SYR-3 C-0 C-1

SYR-3 C-0 C
ANAS ABBASIChOGeneral instructions
Cover sheet

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

IChO
SYR-3 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

SYR-3 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
SYR-3 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
SYR-3 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

SYR-3 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
SYR-3 C-0 G-6

Periodic Table

G0-6
English (Official)IChO
SYR-3 C-0 G-7

1

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

G0-7
English (Official)IChO
SYR-3 C-0 G-1

G0-1
arabic (Syria)

Japan 2021 Olympiad Chemistry International

Japan IChO2021 53rd

2021 August, 2nd – July 25th
https://www.icho2021.org

‫‪IChO‬‬‫‪G0-2‬‬‫‪SYR-3 C-0 G-2‬‬‫)‪arabic (Syria‬‬‫ﺍﻟﺘﻌﻠﻴﻤﺎﺕ ﺍﻟﻌﺎﻣﺔ‬‫• ﻣﺴﻤﻮﺡ ﻟﻚ ﺑﺎﺳﺘﻌﻤﺎﻝ ﻗﻠﻢ ﺍﻟﺤﺒﺮ ﺍﻟﻨﺎﺷﻒ ﻓﻘﻂ ﻟﻜﺘﺎﺑﺔ ﺍﻹﺟﺎﺑﺔ‬‫• ﻳﺠﺐ ﺃﻥ ﺗﻜﻮﻥ ﺁﻟﺘﻚ ﺍﻟﺤﺎﺳﺒﺔ ﻏﻴﺮ ﻗﺎﺑﻠﺔ ﻟﻠﺒﺮﻣﺠﺔ‬‫• ﻳﺘﺄﻟﻒ ﻫﺬﺍ ﺍﻻﻣﺘﺤﺎﻥ ﻣﻦ ﺗﺴﻊ ﻣﺴﺎﺋﻞ‪.‬‬‫• ﻳﻤﻜﻨﻚ ﺣﻞ ﺍﻟﻤﺴﺎﺋﻞ ﻭﻓﻖ ﺃﻱ ﺗﺮﺗﻴﺐ ﺗﺮﻳﺪﻩ‪.‬‬‫• ﻟﺪﻳﻚ ﺧﻤﺲ ﺳﺎﻋﺎﺕ ﻟﺤﻞ ﻛﻞ ﺍﻟﻤﺴﺎﺋﻞ‪.‬‬‫• ﻳﻤﻜﻨﻚ ﺍﻟﺒﺪﺀ ﺑﺎﻟﻌﻤﻞ ﻓﻘﻂ ﺑﻌﺪ ﺃﻥ ﻳﻌﻄﻰ ﺃﻣﺮ ﺍﻟﺒﺪﺀ‪.‬‬‫• ﻳﺠﺐ ﻛﺘﺎﺑﺔ ﻛﻞ ﺍﻟﻨﺘﺎﺋﺞ ﻓﻲ ﺻﻨﺎﺩﻳﻖ ﺍﻹﺟﺎﺑﺔ ﺍﻟﻤﻨﺎﺳﺒﺔ ﺑﺎﺳﺘﺨﺪﺍﻡ ﻗﻠﻢ ﺍﻟﺤﺒﺮ ﺍﻟﻨﺎﺷﻒ ﻋﻠﻰ ﺃﻭﺭﺍﻕ ﺍﻹﺟﺎﺑﺔ‪ .‬ﺍﺳﺘﺨﺪﻡ‬‫ﺃﻥ ﺍﻹﺟﺎﺑﺎﺕ ﺍﻟﻤﻜﺘﻮﺑﺔ ﺧﺎﺭﺝ ﺻﻨﺎﺩﻳﻖ ﺍﻹﺟﺎﺑﺔ ﻟﻦ‬‫ﺧﻠﻔﻴﺔ ﺃﻭﺭﺍﻕ ﺍﻷﺳﺌﻠﺔ ﻓﻲ ﺣﺎﻝ ﺣﺎﺟﺘﻚ ﻟﻮﺭﻕ ﻣﺴﻮﺩﺓ‪ .‬ﻭﺗﺬﻛﺮ ّ‬‫ﺗﺤﺘﺴﺐ‪.‬‬‫• ﺍﻛﺘﺐ ﺍﻟﺤﺴﺎﺑﺎﺕ ﺍﻟﺘﻮﺿﻴﺤﻴﺔ ﺫﺍﺕ ﺍﻟﺼﻠﺔ ﻋﻨﺪ ﺍﻟﻀﺮﻭﺭﺓ ﻓﻲ ﺍﻟﺼﻨﺎﺩﻳﻖ ﺍﻟﻤﻨﺎﺳﺒﺔ‪ .‬ﺳﺘﻌﻄﻰ ﺍﻟﻌﻼﻣﺎﺕ ﺍﻟﻜﺎﻣﻠﺔ ﻟﻺﺟﺎﺑﺎﺕ‬‫ﺍﻟﺼﺤﻴﺤﺔ ﻓﻘﻂ ﻋﻨﺪﻣﺎ ﺗﻘﻮﻡ ﺑﺈﻇﻬﺎﺭ ﻋﻤﻠﻚ‪.‬‬‫• ﺳﻴﻘﻮﻡ ﺍﻟﻤﺮﺍﻗﺒﻮﻥ ﺑﺈﻋﻄﺎﺀ ﺗﻨﺒﻴﻪ ﻗﺒﻞ ‪ 30‬ﺩﻗﻴﻘﺔ ﻣﻦ ﺃﻣﺮ ﺍﻟﺘﻮﻗﻒ‪.‬‬‫• ﻳﺠﺐ ﻋﻠﻴﻚ ﺍﻟﺘﻮﻗﻒ ﻋﻨﺪﻣﺎ ُﻳﻌﻄﻰ ﺃﻣﺮ ﺍﻟﺘﻮﻗﻒ‪ .‬ﺳﻴﻘﻮﺩ ﺍﻹﺧﻔﺎﻕ ﻓﻲ ﺍﻟﺘﻮﻗﻒ ﻋﻦ ﺍﻟﻜﺘﺎﺑﺔ ﺇﻟﻰ ﺇﻟﻐﺎﺀ ﺍﻣﺘﺤﺎﻧﻚ‪.‬‬‫• ﺗﺘﻮﺍﻓﺮ ﺍﻟﻨﺴﺨﺔ ﺍﻹﻧﻜﻠﻴﺰﻳﺔ ﺍﻟﺮﺳﻤﻴﺔ ﻟﻬﺬﺍ ﺍﻻﻣﺘﺤﺎﻥ ﻋﻨﺪ ﺍﻟﻄﻠﺐ ﻓﻘﻂ ﻟﻠﺘﻮﺿﻴﺢ‪.‬‬‫• ﻻ ﻳﺴﻤﺢ ﻟﻚ ﺗﺮﻙ ﻣﻜﺎﻥ ﻋﻤﻠﻚ ﻭﺍﻣﺘﺤﺎﻧﻚ ﻣﻦ ﺩﻭﻥ ﺇﺫﻥ‪ .‬ﻓﻲ ﺣﺎﻝ ﺣﺎﺟﺘﻚ ﻷﻱ ﻣﺴﺎﻋﺪﺓ ) ﺗﻌﻄﻞ ﺁﻟﺔ ﺣﺎﺳﺒﺔ‪ ،‬ﺍﻟﺬﻫﺎﺏ‬‫ﻟﻠﺤﻤﺎﻡ‪ ،‬ﻭﻏﻴﺮﻩ( ‪ ،‬ﺍﺭﻓﻊ ﻳﺪﻙ ﻭﺍﻧﺘﻈﺮ ﻭﺻﻮﻝ ﺍﻟﻤﺮﺍﻗﺐ‪.‬‬‫ﺣﻈﺎً ﻣﻮﻓﻘﺎً‬‫ﻣﻌﻠﻮﻣﺎﺕ ﻋﻦ ﺍﻟﻤﺴﺎﺋﻞ ﻭﺗﻮﺯﻳﻊ ﺍﻟﺪﺭﺟﺎﺕ‬‫ﺍﻟﻌﻨﻮﺍﻥ‬‫ﺍﻟﻌﻼﻣﺔ ﺍﻟﻜﻠﻴﺔ‬‫ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ‬‫‪35‬‬‫‪11‬‬‫‪1‬‬‫ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻋﻠﻰ ﺳﻄﺢ ﻣﻌﺪﻥ‬‫‪24‬‬‫‪3‬‬‫ﻗﺎﻧﻮﻥ ﺑﻴﺮ ‪ -‬ﻻﻣﺒﻴﺮ ‪Law? Lambert–Beer‬‬‫‪22‬‬‫‪2‬‬‫ﻛﺒﺴﻮﻟﺔ ﺍﻟﺰﻣﻦ ﺍﻟﻨﻈﻴﺮﻳﺔ ‪Capsule Time Isotope‬‬

‫‪4‬‬
‫‪5‬‬
‫‪6‬‬
‫‪7‬‬
‫‪8‬‬
‫‪9‬‬

‫‪Total‬‬‫ﻛﻴﻤﻴﺎﺀ ﺍﻷﻛﺴﺪﺓ‪ -‬ﺇﺭﺟﺎﻉ ﻟﻠﺘﻮﺗﻴﺎﺀ‬‫ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﺍﻟﻐﺎﻣﺾ‬‫ﻛﻴﻤﻴﺎﺀ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺼﻠﺒﺔ ﻟﻠﻤﻌﺎﺩﻥ ﺍﻻﻧﺘﻘﺎﻟﻴﺔ‬‫ﺍﻟﻠﻌﺐ ﻣﻊ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﻌﻄﺮﻳﺔ ﺍﻟﻼﺑﻨﺰﻭﺋﻴﺔ‬‫ﺍﻟﺠﺰﻳﺌﺎﺕ ﺍﻟﻌﻀﻮﻳﺔ ﺍﻟﺪﻳﻨﺎﻣﻴﻜﻴﺔ ﻭﻛﻴﺮﺍﻟﻴﺘﻬﺎ‬‫ﻣﺎ ﺗﺤﺒﻪ ﻭﻣﺎ ﺗﻜﺮﻫﻪ ﺍﻟﻜﺒﺴﻮﻟﺔ‬‫‪32‬‬‫‪60‬‬‫‪45‬‬‫‪36‬‬‫‪26‬‬‫‪23‬‬‫‪11‬‬‫‪8‬‬‫‪11‬‬‫‪12‬‬‫‪13‬‬‫‪13‬‬‫‪11‬‬‫‪10‬‬‫‪100‬‬IChO
SYR-3 C-0 G-3

