylenol ‫ ﺑﺎﻟﺘﻨﻘﻴﻂ ﻓﻲ ﻭﺟﻮﺩ‬G ‫ ﻋﻦ ﻃﺮﻳﻖ ﺇﺿﺎﻓﺔ ﻣﺤﻠﻮﻝ‬50.0 mL F ‫ﻣﻌﺎﻳﺮﺓ ﺍﻟﺘﺮﺳﻴﺐ ﻟﻤﺤﻠﻮﻝ‬‫ ﺍﻟﺮﺍﺳﺐ ﺍﻟﻤﺘﻜﻮﻥ‬.‫ ﻳﺘﻐﻴﺮ ﻟﻮﻥ ﺍﻟﻤﺤﻠﻮﻝ ﻣﻦ ﺍﻷﺻﻔﺮ ﺇﻟﻰ ﺍﻷﺭﺟﻮﺍﻧﻲ‬، G ‫ ﻣﻦ ﺍﻟﻤﺤﻠﻮﻝ‬18.8 mL ‫ ﺑﻌﺪ ﺇﺿﺎﻓﺔ‬.‫ ﻛﺪﻟﻴﻞ‬Ce3+ ‫ﺇﻟﻰ‬
Si(OH)4 ‫ ﻭﻣﺮﻛﺐ ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﺍﻟﻨﺎﺗﺞ ﺍﻟﻮﺣﻴﺪ ﻫﻮ‬Ce3+ ‫ﻫﻮ ﻣﺮﻛﺐ ﺛﻨﺎﺋﻲ ﻳﺤﺘﻮﻱ ﻋﻠﻰ‬5pt

Write the balanced equation for the reaction of Na2 SiF6 with Ce2 (SO4 )3 .
Ce2 (SO4 )3 ‫ ﻣﻊ‬Na2 SiF6 ‫ﺃﻛﺘﺐ ﻣﻌﺎﺩﻟﺔ ﻣﻮﺯﻭﻧﺔ ﻟﺘﻔﺎﻋﻞ‬B.1•:Na2 SiF6 ‫ ﻣﻊ ﺍﻝ‬CCl4 ‫• ﺗﻔﺎﻋﻞ ﺍﻝ‬(Substance losses by e.g. evaporation are negligible during the following operations.)(.‫ ﺍﻟﺘﺒﺨﺮ ﻻ ﻳﻜﺎﺩ ﻳﺬﻛﺮ ﺃﺛﻨﺎﺀ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ‬، ‫)ﺧﺴﺎﺋﺮ ﺍﻟﻤﺎﺩﺓ ﻋﻠﻰ ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ‬Na2 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.
‫ ﺩﺭﺟﺔ ﻣﺌﻮﻳﺔ ﻓﻲ ﻭﻋﺎﺀ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻤﺤﻜﻢ‬300 ‫ ﻭﺗﻢ ﺗﺴﺨﻴﻨﻬﺎ ﺇﻟﻰ‬500.0 g CCl4 ‫ ﺇﻟﻰ‬Na2 SiF6 ( 𝑥 [g] ) ‫ﺗﻤﺖ ﺇﺿﺎﻓﺔ‬‫ ﻭﺗﻢ ﺗﺨﻔﻴﻒ ﺍﻟﻤﺎﺩﺓ‬. ‫ ﺍﻟﻤﺘﻮﻟﺪ ﻋﻦ ﻃﺮﻳﻖ ﺍﻟﻔﻠﺘﺮﺓ‬NaCl ‫ ﺍﻟﻐﻴﺮ ﻣﺘﻔﺎﻋﻞ ﻭ‬Na2 SiF6 ‫ ﺗﻤﺖ ﺇﺯﺍﻟﺔ‬.‫ﺍﻟﻐﻠﻖ ﻭﺍﻟﻤﻘﺎﻭﻡ ﻟﻠﻀﻐﻂ‬‫ ﻛﻤﺮﻛﺐ‬SiF4 H ‫ﻟﻤﺤﻠﻮﻝ‬19 F NMR ‫ﻭ‬29 Si ‫( ﻭﺃﻇﻬﺮﺕ ﺃﻃﻴﺎﻑ‬H ‫ )ﻣﺤﻠﻮﻝ‬CCl4 ‫ ﻣﻊ‬L 1.00 ‫ﺍﻟﻤﻔﻠﺘﺮﺓ ﺇﻟﻰ ﺣﺠﻢ ﺇﺟﻤﺎﻟﻲ ﻗﺪﺭﻩ‬CF3 Cl , CF2 Cl2 , CFCl3 ‫ ﻟﻮﺣﻈﺖ ﺍﻹﺷﺎﺭﺍﺕ ﺍﻟﻤﻘﺎﺑﻠﺔ ﻟـ‬، SiF4 ‫ ﺑﺎﻹﺿﺎﻓﺔ ﺇﻟﻰ‬، 19 F NMR ‫ ﻓﻲ ”ﻃﻴﻒ‬.‫ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﺍﻟﻮﺣﻴﺪ‬.‫ ﻣﺘﻨﺎﺳﺒﺔ ﻣﻊ ﻋﺪﺩ ﻣﻮﻻﺕ ﻧﻮﻯ ﺍﻟﻔﻠﻮﺭ‬19 F NMR ‫( ﻧﺴﺐ ﺍﻟﺘﻜﺎﻣﻞ ﻓﻲ ﻃﻴﻒ‬1 ‫ )ﺍﻟﺠﺪﻭﻝ‬CF4 ‫ﻭ‬

CF4

2.0

CF3 Cl18.0

1 TableCF2 Cl2

65.0

CFCl3

45.0

NMR data 19 F

Integration ratio

SiF4 is hydrolyzed to form H2 SiF6 according to the following eq. 8::‫ﻭﻓﻘﺎ ﻟﻤﺎ ﻳﻠﻲ‬H2 SiF6 ‫ ﻳﺘﻢ ﺗﺤﻠﻴﻠﻪ ﻟﺘﺸﻜﻴﻞ‬SiF4
ً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).

IChO
SAU-1 C-5 Q-6

Q5-6
Arabic (SAU) (Saudi Arabia)‫ ﺗﻢ‬، ‫ ﺑﻌﺪ ﺍﻟﻔﺼﻞ‬.SiF4 ‫ ﻣﻤﺎ ﺃﺩﻯ ﺇﻟﻰ ﺍﻟﺘﺤﻠﻞ ﺍﻟﻤﺎﺋﻲ ﺍﻟﻜﺎﻣﻞ ﻟـ‬, ‫ ﺗﻤﺖ ﺇﺿﺎﻓﺘﻪ ﻟﻜﻤﻴﺔ ﺯﺍﺋﺪﺓ ﻣﻦ ﺍﻟﻤﺎﺀ‬10 mL H ‫ﻣﺤﻠﻮﻝ‬( J ‫ )ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ‬Na2 SiF6 ‫ ﺍﻟﻤﺘﻮﻟﺪ ﻣﻦ ﺍﻟﺘﺤﻠﻞ ﺍﻟﻤﺎﺋﻲ ﻓﻲ ﺍﻟﻤﺤﻠﻮﻝ ﺍﻟﻤﺎﺋﻲ ﻭﺗﺤﻮﻳﻠﻪ ﺑﺎﻟﻜﺎﻣﻞ ﺇﻟﻰ‬، H2 SiF6 ‫ﻣﻌﺎﺩﻟﺔ‬The 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).
‫ﺗﻤﺎﻣﺎ‬، (‫ ﻭﺍﻟﺬﻱ ﺗﻤﺖ ﺇﺯﺍﻟﺘﻪ ﺑﺎﻟﻔﻠﺘﺮﺓ ﻓﻲ ﺍﻟﺨﻄﻮﺓ ﺍﻷﻭﻟﻴﺔ )ﺗﺤﺘﻪ ﺧﻂ‬، NaCl ‫ ﺍﻟﻐﻴﺮ ﻣﺘﻔﺎﻋﻞ ﻭ‬Na2 SiF6 ‫ﺗﻢ ﺇﺫﺍﺑﺔ ﺍﻟﺮﺍﺳﺐ‬ً
.(10.0 L :K ‫ﻓﻲ ﺍﻟﻤﺎﺀ ﻹﻋﻄﺎﺀ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ )ﻣﺤﻠﻮﻝ‬Then, additional precipitation titrations using solution G were carried out, and the endpoints of the titrations with G were as follows::‫ ﻛﺎﻧﺖ ﻛﻤﺎ ﻳﻠﻲ‬G ‫ ﻭﻧﻘﻄﺔ ﻧﻬﺎﻳﺔ ﺍﻟﻤﻌﺎﻳﺮﺓ ﻣﻊ‬,‫ ﺗﻢ ﺗﻨﻔﻴﺬﻫﺎ‬G ‫ ﻣﻌﺎﻳﺮﺓ ﺗﺮﺳﻴﺐ ﺇﺿﺎﻓﻴﺔ ﺑﺎﺳﺘﺨﺪﺍﻡ ﻣﺤﻠﻮﻝ‬,‫ﺑﻌﺪ ﺫﻟﻚ‬·For solution J (entire amount): 61.6 mL .
·For 100 mL of solution K: 44.4 mL .
.61.6 mL :(‫ )ﺍﻟﻜﻤﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ‬J ‫ﻟﻠﻤﺤﻠﻮﻝ‬. 44.4 mL :K ‫ ﻣﻦ ﻣﺤﻠﻮﻝ‬100 mL ‫ﻟﻜﻞ‬It should be noted here that the coexistence of NaCl or SiO2 has no effect on the precipitation titration.

.‫ ﻟﻴﺲ ﻟﻪ ﺃﻱ ﺗﺄﺛﻴﺮ ﻋﻠﻰ ﻣﻌﺎﻳﺮﺓ ﺍﻟﺘﺮﺳﻴﺐ‬SiO2 ‫ ﺃﻭ‬NaCl ‫ﻭﺗﺠﺪﺭ ﺍﻹﺷﺎﺭﺓ ﻫﻨﺎ ﺇﻟﻰ ﺃﻥ ﺗﻌﺎﻳﺶ‬15pt

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.
)‫ ﻭﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ‬.(‫ ﺍﻟﻤﻨﺘﺠﺔ ﻓﻲ ﻭﻋﺎﺀ ﺍﻟﺘﻔﺎﻋﻞ )ﺍﻟﻤﻌﻠﻮﻣﺎﺕ ﺗﺤﺘﻬﺎ ﺧﻂ‬NaCl ‫ﺍﺣﺴﺐ ﻛﺘﻠﺔ‬B.2.‫ ﺍﻟﻤﺴﺘﺨﺪﻣﺔ ﻛﻤﻮﺍﺩ ﺃﻭﻟﻴﺔ‬Na2 SiF6 ‫[ 𝑥 ( ﻣﻦ‬g]

8pt

77.8% of the CCl4 used as a starting material was unreacted. Calculate the massof CF3 Cl generated.
.‫ ﺍﻟﻤﺘﻮﻟﺪﺓ‬CF3 Cl ‫ ﺍﺣﺴﺐ ﻛﺘﻠﺔ‬.‫ ﺍﻟﻤﺴﺘﺨﺪﻣﺔ ﻛﻤﻮﺍﺩ ﺃﻭﻟﻴﺔ ﻟﻢ ﺗﺘﻔﺎﻋﻞ‬، CCl4 ‫ ﻣﻦ‬77.8 %

B.3

IChO

A5-1

SAU-1 C-5 A-1

English (Official)

Mysterious Silicon
Part 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
SAU-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
English (Official)IChO
SAU-1 C-5 A-3

Part B
B.1 (5 pt)

B.2 (15 pt)(Continued on the next page)

A5-3
English (Official)IChO
SAU-1 C-5 A-4

B.2 (cont.)