G0-3
arabic (Syria)

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

−34

𝑒 = 1.602176634 × 10𝑚e = 9.10938370 × 10
𝜀0 = 8.85418781 × 10

−12

Js

charge Elementary−1

constant Electricvacuum) of (permittivity

kg

Fm

𝑁A = 6.02214076 × 1023 mol−1
𝑘B = 1.380649 × 10

−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
−1

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 = 10
1 ppb = 10

1 ppt = 10

constant Planck

C

−19

−31

vacuum in light of Speed

J

−6
−9

−12

𝜋 = 3.141592653589793
𝑒 = 2.718281828459045

mass Electron

constant Avogadro

constant Boltzmann
constant Faraday

constant Gas
unit mass atomic Unified
pressure Standard

pressure Atmospheric

Celsius degree ZeroÅngstrom

Picometer

Electronvolt

Part-per-million
Part-per-billion

Part-per-trillion
pilogarithm natural the of base The
number) (Euler’s

IChO

G0-4

SYR-3 C-0 G-4

arabic (Syria)

Equations
𝑃 𝑉 = 𝑛𝑅𝑇
substance, of amount the is 𝑛 volume, the is 𝑉 pressure, the is 𝑃 where ,gas. ideal of temperature absolute the is 𝑇𝑞 𝑞
𝐹 = 𝑘e 1 2 2𝑟
Coulomb’s is 𝑘e (≃ 9.0 × 109 N m2 C−2 ) force, electrostatic the is 𝐹 where ,the is 𝑟 and charges, the of magnitudes the are 𝑞2 and 𝑞1 constant,charges. the between distance
Δ𝑈 = 𝑞 + 𝑤
supplied, heat the is 𝑞 energy, internal the in change the is Δ𝑈 where ,done. work the is 𝑤
𝐻 = 𝑈 + 𝑃𝑉
𝑆 = 𝑘B ln 𝑊
microstates. of number the is 𝑊 where ,𝑞
Δ𝑆 = rev𝑇
process. reversible the for heat the is 𝑞rev where ,𝐺 = 𝐻 − 𝑇𝑆
Δr 𝐺∘ = −𝑅𝑇 ln 𝐾 = −𝑧𝐹 𝐸 ∘𝐸 ∘ electrons, of number the is 𝑧 constant, equilibrium the is 𝐾 where ,potential. electrode standard the is

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

of

law
first
The
thermodynamics
𝐻 Enthalpy

on
based
Entropy
𝑆 principle Boltzmann’s
entropy of change The
Δ𝑆
𝐺 energy free Gibbs

𝑄 quotient Reaction

IChO

G0-5

SYR-3 C-0 G-5

arabic (Syria)

Δ𝑞 = 𝑛𝑐m Δ𝑇
capacity. heat molar temperature-independent the is 𝑐m where ,
𝐶
𝑅𝑇
ln ox
𝐸 = 𝐸∘ +𝑧𝐹
𝐶red
the is 𝐶red substance, oxidized of concentration the is 𝐶ox where ,substance. reduced of concentration𝐸
𝑘 = 𝐴 exp (− 𝑎 )𝑅𝑇the is 𝐸𝑎 factor, pre-exponential the is 𝐴 constant, rate the is 𝑘 where ,energy. activation
exp(𝑥) = 𝑒𝑥
𝐴 = 𝜀𝑙𝑐
is 𝑙 coefficient, absorption molar the is 𝜀 absorbance, the is 𝐴 where ,solution. the of concentration the is 𝑐 length, path optical theequilibrium an For
HA ⇌ H+ + A−
,𝐾a is constant equilibrium where ,[A− ]pH = p𝐾a + log ()[HA]𝑐
𝐸 = ℎ𝜈 = ℎ
𝜆light. the of wavelength the is 𝜆 frequency, the is 𝜈 where ,2

1+𝑥+𝑥 +⋯+𝑥𝑛Δ𝑞 change Heat
forequation Nernst
reaction redox
equation Arrhenius

equation Lambert–BeerHenderson–Hasselbalch
equation

photon a of Energy,𝑥 ≠ 1 Whengeometric a of sum The
series,𝑥 ≪ 1 When
1
≃1+𝑥
1−𝑥

Approximation
be can that equation
problems solve to used

1 − 𝑥𝑛+1𝑛= ∑𝑖=0 𝑥𝑖 =
1−𝑥

IChO
SYR-3 C-0 G-6

G0-6
arabic (Syria)

Table Periodic

IChO
SYR-3 C-0 G-7

G0-7
arabic (Syria)

Shifts Chemical NMR H1

ppm 4.0+ ca. group-substitution: alkyl one for Δ𝛿SYR-3 C-1 C-1

SYR-3 C-1 C
ANAS ABBAS

IChO
Problem 1
Cover sheet

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

Q1-1

IChO
SYR-3 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
SYR-3 C-1 Q-2

Q1-2

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

Fig.2

English (Official)

Q1-3

IChO
SYR-3 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
SYR-3 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
SYR-3 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‫‪Q1-1‬‬‫‪IChO‬‬‫‪SYR-3 C-1 Q-1‬‬‫)‪arabic (Syria‬‬‫ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻋﻠﻰ ﺳﻄﺢ ﺍﻟﻤﻌﺪﻥ‬‫‪Question‬‬‫‪Points‬‬‫‪Score‬‬‫‪A.1‬‬‫‪6‬‬‫‪A.2‬‬‫‪4‬‬‫‪total the of % 11‬‬‫‪B.1‬‬‫‪5‬‬‫‪B.2‬‬‫‪3‬‬‫‪B.3‬‬‫‪3‬‬‫‪B.4‬‬‫‪3‬‬‫‪Total‬‬‫‪24‬‬‫ﻳﺘﻮﻗﻊ ﺃﻥ ﻳﻜﻮﻥ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻣﺼﺪﺭ ﻃﺎﻗﺔ ﺍﻟﻤﺴﺘﻘﺒﻞ ﺍﻟﺘﻲ ﻻ ﺗﻌﺘﻤﺪ ﻋﻠﻰ ﺍﻟﻮﻗﻮﺩ ﺍﻟﺤﺠﺮﻱ‪ .‬ﻓﻲ ﻫﺬﻩ ﺍﻟﻤﺴﺎﻟﺔ ﺳﻨﺪﺭﺱ‬‫ﺗﺨﺰﻳﻦ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻓﻲ ﺍﻟﻤﻌﺪﻥ ﻭﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﺗﺮﺗﺒﻂ ﺑﺘﻨﻜﻨﻮﻟﻮﺟﻴﺎ ﻧﻘﻞ ﻭﺗﺨﺰﻳﻦ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‬‫ﺍﻟﺠﺰﺀ ‪A‬‬

‫ﺃﻭﻻ ﻋﻤﻠﻴﺔ ﺍﺩﻣﺼﺎﺹ )ﺍﻣﺘﺰﺍﺯ( ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‬‫ﺑﻤﺎ ﺃﻥ ﺍﻣﺘﺼﺎﺹ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺇﻟﻰ ﺩﺍﺧﻞ ﺍﻟﻤﻌﺪﻥ ﻳﺘﻢ ﻋﺒﺮ ﺳﻄﺤﻪ‪ ،‬ﻟﺬﻟﻚ ﻧﻌﺘﺒﺮ‬‫ً‬‫ﻋﻠﻰ ﺳﻄﺢ ﺍﻟﻤﻌﺪﻥ‪:‬‬‫)‪H2 (g) → 2H(ad‬‬‫ﺣﻴﺚ ﻳﺘﻢ ﺗﺮﻣﻴﺰ ﺣﺎﻟﺔ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﻐﺎﺯﻱ ﺑـ )‪ (g‬ﻭﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﻤﻤﺘﺰ ﺑـ )‪. (ad‬‬‫ﻋﻨﺪﻣﺎ ﻳﺼﻞ ﺟﺰﻱﺀ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ) ‪ (H2‬ﺇﻟﻰ ﺳﻄﺢ ﺍﻟﻤﻌﺪﻥ )‪ (M‬ﻳﺘﻔﻜﻚ ﻋﻠﻰ ﺍﻟﺴﻄﺢ ﻭﻳﺪﻣﺺ )ﻳﻤﺘﺰ( ﻋﻠﻰ ﺷﻜﻞ ﺫﺭﺍﺕ‬‫ﻫﻴﺪﺭﻭﺟﻴﻦ )‪ (H‬ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﻓﻲ ﺍﻟﺸﻜﻞ ‪.1‬‬‫ﻳﺘﻢ ﺗﻤﺜﻴﻞ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﺎﻣﻨﺔ ﻟﻠﻬﻴﺪﺭﻭﺟﻴﻦ ) ‪ (H2‬ﺑﻤﺘﻐﻴﺮﻳﻦ‪ :‬ﺍﻻﻭﻝ ﻫﻮ ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺍﻟﺬﺭﺍﺕ )𝑑(‪ ،‬ﻭﺍﻟﺜﺎﻧﻲ ﺍﻻﺭﺗﻔﺎﻉ ﺍﻟﻨﺴﺒﻲ‬‫ﻋﻦ ﺫﺭﺍﺕ ﺳﻄﺢ ﺍﻟﻤﻌﺪﻥ )𝑧(‪.‬‬‫ﻧﻔﺘﺮﺽ ﺍﻥ ﺍﻟﻤﺤﻮﺭ ﺍﻟﻮﺍﺻﻞ ﺑﻴﻦ ﺫﺭﺗﻲ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻣﻮﺍﺯﻱ ﻟﻠﺴﻄﺢ ﻭﺃﻥ ﻣﺮﻛﺰ ﺍﻟﺜﻘﻞ ﻳﻘﻊ ﺩﺍﺋﻤﺎً ﻋﻠﻰ ﺍﻟﻌﻤﻮﺩ ﺍﻟﻤﻨﻘﻂ ﺍﻟﻤﻮﺿﺢ‬‫ﺑﺎﻟﺸﻜﻞ ‪.1‬‬‫ﻳﻈﻬﺮ ﺍﻟﺸﻜﻞ ‪ 2‬ﺧﺮﻳﻄﺔ ﺗﺴﺎﻭﻱ ﺍﻟﻜﻤﻮﻥ )‪ (potential energy contour plot‬ﻟﻠﺘﻔﻜﻚ ﻋﻠﻰ ﺍﻟﺴﻄﺢ‪ .‬ﺗﻤﺜﻞ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﺪﺩﻳﺔ‬‫ﻋﻠﻰ ﺍﻟﻤﺨﻄﻂ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﺎﻣﻨﺔ ﻣﻘﺪﺭﺓ ﺑﻮﺍﺣﺪﺓ ﺍﻟﻜﻴﻠﻮﺟﻮﻝ )‪ (kJ‬ﻟﻜﻞ ﻭﺍﺣﺪ ﻣﻮﻝ ﻣﻦ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ) ‪ .(H2‬ﻳﻤﺜﻞ ﺍﻟﻔﺮﺍﻍ ﻣﺎ ﺑﻴﻦ‬

‫‪Q1-2‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-1 Q-2‬‬‫ﺧﻄﻴﻦ ﻣﺘﺼﻠﻴﻦ ﻣﺎﻗﻴﻤﺘﻪ ‪ 20‬ﻛﻴﻠﻮﺟﻮﻝ ﻟﻜﻞ ﻣﻮﻝ ) ‪ ،(20 kJ mol−1‬ﻭﺍﻟﻔﺮﺍﻍ ﻣﺎﺑﻴﻦ ﺍﻟﺨﻄﻮﻁ ﺍﻟﻤﺘﻘﻄﻌﺔ ﻫﻮ ‪ 100‬ﻛﻴﻠﻮﺟﻮﻝ‬‫ﻟﻜﻞ ﻣﻮﻝ ) ‪ (100 kJ mol−1‬ﻭ ﺍﻟﻔﺮﺍﻍ ﺑﻴﻦ ﺧﻂ ﻣﺘﺼﻞ ﻭﺧﻂ ﻣﺘﻘﻄﻊ ‪ 80‬ﻛﻴﻠﻮﺟﻮﻝ ﻟﻜﻞ ﻣﻮﻝ‪.‬‬‫ﻧﻬﻤﻞ ﻃﺎﻗﺔ ﻧﻘﻄﺔ ﺍﻻﻫﺘﺰﺍﺯ ﺍﻟﺼﻔﺮﻱ‪.‬‬‫�������� ������� ������� ��� ����� ���������‪Fig.1 ����� .‬‬‫‪Q1-3‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-1 Q-3‬‬‫‪Fig.2‬‬‫‪A.1‬‬‫ﺍﺧﺘﺮ ﺍﻗﺮﺏ ﻗﻴﻤﺔ ﻣﻦ ﺑﻴﻦ ﺍﻟﻘﻴﻢ ‪ A‬ﺣﺘﻰ ‪ G‬ﻭﺍﻟﺘﻲ ﺗﻨﺎﺳﺐ ﻛﻞ ﻣﻦ ﺍﻟﻌﺒﺎﺭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬‫)‪ (i‬ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺫﺭﺗﻲ ﺟﺰﻱﺀ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﻐﺎﺯﻱ ) ‪(H2‬‬‫)‪ (ii‬ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺫﺭﺍﺕ ﺍﻟﻤﻌﺪﻥ ) ‪ (𝑑M‬ﺍﻟﻤﻮﺿﺤﺔ ﻓﻲ ﺍﻟﺸﻜﻞ ‪.1‬‬‫)‪ (iii‬ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺫﺭﺓ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﻤﺪﻣﺼﺔ )ﺍﻟﻤﻤﺘﺰﺓ( ﻭﺳﻄﺢ ﺍﻟﻤﻌﺪﻥ ) ‪ (ℎad‬ﺍﻟﻤﻮﺿﺤﺔ ﻓﻲ‬‫ﺍﻟﺸﻜﻞ ‪.1‬‬‫‪A. 0.03 nm‬‬‫‪B. 0.07 nm‬‬‫‪C. 0.11 nm‬‬‫‪E. 0.19 nm‬‬‫‪F. 0.23 nm‬‬‫‪G. 0.27 nm‬‬‫‪D. 0.15 nm‬‬‫‪6pt‬‬

‫‪Q1-4‬‬‫)‪arabic (Syria‬‬‫‪A.2‬‬‫‪IChO‬‬‫‪SYR-3 C-1 Q-4‬‬‫ﺍﺧﺘﺮ ﺍﻗﺮﺏ ﻗﻴﻤﺔ ﻣﻦ ﺑﻴﻦ ﺍﻟﻘﻴﻢ ‪ A‬ﺣﺘﻰ ‪ H‬ﻭﺍﻟﺘﻲ ﺗﻨﺎﺳﺐ ﻛﻞ ﻣﻦ ﺍﻟﻌﺒﺎﺭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ )‪ (i‬ﻭ ‪(ii):‬‬‫)‪ (i‬ﺍﻟﻄﺎﻗﺔ ﺍﻟﻼﺯﻣﺔ ﻟﺘﻔﻜﻚ ﺟﺰﻱﺀ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﻐﺎﺯﻱ )‪ ( H2 gaseous‬ﺇﻟﻰ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‬‫ﺍﻟﻐﺎﺯﻳﺔ )‪[H2 (g) → 2H(g)] ( 𝐻 gaseous‬‬‫)‪ (ii‬ﺍﻟﻄﺎﻗﺔ ﺍﻟﻤﺘﺤﺮﺭﺓ ﺃﺛﻨﺎﺀ ﺍﺩﻣﺼﺎﺹ )ﺍﻣﺘﺰﺍﺯ( ﺟﺰﻱﺀ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﻐﺎﺯﻱ ) ‪(H2 gaseous‬‬‫])‪[H2 (g) → 2H(ad‬‬‫‪A. 20 kJ mol−1‬‬‫‪E. 150 kJ mol−1‬‬‫‪B. 40 kJ mol−1‬‬‫‪F. 200 kJ mol−1‬‬‫‪C. 60 kJ mol−1‬‬‫‪G. 300 kJ mol−1‬‬‫‪D. 100 kJ mol−1‬‬‫‪H. 400 kJ mol−1‬‬‫‪4pt‬‬

‫‪Q1-5‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-1 Q-5‬‬‫ﺍﻟﺠﺰﺀ ‪B‬‬‫ﺗﺴﻠﻚ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﻤﺪﻣﺼﺔ )ﺍﻟﻤﻤﺘﺰﺓ( ﺑﻌﺪ ﺫﻟﻚ ﺇﻣﺎ ﻃﺮﻳﻖ ﺍﻻﻣﺘﺼﺎﺹ ﻧﺤﻮ ﺩﺍﺧﻞ ﺍﻟﻤﻌﺪﻥ ﺍﻟﻤﻮﺿﺢ ﻓﻲ ﺍﻟﺘﻔﺎﻋﻞ ‪1b‬‬‫ﺍﻭ ﻃﺮﻳﻖ ﺍﻻﺗﺤﺎﺩ ﻭﺍﻟﺘﺤﺮﺭ ﺇﻟﻰ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻐﺎﺯﻳﺔ ﺍﻟﻤﻮﺿﺢ ﻓﻲ ﺍﻟﺘﻔﺎﻋﻞ ‪ 1a.‬ﺗﻤﺜﻞ ﺍﻟﺮﻣﺰ )‪ H(ab‬ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﻤﻤﺘﺼﺔ‬‫ﻧﺤﻮ ﺩﺍﺧﻞ ﺍﻟﻤﻌﺪﻥ‪.‬‬‫ﺗﺮﻣﺰ ﻣﻌﺪﻻﺕ ﺍﻟﺘﻔﺎﻋﻞ ﺑﻮﺍﺣﺪﺓ ﺍﻟﻤﻮﻗﻊ ﺍﻟﺴﻄﺤﻲ ﻟﻼﻣﺘﺰﺍﺯ ‪ r1‬ﻭﻧﺰﻉ ﺍﻻﻣﺘﺰﺍﺯ ‪ r2‬ﻭﺍﻻﻣﺘﺼﺎﺹ ‪ r3‬ﻛﻤﺎ ﻳﻠﻲ‪𝑟1 [s−1 ], 𝑟2 [s−1 ] :‬‬‫ﻭ ] ‪ 𝑟3 [s−1‬ﻭﻓﻖ ﺍﻟﻌﻼﻗﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬‫ﺣﻴﺚ‪:‬‬‫] ‪Pa ], 𝑘2 [s−1‬‬‫‪−1‬‬‫‪ 𝑘3 [s ] ,𝑘1 [s‬ﻫﻲ ﺛﻮﺍﺑﺖ ﺍﻟﺘﻔﺎﻋﻼﺕ‬‫‪−1‬‬‫‪−1‬‬‫‪ 𝑃H2‬ﺿﻐﻂ ﻏﺎﺯ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ‪H2‬‬‫ﺗﻤﺜﻞ 𝜃 )‪ (0 ≤ 𝜃 ≤ 1‬ﻧﺴﺒﺔ ﻋﺪﺩ ﺍﻟﻤﻮﺍﻗﻊ ﺍﻟﻤﺸﻐﻮﻟﺔ ﺑﺬﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺇﻟﻰ ﻋﺪﺩ ﺍﻟﻤﻮﺍﻗﻊ ﺍﻟﻤﺘﻮﻓﺮﺓ ﻋﻠﻰ ﺍﻟﺴﻄﺢ‪.‬‬‫ﻧﻔﺘﺮﺽ ﺃﻥ ﺍﻻﻣﺘﺰﺍﺯ ﻭﻧﺰﻉ ﺍﻻﻣﺘﺰﺍﺯ ﺍﺳﺮﻉ ﻣﻦ ﺍﻻﻣﺘﺼﺎﺹ ) ‪ (𝑟1 , 𝑟2 ≫ 𝑟3‬ﻭﺃﻥ ﻧﺴﺒﺔ ﺍﻟﻤﻮﺍﻗﻊ ﺍﻟﻤﺸﻐﻮﻟﺔ 𝜃 ﺗﺒﻘﻰ ﺛﺎﺑﺘﺔ‪.‬‬‫‪B.1‬‬‫ﻳﻤﻜﻦ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻦ ‪ 𝑟3‬ﺑﺎﻟﻌﻼﻗﺔ‪:‬‬‫ﻋﺒﺮ ﻋﻦ ‪ C‬ﺑﺪﻻﻟﺔ ﻛﻞ ﻣﻦ ‪ 𝑘1‬ﻭ ‪𝑘2‬‬‫‪5pt‬‬