NaCl ∶

g, Na2 SiF6 ∶g

A5-4
English (Official)IChO
SAU-1 C-5 A-5

B.3 (8 pt)

CF3 Cl ∶g

A5-5English (Official)SAU-1 C-6 C-1

SAU-1 C-6 C
Mohammad AlHudaithi

IChO
Problem 6
Cover sheet

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

Q6-1

IChO
SAU-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
SAU-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
SAU-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
SAU-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
SAU-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
SAU-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
SAU-1 C-6 Q-1

Arabic (SAU) (Saudi Arabia)‫ﻛﻴﻤﻴﺎﺀ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺼﻠﺒﺔ ﻟﻠﻔﻠﺰﺍﺕ ﺍﻻﻧﺘﻘﺎﻟﻴﺔ‬13 % of the total
Total
45

C.3
5

C.2
5

C.1
5

B.4
4

B.3
4

B.2
4

B.1
6

A.3
3

A.2
3

A.1
6

Question
PointsScore

Volcano at Sakurajima island

A PartJapan 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:‫ ﻋﻨﺪﻣﺎ ﺗﺘﺒﻠﻮﺭ ﻣﻌﺎﺩﻥ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ‬.‫ﺍﻟﻴﺎﺑﺎﻥ ﻫﻲ ﻭﺍﺣﺪﺓ ﻣﻦ ﺍﻟﺪﻭﻝ ﺍﻟﺘﻲ ﻟﺪﻳﻬﺎ ﺃﻛﺒﺮ ﻋﺪﺩ ﻣﻦ ﺍﻟﺒﺮﺍﻛﻴﻦ ﻓﻲ ﺟﻤﻴﻊ ﺃﻧﺤﺎﺀ ﺍﻟﻌﺎﻟﻢ‬Mn+ ‫ ﺗﻜﻮﻥ‬.‫( ﻓﻲ ﺍﻟﺼﻬﺎﺭﺓ ﻓﻲ ﻣﻌﺎﺩﻥ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ‬Mn+ ) ‫ ﻳﺤﺼﻞ ﺍﻧﺪﻣﺎﺝ ﺟﺰﺀ ﻣﻦ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻤﻌﺎﺩﻥ ﺍﻻﻧﺘﻘﺎﻟﻴﺔ‬،‫ﻣﻦ ﺍﻟﺼﻬﺎﺭﺓ‬‫( ﻭﺗﺸﻜﻞ ﻋﺪﺩ ﺗﻨﺎﺳﻘﻲ ﺃﺭﺑﻌﺔ ﻓﻲ ﺷﻜﻞ ﺭﺑﺎﻋﻲ‬O2− ) ‫ﺍﻟﻤﺪﺭﻭﺳﺔ ﻓﻲ ﻫﺬﻩ ﺍﻟﻤﺴﺄﻟﺔ ﻣﺮﺗﺒﻄﺔ ﺗﻨﺎﺳﻘﻴﺎ ﻣﻊ ﺃﻳﻮﻧﺎﺕ ﺍﻷﻛﺴﻴﺪ‬‫ ﻭﻛﻼﻫﻤﺎ‬،‫𝑂( ﻓﻲ ﻣﻌﺎﺩﻥ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ‬h ) ‫𝑇( ﻓﻲ ﺍﻟﺼﻬﺎﺭﺓ ﻭﺗﺸﻜﻞ ﻋﺪﺩ ﺗﻨﺎﺳﻘﻲ ﺳﺘﺔ ﻓﻲ ﺷﻜﻞ ﺛﻤﺎﻧﻲ ﺍﻟﻮﺟﻮﻩ‬d ) ‫ﺍﻷﻭﺟﻪ‬،𝐷 ،‫ ﺑﻴﻦ ﻣﻌﺎﺩﻥ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ ﻭﺍﻟﺼﻬﺎﺭﺓ‬Mn+ ‫ ﻳﻤﻜﻦ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻦ ﻣﻌﺎﻣﻞ ﺍﻟﺘﻮﺯﻳﻊ ﻟﻼﻳﻮﻥ‬.‫ﻳﻈﻬﺮ ﺗﻜﻮﻳﻦ ﻏﺰﻝ ﺍﻟﻜﺘﺮﻭﻧﻲ ﻣﺮﺗﻔﻊ‬:‫ﺑﺎﻟﻌﻼﻗﺔ‬IChO
SAU-1 C-6 Q-2

Q6-2
Arabic (SAU) (Saudi Arabia)

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.

‫ ﻳﻮﺿﺢ ﺍﻟﺠﺪﻭﻝ ﺃﺩﻧﺎﻩ ﻗﻴﻢ‬.‫ ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‬،‫ ﻓﻲ ﻛﻞ ﻣﻦ ﻣﻌﺎﺩﻥ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ ﻭﺍﻟﺼﻬﺎﺭﺓ‬Mn+ ‫[ ﺗﻤﺜﻞ ﺗﺮﺍﻛﻴﺰ‬M]l ‫[ ﻭ‬M]s ‫ﺣﻴﺚ‬.‫ ﻛﺄﻣﺜﻠﺔ‬Mn2+ ‫ ﻭ‬Cr2+ ‫𝐷 ﻟـ‬

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.

،𝑂h ‫ ﻭﻃﺎﻗﺔ ﺍﺳﺘﻘﺮﺍﺭ ﺍﻟﻤﺠﺎﻝ ﺍﻟﺒﻠﻮﺭﻱ ﻓﻲ ﻣﺠﺎﻝ‬Mn+ ‫ ﻟﻸﻳﻮﻥ‬d -‫ ﻳﻤﺜﻼﻥ ﻃﺎﻗﺔ ﺍﻻﻧﻘﺴﺎﻡ ﻟﻤﺪﺍﺭﺍﺕ‬CFSEO ‫ ﻭ‬ΔO ‫ﺑﻔﺮﺽ‬. 𝑇d ‫ ﻳﻤﺜﻼﻧﻬﻤﺎ ﻓﻲ ﻣﺠﺎﻝ‬CFSET ‫ ﻭ‬ΔT ‫ ﻭﺑﻔﺮﺽ‬.‫ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‬
6pt

3pt

Calculate |CFSEO −CFSET | = ΔCFSE in terms of ΔO for Cr2+ , Mn2+ , and Co2+ ;assume ΔT = 4/9ΔO .
‫؛‬Co2+ ‫ ﻭ‬Mn2+ ‫ ﻭ‬Cr2+ ‫ ﻟﻜﻞ ﻣﻦ‬ΔO ‫| ﺑﺪﻻﻟﺔ ﺍﻟﻤﺼﻄﻠﺤﺎﺕ‬CFSEO −CFSET | = ΔCFSE ‫ﺍﺣﺴﺐ‬.ΔT = 4/9ΔO ‫ﺍﻓﺘﺮﺽ‬A linear relationship is observed by plotting ln𝐷 against ΔCFSE / ΔO in the Cartesian coordinate system shown below.
Estimate 𝐷 for Co2+ .
‫ ﻓﻲ ﻧﻈﺎﻡ ﺍﻹﺣﺪﺍﺛﻴﺎﺕ‬ΔCFSE/ΔO ‫ ﻣﻘﺎﺑﻞ‬ln𝐷 ‫ﻟﻮﺣﻆ ﻭﺟﻮﺩ ﻋﻼﻗﺔ ﺧﻄﻴﺔ ﻣﻦ ﺧﻼﻝ ﺭﺳﻢ‬.‫ﺍﻟﺪﻳﻜﺎﺭﺗﻴﺔ ﺍﻟﻤﻮﺿﺢ ﺃﺩﻧﺎﻩ‬.Co2+ ‫ﻋﻴﻦ 𝐷 ﻷﻳﻮﻥ‬
ِ

A.1

A.2

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 from‫‪Q6-3‬‬‫)‪Arabic (SAU) (Saudi Arabia‬‬‫‪IChO‬‬‫‪SAU-1 C-6 Q-3‬‬‫‪the CFSE of Mn+ in the 𝑂h field.‬‬‫ﺗﺘﺒﻠﻮﺭ ﺃﻛﺎﺳﻴﺪ ﺍﻟﻤﻌﺎﺩﻥ ‪) MO‬ﺣﻴﺚ ﺍﻟﻤﻌﺪﻥ ﻳﻤﺜﻞ ﺃﻱ ﻣﻦ ﺍﻟﻤﻌﺎﺩﻥ ﺍﻟﺘﺎﻟﻴﺔ ‪ Mn V، Ti، ،Ca،‬ﺃﻭ ‪ (Co‬ﻋﻠﻰ ﻫﻴﺌﺔ ﺑﻠﻮﺭﺓ ﺍﻟﻤﻠﺢ‬‫ﺍﻟﺼﺨﺮﻱ ﺣﻴﺚ ﻳﺘﺒﻨﻰ ‪ Mn+‬ﺍﻟﺸﻜﻞ ﺍﻟﻬﻨﺪﺳﻲ ‪ 𝑂h‬ﻣﻊ ﺗﺮﻛﻴﺐ ﺇﻟﻜﺘﺮﻭﻧﻲ ﻣﺮﺗﻔﻊ ﺍﻟﻐﺰﻝ‪ .‬ﺗﺮﺗﺒﻂ ﺍﻧﺜﺎﻟﺒﻴﺔ ﺍﻟﺸﺒﻜﺔ ﺍﻟﺒﻠﻮﺭﻳﺔ‬‫ﻟﻬﺬﻩ ﺍﻷﻛﺎﺳﻴﺪ ﺑﺸﻜﻞ ﺃﺳﺎﺳﻲ ﺑﺘﺄﺛﻴﺮﺍﺕ ﻛﻮﻟﻮﻡ ﺑﻴﻦ ﺍﻟﺸﺤﻨﺎﺕ ﺍﻟﺘﻲ ﺗﻌﺘﻤﺪ ﻋﻠﻰ ﺃﺳﺎﺱ ﻧﺼﻒ ﻗﻄﺮ ﻭﺷﺤﻨﺔ ﺍﻷﻳﻮﻧﺎﺕ‬‫ﻭﺑﻌﺾ ﺍﻟﻤﺴﺎﻫﻤﺎﺕ ﻣﻦ ‪ CFSE‬ﻟﻸﻳﻮﻥ ‪ Mn+‬ﻓﻲ ﻣﺠﺎﻝ ‪.𝑂h‬‬‫‪A.3‬‬‫‪Choose the appropriate set of lattice enthalpies [kJ mol−1 ] from one of the op‬‬‫‪tions (a) to (f).‬‬‫ﺍﺧﺘﺮ ﺍﻟﻤﺠﻤﻮﻋﺔ ﺍﻟﻤﻨﺎﺳﺒﺔ ﻣﻦ ﺍﻧﺜﺎﻟﺒﻴﺎﺕ ﺍﻟﺸﺒﻜﺔ ﺍﻟﺒﻠﻮﺭﻳﺔ ] ‪ [kJ mol−1‬ﻣﻦ ﺃﺣﺪ ﺍﻟﺨﻴﺎﺭﺍﺕ )‪(a‬‬‫ﺇﻟﻰ ‪(f).‬‬‫‪3pt‬‬IChO
SAU-1 C-6 Q-4

Q6-4
Arabic (SAU) (Saudi Arabia)

B PartA 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 ).
‫ ﻓﻲ ﺍﻟﺸﻜﻞ‬.Fig.1 ‫ ﻳﺘﺒﻠﻮﺭ ﻓﻲ ﺧﻠﻴﺔ ﻭﺣﺪﺓ ﺭﺑﺎﻋﻴﺔ ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﻓﻲ‬،Cu2+ ‫ ﻭ‬La3+ ‫ ﻳﺤﺘﻮﻱ ﻋﻠﻰ‬، A ‫ﻣﺰﻳﺞ ﺃﻛﺴﻴﺪ‬‫[ ﻣﻦ‬CuO6 ] ‫ ﻭﻳﺘﺸﻮﻩ‬، (𝑙𝑥 ) x ‫ ﺃﻃﻮﻝ ﻣﻦ ﺗﻠﻚ ﻋﻠﻰ ﺍﻟﻤﺤﻮﺭ‬، (𝑙𝑧 ) z ‫ ﻋﻠﻰ ﻃﻮﻝ ﺍﻟﻤﺤﻮﺭ‬Cu–O ‫ ﻳﻜﻮﻥ ﻃﻮﻝ‬،[CuO6 ] ‫ﺍﻟﺜﻤﺎﻧﻲ‬( d𝑧2 ‫ ﻭ‬d𝑥2 −𝑦2 ) eg ‫ ﻫﺬﻩ ﺍﻟﺘﺸﻮﻫﺎﺕ ﺗﺆﺩﻱ ﺍﻟﻰ ﺍﺯﺍﻟﺔ ﺍﻟﺘﺴﺎﻭﻱ ﺍﻟﻄﺎﻗﻲ ﻟﻤﺪﺍﺭﺍﺕ‬.𝑂h ‫ﺍﻟﺸﻜﻞ ﺍﻟﻤﻨﺘﻈﻢ‬Fig. 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.
‫ ﻭﺍﻟﺬﻱ ﻳﺘﻜﻮﻥ ﻋﻦ ﻃﺮﻳﻖ ﺧﻠﻂ ﻛﻠﻮﺭﻳﺪﺍﺕ ﺍﻟﻔﻠﺰﺍﺕ‬، B ‫ ﻋﻦ ﻃﺮﻳﻖ ﺍﻟﺘﺤﻠﻞ ﺍﻟﺤﺮﺍﺭﻱ )ﺍﻻﻧﺤﻼﻝ ﺍﻟﺤﺮﺍﺭﻱ( ﻟﻠﻤﻌﻘﺪ‬A ‫ﻳﻤﻜﻦ ﺗﺸﻴﻴﺪ‬‫ ُﻳﻈﻬﺮ ﺳﻠﻮﻙ‬.‫ ﺃﻱ ﺣﻤﺾ ﺛﻨﺎﺋﻲ‬،C4 H2 O4 (squaric acid) ‫ﻓﻲ ﻣﺤﻠﻮﻝ ﺃﻣﻮﻧﻴﺎ ﻣﺎﺋﻲ ﻣﺨﻔﻒ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺣﻤﺾ ﺳﻜﻮﺍﺭﻳﻚ‬، ‫ ﺩﺭﺟﺔ ﻣﺌﻮﻳﺔ ﺑﺴﺒﺐ ﻓﻘﺪ ﻣﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ‬200 ‫ ﺣﺘﻰ‬29.1% ‫ﻓﻘﺪﺍ ﻓﻲ ﺍﻟﻮﺯﻥ ﺑﻨﺴﺒﺔ‬‫ ﻓﻲ ﺍﻟﻬﻮﺍﺀ ﺍﻟﺠﺎﻑ‬B ‫ﺍﻻﻧﺤﻼﻝ ﺍﻟﺤﺮﺍﺭﻱ ﻟـ‬ًA ‫ ﺇﺟﻤﺎﻟﻲ ﻓﻘﺪﺍﻥ ﺍﻟﻮﺯﻥ ﺃﺛﻨﺎﺀ ﺗﻜﻮﻳﻦ‬.CO2 ‫ ﺩﺭﺟﺔ ﻣﺌﻮﻳﺔ ﺑﺴﺒﺐ ﺇﻃﻼﻕ‬700 ‫ﻣﺘﺒﻮﻋﺎ ﺑﻔﻘﺪﺍﻥ ﺁﺧﺮ ﻟﻠﻮﺯﻥ ﺣﺘﻰ ﺍﻟﻮﺻﻮﻝ ﺇﻟﻰ‬ً.‫ ﻓﻘﻂ ﻓﻲ ﺗﻔﺎﻋﻞ ﺍﻻﻧﺤﻼﻝ ﺍﻟﺤﺮﺍﺭﻱ‬CO2 ‫ ﻭﺗﺠﺪﺭ ﺍﻹﺷﺎﺭﺓ ﺇﻟﻰ ﺃﻧﻪ ﻳﺘﻢ ﺇﻃﻼﻕ ﺍﻟﻤﺎﺀ ﻭ‬. 63.6% ‫ ﻫﻮ‬B ‫ﻣﻦ‬
6pt

Write the chemical formulae for A and B.
. B ‫ ﻭ‬A ‫ﺍﻛﺘﺐ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻜﻞ ﻣﻦ‬

B.1

IChO
SAU-1 C-6 Q-5

4pt

4pt

Calculate 𝑙𝑥 and 𝑙𝑧 using Fig. 1.

Q6-5
Arabic (SAU) (Saudi Arabia)

B.2
Fig.1. ‫ﺍﺣﺴﺐ 𝑥𝑙 ﻭ 𝑧𝑙 ﺑﺎﺳﺘﺨﺪﺍﻡ‬For 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.
‫ ﺍﻛﺘﺐ ﺃﺳﻤﺎﺀ‬، Fig.1 ‫ ﻣﻦ‬A ‫[ ﻓﻲ‬CuO6 ] ‫ ﻓﻲ ﺍﻟﺸﻜﻞ ﺍﻟﺜﻤﺎﻧﻲ ﺍﻟﻤﺸﻮﻩ‬Cu2+ ‫ﻣﻦ ﺍﺟﻞ‬‫ ﻭﺍﺭﺳﻢ ﺍﻟﺘﻮﺯﻳﻊ ﺍﻻﻟﻜﺘﺮﻭﻧﻲ ﻓﻲ‬، (ii) ‫( ﻭ‬i) ‫ ( ﻓﻲ‬d𝑧2 ‫ ﻭ‬d𝑥2 −𝑦2 ) eg ‫ﺍﻟﻤﺪﺍﺭﺍﺕ ﺍﻟﻤﻨﻘﺴﻤﺔ‬.‫ﺍﻟﻤﺮﺑﻊ ﺍﻟﻤﻨﻘﻂ ﻓﻲ ﻭﺭﻗﺔ ﺇﺟﺎﺑﺘﻚ‬

B.3

Q6-6

IChO
SAU-1 C-6 Q-6Arabic (SAU) (Saudi Arabia)

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.
‫ ﻳﻈﻬﺮ ﺑﺎﻟﻤﻘﺎﺑﻞ ﺛﻘﺐ ﻭﺍﺣﺪ ﻓﻲ ﺍﻟﺸﺒﻜﺔ‬Sr2+ ‫ ﺑﺄﻳﻮﻥ ﻭﺍﺣﺪ ﻣﻦ‬La3+ ‫ ﻭﻟﺪﻯ ﺍﺳﺘﺒﺪﺍﻝ ﺃﻳﻮﻥ ﻭﺍﺣﺪ ﻣﻦ‬.‫ ﻣﺎﺩﺓ ﻋﺎﺯﻟﺔ‬A ‫ﻳﻌﺘﺒﺮ‬‫ ﻣﻮﺻﻠﻴﺔ ﻓﺎﺋﻘﺔ ﺗﺤﺖ‬Sr2+ -‫ ﺍﻟﺬﻱ ﺗﻢ ﺗﻄﻌﻴﻤﻪ ﺑﺄﻳﻮﻥ‬A ‫ ﺃﻇﻬﺮ‬،‫ ﻧﺘﻴﺠﺔ ﻟﺬﻟﻚ‬.‫ﺍﻟﺒﻠﻮﺭﻳﺔ ﺍﻟﺘﻲ ﻳﻤﻜﻨﻬﺎ ﻣﻦ ﺗﻮﺻﻴﻞ ﺍﻟﻜﻬﺮﺑﺎﺀ‬.‫ ﻣﻦ ﺍﻟﺜﻘﻮﺏ ﻟﻜﻞ ﻣﺘﺮ ﻣﻜﻌﺐ‬2.05 × 1027 ‫ ﻳﺘﻜﻮﻥ‬، A ‫ ﻋﻨﺪﻣﺎ ﺣﺪﺙ ﺗﻔﺎﻋﻞ ﺍﺳﺘﺒﺪﺍﻝ ﻟـ‬. 38 K
4pt

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.
‫ ﺑﻨﺴﺒﺔ ﻣﻮﻟﻴﺔ ﻓﻲ‬La3+ ‫ ﺍﻟﻤﺴﺘﺒﺪﻝ ﻓﻲ ﻣﻮﺍﻗﻊ ﺃﻳﻮﻥ‬Sr2+ ‫ﺍﺣﺴﺐ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﻷﻳﻮﻥ‬B.4‫ ﻻﺣﻆ ﺃﻥ ﺗﻜﺎﻓﺆﺍﺕ ﺍﻷﻳﻮﻧﺎﺕ ﺍﻟﻤﻜﻮﻧﺔ ﻭﺍﻟﺒﻨﻴﺔ ﺍﻟﺒﻠﻮﺭﻳﺔ ﻟﻢ ﺗﺘﻐﻴﺮ ﺃﺛﻨﺎﺀ ﺗﻔﺎﻋﻞ‬.‫ﺗﻔﺎﻋﻞ ﺍﻻﺳﺘﺒﺪﺍﻝ‬.‫ﺍﻻﺳﺘﺒﺪﺍﻝ‬C PartCu2 (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.
−2+‫ ُﻳﻈﻬﺮ‬.Fig.2A ‫ ﺍﻟﺸﻜﻞ‬Cu ‫ ﻣﺘﺮﺍﺑﻄﺔ ﺗﻨﺎﺳﻘﻴﺎ ﺇﻟﻰ ﺍﺛﻨﻴﻦ ﻣﻦ‬CH3 CO2 ‫ ﻣﻦ ﺃﺭﺑﻊ ﻟﻴﺠﺎﻧﺪﺍﺕ‬Cu2 (CH3 CO2 )4 ‫ﻳﺘﻜﻮﻥ‬‫ ﻣﻊ ﻣﺤﻮﺭﻳﻦ ﻳﻤﺮﺍﻥ ﻋﺒﺮ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ﻓﻲ ﺍﻟﻠﻴﺠﺎﻧﺪﺍﺕ ﺍﻻﺭﺑﻌﺔ‬،‫ ﻣﺴﺘﻮﻳﺎﺕ ﻋﺎﻟﻴﺔ ﻣﻦ ﺍﻟﺘﻨﺎﻇﺮ ﺍﻟﻬﻴﻜﻠﻲ‬Cu2 (CH3 CO2 )4−2+‫ ﻋﻨﺪ‬.‫ ﻭﺟﻤﻴﻊ ﺍﻟﻤﺤﺎﻭﺭ ﻣﻮﺟﻬﺔ ﺑﺸﻜﻞ ﻣﺘﻌﺎﻣﺪ ﺑﺎﻟﻨﺴﺒﺔ ﻟﺒﻌﻀﻬﺎ ﺍﻟﺒﻌﺾ‬،Cu ‫ ﻭﻣﺤﻮﺭ ﻳﻤﺮ ﻋﺒﺮ ﺃﻳﻮﻧﻴﻦ ﻣﻦ‬CH3 CO2−‫ ﻳﺘﻜﻮﻥ‬.”‫ ﻳﺘﻜﻮﻥ ”ﻣﻌﻘﺪ ﺫﻭ ﻓﺠﻮﺍﺕ ﻗﻔﺺ‬،CH3 CO2 ‫ﺑﺪﻻ ﻣﻦ‬(dicarboxylate) ‫ﺍﺳﺘﺨﺪﺍﻡ ﻟﻴﺠﺎﻧﺪ ﺛﻨﺎﺋﻲ ﺍﻟﻜﺮﺑﻮﻛﺴﻴﻞ‬
ً
2+

−

L1
‫ ﺗﺤﺪﺩ ﺍﻟﺰﺍﻭﻳﺔ‬.Fig.2C ‫ ﺍﻟﺸﻜﻞ‬Cu ‫ ﻭ‬Fig.2B ‫ ﺍﻟﺸﻜﻞ‬L1 ‫ ﻣﻦ ﻣﺴﺘﻮﻱ ﻣﻦ ﺛﻨﺎﺋﻲ ﻛﺮﺑﻮﻛﺴﻴﻼﺕ‬Cu4 (L1L1)4 ‫ﻣﻌﻘﺪ ﺍﻟﻘﻔﺺ‬‫ ﺗﺮﻛﻴﺐ ﺑﻨﻴﺔ‬، Fig.2B ‫ ﺍﻟﻤﺸﺎﺭ ﺇﻟﻴﻬﺎ ﺑﻮﺍﺳﻄﺔ ﺍﻷﺳﻬﻢ ﻓﻲ ﺍﻟﺸﻜﻞ‬،‫ ﺑﻴﻦ ﺍﺗﺠﺎﻫﺎﺕ ﺍﻟﺘﻨﺎﺳﻖ ﻻﺛﻨﻴﻦ ﻣﻦ ﺍﻟﻜﺮﺑﻮﻛﺴﻴﻼﺕ‬θ
Fig.2. ‫ ﻻﺣﻆ ﺃﻥ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻏﻴﺮ ﻣﻮﺿﺤﺔ ﻓﻲ ﺍﻟﺸﻜﻞ‬.L1 ‫ ﺩﺭﺟﺔ ﻟـ‬0 ‫ ﺗﺴﺎﻭﻱ‬θ .‫ﻣﻌﻘﺪ ﺍﻟﻘﻔﺺ‬
2+