‫‪Q1-6‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-1 Q-6‬‬‫ﺃﺧﺬ ﻋﻴﻨﺔ ﻣﻌﺪﻧﻴﺔ ﺫﺍﺕ ﺳﻄﺢ ﻣﺴﺎﺣﺘﻪ ) ‪ (𝑆 = 1.0 × 10−3 m2‬ﻭﻧﻀﻌﻪ ﻓﻲ ﺣﺎﻭﻳﺔ ﺣﺠﻤﻬﺎ ‪ 1‬ﻟﻴﺘﺮ ) ‪ (1.0 × 10−3 m3‬ﺗﺤﻮﻱ‬‫ﻋﻠﻰ ﻏﺎﺯ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ )‪ .(𝑃H2 = 1.0 × 102 Pa‬ﺇﻥ ﻛﺜﺎﻓﺔ ﻣﻮﺍﻗﻊ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﻤﻤﺘﺰﺓ ﻋﻠﻰ ﺍﻟﺴﻄﺢ ﻫﻲ )= 𝑁‬‫‪ .(1.3 × 1018 m−2‬ﺗﺘﺮﻙ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﺍﻟﺴﻄﺢ ﻋﻨﺪ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ )‪.(𝑇 = 400 K‬‬‫ﻋﻨﺪ ﺑﺪﺀ ﺍﻟﺘﻔﺎﻋﻞ ‪ ،1‬ﻳﺘﻨﺎﻗﺺ ﺍﻟﻀﻐﻂ ‪ 𝑃H2‬ﺑﻤﻌﺪﻝ ﺛﺎﺑﺖ ) ‪ .(𝑣 = 4.0 × 10−4 Pa s−1‬ﻧﻔﺘﺮﺽ ﺃﻥ ﻏﺎﺯ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻫﻮ ﻏﺎﺯ‬‫ﻣﺜﺎﻟﻲ ﻭﺃﻥ ﺣﺠﻢ ﻋﻴﻨﺔ ﺍﻟﻤﻌﺪﻥ ﻣﻬﻤﻠﺔ‪.‬‬‫‪B.2‬‬‫ﺍﺣﺴﺐ ﻛﻤﻴﺔ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﻤﻤﺘﺰﺓ ﻣﻘﺪﺭﺓ ﺑﺎﻟﻤﻮﻝ ﺑﻮﺍﺣﺪﺓ ﺍﻟﺴﻄﺢ ﻭﻭﺍﺣﺪﺓ ﺍﻟﺰﻣﻦ‪.‬‬‫)] ‪(𝐴 [mol s−1 m−2‬‬‫‪3pt‬‬‫‪B.3‬‬‫ﻋﻨﺪ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ )‪ (𝑇 = 400 K‬ﻛﺎﻧﺖ ‪ C‬ﺗﺴﺎﻭﻱ ) ‪ .(1.0 × 102 Pa−1‬ﺍﺣﺴﺐ ﻗﻴﻤﺔ ‪ 𝑘3‬ﻋﻨﺪ ﺩﺭﺟﺔ‬‫ﺣﺮﺍﺭﺓ ‪ 400‬ﻛﻴﻠﻔﻦ‪.‬‬‫ﺇﺫﺍ ﻟﻢ ﺗﺤﺼﻞ ﻋﻠﻰ ﺟﻮﺍﺏ ﻟﻠﻄﻠﺐ ‪ B.2‬ﺍﺳﺘﺨﺪﻡ ‪𝐴 = 3.6 × 10−7 mol s−1 m−2‬‬‫‪3pt‬‬‫‪B.4‬‬‫ﻋﻨﺪ ﺩﺭﺟﺎﺕ ﺣﺮﺍﺭﺓ ﻣﺨﺘﻠﻔﺔ ‪ T،‬ﻟﺪﻳﻨﺎ ‪ .𝑘3 = 4.8 × 10−2 s−1 ،𝐶 = 2.5 × 103 Pa−1‬ﺗﻌﻄﻲ‬‫ﺍﻻﺷﻜﺎﻝ ﺍﻟﺘﺎﻟﻴﺔ ﻣﻨﺤﻨﻴﺎﺕ ﻋﻼﻗﺔ ‪ 𝑟3‬ﺑﺪﻻﻟﺔ ‪ 𝑃H2‬ﻋﻨﺪ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‪ .‬ﺍﺧﺘﺮ ﺍﻟﻤﻨﺤﻨﻲ ﺍﻟﻤﻨﺎﺳﺐ‬‫ﻟﻠﻤﻌﻄﻴﺎﺕ ﺍﻟﺴﺎﺑﻘﺔ ﻣﻦ ﺍﻻﺷﻜﺎﻝ )‪ (a‬ﺣﺘﻰ )‪(h‬‬‫‪3pt‬‬IChO
SYR-3 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
SYR-3 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
SYR-3 C-1 A-3

B.3 (3 pt)

𝑘3 =B.4 (3 pt)

s−1

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

SYR-3 C-2 C
ANAS ABBAS

IChO
Problem 2
Cover sheet

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

Q2-1

IChO
SYR-3 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
SYR-3 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
SYR-3 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
SYR-3 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

‫‪Q2-1‬‬‫‪IChO‬‬‫‪SYR-3 C-2 Q-1‬‬‫)‪arabic (Syria‬‬‫‪Capsule Time Isotope‬‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪8‬‬‫‪total the of % 11‬‬‫‪A.2‬‬‫‪8‬‬‫‪A.3‬‬‫‪10‬‬‫‪A.4‬‬‫‪9‬‬‫‪Total‬‬‫‪35‬‬‫‪Score‬‬‫ﺗﺴﻤﻰ ﺍﻟﺠﺰﻳﺌﺎﺕ ﺍﻟﺘﻲ ﺗﺨﺘﻠﻒ ﻋﻦ ﺑﻌﻀﻬﺎ ﻓﻘﻂ ﺑﺎﻟﻨﻈﺎﺋﺮ ﻣﺜﻞ ) ‪ (CH4‬ﻭ )‪ (CH3 D‬ﺑﺎﻟﺠﺰﻳﺌﺎﺕ ﺍﻻﻳﺰﻭﺗﺒﻮﻟﻮﻍ )‪(isotopologues‬‬‫‪ .‬ﺗﻤﺘﺎﺯ ﺟﺰﻳﺌﺎﺕ ﺍﻻﻳﺰﻭﺗﺒﻮﻟﻮﻍ ﺑﺎﻥ ﻟﻬﺎ ﺍﻟﺨﻮﺍﺹ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻧﻔﺴﻬﺎ‪ .‬ﻟﻜﻦ ﻓﻲ ﺍﻟﻄﺒﻴﻌﺔ‪ ،‬ﻳﻮﺟﺪ ﺍﺧﺘﻼﻑ ﺑﺴﻴﻂ‪.‬‬‫ﻧﻔﺘﺮﺽ ﺍﻥ ﻛﻞ ﺍﻟﻤﺮﻛﺒﺎﺕ ﻓﻲ ﻫﺬﺍ ﺍﻟﺴﺆﺍﻝ ﻫﻲ ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻐﺎﺯﻳﺔ‪.‬‬‫ﻟﻨﺎﺧﺬ ﺗﻔﺎﻋﻞ ﺍﻟﺘﻮﺍﺯﻥ ﺍﻟﺘﺎﻟﻲ‪:‬‬‫ﺗﺰﺩﺍﺩ ﺍﻻﻧﺘﺮﻭﺑﻴﻪ 𝑆 ﺑﺎﺯﺩﻳﺎﺩ ﻋﺪﺩ ﺍﻟﺤﺎﻻﺕ ﺍﻟﻤﻴﻜﺮﻭﺳﻜﻮﺑﻴﺔ ﺍﻟﻤﺤﺘﻤﻠﺔ ﻟﻠﺠﻤﻠﺔ ‪: W‬‬‫ﻟﺪﻳﻨﺎ ‪ 𝑊 = 1‬ﻣﻦ ﺃﺟﻞ ‪ 12 C16 O2‬ﻭ ‪ .12 C18 O2‬ﺑﺎﻟﻤﻘﺎﺑﻞ ‪ 𝑊 = 2‬ﻣﻦ ﺃﺟﻞ ﺍﻟﺠﺰﻱﺀ ‪ 12 C16 O18 O‬ﻻﻥ ﺫﺭﺍﺕ ﺍﻷﻭﻛﺴﺠﻴﻦ ﻣﺘﻤﺎﻳﺰﺓ‬‫ﻓﻲ ﻫﺬﺍ ﺍﻟﺠﺰﻱﺀ‪ .‬ﺑﻤﺎ ﺃﻥ ﺍﻟﺠﺰﺀ ﺍﻟﻴﻤﻴﻨﻲ ﻣﻦ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻤﺘﻮﺍﺯﻥ ﺍﻟﺴﺎﺑﻖ )‪ (eq. 1‬ﻳﻈﻬﺮ ﺟﺰﻳﺌﺘﻴﻦ ﻣﻦ ‪ 12 C16 O18 O‬ﻓﺈﻥ‪:‬‬‫‪𝑊 = 22 = 4‬‬

‫‪Q2-2‬‬‫)‪arabic (Syria‬‬‫‪A.1‬‬‫‪IChO‬‬‫‪SYR-3 C-2 Q-2‬‬‫ﺗﻐﻴﺮ ﺍﻻﻧﺘﺎﻟﺒﻴﺔ 𝐻‪ Δ‬ﻓﻲ ﻣﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻋﻞ ‪ (eq. 3) 3‬ﻫﻲ ﻣﻮﺟﺒﺔ ﺑﻐﺾ ﺍﻟﻨﻈﺮ ﻋﻦ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‬‫‪8pt‬‬‫ﺍﺣﺴﺐ ﺛﺎﺑﺖ ﺍﻟﺘﻮﺍﺯﻥ 𝐾 ﻟﻠﺘﻔﺎﻋﻞ ‪ (eq. 3) 3‬ﻋﻨﺪ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﻣﻨﺨﻔﻀﺔ )ﺍﻋﺘﺒﺮ ‪ (𝑇 → 0‬ﻭﻋﻨﺪ‬‫ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﻣﺮﺗﻔﻌﺔ ) ﺍﻋﺘﺒﺮ ∞‪.(𝑇 → +‬‬‫ﺍﻓﺘﺮﺽ ﺃﻥ ﺍﻟﺘﻔﺎﻋﻞ ﻳﺒﻘﻰ ﺛﺎﺑﺘﺎً ﻭﻻﻳﺘﻐﻴﺮ ﻋﻨﺪ ﺩﺭﺟﺎﺕ ﺍﻟﺤﺮﺍﺭﺓ ﻫﺬﻩ ﻭﺍﻥ ﺗﻐﻴﺮ ﺍﻻﻧﺘﺎﻟﺒﻴﺔ 𝐻‪Δ‬‬‫ﻳﺘﻘﺎﺭﺏ ﺇﻟﻰ ﻗﻴﻤﺔ ﺛﺎﺑﺘﺔ ﻋﻨﺪ ﺩﺭﺟﺎﺕ ﺍﻟﺤﺮﺍﺭﺓ ﺍﻟﻤﺮﺗﻔﻌﺔ‪.‬‬‫ﻳﻤﻜﻦ ﺷﺮﺡ ﺗﻐﻴﺮ ﺍﻻﻧﺘﺎﻟﺒﻴﺔ 𝐻‪ Δ‬ﻟﻠﻌﻤﻠﻴﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ﻋﻦ ﻃﺮﻳﻖ ﺍﻫﺘﺰﺍﺯﺍﺕ ﺍﻟﺠﺰﻳﺌﺎﺕ‪.‬‬‫ﻋﻨﺪ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ 𝑇 = ‪ 0‬ﻛﻴﻠﻔﻦ‪ ،‬ﻳﻌﺒﺮ ﻋﻦ ﻃﺎﻗﺔ ﺍﻻﻫﺘﺰﺍﺯ ﻟﻠﺠﺰﻳﺌﺎﺕ ﺍﻟﺜﻨﺎﺋﻴﺔ ﻭﺍﻟﺘﻲ ﺗﻮﺍﺗﺮ ﺍﻻﻫﺘﺰﺍﺯ ﻫﻮ 𝜈 ] ‪ [s−1‬ﻛﻤﺎﻳﻠﻲ‪:‬‬‫ﺣﻴﺚ ﺍﻥ 𝑘 ﻫﻮ ﺗﺎﺑﺖ ﺍﻟﻘﻮﺓ ﻭ ﺃﻥ 𝜇 ﻫﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﺨﺘﺰﻟﺔ ﻭﺍﻟﺘﻲ ﻳﻌﺒﺮ ﻋﻨﻬﺎ ﺑﻜﺘﻠﺔ ﺍﻟﺬﺭﺗﻴﻦ ﻓﻲ ﺍﻟﺠﺰﻳﺌﺎﺕ ﺍﻟﺜﻨﺎﺋﻴﺔ ﺍﻟﺬﺭﺓ ‪𝑚1‬‬‫ﻭ ‪ 𝑚2‬ﻭﻓﻖ ﻣﺎﻳﻠﻲ‪:‬‬‫‪A.2‬‬‫ﻳﻜﻮﻥ ﺍﻫﺘﺰﺍﺯ ﺟﺰﻳﺌﺔ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ‪ H2‬ﻋﻨﺪ ) ‪ ( 4161.0 cm−1‬ﻭﺫﻟﻚ ﺑﺎﺳﺘﻌﻤﺎﻝ ﺍﻟﻌﺪﺩ ﺍﻟﻤﻮﺟﻲ‪.‬‬‫ﺍﺣﺴﺐ ﺗﻐﻴﺮ ﺍﻻﻧﺘﺎﻟﺒﻴﺔ 𝐻‪ Δ‬ﻟﻠﺘﻔﺎﻋﻞ ﺍﻟﺘﺎﻟﻲ ﻋﻨﺪ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ 𝑇 = ‪ 0‬ﻛﻴﻠﻔﻦ ﻣﻘﺪﺭﺍ ﺑﺎﻟﻮﺍﺣﺪﺓ‬‫) ‪(J mol−1‬‬‫ﻧﻔﺘﺮﺽ ﺃﻥ‪:‬‬‫• ﺃﻧﻪ ﻃﺎﻗﺔ ﺍﻻﻫﺘﺰﺍﺯ ﻓﻘﻂ ﻫﻲ ﺍﻟﺘﻲ ﺗﺴﺎﻫﻢ ﻓﻲ 𝐻‪Δ‬‬‫• ﻗﻴﻤﺔ 𝑘 ﻟﻜﻞ ﻣﻦ ‪ HD‬ﻭ ‪ H2‬ﻭ ‪ D2‬ﻣﺘﻤﺎﺛﻠﺔ‬‫• ﻛﺘﻠﺔ ﺫﺭﺓ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ‪ H‬ﺗﺴﺎﻭﻱ )‪ (1 Da‬ﻭﻛﺘﻠﺔ ﺫﺭﺓ ﺍﻟﺪﻳﺘﺮﻳﻮﻡ ‪ D‬ﺗﺴﺎﻭﻱ )‪(2 Da‬‬‫‪8pt‬‬

‫‪Q2-3‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-2 Q-3‬‬‫ﺗﺮﺗﺒﻂ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟـ ‪ D2 ،HD ،H2‬ﺑﺪﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻓﻲ ﺍﻟﺠﻤﻠﺔ ﻓﻲ ﺣﺎﻟﺔ ﺍﻟﺘﻮﺍﺯﻥ‪ .‬ﻳﻌﺒﺮ ﺍﻟﺮﻣﺰ ‪ ΔD2‬ﻋﻦ ﺍﻟﺘﻐﻴﺮ ﺑﺎﻟﻨﺴﺒﺔ‬‫ﺍﻟﻤﻮﻟﻴﺔ ﺑـ ‪.D2‬‬‫] ‪[D2‬‬‫] ‪[D‬‬‫ﻟﺪﻳﻨﺎ ‪ 𝑅D2‬ﺗﺸﻴﺮ ﺇﻟﻰ ‪ 2‬ﻓﻲ ﺍﻟﻌﻴﻨﺔ ﻭ ‪ 𝑅D∗ 2‬ﺇﻟﻰ‬‫] ‪[H2‬‬‫] ‪[H2‬‬‫ﻋﻨﺪ )∞‪(T→+‬‬‫‪A.3‬‬‫ﻋﻨﺪ )∞‪ .(T→+‬ﻳﺠﺐ ﺃﻥ ﻧﻼﺣﻆ ﻫﻨﺎ ﺃﻥ ﺗﻮﺯﻉ ﺍﻟﻨﻈﺎﺋﺮ ﻳﺼﺒﺢ ﻋﺸﻮﺍﺋﻴﺎً‬‫ﺍﺣﺴﺐ ‪ ΔD2‬ﻣﻊ ﻭﻓﺮﺓ ‪ D‬ﺍﻟﻄﺒﻴﻌﻴﺔ ﻋﻨﺪﻣﺎ ﻳﻜﻮﻥ ﺍﻟﺘﺒﺎﺩﻝ ﺍﻟﻨﻈﻴﺮﻱ ﻓﻲ ﺣﺎﻟﺔ ﺗﻮﺍﺯﻥ ﻋﻨﺪ ﺩﺭﺟﺔ‬‫ﺍﻟﺤﺮﺍﺭﺓ ﺣﻴﺚ 𝐾 ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ )‪ (eq. 4‬ﻳﺴﺎﻭﻱ )‪.(0.300‬‬‫ﺍﻓﺘﺮﺽ ﺍﻥ ﻧﺴﺒﺔ ﺍﻟﻮﻓﺮﺓ ﺍﻟﻄﺒﻴﻌﻴﺔ ﻟـ ‪ D‬ﻭ ‪ H‬ﻫﻲ ) ‪ (1.5576 × 10–4‬ﻭ ) ‪(1 − 1.5576 × 10–4‬‬‫ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‪.‬‬‫‪10pt‬‬‫‪Q2-4‬‬‫‪IChO‬‬‫‪SYR-3 C-2 Q-4‬‬‫)‪arabic (Syria‬‬‫ﺑﺸﻜﻞ ﻋﺎﻡ‪ ،‬ﺗﺰﺩﺍﺩ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻣﻊ ﺍﺯﺩﻳﺎﺩ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻟﻼﻳﺰﻭﺗﺒﻮﻟﻮﻍ ﺍﻟﻤﺴﺘﺒﺪﻝ ﻣﺮﺗﻴﻦ ﻭﺍﻟﺬﻱ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺫﺭﺗﻴﻦ‬‫ﻧﻈﻴﺮﺗﻴﻦ ﺛﻘﻴﻠﺘﻴﻦ ﻓﻲ ﺟﺰﻳﺌﺔ ﻭﺍﺣﺪﺓ‪.‬‬‫ﻟﻨﺄﺧﺬ ﺑﻌﻴﻦ ﺍﻻﻋﺘﺒﺎﺭ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﺠﺰﺋﻴﺎﺕ ‪ CO2‬ﺍﻟﺘﻲ ﻟﻬﺎ ﻭﺯﻥ ﺟﺰﻳﺌﻲ ‪ 44‬ﻭ ‪ 47‬ﻭﺍﻟﺬﻱ ﻳﻮﺻﻒ ﻛـ ]‪ CO2 [44‬ﻭ ]‪.CO2 [47‬‬‫ﺗﻌﺮﻑ ﺍﻟﻜﻤﻴﺔ ‪ Δ47‬ﻛﻤﺎ ﻳﻠﻲ‪:‬‬‫]]‪[CO2 [47‬‬‫]]‪[CO2 [47‬‬‫∗‬‫‪ 𝑅47‬ﺇﻟﻰ‬‫ﻓﻲ ﺍﻟﻌﻴﻨﺔ ﻭ‬‫ﺗﺸﻴﺮ ‪ 𝑅47‬ﺇﻟﻰ‬‫]]‪[CO2 [44‬‬‫]]‪[CO2 [44‬‬‫ﻟﺬﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ﻭﺍﻻﻭﻛﺴﺠﻴﻦ‪ ،‬ﻧﻬﻤﻞ ﺍﻟﻨﻈﺎﺋﺮ ﻏﻴﺮ ﺍﻟﻤﻮﺟﻮﺩﺓ ﻫﻨﺎ‪.‬‬‫‪abundance natural‬‬‫ﺍﻟﻮﻓﺮﺓ ﺍﻟﻄﺒﻴﻌﻴﺔ‬‫‪O‬‬‫‪16‬‬‫‪0.997621‬‬‫ﻋﻨﺪﻣﺎ )∞‪ .(T→+‬ﻳﻈﻬﺮ ﺍﻟﺠﺪﻭﻝ ﺍﻟﺘﺎﻟﻲ ﺍﻟﻮﻓﺮﺓ ﺍﻟﻄﺒﻴﻌﻴﺔ‬‫‪C‬‬‫‪12‬‬‫‪C‬‬‫‪13‬‬‫‪0.988888‬‬‫‪0.011112‬‬‫‪17‬‬