IChO
SAU-1 C-6 Q-7

Q6-7
Arabic (SAU) (Saudi Arabia)

Fig. 25pt

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.
.‫ ﺩﺭﺟﺔ‬90 ‫ ﺍﻟﻤﻮﺿﺢ ﺃﺩﻧﺎﻩ ﻋﻨﺪ ﺯﺍﻭﻳﺔ‬L2 ‫ ﻟﺜﻨﺎﺋﻲ ﻛﺮﺑﻮﻛﺴﻴﻼﺕ ﺍﻟﻤﺴﺘﻮﻱ‬θ ‫ﺗﻢ ﺗﺜﺒﻴﺖ ﺍﻟﺰﺍﻭﻳﺔ‬L2
L2)𝑚 ‫ ﻫﻮ‬Cu ‫ ﻭ‬L2 ‫ﺇﺫﺍ ﻛﺎﻥ ﺗﺮﻛﻴﺐ ﻣﻌﻘﺪ ﺍﻟﻘﻔﺺ ﺍﻟﻤﺘﻜﻮﻥ ﻣﻦ‬‫ ﺍﺫﻛﺮ ﺃﺻﻐﺮ ﻣﺠﻤﻮﻋﺔ‬، Cu𝑛 (L2−‫ ﻫﻲ ﺍﻟﺘﻲ‬L2 ‫ ﻟﻠﻴﺠﺎﻧﺪ‬CO2 ‫ ﺍﻓﺘﺮﺽ ﺃﻥ ﻣﺠﻤﻮﻋﺎﺕ‬. m ‫ ﻭ‬n ‫ﻣﻦ ﺍﻻﻋﺪﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ ﻟﻜﻞ ﻣﻦ‬
2+

.Cu

2+

‫ﺗﺮﺗﺒﻂ ﻓﻘﻂ ﺑﺮﺍﺑﻄﺔ ﺗﻨﺎﺳﻘﻴﺔ ﺍﻟﻰ ﺃﻳﻮﻧﺎﺕ‬

C.1

Q6-8

IChO
SAU-1 C-6 Q-8Arabic (SAU) (Saudi Arabia)

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 dark gray polyhedron in Fig. 3C and 3D. Note that hydrogenatoms are not shown in Fig. 3.
2+−‫ ﺍﻟﺸﻜﻞ‬O2− ‫ ﻭﻭﺍﺣﺪ ﻣﻦ‬CH3 CO2 ‫ ﻭﺳﺘﺔ ﻣﻦ‬Zn ‫ ﻋﻠﻰ ﺃﺭﺑﻊ ﺭﺑﺎﻋﻴﺎﺕ ﺍﻟﺴﻄﻮﺡ ﻣﻦ‬،Zn4 O(CH3 CO2 )6 ‫ﻳﺤﺘﻮﻱ ﻣﻌﻘﺪ ﺍﻟﺰﻧﻚ‬‫ ﻭﺍﻟﻤﺤﺎﻭﺭ ﺍﻟﺜﻼﺛﺔ ﺍﻟﺘﻲ ﺗﻤﺮ ﻋﺒﺮ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ‬،‫ ﻋﻨﺪ ﻣﺮﻛﺰ ﻧﻘﻄﺔ ﺍﻷﺻﻞ‬O2− ‫ ﻳﻘﻊ‬Zn4 O(CH3 CO2 )6 ‫ ﻓﻲ ﺍﻟﻤﻌﻘﺪ‬. Fig.3A−
، Fig.3B ‫ )ﺍﻟﺸﻜﻞ‬p-benzenedicarboxylate ‫ ﻋﻨﺪﻣﺎ ﻳﺘﻢ ﺍﺳﺘﺨﺪﺍﻡ‬.‫ ﻣﺘﻌﺎﻣﺪﺓ ﺑﺎﻟﻨﺴﺒﺔ ﻟﺒﻌﻀﻬﺎ ﺍﻟﺒﻌﺾ‬CH3 CO2 ‫ﻓﻲ‬−2+‫ )ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﻌﻨﻘﻮﺩﻳﺔ( ﺑﺒﻌﻀﻬﺎ ﺍﻟﺒﻌﺾ ﻟﺘﺸﻜﻴﻞ ﻣﺎﺩﺓ ﺻﻠﺒﺔ‬Zn ‫ ﺗﺮﺗﺒﻂ ﻛﻼﺳﺘﺮﺍﺕ‬،CH3 CO2 ‫ﺑﺪﻻ ﻣﻦ‬( 180 = θ ، L3
ًL3
‫ ﻭﻟﻪ ﻫﻴﻜﻞ‬،[Zn4 O(L3L3)3 ]𝑛 ‫ ﻫﻮ‬X ‫ ﻳﻜﻮﻥ ﺗﺮﻛﻴﺐ‬. Fig3C ‫( ﻭﺍﻟﺘﻲ ﺗﺴﻤﻰ ”ﺑﻮﻟﻴﻤﺮ ﺗﻨﺎﺳﻘﻲ ﻣﺴﺎﻣﻲ” ﺍﻟﺸﻜﻞ‬X) ‫ﺑﻠﻮﺭﻳﺔ‬‫ ﻭﻳﺘﻢ ﺗﻤﺜﻴﻞ ﻛﻞ‬، Fig.3D ‫ ﻳﺘﻢ ﺗﻤﺜﻴﻞ ﺃﺣﺪ ﺍﻟﻤﺴﺎﻡ ﻋﻠﻰ ﺷﻜﻞ ﻛﺮﺓ ﻓﻲ ﺍﻟﺸﻜﻞ‬.‫ﺑﻠﻮﺭﻱ ﻣﻜﻌﺒﻲ ﻣﻊ ﻣﺴﺎﻡ ﺑﺤﺠﻢ ﺍﻟﻨﺎﻧﻮ‬2+−‫ ﻻﺣﻆ ﺃﻥ ﺫﺭﺍﺕ‬. Fig.3D ‫ ﻭ‬Fig.3C ‫ ﻋﻠﻰ ﺷﻜﻞ ﻣﺘﻌﺪﺩ ﺍﻟﺴﻄﻮﺡ ﺭﻣﺎﺩﻱ ﻏﺎﻣﻖ ﻓﻲ ﺍﻟﺸﻜﻞ‬Zn ‫ﺭﺑﺎﻋﻲ ﺳﻄﻮﺡ ﻛﻼﺳﺘﺮ‬.Fig.3 ‫ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻏﻴﺮ ﻣﻮﺿﺤﺔ ﻓﻲ ﺍﻟﺸﻜﻞ‬2+

Fig. 35pt

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].
.0.592 g cm−3 ‫ ﻭﻛﺜﺎﻓﺔ‬Fig.3C ‫ ﻋﻠﻰ ﺧﻠﻴﺔ ﻭﺣﺪﺓ ﻣﻜﻌﺒﻴﺔ ﻃﻮﻝ ﺿﻠﻌﻬﺎ 𝑎 ﺍﻟﺸﻜﻞ‬X ‫ﻳﺤﺘﻮﻱ‬. [cm] ‫ﺍﺣﺴﺐ 𝑎 ﺑﻮﺣﺪﺓ‬

C.2

IChO
SAU-1 C-6 Q-9

5pt

Q6-9
Arabic (SAU) (Saudi Arabia)

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.
‫ ﻣﻦ‬3.0 × 102 𝑚𝐿 ‫ ﺍﺳﺘﻴﻌﺎﺏ‬X ‫ ﻣﻦ‬1g ‫ ﻭﻳﻤﻜﻦ ﻟـ‬، ‫ ﻋﻠﻰ ﻋﺪﺩ ﻛﺒﻴﺮ ﻣﻦ ﺍﻟﻤﺴﺎﻣﺎﺕ‬X ‫ﻳﺤﺘﻮﻱ‬CO2 ‫ ﺍﺣﺴﺐ ﻣﺘﻮﺳﻂ ﻋﺪﺩ ﺍﻟﺠﺰﻳﺌﺎﺕ ﻣﻦ‬.‫ ﺩﺭﺟﺔ ﻣﺌﻮﻳﺔ‬25 ‫ ﻭ‬1 bar ‫ ﻓﻲ ﺍﻟﻤﺴﺎﻡ ﻋﻨﺪ‬CO2 ‫ﻏﺎﺯ‬.‫ﻟﻜﻞ ﻣﺴﺎﻡ ﻭﺍﺣﺪ‬

C.3

IChO
SAU-1 C-6 A-1

A6-1
English (Official)

The Solid-State Chemistry of Transition Metals
Part A
A.1 (6 pt)

Cr2+ ∶ΔO , Mn2+ ∶ΔO , Co2+ ∶

ΔO

IChO
SAU-1 C-6 A-2

A.2 (3 pt)

𝐷∶

A.3 (3 pt)

A6-2
English (Official)

IChO

A6-3

SAU-1 C-6 A-3

English (Official)Part B
B.1 (6 pt)

A∶

,B∶

B.2 (4 pt)

𝑙𝑥 =nm, 𝑙𝑧 =nm

IChO
SAU-1 C-6 A-4

B.3 (4 pt), (ii) ∶(i) ∶B.4 (4 pt)%

A6-4
English (Official)IChO
SAU-1 C-6 A-5Part C
C.1 (5 pt)𝑛=,𝑚=

C.2 (5 pt)

𝑎=

cm

A6-5
English (Official)IChO
SAU-1 C-6 A-6

C.3 (5 pt)

A6-6
English (Official)SAU-1 C-7 C-1

SAU-1 C-7 C
Mohammad AlHudaithi

IChO
Problem 7
Cover sheet

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

Q7-1

IChO
SAU-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
SAU-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
SAU-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
SAU-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
SAU-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
SAU-1 C-7 Q-1

Arabic (SAU) (Saudi Arabia)‫ﺍﻟﻠﻌﺐ ﻣﻊ ﺍﻷﺭﻭﻣﺎﺗﻴﺔ ﻏﻴﺮ ﺍﻟﺒﻨﺰﻳﻨﻮﺩﻳﺔ‬
Total
36

B.1
10

total the of % 13
A.3
19

A.2
2

A.1
5

Question
PointsScore

Prof. Nozoe (1902–1996) opened the research field of non-benzenoid aromatic compounds, which are
now ubiquitous in organic chemistry.
‫ ﻭﻫﻮ ﺍﻵﻥ ﻣﺘﻮﻓﺮ ﻓﻲ‬non-benzenoid aromatic ‫( ﻣﺠﺎﻝ ﺍﻟﺒﺤﺚ ﻋﻦ ﺍﻟﻤﺮﻛﺒﺎﺕ‬1996-1902) Nozoe ‫ﺍﻓﺘﺘﺢ ﺍﻟﺒﺮﻭﻓﻴﺴﻮﺭ‬.‫ﻛﻞ ﻣﻜﺎﻥ ﻓﻲ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻌﻀﻮﻳﺔ‬Photo courtesy: Tohoku Univ.

A PartLineariifolianone 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.
‫ ﺗﻢ ﻋﺰﻟﻪ ﻣﻦ‬، ‫ ﻫﻮ ﻣﻨﺘﺞ ﻃﺒﻴﻌﻲ ﻟﻪ ﻫﻴﻜﻞ ﻓﺮﻳﺪ‬Lineariifolianone‫ ﻗﺒﻞ ﺍﻟﺘﺤﻮﻝ ﻋﻦ ﻃﺮﻳﻖ‬، 2 ‫( ﺑﺎﻟﺘﺤﻮﻝ ﻋﻦ ﻃﺮﻳﻖ ﺧﻄﻮﺓ ﻭﺍﺣﺪﺓ ﺍﻟﻤﺮﻛﺐ‬1) valencene ‫ ﻳﻨﺘﺞ ﻣﻦ‬. Inula linariifolia‫ ﻋﻦ ﻃﺮﻳﻖ ﺇﺟﺮﺍﺀ ﻧﻔﺲ ﺍﻟﺘﺤﻮﻝ‬6 ‫( ﺇﻟﻰ‬5) Eremophilene ‫ ﻳﺘﻢ ﺗﺤﻮﻳﻞ‬.4 ‫ ﺍﻟﺬﻱ ﻳﻨﺘﺞ ﻋﻨﻪ ﺍﻟﻜﻴﺘﻮﻥ‬3 ‫ﺛﻼﺙ ﺧﻄﻮﺍﺕ ﻟﻠﻤﺮﻛﺐ‬.‫ﺍﻟﻤﻜﻮﻥ ﻣﻦ ﺃﺭﺑﻊ ﺧﻄﻮﺍﺕ‬IChO
SAU-1 C-7 Q-2Inula linariifolia

Q7-2
Arabic (SAU) (Saudi Arabia)

Q7-3

IChO
SAU-1 C-7 Q-3

5pt

Arabic (SAU) (Saudi Arabia)

Draw the structures of 2 and 6 and clearly identify the stereochemistry where
necessary.
.‫ ﻭﺣﺪﺩ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺑﻮﺿﻮﺡ ﻋﻨﺪ ﺍﻟﻀﺮﻭﺭﺓ‬6 ‫ ﻭ‬2 ‫ﺍﺭﺳﻢ ﺗﺮﺍﻛﻴﺐ‬A.1

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
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.
‫( ﺑﺠﻤﻴﻊ ﺍﻟﻤﺮﺍﻛﺰ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺍﻟﻤﻮﺟﻮﺩﺓ‬188 :‫ )ﺍﻟﻮﺯﻥ ﺍﻟﺠﺰﻳﺌﻲ‬8 ‫ ﻳﺤﺘﻔﻆ ﺍﻟﻤﺮﻛﺐ‬.15 ‫ ﺇﻟﻰ ﺍﺳﺘﺮ‬4 ‫ ﻳﺘﻢ ﺗﺤﻮﻳﻞ ﺍﻟﻜﻴﺘﻮﻥ‬، ‫ﺑﻌﺪ ﺫﻟﻚ‬‫ ﺃﻓﺘﺮﺽ ﺃﻥ‬.‫ ﻋﻠﻰ ﺧﻤﺴﺔ ﻣﺮﺍﻛﺰ ﻓﺮﺍﻏﻴﺔ ﻭﻻ ﺗﻮﺟﺪ ﺭﻭﺍﺑﻂ ﺛﻨﺎﺋﻴﺔ ﺑﻴﻦ ﺍﻟﻜﺮﺑﻮﻥ ﻭﺍﻟﻜﺮﺑﻮﻥ‬10 ‫ ﻭ‬9 ‫ ﺗﺤﺘﻮﻱ ﺍﻟﻤﺮﻛﺒﺎﺕ‬.7 ‫ﻓﻲ‬ً ‫ ﻳﺴﺘﺨﺪﻡ‬H2 18 O
‫ ﻟﺘﺸﻴﻴﺪ‬H2 16 O ‫ﺑﺪﻻ ﻣﻦ‬18

O labelled-lineariifolianones-.‫ ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‬,12 ‫ ﻭ‬11 ‫ ﻣﻦ‬14 ‫ ﻭ‬13
18

O - labelled isotopomers ‫ ﻫﻲ‬14 ‫ ﻭ‬13 ‫ﺍﻟﻤﺮﻛﺒﺎﺕ‬isotopic labelling ‫ﺗﺠﺎﻫﻞ‬.‫ ﻣﻊ ﻛﻴﻤﻴﺎﺀ ﻓﺮﺍﻏﻴﺔ ﻣﺘﻄﺎﺑﻘﺔ‬15 ‫ ﺗﻌﻄﻲ ﻧﻔﺲ ﺍﻟﻨﺎﺗﺞ‬14 ‫ ﻭ‬13 ‫ﺍﻟﻤﺮﻛﺒﺎﺕ‬IChO
SAU-1 C-7 Q-4

Q7-4
Arabic (SAU) (Saudi Arabia)

IChO
SAU-1 C-7 Q-5

2pt

19pt

Choose the appropriate structure for A.

Q7-5
Arabic (SAU) (Saudi Arabia)

A.2. A ‫ﺍﺧﺘﺮ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻬﻴﻜﻠﻲ ﺍﻟﻤﻨﺎﺳﺐ ﻟـ‬

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.
‫ﺃﻳﻀﺎ ﺇﻟﻰ‬ً ‫ ﺃﺷﺮ‬.‫ ﻭﺣﺪﺩ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺑﻮﺿﻮﺡ ﻋﻨﺪ ﺍﻟﻀﺮﻭﺭﺓ‬14-8 ‫ﺍﺭﺳﻢ ﺗﺮﺍﻛﻴﺐ ﻫﻴﻜﻠﻴﺔ‬.‫ ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﻓﻲ ﺍﻟﻤﺜﺎﻝ ﺃﺩﻧﺎﻩ‬14 ‫ ﻭ‬13 ‫ ﻓﻲ‬18 O ‫ﺫﺭﺍﺕ‬

A.3

IChO
SAU-1 C-7 Q-6

Q7-6
Arabic (SAU) (Saudi Arabia)

B PartCompound 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.
‫ ﻛﻤﻨﺸﻂ‬19 ‫ ﻳﻤﻜﻦ ﺍﺳﺘﺨﺪﺍﻡ‬، ‫ ﻓﻴﻤﺎ ﻳﺘﻌﻠﻖ ﺑﺎﻷﺭﻭﻣﺎﺗﻴﺔ ﻏﻴﺮ ﺍﻟﺒﻴﻨﺰﻳﻨﻮﺩﻳﺔ‬.‫ ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺃﺩﻧﺎﻩ‬19 ‫ﻳﺘﻢ ﺗﺸﻴﻴﺪ ﺍﻟﻤﺮﻛﺐ‬‫ ﻗﺪ ﻟﻮﺣﻆ ﺑﻮﺍﺳﻄﺔ‬21 ‫ ﻋﻠﻰ ﺍﻟﺮﻏﻢ ﻣﻦ ﺃﻥ ﺗﻜﻮﻳﻦ‬.21 ‫ ﻋﺒﺮ ﻣﺮﻛﺐ ﻭﺳﻄﻲ ﺯﻭﺝ ﺃﻳﻮﻧﻲ‬22 ‫ ﺇﻟﻰ‬20 ‫ ﻭﺗﻢ ﺗﺤﻮﻳﻞ‬، ‫ﻟﻠﻜﺤﻮﻻﺕ‬. 22 ‫ ﻭ‬18 ‫ﺗﺪﺭﻳﺠﻴﺎ ﻹﻋﻄﺎﺀ‬‫ ﻳﺘﺤﻠﻞ‬21 ‫ ﻓﺈﻥ‬NMRً

10pt

Draw the structures of 17–19 and 21. Identifying the stereochemistry is notnecessary.
.‫ﺿﺮﻭﺭﻳﺎ‬‫ ﺗﺤﺪﻳﺪ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ ﻟﻴﺲ‬.21 ‫ ﻭ‬19-17 ‫ﺍﺭﺳﻢ ﺗﺮﺍﻛﻴﺐ ﻟـ‬
ً

B.1

IChO
SAU-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
English (Official)IChO
SAU-1 C-7 A-2

A.3 (19 pt)

A7-2
English (Official)IChO
SAU-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
English (Official)SAU-1 C-8 C-1

SAU-1 C-8 C
Mohammad AlHudaithi

IChO
Problem 8
Cover sheet

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

Q8-1

IChO
SAU-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
SAU-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
SAU-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
SAU-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
SAU-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
SAU-1 C-8 Q-1

Arabic (SAU) (Saudi Arabia)‫ﺍﻟﺠﺰﻳﺌﺎﺕ ﺍﻟﻌﻀﻮﻳﺔ ﺍﻟﺪﻳﻨﺎﻣﻴﻜﻴﺔ ﻭ ﻛﻴﺮﺍﻟﻴﺘﻬﺎ‬
Total
26

B.2
4

total the of % 11

B.1
3

A.3
7

A.2
3

A.1
9

Question
PointsScore

A PartPolycyclic 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.
‫ ( ﺍﻟﺘﻲ ﺗﺮﺗﺒﻂ ﺑﺸﻜﻞ ﻣﺘﻌﺎﻗﺐ‬Polycyclic aromatic hydrocarbons ) ‫ﺍﻟﻬﻴﺪﺭﻭﻛﺮﺑﻮﻧﺎﺕ ﺍﻻﺭﻭﻣﺎﺗﻴﺔ ﻣﺘﻌﺪﺩﺓ ﺍﻟﺤﻠﻘﺎﺕ‬‫ ( ﻳﺮﻣﺰ ﺍﻟﻰ ﻋﺪﺩ ﺍﻟﺤﻠﻘﺎﺕ ﺍﻟﺴﺪﺍﺳﻴﺔ ﻓﻲ ﺍﻟﻤﺮﻛﺐ )ﺍﻧﻈﺮ‬n ) ‫[ ( ﻭﻫﻨﺎ ﺍﻟﺤﺮﻑ‬n]carbohelicenes ) ‫ﺑﻤﻮﻗﻊ ﺍﻻﻭﺭﺛﻮ ﺗﺴﻤﻰ‬‫[( ﺗﻢ ﺗﺤﻀﻴﺮﻩ ﺑﻔﻌﺎﻟﻴﺔ ﻋﻦ ﻃﺮﻳﻖ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻤﺘﻌﺎﻗﺒﺔ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻀﻮﺋﻴﺔ‬4]C) ‫[ ﺍﻭ‬4]carbohelicene ,(‫ﺑﺎﻻﺳﻔﻞ‬. (iodine) ‫(( ﺍﻟﺬﻱ ﻳﺘﺄﻛﺴﺪ ﺑﺴﻬﻮﻟﺔ ﺑﺎﻟﻴﻮﺩ‬Int) intermediate ) ‫ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺑﺎﻻﺳﻔﻞ ﻋﻦ ﻃﺮﻳﻖ ﻣﺮﻛﺐ ﻭﺳﻄﻲ‬IChO
SAU-1 C-8 Q-2

Q8-2
Arabic (SAU) (Saudi Arabia)

The photoreaction proceeds in a manner similar to the following example.
.‫ﺍﻟﺘﻔﺎﻋﻼﺕ ﺍﻟﻀﻮﺋﻴﺔ ﺗﺘﻢ ﺑﻨﻔﺲ ﺍﻟﻄﺮﻳﻘﺔ ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺑﺎﻟﻤﺜﺎﻝ ﺍﻟﺘﺎﻟﻲ‬Note: 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.
‫ ﻣﻦ ﻓﻀﻠﻚ ﺍﺭﺳﻢ ﺭﻭﺍﺑﻂ ﺍﺣﺎﺩﻳﺔ ﻭ ﺛﻨﺎﺋﻴﺔ ﺑﺸﻜﻞ ﻣﺘﻨﺎﻭﺏ ﻓﻲ ﺍﺟﺎﺑﺎﺗﻚ‬,8 ‫ ﻓﻲ ﻛﻞ ﻓﻘﺮﺍﺕ ﺍﻟﺴﺆﺍﻝ‬:‫ﻣﻼﺣﻀﺔ‬𝜋 ‫( ﻻ ﺗﻀﻊ ﺩﻭﺍﺋﺮ ﻷﻧﻈﻤﺔ ﻣﺪﺍﺭﺍﺕ‬carbohelicene), ‫ﻋﻠﻰ ﺍﻟﺴﺆﺍﻝ ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﻓﻲ ﻣﺜﺎﻝ ﺍﻟﻜﺮﺑﻮﻫﻴﻠﻴﺴﻴﻨﺎﺕ‬.‫ﺍﻟﻤﺘﺮﺍﻓﻘﺔ‬
9pt

3pt

Draw the structures of A–C. Stereoisomers should be distinguished.
. ‫ ﻳﺠﺐ ﺗﻤﻴﻴﺰ ﺍﻟﻤﺘﻤﺎﻛﺒﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ‬,C ‫ ﺍﻟﻰ‬A ‫ﺍﺭﺳﻢ ﺗﺮﺍﻛﻴﺐ ﻣﻦ‬