‫‪O‬‬

‫‪O‬‬

‫‪0.0003790‬‬‫‪18‬‬‫‪0.0020000‬‬‫ﻳﺤﺪﺩ ﺍﺭﺗﺒﺎﻁ ‪ Δ47‬ﺑﺪﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻛﻤﺎ ﻳﻠﻲ‪ ،‬ﺣﻴﺚ ﺗﻌﻄﻰ 𝑇 ﺑﺪﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﺍﻟﻤﻄﻠﻘﺔ ﺑﻮﺍﺣﺪﺓ ﺍﻟﻜﻠﻔﻦ ‪k.‬‬‫‪A.4‬‬‫ﺗﻢ ﺍﻟﺤﺼﻮﻝ ﻋﻠﻰ ﻗﻴﻤﺔ ‪ 𝑅47‬ﻟﻤﺴﺘﺤﺎﺛﺔ ﺍﻟﺒﻼﻧﻜﺘﻮﻥ ﻫﺬﻩ ﻣﻦ ﻗﺎﻉ ﺍﻟﻤﺤﻴﻂ ﺍﻟﻤﺘﺠﻤﺪﺓ ﻭﺗﺴﺎﻭﻱ‬‫) ‪.(4.50865 × 10–5‬‬‫ﻗﺪﺭ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻣﺴﺘﻌﻤﻼ ﻗﻴﻤﺔ ‪ . 𝑅47‬ﺗﻔﺴﺮ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻫﺬﻩ ﺑﺄﻧﻬﺎ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‬‫ﺍﻟﺴﺎﺋﺪﺓ ﺃﺛﻨﺎﺀ ﺍﻟﻔﺘﺮﺓ ﺍﻟﺘﻲ ﻋﺎﺵ ﻓﻴﻬﺎ ﺍﻟﺒﻼﻧﻜﺘﻮﻥ‪ .‬ﺍﻋﺘﺒﺮ ﻓﻘﻂ ﺍﻻﻳﺰﻭﺗﺒﻮﻟﻮﻍ ﺍﻻﻛﺜﺮ ﺷﻬﺮﺓ ﻟـ‬‫]‪ CO2 [47‬ﻓﻲ ﺍﻟﺤﺴﺎﺑﺎﺕ‪.‬‬‫‪9pt‬‬IChO
SYR-3 C-2 A-1

Isotope Time Capsule
A.1 (8 pt)

𝑇 →0∶𝐾=, 𝑇 → +∞ ∶ 𝐾 =A2-1
English (Official)IChO
SYR-3 C-2 A-2

A.2 (8 pt)

Δ𝐻 =J mol−1

A2-2
English (Official)IChO
SYR-3 C-2 A-3

A.3 (10 pt)

ΔD2 =

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

A.4 (9 pt)

𝑇 =

K

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

SYR-3 C-3 C
ANAS ABBAS

IChO
Problem 3
Cover sheet

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

Q3-1

IChO
SYR-3 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
SYR-3 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

‫‪Q3-1‬‬‫‪IChO‬‬‫‪SYR-3 C-3 Q-1‬‬‫)‪arabic (Syria‬‬‫ﻗﺎﻧﻮﻥ ﺑﻴﻴﺮ ﻻﻣﺒﻴﺮ؟‬‫‪Question‬‬‫‪Points‬‬‫‪total the of % 8‬‬‫‪A.1‬‬‫‪10‬‬‫‪B.1‬‬‫‪6‬‬‫‪B.2‬‬‫‪6‬‬‫‪Total‬‬‫‪22‬‬‫‪Score‬‬‫ﻓﻲ ﻫﺬﻩ ﺍﻟﻤﺴﺄﻟﺔ‪ ،‬ﻧﻬﻤﻞ ﺍﻣﺘﺼﺎﺹ ﺍﻟﺨﻠﻴﺔ ﻭﺍﻟﻤﺬﻳﺐ‪ .‬ﻭﻧﻌﺘﺒﺮ ﺍﻥ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﻛﻞ ﺍﻟﻤﺤﺎﻟﻴﻞ ﻭﺍﻟﻐﺎﺯﺍﺕ ﺗﺒﻘﻰ ﺛﺎﺑﺘﺔ ﻋﻨﺪ ‪25‬‬‫ﺩﺭﺟﺔ ﺳﻴﻠﺴﻴﻮﺱ‬‫ﺍﻟﺠﺰﺀ ‪A‬‬‫ﺗﻢ ﺗﺤﻀﻴﺮ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ‪ X‬ﺑﺎﺳﺘﺨﺪﺍﻡ ‪ HA‬ﻭ ‪ NaA,‬ﺗﺮﻛﻴﺰ ﻛﻞ ﻣﻦ ] ‪ [H+ ] ،[HA] ،[A−‬ﻓﻲ ﺍﻟﻤﺤﻠﻮﻝ ‪ X‬ﻫﻮ ﻋﻠﻰ ﺍﻟﺘﺘﺎﻟﻲ‪:‬‬‫‪ .1.00 × 10−4 mol L−1 ،1.00 × 10−3 mol L−1 ،1.00 × 10−2 mol L−1‬ﻭﺍﻟﺘﻲ ﺗﺮﺗﺒﻂ ﻭﻓﻖ ﺗﻔﺎﻋﻞ ﺍﻟﺘﻮﺍﺯﻥ ﺣﻤﺾ ‪-‬‬
‫ﺍﺳﺎﺱ ﺍﻟﺘﺎﻟﻲ‪:‬‬‫] ‪[A− ][H+‬‬‫]‪[HA‬‬‫)‪(1‬‬‫=𝐾‬‫‪HA ⇌ A− + H+‬‬‫ﺍﻟﻤﺴﺎﺭ ﺍﻟﻀﻮﺋﻲ ﻓﻲ ﺍﻟﺠﺰﺀ ‪ A‬ﻫﻮ 𝑙‪ .‬ﻧﻬﻤﻞ ﺗﻐﻴﺮ ﺍﻟﻜﺜﺎﻓﺔ ﺑﺎﻟﺘﻤﺪﻳﺪ‪ .‬ﻧﻔﺘﺮﺽ ﺃﻧﻪ ﻻﻳﻮﺟﺪ ﺗﻔﺎﻋﻞ ﻛﻴﻤﻴﺎﺋﻲ ﺍﺧﺮ ﻏﻴﺮ ﺍﻟﻤﻮﺿﺢ‬‫ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ )‪(eq 1‬‬‫‪A.1‬‬‫ﻛﺎﻥ ﺍﻣﺘﺼﺎﺹ ﺍﻟﻤﺮﻛﺐ ‪ X‬ﻫﻮ ‪ 𝐴1‬ﻋﻨﺪ ﻃﻮﻝ ﺍﻟﻤﻮﺟﺔ ‪ .𝜆1‬ﺛﻢ ﺗﻢ ﺗﻤﺪﻳﺪ ﺍﻟﻤﺤﻠﻮﻝ ‪ X‬ﺇﻟﻰ ﺿﻌﻔﻲ‬‫ﺍﻟﺤﺠﻢ ﺍﻷﻭﻟﻲ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺣﻤﺾ ﻛﻠﻮﺭ ﺍﻟﻤﺎﺀ ﺫﻱ ﺩﺭﺟﺔ ﺍﻟﺤﻤﻮﺿﺔ ‪ (pH=2.500).‬ﺑﻌﺪ ﺍﻟﺘﻤﺪﻳﺪ‪،‬‬‫ﻳﺒﻘﻰ ﺍﻻﻣﺘﺼﺎﺹ ‪ 𝐴1‬ﻋﻨﺪ ﻃﻮﻝ ﺍﻟﻤﻮﺟﺔ ‪.𝜆1‬‬‫ﺣﺪﺩ ﺍﻟﻨﺴﺒﺔ ) ‪ (𝜀HA /𝜀A−‬ﺣﻴﺚ ﺗﻤﺜﻞ ﻛﻞ ﻣﻦ ‪ 𝜀HA‬ﻭ ‪ 𝜀A−‬ﻣﻌﺎﻣﻞ ﺍﻻﻣﺘﺼﺎﺹ ﻟﻜﻞ ﻣﻦ ‪ HA‬ﻭ ‪A−‬‬‫ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ﻋﻨﺪ ﻃﻮﻝ ﺍﻟﻤﻮﺟﺔ ‪.𝜆1‬‬‫‪10pt‬‬‫‪Q3-2‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-3 Q-2‬‬‫ﺍﻟﺠﺰﺀ ‪B‬‬‫ﻟﻨﺄﺧﺬ ﺍﻟﺘﻮﺍﺯﻥ ﺍﻟﺘﺎﻟﻲ ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻐﺎﺯﻳﺔ‪.‬‬‫‪D ⇌ 2M‬‬‫(‪)2‬‬‫ﻧﻤﻸ ﻏﺎﺯ ﻧﻘﻲ ‪ D‬ﻋﻨﺪ ﺿﻐﻂ 𝑃 ﻓﻲ ﺍﻟﺤﺎﻭﻳﺔ ﺍﻟﻤﻮﺿﺤﺔ ﺑﺎﻟﺸﻜﻞ ﻭﺍﻟﺘﻲ ﻟﻬﺎ ﺟﺪﺍﺭ ﺷﻔﺎﻑ ﻣﺘﺤﺮﻙ ﻭﻣﻘﻄﻊ ﻣﺴﺎﺣﺘﻪ 𝑆‪ ،‬ﻳﺤﺪﺙ‬‫ﺍﻟﺘﻮﺍﺯﻥ ﻣﻊ ﺑﻘﺎﺀ ﺍﻟﻀﻐﻂ ﺍﻟﻜﻠﻲ ﺛﺎﺑﺘﺎً ﻋﻨﺪ 𝑃‪.‬‬‫ﺗﻌﻄﻰ ﺍﻣﺘﺼﺎﺻﻴﺔ ﺍﻟﻐﺎﺯ ﺑﺎﻟﻌﻼﻗﺔ )𝑙) 𝑉‪(𝐴 = 𝜀(𝑛/‬‬‫ﺣﻴﺚ 𝜀‪ 𝑙 ،𝑉 ،𝑛 ،‬ﻫﻲ ﻣﻌﺎﻣﻞ ﺍﻻﻣﺘﺼﺎﺻﻴﺔ‪ ،‬ﻋﺪﺩ ﻣﻮﻻﺕ ﺍﻟﻐﺎﺯ‪ ،‬ﺣﺠﻢ ﺍﻟﻐﺎﺯ‪ ،‬ﻃﻮﻝ ﺍﻟﻤﺴﺎﺭ ﺍﻟﻀﻮﺋﻲ ﻋﻠﻰ ﺍﻟﺘﺘﺎﻟﻲ‪.‬‬‫ﻧﻔﺘﺮﺽ ﺃﻥ ﻛﻞ ﻣﻜﻮﻧﺎﺕ ﺍﻟﻐﺎﺯ ﺗﺘﺼﺮﻑ ﺗﺼﺮﻑ ﺍﻟﻐﺎﺯ ﺍﻟﻤﺜﺎﻟﻲ‪.‬‬‫ﺍﺳﺘﺨﺪﻡ ﺍﻟﺘﻌﺮﻳﻔﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ﻓﻲ ﺣﺎﻝ ﺍﻟﻀﺮﻭﺭﺓ‪.‬‬‫ﺍﻟﺤﺎﻟﺔ ﺍﻻﺑﺘﺪﺍﺋﻴﺔ‬‫ﺍﻟﻀﻐﻂ ﺍﻟﺠﺰﺋﻲ‬‫ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ‬‫ﺍﻟﺤﺠﻢ‬‫ﺑﻌﺪ ﺍﻟﺘﻮﺗﺰﻥ‬