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)]

A.1

A.2

phosphonium salt ) ‫[ ( ﻣﻦ ﻣﻠﺢ ﻓﻮﺳﻔﻮﻧﻴﻮﻡ ﻣﺸﺎﺑﻪ‬5]carbohelicene ) ‫ﻣﺤﺎﻭﻻﺕ ﻟﺘﺸﻴﻴﺪ‬[5]carbohelicene ‫( ﻭ ﻧﺘﺞ ﺍﺛﺎﺭ ﻣﻦ ﻣﺮﻛﺐ ﻣﻨﺎﺳﺐ ﻓﻲ ﺍﻟﺒﺪﺍﻳﺔ ﺑﻜﻤﻴﺔ ﻗﻠﻴﻠﺔ ﺟﺪﺍ ﻣﻦ‬‫ ﻣﻦ‬2Da ‫ ﺍﻟﺬﻱ ﻟﻪ ﻭﺯﻥ ﺟﺰﻳﺌﻲ ﺗﻌﺘﺒﺮ ﺍﻗﻞ ﺑﻤﻘﺪﺍﺭ‬D ‫ ﻭﻟﻜﻦ ﺍﻟﺘﺸﻴﻴﺪ ﻧﺘﺞ ﻋﻨﻪ ﺍﻟﻨﺎﺗﺞ‬،‫ ( ﻟﻠﺮﻧﻴﻦ ﺍﻟﻨﻮﻭﻱ‬chemical shifts ) ‫ ﺍﻻﻧﺰﻳﺎﺣﺎﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬, [5]carbohelicene ‫ﺗﻠﻚ ﻟـ‬‫ ﺃﺭﺳﻢ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‬,‫ ﻣﺪﺭﺟﺔ ﺑﺎﻻﺳﻔﻞ‬D ‫ ( ﻟـ‬1 H NMR ) ‫ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﺍﻟﺒﺮﻭﺗﻮﻧﻲ‬. D ‫ﻟـ‬[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)]
The [5]- and larger [n]carbohelicenes have helical chirality and interconversion between enantiomers ofthese helicenes is significantly slow at room temperature. The chirality of [n]carbohelicenes is definedas (M) or (P) as shown below.
‫ ﻛﻴﺮﺍﻟﻴﺔ ﻣﻠﺘﻔﺔ ﺑﺸﻜﻞ ﺣﻠﺰﻭﻧﻲ ﻭ ﺍﻟﺘﺤﻮﻝ ﺍﻟﺒﻴﻨﻲ‬helical chirality ‫[ ﺍﻻﻛﺒﺮ ﻋﻠﻰ‬n]carbohelicenes ‫ ﻭ‬-[5] ‫ﺗﺤﺘﻮﻱ‬.‫( ﺑﻄﻲﺀ ﺑﺸﻜﻞ ﻛﺒﻴﺮ ﻓﻲ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﺍﻟﻐﺮﻓﺔ‬helicenes) ‫( ﺑﻴﻦ ﺍﺍﻻﻳﻨﺎﻧﺘﻮﻣﻴﺮﻳﻦ ﻟﻬﺬﺓ ﺍﻟﻜﻴﺮﺍﻟﻴﺘﺎﺕ‬interconversion).‫( ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺑﺎﻻﺳﻔﻞ‬P) ‫( ﺍﻭ‬M) ‫[ ﻣﺸﺎﺭ ﺍﻟﻴﻬﺎ ﺑﺎﻟﺮﻣﻮﺯ‬n]carbohelicenes ‫ﺍﻟﻜﻴﺮﺍﻟﻴﺔ ﻟـ‬

IChO
SAU-1 C-8 Q-3

Q8-3
Arabic (SAU) (Saudi Arabia)

The [n]Carbohelicenes with n larger than 4 can be enantiomerically separated by a chiral column chromatography, which was developed by Prof. Yoshio Okamoto.
) ‫ ﻳﻤﻜﻦ ﻓﺼﻠﻬﺎ ﺍﻳﻨﺎﻧﺘﻮﻣﻴﺮﻳﺎ ﻋﻦ ﻃﺮﻳﻖ ﻛﺮﻭﻣﺎﺗﻮﻗﺮﺍﻓﻴﺎ ﺍﻟﻌﻤﻮﺩ ﺍﻟﻜﻴﺮﺍﻟﻲ‬4 ‫ ﺍﻛﺒﺮ ﻣﻦ‬n ‫[ ﺫﻭ‬n]Carbohelicenes ‫ﺍﻝ‬‫ ﺍﻟﺬﻱ ﺗﻢ ﺗﻄﻮﻳﺮﻩ ﺑﻮﺍﺳﻄﺔ ﺍﻟﺒﺮﻭﻓﻴﺴﻮﺭ ﻳﻮﺷﻲ ﺍﻭﻛﺎﻣﻮﺗﻮ‬,( chiral column chromatography

Foundation Prize Japan The courtesy: PhotoIChO
SAU-1 C-8 Q-4

Q8-4
Arabic (SAU) (Saudi Arabia)

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.
‫ ﺍﺫﺍ ﺍﺧﺬﻧﺎ ﺑﻌﻴﻦ ﺍﻻﻋﺘﺒﺎﺭ‬,‫ﺍﻟﻬﻴﻠﻴﺴﻴﻨﺎﺕ ﺍﻟﻤﺘﻌﺪﺩﺓ ﻫﻲ ﻣﺮﻛﺒﺎﺕ ﺗﺤﺘﻮﻱ ﻋﻠﻰ ﺍﺛﻨﻴﻦ ﺍﻭ ﺍﻛﺜﺮ ﻣﻦ ﺗﺮﺍﻛﻴﺐ ﺍﺷﺒﺎﻩ ﺍﻟﻬﻴﻠﻴﺴﻴﻦ‬,‫( ﻣﺘﻌﺪﺩﺓ ﻓﻲ ﺍﻟﻬﻴﻠﻴﺴﻴﻨﺎﺕ ﺍﻟﻤﺘﻌﺪﺩﺓ‬stereoisomers) ‫ﺍﻟﻜﻴﺮﺍﻟﻴﺔ ﺍﻟﻤﻠﺘﻔﺔ ﻓﺎﻥ ﺫﻟﻚ ﻳﻌﻨﻲ ﻭﺟﻮﺩ ﺍﻳﺰﻭﻣﻴﺮﺍﺕ ﻓﺮﺍﻏﻴﺔ‬‫ ﻭﺍﺣﺪﺓ ﻣﻦ‬.‫[ ﻓﻲ ﺟﺰﻱﺀ ﻭﺍﺣﺪ‬5]carbohelicene ‫ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺛﻼﺛﺔ ﻣﺜﻴﻼﺕ ﻟﺘﺮﻛﻴﺐ ﺍﻝ‬E ‫ ﺍﻟﻤﺮﻛﺐ‬:‫ﻋﻠﻰ ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ‬.‫ ( ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺑﺎﻻﺳﻔﻞ‬P , P , P) ‫ﺍﻟﻤﺘﻤﺎﻛﺒﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺗﻮﺻﻒ ﻛﺎﻻﺗﻲ‬7pt

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).
‫ ﻋﻨﺪﻣﺎ‬, triphenylene ‫ ( ﻳﻨﺘﺞ ﻋﻨﻪ‬1,2-dibromobenzene ) ‫ﺍﻟﺘﺮﺍﻳﻤﺮﺓ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻨﻴﻜﻞ ﻟـ‬,(C66 H36 ) G ‫( ﻟﻨﻔﺲ ﺍﻟﺘﻔﺎﻋﻞ ﻳﻨﺘﺞ ﻋﻨﺔ ﺍﻟﻬﻴﻠﻴﺴﻴﻦ ﺍﻟﻤﺘﻌﺪﺩ‬P)- F‫ ﻭ‬F ‫ﻳﺘﻌﺮﺽ ﺍﻻﻳﻨﺎﻧﺘﻮﻣﻴﺮ‬،‫ ﻻ ﻳﺤﺪﺙ ﺍﺛﻨﺎﺀ ﺍﻟﺘﻔﺎﻋﻞ‬interconversion) ‫ﻋﻠﻤﺎ ﺑﺎﻥ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﺒﻴﻨﻲ ﺑﻴﻦ ﺍﻟﻤﺘﻤﺎﻛﺒﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ‬‫ ﺍﻟﻨﺎﺗﺠﺔ ﻋﻦ ﻫﺬﺓ‬G ‫( ﺍﻟﻤﻤﻜﻨﺔ ﻟﻠﻤﺮﻛﺐ‬stereoisomers) ‫ﺣﺪﺩ ﺟﻤﻴﻊ ﺍﻻﻳﺰﻭﻣﻴﺮﺍﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ‬‫ ﻳﺠﺐ ﺍﻥ ﻳﺮﺳﻢ ﺍﻳﺰﻭﻣﻴﺮ ﻭﺍﺣﺪ ﺑﺎﻟﻜﺎﻣﻞ ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺑﺎﻻﻋﻠﻰ‬,‫ ﻛﻤﺮﺟﻊ‬.‫ﺍﻟﻌﻤﻠﻴﺔ ﺑﺪﻭﻥ ﺗﻜﺮﺍﺭ‬‫ ﻛﺬﻟﻚ ﻳﺠﺐ ﺍﺩﺭﺍﺝ ﺍﻻﻳﺰﻭﻣﻴﺮﺍﺕ‬,‫ﻣﻊ ﻛﻴﺮﺍﻟﻴﺔ ﻣﺤﺪﺩﺓ ﻛﻤﺎ ﻓﻲ ﺍﻟﻤﺜﺎﻝ ﺍﻋﻼﻩ ﻣﻊ ﺍﺳﺘﺨﺪﺍﻡ ﺍﻻﺭﻗﺎﻡ‬‫ ﻓﻲ ﺍﻣﺎﻛﻦ‬P ‫ ﻭ‬M ‫ﺍﻟﻔﺮﺍﻏﻴﺔ ﺍﻻﺧﺮﻯ ﻋﻦ ﻃﺮﻳﻖ ﺍﻟﺘﻌﺮﻳﻒ ﺑﺎﻻﺭﻗﺎﻡ ﺣﺴﺐ ﻣﻮﺍﻗﻊ ﺍﻻﺭﻗﺎﻡ ﺑـﺎﻟﺮﻣﻮﺯ‬:‫ ﻳﺠﺐ ﺍﺩﺭﺍﺟﻬﺎ ﻛﺎﻟﺘﺎﻟﻲ‬E ‫ ﻟﻠﻤﺮﻛﺐ‬:‫ ﻋﻠﻰ ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ‬.‫ﺍﻻﺭﻗﺎﻡ‬(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).