‫‪D‬‬

‫‪M‬‬

‫‪D‬‬

‫𝑃‬

‫‪0‬‬

‫‪𝑝D‬‬‫‪𝑝M‬‬‫‪𝑛0‬‬‫‪0‬‬‫‪𝑛D‬‬‫‪𝑛M‬‬‫‪𝑉0‬‬‫‪M‬‬‫𝑉‬‫‪B.1‬‬‫ﺍﻣﺘﺼﺎﺻﻴﺔ ﺍﻟﻐﺎﺯ ﻋﻨﺪ ﻃﻮﻝ ﺍﻟﻤﻮﺟﺔ ‪ 𝜆B1‬ﻭﺍﻟﻤﻘﺎﺳﺔ ﺑﺎﻻﺗﺠﺎﻩ 𝑥 ) 𝑥𝑙 = 𝑙( ﻫﻲ ‪ 𝐴B1‬ﻓﻲ ﺍﻟﺤﺎﻟﺘﻴﻦ‬‫ﺍﻻﺑﺘﺪﺍﺋﻴﺔ ﻭﺑﻌﺪ ﺍﻟﺘﻮﺍﺯﻥ‪.‬‬‫ﺣﺪﺩ ﺍﻟﻨﺴﺒﺔ ) ‪ (𝜀D /𝜀M‬ﻋﻨﺪ ﻃﻮﻝ ﺍﻟﻤﻮﺟﺔ ‪ 𝜆B1‬ﺣﻴﺚ ‪ 𝜀D‬ﻭ ‪ 𝜀M‬ﺗﻤﺜﻼﻥ ﻣﻌﺎﻣﻞ ﺍﻻﻣﺘﺼﺎﺻﻴﺔ ﻟـ ‪D‬‬‫ﻭ ‪ M‬ﻋﻠﻰ ﺍﻟﺘﺘﺎﻟﻲ‪.‬‬‫‪6pt‬‬‫‪B.2‬‬‫ﺍﻣﺘﺼﺎﺻﻴﺔ ﺍﻟﻐﺎﺯ ﻋﻨﺪ ﻃﻮﻝ ﺍﻟﻤﻮﺟﺔ ‪ 𝜆B2‬ﻭﺍﻟﻤﻘﺎﺳﺔ ﺑﺎﻻﺗﺠﺎﻩ 𝑦 ﻫﻲ ‪ 𝐴B2‬ﻓﻲ ﺍﻟﺤﺎﻟﺘﻴﻦ‬‫ﺍﻻﺑﺘﺪﺍﺋﻴﺔ ) ‪ (𝑙 = 𝑙𝑦0‬ﻭﺑﻌﺪ ﺍﻟﺘﻮﺍﺯﻥ ) 𝑦𝑙 = 𝑙(‪.‬‬‫ﺣﺪﺩ ﺍﻟﻨﺴﺒﺔ ) ‪ (𝜀D /𝜀M‬ﻋﻨﺪ ﻃﻮﻝ ﺍﻟﻤﻮﺟﺔ ‪𝜆B2‬‬‫‪6pt‬‬IChO
SYR-3 C-3 A-1

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

A.1 (cont.)

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

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

B.2 (6 pt)

𝜀D /𝜀M =A3-4
English (Official)SYR-3 C-4 C-1

SYR-3 C-4 C
ANAS ABBAS

IChO
Problem 4
Cover sheet

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

Q4-1

IChO
SYR-3 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
SYR-3 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
SYR-3 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

‫‪Q4-1‬‬‫‪IChO‬‬‫‪SYR-3 C-4 Q-1‬‬‫)‪arabic (Syria‬‬‫ﻛﻴﻤﻴﺎﺀ ﺍﻷﻛﺴﺪﺓ ﻭﺍﻹﺭﺟﺎﻉ ﻟﻠﺘﻮﺗﻴﺎﺀ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪6‬‬‫‪A.2‬‬‫‪5‬‬‫‪total the of % 11‬‬‫‪B.1‬‬‫‪4‬‬‫‪B.2‬‬‫‪3‬‬‫‪B.3‬‬‫‪5‬‬‫‪B.4‬‬‫‪9‬‬‫‪Total‬‬‫‪32‬‬‫‪Score‬‬‫ﺍﻟﺼﻔﺮ ﻭﻣﻮﺍﺩ ﺍﻟﻔﻮﻻﺫ‪ .‬ﻳﺠﺮﻱ ﻓﺼﻞ ﺍﻟﺘﻮﺗﻴﺎﺀ ﺍﻟﻤﻮﺟﻮﺩﺓ ﻓﻲ ﻣﻴﺎﻩ‬‫ُﻳﺴﺘﻌﻤﻞ ﺍﻟﺘﻮﺗﻴﺎﺀ ﻣﻨﺬ ﺯﻣﻦ ﺑﻌﻴﺪ ﻓﻲ ﺍﻟﺨﻼﺋﻂ ﻣﺜﻞ ﱡ‬‫ﺍﻟﺼﺮﻑ ﺍﻟﺼﻨﺎﻋﻴﺔ ﺑﺎﻟﺘﺮﺳﻴﺐ ﻭﺫﻟﻚ ﻹﺯﺍﻟﺔ ﺍﻟﺴﻤﻴﺔ ﻣﻦ ﺍﻟﻤﻴﺎﻩ‪ ،‬ﻭﻳﺠﺮﻱ ﺑﻌﺪ ﺫﻟﻚ ﺇﺭﺟﺎﻉ ﺍﻟﺮﺍﺳﺐ ﻻﺳﺘﻌﺎﺩﺓ ﺍﻟﺘﻮﺗﻴﺎﺀ ﻭﺇﻋﺎﺩﺓ‬‫ﺍﺳﺘﻌﻤﺎﻟﻬﺎ‪.‬‬‫ﺍﻟﺠﺰﺀ ‪A‬‬‫ﺇﻥ ﺗﻮﺍﺯﻥ ﺍﻧﺤﻼﻝ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺘﻮﺗﻴﺎﺀ )‪ Zn(OH)2 (s‬ﻋﻨﺪ ‪ 25 ∘ C‬ﻭﺛﻮﺍﺑﺖ ﺍﻟﺘﻮﺍﺯﻧﺎﺕ ﺍﻟﻤﺘﻌﻠﻘﺔ ﺑﺬﻟﻚ ﻣﻌﻄﺎﺓ ﺑﺎﻟﻤﻌﺎﺩﻻﺕ ‪eq.‬‬‫‪.4-1‬‬