A.3

IChO
SAU-1 C-8 Q-5

Q8-5
Arabic (SAU) (Saudi Arabia)

IChO
SAU-1 C-8 Q-6

Q8-6
Arabic (SAU) (Saudi Arabia)

B PartSumanene 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 was achieved by a reaction sequence that consists of a ring-opening and aring-closing metathesis.
‫ ﺍﻻﺳﻢ‬,2003 ‫( ﻣﺮﻛﺐ ﻫﻴﺪﺭﻭﻛﺎﺭﺑﻮﻧﻲ ﻳﺸﺒﻪ ﺍﻟﻮﻋﺎﺀ ﻭﺍﻟﺬﻱ ﺍﻋﻠﻦ ﻋﻨﻪ ﺑﺎﻟﺒﺪﺍﻳﺔ ﻓﻲ ﺍﻟﻴﺎﺑﺎﻥ ﻓﻲ ﻋﺎﻡ‬Sumanene) ‫ﺍﻟﺴﻮﻣﺎﻧﻴﻦ‬.‫” ﻭﺍﻟﺘﻲ ﺗﻌﻨﻲ ﻭﺭﺩﺓ ﺩﻭﺍﺭ ﺍﻟﺸﻤﺲ‬suman” ‫ ” ﻣﺸﺘﻖ ﻣﻦ ﺍﻟﻜﻠﻤﺔ ﺍﻟﺴﻨﺴﻜﺮﻳﺘﻴﺔ ﺍﻟﻬﻨﺪﻳﺔ‬sumanene ”‫ ﻋﻦ ﻃﺮﻳﻖ ﺗﻔﺎﻋﻼﺕ ﻣﺘﻌﺎﻗﺒﺔ ﺗﺘﻜﻮﻥ ﻣﻦ ﻓﺘﺢ ﺍﻟﺤﻠﻘﺔ ﻭ ﺍﻏﻼﻕ ﺍﻟﺤﻠﻘﺔ ﻭﺗﻐﻴﻴﺮ ﺍﻟﻤﻮﺍﺿﻊ ﺗﻌﺮﻑ‬Sumanene ‫ﺍﻣﻜﻦ ﺗﺸﻴﻴﺪ‬metathesis:(ring-opening and a ring-closing metathesis)

Representative metathesis reactions catalyzed by a ruthenium catalyst (Ru*) are shown below.
‫ ( ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ‬Ru* ) ‫ ( ﺣﺎﻓﺰ ﺍﻟﺮﻭﺛﻴﻨﻴﻮﻡ‬Representative metathesis ) ‫ ﺑﺎﺳﺘﺨﺪﺍﻡ‬metathesis ‫ﻳﺘﻢ ﺗﺤﻔﻴﺰ ﺗﻔﺎﻋﻼﺕ‬.‫ﺑﺎﻻﺳﻔﻞ‬IChO
SAU-1 C-8 Q-7

3pt

Q8-7
Arabic (SAU) (Saudi Arabia)

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

B.1(‫ ) ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ ﻟﻬﺬﺍ ﺍﻟﻤﺮﻛﺐ ﻏﻴﺮ ﻣﻄﻠﻮﺏ‬I ‫ﺍﺭﺳﻢ ﺗﺮﻛﻴﺐ ﺍﻟﻨﺎﺗﺞ ﺍﻟﻮﺳﻄﻲ‬4pt

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.
‫ ﺍﻟﻔﻌﺎﻝ ﺿﻮﺋﻴﺎ ﻳﺘﻜﻮﻥ ﺑﻨﻔﺲ ﺗﺮﺗﻴﺐ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻤﺮﻛﺐ ﺍﻟﻔﻌﺎﻝ‬J ‫ﺑﺪﺀﺍ ﻣﻦ ﺍﻟﻤﺮﻛﺐ ﺍﻻﺑﺘﺪﺍﺋﻲ‬‫ ﻻ‬J ‫ ( ﻓﻲ‬stereocenters ) ‫ ﺍﻟﻤﺮﺍﻛﺰ ﺍﻟﻔﺮﺍﻏﻴﺔ‬. K derivative sumanene ‫ﺿﻮﺋﻴﺎ ﻣﺸﺘﻖ‬.( metathesis reaction ) ‫ ( ﺧﻼﻝ ﺗﻔﺎﻋﻞ‬inversion ) ‫ﺗﻌﺎﻧﻲ ﻣﻦ ﺍﻧﻌﻜﺎﺱ‬.‫ ﻣﻊ ﻣﺮﺍﻋﺎﺓ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ ﻟﻠﻤﺮﻛﺐ‬K ‫ﺍﺭﺳﻢ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻟﻠﻤﺮﻛﺐ‬

B.2

IChO

A8-1

SAU-1 C-8 A-1

English (Official)

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
SAU-1 C-8 A-2

A.3 (7 pt)

A8-2
English (Official)IChO
SAU-1 C-8 A-3

Part B
B.1 (3 pt)

B.2 (4 pt)

A8-3
English (Official)SAU-1 C-9 C-1

SAU-1 C-9 C
Mohammad AlHudaithi

IChO
Problem 9
Cover sheet

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

Q9-1

IChO
SAU-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
SAU-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
SAU-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
SAU-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

IChO
SAU-1 C-9 Q-5

Q9-5
English (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
SAU-1 C-9 Q-1

Arabic (SAU) (Saudi Arabia)‫ﻳﺤﺐ ﻭﻳﻜﺮﻩ ﺍﻟﻜﺒﺴﻮﻟﺔ‬
Total
23

A.5
3

total the of % 10

A.4
3

A.3
2

A.2
2

A.1
13

Question
PointsScore‫ﺟﺰﺀ‬Good kids don’t do this, but if you unseam a tennis ball, you can disassemble it into two U-shaped pieces.
(U) ‫ ﻟﻜﻦ ﺍﺫﺍ ﻗﻤﺖ ﺑﻘﺺ ﻛﺮﺓ ﺍﻟﺘﻨﺲ ﻳﻤﻜﻨﻚ ﻓﺼﻞ ﻗﻄﻌﺘﻴﻦ ﻋﻠﻰ ﺷﻜﻞ ﺣﺮﻑ ﺍﻝ‬,‫ﺍﻷﻭﻻﺩ ﺍﻟﺠﻴﺪﻭﻥ ﻻ ﻳﻔﻌﻠﻮﻥ ﺫﻟﻚ‬.‫ﺑﺎﻻﻧﺠﻠﻴﺰﻳﺔ‬Based 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.
3 ‫ ﺗﻢ ﺗﺤﻀﻴﺮ ﻣﺮﻛﺐ‬,‫” ﺟﺰﻳﺌﺎﺕ ﺑﺄﺣﺠﺎﻡ ﻣﺨﺘﻠﻔﺔ‬U” ‫ ﻋﻠﻰ ﺷﻜﻞ ﺣﺮﻑ‬2 ‫ ﻭ ﻣﺮﻛﺐ‬1 ‫ ﺗﻢ ﺗﺤﻀﻴﺮ ﻣﺮﻛﺐ‬:‫ﻃﺒﻘﺎ ﻟﻬﺬﺓ ﺍﻟﻔﻜﺮﺓ‬” encapsulation ” ‫ ﻭﺗﻤﺖ ﺩﺭﺍﺳﺔ ﻣﺪﻯ ﻗﺪﺭﺓ ﺍﻟﻤﺮﻛﺒﺎﺕ ﻋﻠﻰ ﺍﻟﻜﺒﺴﻠﺔ‬1 ‫ﻟﻠﻤﻘﺎﺭﻧﺔ ﻣﻊ ﻣﺮﻛﺐ‬IChO
SAU-1 C-9 Q-2

Q9-2
Arabic (SAU) (Saudi Arabia)

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.
‫ ﻭ‬40.49 % ‫ ﺍﻟﻜﺎﺭﺑﻮﻥ ﺑﻨﺴﺒﺔ‬:‫ ﻫﻮ ﻛﺎﻟﺘﺎﻟﻲ‬9 ‫ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻌﻨﺼﺮﻱ ﻟﻠﻤﺮﻛﺐ‬,‫ ﻣﻮﺿﺢ ﺑﺎﻻﺳﻔﻞ‬2 ‫ﺍﻟﺘﺸﻴﻴﺪ ﺍﻟﻜﻠﻲ ﻟﻠﻤﺮﻛﺐ‬17.98 % ‫ ﻭ ﺍﻷﻛﺴﺠﻴﻦ ﺑﻨﺴﺒﺔ‬1.70 % ‫ﺍﻟﻬﺎﻳﺪﺭﻭﺟﻴﻦ ﺑﻨﺴﺔ‬IChO
SAU-1 C-9 Q-3

Q9-3
Arabic (SAU) (Saudi Arabia)

IChO
SAU-1 C-9 Q-4

13pt

Q9-4
Arabic (SAU) (Saudi Arabia)

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.
,(‫ ﻟﻴﺲ ﻣﻦ ﺍﻟﻀﺮﻭﺭﻱ ﺗﻮﺿﻴﺢ ﺍﻟﻤﺘﻤﺎﻛﺒﺎﺕ )ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ‬, 9 - 4 ‫ﺃﺭﺳﻢ ﺍﻟﻤﺮﻛﺒﺎﺕ ﻣﻦ‬
p-methoxybenzyl ‫” ﻛﺈﺧﺘﺼﺎﺭ ﺑﺪﻻ ﻣﻦ ﺭﺳﻢ ﺗﺮﻛﻴﺒﺔ ﺍﻟﻤﺮﻛﺐ ﻛﺎﻣﻠﺔ ﻟﻞ‬PMB” ‫ﺍﺳﺘﺨﺪﻡ ﺍﻟﺮﻣﺰ‬‫ﺍﻟﻤﻮﺿﺢ ﻓﻲ ﺍﻟﻤﺨﻄﻂ ﺑﺎﻷﻋﻠﻰ‬A.1

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.
‫ ﺑﻴﻨﻤﺎ ﻟﻢ ﺗﺘﻢ ﻣﻼﺣﻈﺔ‬,‫ ( ﻛﺎﻥ ﻭﺍﺿﺤﺎ‬12 ) ‫ ﻗﻤﺔ ﺍﻻﻳﻮﻥ ﺍﻟﻤﺘﻮﺍﻓﻖ ﻣﻊ ﺛﻨﺎﺋﻲ ﺍﻟﻤﺮﻛﺐ‬,1 ‫ﻓﻲ ﺍﻟﻄﻴﻒ ﺍﻟﻜﺘﻠﻲ ﻟﻠﻤﺮﻛﺐ‬1
𝐻 𝑁 𝑀 𝑅 ‫ ﻓﻲ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻨﻮﻭﻱ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﺍﻟﺒﺮﻭﺗﻮﻧﻲ‬.3 ‫( ﻓﻲ ﺍﻟﻄﻴﻒ ﺍﻟﻜﺘﻠﻲ ﻟﻠﻤﺮﻛﺐ‬32 )‫ﺍﻟﻘﻤﺔ ﻷﻱ ﺃﻳﻮﻥ ﻟﻠﺜﻨﺎﺋﻲ ﻝ‬
chemical shift ‫ ﻣﺘﻄﺎﺑﻘﺔ ﻣﻐﻨﺎﻃﻴﺴﻴﺎ ﻭ ﺍﻻﻧﺰﻳﺎﺡ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‬1 ‫( ﻣﻦ ﺍﻟﻤﺮﻛﺐ‬NH) ‫ ﺟﻤﻴﻊ ﺍﻟﺒﺮﻭﺗﻮﻧﺎﺕ ﻣﻦ‬, 12 ‫ﻟﻤﺤﻠﻮﻝ ﻣﻦ‬‫ ﻫﺬﻩ ﺍﻟﻤﻌﻄﻴﺎﺕ ﺗﺪﻝ ﻋﻠﻰ ﻭﺟﻮﺩ ﺭﻭﺍﺑﻂ‬.3 ‫( ﻓﻲ ﺍﻟﻤﺮﻛﺐ‬NH)‫ﻛﺎﻥ ﻣﺨﺘﻠﻒ ﻛﺜﻴﺮﺍ ﻋﻦ ﺍﻻﻧﺰﻳﺎﺡ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻟﺒﺮﻭﺗﻮﻧﺎﺕ ﺍﻝ‬‫ ﻻﻧﺘﺎﺝ ﻣﺎ ﻳﻌﺮﻑ ﺑﺎﻟﻜﺒﺴﻮﻟﺔ ﺍﻟﺜﻨﺎﺋﻴﺔ‬1 ‫ ﻭ ﺍﻟﺬﺭﺓ ﺍﻛﺲ ﻟﺠﺰﻱﺀ ﺍﺧﺮ ﻣﻦ ﺍﻟﻤﺮﻛﺐ‬1 ‫( ﻟﻠﻤﺮﻛﺐ‬NH) ‫ﻫﻴﺪﺭﻭﺟﻴﻨﻴﺔ ﺑﻴﻦ ﺍﻝ‬
dimeric capsule

2pt

2pt

Circle all the appropriate atom(s) X in 1.
1 ‫ ﺍﻟﻤﻤﻜﻨﺔ ﻓﻲ‬X (‫ﺿﻊ ﺩﺍﺋﺮﺓ ﻋﻠﻰ ﺟﻤﻴﻊ ﺫﺭﺓ)ﺍﺕ‬Give the number of the hydrogen bonds in the dimeric capsule (112 ).
(112 ) ‫ﺃﻋﻄﻲ ﻋﺪﺩ ﺍﻟﺮﻭﺍﺑﻂ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻨﻴﺔ ﻓﻲ ﺍﻟﻜﺒﺴﻮﻟﺔ ﺍﻟﺜﻨﺎﺋﻴﺔ‬

A.2

A.3

IChO
SAU-1 C-9 Q-5

Q9-5
Arabic (SAU) (Saudi Arabia)

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:‫ ﻫﺬﺓ ﺍﻟﻈﺎﻫﺮﺓ‬,‫ ﺻﻐﻴﺮ ﻛﻔﺎﻳﺔ ﺍﻟﺘﻜﺒﺴﻞ ﺑﺪﺍﺧﻠﻪ‬Z ‫( ﺗﺤﺘﻮﻱ ﻋﻠﻰ ﻓﺮﺍﻍ ﺩﺍﺧﻠﻲ ﻳﻤﻜﻦ ﻟﻤﺮﻛﺐ‬112 ) 1 ‫ﺍﻟﻜﺒﺴﻮﻟﺔ ﺍﻟﺜﻨﺎﺋﻴﺔ ﻟﻠﻤﺮﻛﺐ‬:‫ﻳﻤﻜﻦ ﺗﻔﺴﻴﺮﻫﺎ ﺑﺎﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ‬The equilibrium constant of the encapsulation of Z into 1 2 is given as below::‫ ﻣﻌﻄﻰ ﺑﺎﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ‬1 2 ‫ ﺑﺪﺍﺧﻞ‬Z ‫ﺛﺎﺑﺖ ﺍﻟﺘﻮﺍﺯﻥ ﻟﻠﺘﻜﺒﺴﻞ ﻝ‬

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 .

1 2 ‫( ﻋﻠﻰ ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ‬NMR), ‫ﺍﻟﺘﻜﺒﺴﻞ ﻟﻠﻤﺮﻛﺐ ﺑﺪﺍﺧﻞ ﺍﻟﻜﺒﺴﻮﻟﺔ ﻳﻤﻜﻦ ﻣﺘﺎﺑﻌﺘﻬﺎ ﻋﻦ ﻃﺮﻳﻖ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻨﻮﻱ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ‬CH4 ‫( ﻗﺒﻞ ﻭﺑﻌﺪ ﺍﺿﺎﻓﺔ‬1 H NMR ) ‫ ﻳﻌﻄﻲ ﺇﺷﺎﺭﺍﺕ ﻣﺨﺘﻠﻔﺔ ﻓﻲ ﻃﻴﻒ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﺍﻟﻨﻮﻭﻱ ﺍﻟﺒﺮﻭﺗﻮﻧﻲ‬C6 D6 ‫ﻓﻲ‬Compound 2 also forms a rigid and larger dimeric capsule (222 ). The 1 H NMR spectrum of 2 2 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.
‫ﺗﻢ ﻗﻴﺎﺱ ﺃﻃﻴﺎﻑ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻨﻮﻭﻱ‬,(222 ) ‫ ﺍﻳﻀﺎ ﻳﻤﻜﻦ ﻣﻦ ﺧﻼﻟﻪ ﺗﻜﻮﻳﻦ ﻛﺒﺴﻮﻟﺔ ﺛﻨﺎﺋﻴﺔ ﺃﻛﺒﺮ ﻭﺃﻛﺜﺮ ﺻﻼﺑﺔ‬2 ‫ﺍﻟﻤﺮﻛﺐ‬C6 D6 /C6 D5 F ‫ ﻭ ﺃﻳﻀﺎ ﻓﻲ ﺍﻟﺨﻠﻴﻂ ﻣﻦ ﺍﻟﻤﺬﻳﺒﺎﺕ‬C6 D5 F ‫ ﻭ‬C6 D6 ‫ ﻓﻲ ﺍﻟﻤﺬﻳﺒﺎﺕ‬2 2 ‫ ﻝ‬1 H NMR ‫ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﺍﻟﺒﺮﻭﺗﻮﻧﻲ‬2 ‫ ﻓﻲ ﺍﻟﻤﺮﻛﺐ‬Ha ‫ ﻟﻠﺒﺮﻭﺗﻮﻥ‬chemical shifts ‫ ﺍﻻﻧﺰﻳﺎﺣﺎﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬,‫ﻣﻊ ﻣﺮﺍﻋﺎﺓ ﺇﺑﻘﺎﺀ ﻛﺎﻓﺔ ﺍﻟﻤﻌﻄﻴﺎﺕ ﺍﻷﺧﺮﻯ ﺛﺎﺑﺘﺔ‬‫ﻏﻴﺮ‬, 2 ‫ ﻓﻲ‬Ha ‫ﻓﻲ ﺍﻟﻤﺬﻳﺒﺎﺕ ﺍﻟﻤﺬﻛﻮﺭﺓ ﺑﺎﻷﻋﻠﻰ ﺗﻢ ﺗﻠﺨﻴﺼﻬﺎ ﻓﻲ ﺍﻷﺳﻔﻞ ﻭ ﻻ ﻳﻮﺟﺪ ﺃﻱ ﺇﺷﺎﺭﺓ ﻃﻴﻔﻴﺔ ﺍﺧﺮﻯ ﻟﻠﺒﺮﻭﺗﻮﻥ‬‫ ﺍﻓﺘﺮﺽ ﺃﻥ ﺍﻟﻔﺮﺍﻍ ﺑﺪﺍﺧﻞ ﺍﻟﻜﺒﺴﻮﻟﺔ ﻳﺘﻢ ﻣﻠﺌﻪ ﺩﺍﺋﻤﺎ ﺑﺄﻛﺒﺮ ﻋﺪﺩ ﻣﻤﻜﻦ ﻣﻦ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻤﺬﻳﺐ ﻭﺃﻥ ﻛﻞ‬.‫ﺍﻟﻤﺬﻛﻮﺭﺓ ﺑﺎﻷﺳﻔﻞ‬.‫ﺇﺷﺎﺭﺓ ﺗﺘﻮﺍﻓﻖ ﻣﻊ ﻧﻮﻉ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﻜﺒﺴﻮﻟﺔ ﺍﻟﻤﻤﻠﺆﺓ‬IChO
SAU-1 C-9 Q-6

Arabic (SAU) (Saudi Arabia)

𝛿(ppm)of Hasolvent

4.71

C6 D5 F

4.60

4.60 , 4.71 , 4.82
3pt

Q9-6
C6 D6

C6 D6 / C6 D5 F

Determine the number of C6 D6 and C6 D5 F molecules encapsulated in 2 2 giving
each Ha signal.
‫ ﺗﻌﻄﻲ ﻛﻞ ﻭﺍﺣﺪ‬2 2 ‫ ﺍﻟﻤﺘﻜﺒﺴﻠﺔ ﻓﻲ‬C6 D5 F ‫ ﻭ‬C6 D6 ‫ﺣﺪﺩ ﻋﺪﺩ ﺟﺰﻳﺌﺎﺕ‬Ha ‫ﺍﺷﺎﺭﺓ‬

A.4

IChO
SAU-1 C-9 Q-7

Q9-7
Arabic (SAU) (Saudi Arabia)

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 ﻳﻤﻜﻨﻬﺎ ﺍﻥ ﺗﺤﻮﻱ ﺟﺰﻱﺀ ﻭﺍﺣﺪ ﻣﻦ‬2 2 ‫ ﺍﻥ‬C6 D6 ‫( ﻝ‬1 H NMR ) ‫ﺍﺿﻬﺮﺕ ﻗﻴﺎﺳﺎﺕ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻨﻮﻭﻱ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﺍﻟﺒﺮﻭﺗﻮﻧﻲ‬1-adamantanecarboxylic acid
(AdA) ‫ﺍﻭ‬

‫ ﺍﻟﺮﻣﺰ‬.‫𝐾( ﻳﻤﻜﻦ ﻭﺻﻔﻪ ﺑﺎﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻮﺿﺤﺔ ﺑﺎﻷﺳﻔﻞ ﻭﺗﻢ ﺣﺴﺎﺑﻪ ﻓﻲ ﺩﺭﺟﺎﺕ ﺣﺮﺍﺭﺓ ﻣﺨﺘﻠﻔﺔ‬a ) ‫ﻭ ﺛﺎﺑﺖ ﺍﻟﺘﻜﺒﺴﻞ‬[ solvent@ 2 2 ].‫ﻳﺼﻒ ﺟﺴﻴﻢ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﻭﺍﺣﺪ ﺃﻭ ﺃﻛﺜﺮ ﻣﻦ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻤﺬﻳﺐ‬Similarly, 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.
‫ ﺗﻢ ﺍﺳﺘﻨﺘﺎﺟﻬﺎ ﻋﻦ ﻃﺮﻳﻖ‬C6 D6 ‫( ﻓﻲ ﺩﺭﺟﺎﺕ ﺣﺮﺍﺭﺓ ﻣﺨﺘﻠﻔﺔ ﻓﻲ‬2) ‫ ﺍﻟﻤﻌﻄﺎﺓ ﻣﻦ ﺍﻟﻤﻌﺎﺩﻟﺔ‬1 2 ‫( ﻭ‬CH4 )‫𝐾 ﻝ‬a ‫ ﻗﻴﻢ‬,‫ﺑﺎﻟﻤﺜﻞ‬‫ ﺍﻟﻤﻨﺤﻨﻰ ﻟﺜﺎﺑﺘﻲ ﺍﻟﺘﻜﺒﺴﻞ ﺍﻟﻤﻌﻄﺎﻩ ﻛﺎﻟﺘﺎﻟﻲ‬,1 H NMR ‫ﻗﻴﺎﺳﺎﺕ ﺍﻃﻴﺎﻑ ﺍﻝ‬
as ln 𝐾a vs 1/T.‫ﻣﻮﺿﺤﺔ ﺑﺎﻷﺳﻔﻞ‬No 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 ).
,(2) ‫( ﻫﻮ‬Δ𝐻) ‫( ﻭ ﺍﻟﺘﻐﻴﺮ ﺍﻻﻧﺜﺎﻟﺒﻲ‬1) ‫( ﻫﻮ‬Δ𝑆) ‫ ﺍﻟﺘﻐﻴﺮ ﺍﻻﻧﺘﺮﻭﺑﻲ‬II ‫ ﻓﻲ ﺍﻟﺨﻂ‬.112 ‫ ﻣﺘﻜﺒﺴﻞ ﻓﻲ‬C6 D6 ‫ﻻ ﻭﺟﻮﺩ ﻟﻤﺮﻛﺐ‬‫( ﻭ ﺍﻟﺨﻂ‬4) ‫ ﻳﺘﻮﺍﻓﻖ ﻣﻊ‬I ‫ ﻭﺑﺎﻟﺘﺎﻟﻲ ﻓﺎﻥ ﺍﻟﺨﻂ‬,(3) ‫ ﻫﻲ‬II ‫ﻭﺫﻟﻚ ﺩﻻﻟﺔ ﻋﻠﻰ ﺃﻥ ﺍﻟﻘﻮﺓ ﺍﻟﺘﻲ ﺗﻘﻮﺩ ﺍﻟﻰ ﺍﻟﻜﺒﺴﻠﺔ ﻓﻲ ﺍﻟﺨﻂ‬.(5) ‫ ﻳﺘﻮﺍﻓﻖ ﻣﻊ‬II

Q9-8

IChO
SAU-1 C-9 Q-8

3pt

Arabic (SAU) (Saudi Arabia)

Choose the correct options in gaps (1)–(5) in the following paragraph from A
and B.
B ‫ ﻭ‬A ‫( ﻓﻲ ﺍﻟﻔﻘﺮﺓ ﺍﻟﺘﺎﻟﻴﺔ ﻣﻦ‬5)-(1) ‫ﺍﺧﺘﺎﺭ ﺍﻟﺨﻴﺎﺭ ﺍﻟﺼﺤﻴﺢ ﻓﻲ ﺍﻟﻔﺠﻮﺍﺕ‬B

A

negative

positive

(1)

Δ𝐻

Δ𝑆

2 2 and AdA

1 2 and CH4(3)

negative

2 2 and ADA

positive

1 2 and CH4

(2)
(4)
(5)

A.5

IChO
SAU-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
English (Official)IChO
SAU-1 C-9 A-2

A9-2
English (Official)

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) ∶