‫‪Q4-2‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-4 Q-2‬‬‫ﺇﻥ ﺍﻻﻧﺤﻼﻟﻴﺔ‪ ،𝑆 ،‬ﻟﻠﺘﻮﺗﻴﺎﺀ )ﺗﺮﻛﻴﺰ ﺍﻟﺘﻮﺗﻴﺎﺀ ﻓﻲ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ﻣﺸﺒﻊ( ﻣﻌﻄﺎﺓ ﺑﺎﻟﻤﻌﺎﺩﻟﺔ ‪.5‬‬‫‪A.1‬‬‫ﻋﻨﺪﻣﺎ ﺗﺘﺤﻘﻖ ﺍﻟﺘﻮﺍﺯﻧﺎﺕ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻻﺕ ‪ ،4-1‬ﺍﺣﺴﺐ ﻣﺠﺎﻝ ﺍﻟـ ‪ pH‬ﺍﻟﺬﻱ ﻳﻜﻮﻥ ﻓﻴﻪ ﺗﺮﻛﻴﺰ‬‫‪2+‬‬‫‪.[Zn(OH)2−‬‬‫])‪ [Zn(OH)2 (aq‬ﻫﻮ ﺍﻷﻛﺒﺮ ﻣﻦ ﺑﻴﻦ ])‪ ، [Zn(OH)2 (aq)] ، [Zn (aq‬ﻭ ])‪4 (aq‬‬‫‪6pt‬‬‫‪A.2‬‬‫ﺟﺮﻯ ﺗﺤﻀﻴﺮ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ﻣﺸﺒﻊ ﻣﻦ )‪ Zn(OH)2 (s‬ﻋﻨﺪ ‪ pH= 7.00‬ﻭﺟﺮﻯ ﺗﺮﺷﻴﺤﻪ‪ .‬ﺃﺿﻴﻒ ﺇﻟﻰ‬‫ﺍﻟﺮﺷﺎﺣﺔ ‪ NaOH‬ﻟﺮﻓﻊ ‪ pH‬ﺍﻟﻤﺤﻠﻮﻝ ﺇﻟﻰ ‪ .12.00‬ﺍﺣﺴﺐ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﺘﻮﺗﻴﺎﺀ‬‫‪5pt‬‬‫ﺍﻟﻤﺘﺮﺳﺒﺔ ﻋﻨﺪ ﺭﻓﻊ ﺍﻟـ ‪ pH‬ﻣﻦ ‪ 7.00‬ﺇﻟﻰ ‪ .12.00‬ﺃﻫﻤﻞ ﺍﻟﺘﻐﻴﺮﺍﺕ ﻓﻲ ﺍﻟﺤﺠﻢ ﻭﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‪.‬‬‫ﺍﻟﺠﺰﺀ ‪B‬‬‫ﻓﻴﻤﺎ ﻳﻠﻲ‪ ،‬ﺟﺮﻯ ﺗﺴﺨﻴﻦ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺘﻮﺗﻴﺎﺀ ﺍﻟﻤﺴﺘﻌﺎﺩ ﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ ﺃﻭﻛﺴﻴﺪ ﺍﻟﺘﻮﺗﻴﺎﺀ ﻭﻓﻖ ﺍﻟﺘﻔﺎﻋﻞ ﺃﺩﻧﺎﻩ‪:‬‬‫ﻭﺑﻌﺪ ﺫﻟﻚ‪ ،‬ﺟﺮﻯ ﺇﺭﺟﺎﻉ ﺃﻛﺴﻴﺪ ﺍﻟﺘﻮﺗﻴﺎﺀ ﺇﻟﻰ ﺍﻟﺘﻮﺗﻴﺎﺀ ﺍﻟﻤﻌﺪﻧﻲ ﺑﺎﻟﺘﻔﺎﻋﻞ ﻣﻊ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‪:‬‬‫‪B.1‬‬‫ﻟﻜﻲ ﻳﺠﺮﻱ ﺍﻟﺘﻔﺎﻋﻞ )‪ (7‬ﻋﻨﺪ ﺿﻐﻂ ﻫﻴﺪﺭﻭﺟﻴﻦ ﺛﺎﺑﺖ ﻋﻨﺪ ‪ 1bar،‬ﻣﻦ ﺍﻟﻀﺮﻭﺭﻱ ﺗﺨﻔﻴﺾ ﺍﻟﻀﻐﻂ‬‫ﺍﻟﺠﺰﺋﻲ ﻟﺒﺨﺎﺭ ﺍﻟﻤﺎﺀ ﺍﻟﻨﺎﺗﺞ‪ .‬ﺍﺣﺴﺐ ﺍﻟﺤﺪ ﺍﻷﻋﻠﻰ ﻟﻠﻀﻐﻂ ﺍﻟﺠﺰﺋﻲ ﻟﺒﺨﺎﺭ ﺍﻟﻤﺎﺀ ﻟﻴﺠﺮﻱ ﺍﻟﺘﻔﺎﻋﻞ‬‫‪4pt‬‬‫)‪ (7‬ﻋﻨﺪ ‪ . 300∘ C‬ﻟﺪﻳﻨﺎ ﻃﺎﻗﺘﺎ ﺟﻴﺒﺲ ﺍﻟﻘﻴﺎﺳﻴﺔ ﻟﺘﺸﻜﻞ ﻛﻞ ﻣﻦ ﺃﻛﺴﻴﺪ ﺍﻟﺘﻮﺗﻴﺎﺀ ﻭﺑﺨﺎﺭ ﺍﻟﻤﺎﺀ‬‫ﻋﻨﺪ ‪ 300∘ C‬ﻭ ‪ bar 1‬ﻫﻤﺎ ﺑﺎﻟﺘﺮﺗﻴﺐ‪:‬‬
‫∘‬
‫‪2‬‬
‫‪−1‬‬‫‪ Δ𝐺ZnO (300∘ C) = −2.90 × 102 kJ mol−1‬ﻭ ‪.Δ𝐺H2 O (300 C) = −2.20 × 10 kJ mol‬‬‫ُﻳﺴﺘﻌﻤﻞ ﺍﻟﺘﻮﺗﻴﺎﺀ ﺍﻟﻤﻌﺪﻧﻲ ﺑﻮﺻﻔﻪ ﺇﻟﻜﺘﺮﻭﺩﺍً ﺳﺎﻟﺒﺎً )ﻣﺼﻌﺪﺍً ( ﻓﻲ ﺑﻄﺎﺭﻳﺎﺕ ﻣﻌﺪﻥ‪-‬ﻫﻮﺍﺀ‪ .‬ﻳﺘﺄﻟﻒ ﺍﻹﻟﻜﺘﺮﻭﺩ ﻣﻦ ‪ Zn‬ﻭ ‪ZnO.‬‬‫ﻭﻳﺤﺼﻞ ﺗﻔﺎﻋﻞ ﺍﻷﻛﺴﺪﺓ ﻭﺍﻹﺭﺟﺎﻉ ﺍﻟﺘﺎﻟﻲ ﻟﺘﻮﻟﻴﺪ ﺍﻟﻜﻬﺮﺑﺎﺀ ﺑﻘﻮﺓ ﻣﺤﺮﻛﺔ ﻛﻬﺮﺑﺎﺋﻴﺔ )‪ (.e.m.f‬ﻗﺪﺭﻫﺎ ∘ 𝐸ﻋﻨﺪ ‪ 25∘ C‬ﻭﺍﻟﻀﻐﻂ‬‫‪ 1‬ﺑﺎﺭ‪.‬‬‫‪B.2‬‬‫ﺟﺮﻯ ﺗﻔﺮﻳﻎ ﺑﻄﺎﺭﻳﺔ ﺗﻮﺗﻴﺎﺀ‪-‬ﻫﻮﺍﺀ ﻋﻨﺪ ‪ 20mA‬ﻟﻤﺪﺓ ‪ 24‬ﺳﺎﻋﺔ‪ .‬ﺍﺣﺴﺐ ﺍﻟﺘﻐﻴﺮ ﻓﻲ ﻛﺘﻠﺔ‬‫ﺍﻹﻟﻜﺘﺮﻭﺩ ﺍﻟﺴﺎﻟﺐ )ﺍﻟﻤﺼﻌﺪ( ﻟﻠﺒﻄﺎﺭﻳﺔ‪.‬‬‫‪3pt‬‬

‫‪Q4-3‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-4 Q-3‬‬‫ﺟﺒﻞ ﻓﻮﺟﻲ‬‫‪B.3‬‬‫ﻟﻨﺘﺄﻣﻞ ﺗﻐﻴﺮ ﺍﻟﻘﻮﺓ ﺍﻟﻤﺤﺮﻛﺔ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ ‪ e.m.f‬ﻟﺒﻄﺎﺭﻳﺔ ﺍﻟﺘﻮﺗﻴﺎﺀ‪-‬ﻫﻮﺍﺀ ﺗﺒﻌﺎً ﻟﻈﺮﻭﻑ ﺍﻟﻮﺳﻂ‬‫ﺍﻟﻤﺤﻴﻂ‪ .‬ﺍﺣﺴﺐ ﺍﻟـ ‪ e.m.f.‬ﻋﻨﺪ ﻗﻤﺔ ﺟﺒﻞ ﻓﻮﺟﻲ‪ ،‬ﺣﻴﺚ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ‪) −38 ∘ C‬ﺷﺒﺎﻁ(‪،‬‬‫‪5pt‬‬‫ﺇﻥ ﺍﻟﻀﻐﻂ ﺍﻟﺠﻮﻱ ﻣﻌﻄﻰ ﺑﺎﻟﻌﻼﻗﺔ‪:‬‬‫ﻭﺍﻻﺭﺗﻔﺎﻉ ‪ 3776‬ﻣﺘﺮﺍً ‪ّ .‬‬‫ﻋﻨﺪ ﺍﻻﺭﺗﻔﺎﻉ ]‪ ℎ [m‬ﻭﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ]‪ .𝑇 [∘ C‬ﺇﻥ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻸﻛﺴﺠﻴﻦ ﻓﻲ ﺍﻟﺠﻮ ﻫﻲ‪.21%‬‬‫ﺗﻐﻴﺮ ﻃﺎﻗﺔ ﺟﻴﺒﺲ ﺍﻟﻘﻴﺎﺳﻴﺔ ﻟﻠﺘﻔﺎﻋﻞ )‪ (8‬ﻫﻲ ‪Δ𝐺ZnO (−38∘ C) = −3.26 × 102 kJ mol−1‬‬‫ﻋﻨﺪ ‪ −38 ∘ C‬ﻭ ‪.1 bar‬‬‫‪B.4‬‬‫∘‬‫ﺃﻥ ﻛﻤﻮﻧﺎﺕ ﺍﻹﺭﺟﺎﻉ ﺍﻟﻘﻴﺎﺳﻴﺔ‬‫ﺍﺣﺴﺐ ﺍﻟﺘﻐﻴﺮ ﻓﻲ ﻃﺎﻗﺔ ﺟﻴﺒﺲ ﻟﻠﺘﻔﺎﻋﻞ )‪ (6‬ﻋﻨﺪ ‪ . 25 C‬ﻻﺣﻆ ّ‬‫ﻟـ )‪ 𝐸 ∘ (Zn2+ /Zn‬ﻭ )‪ 𝐸 ∘ (O2 /H2 O‬ﻋﻨﺪ ‪ 25 ∘ C‬ﻭ ‪ 1 bar‬ﻣﻌﻄﺎﺓ ﺑﺎﻟﻤﻌﺎﺩﻟﺘﻴﻦ )‪ (10‬ﻭ )‪(11‬‬‫ﺑﺎﻟﺘﺮﺗﻴﺐ‪.‬‬‫‪9pt‬‬IChO
SYR-3 C-4 A-1

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

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

A.2 (5 pt)%

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

Part B
B.1 (4 pt)

bar

𝑝H2 O=
B.2 (3 pt)

g

A4-3
English (Official)IChO
SYR-3 C-4 A-4

B.3 (5 pt)

V

A4-4
English (Official)IChO
SYR-3 C-4 A-5

B.4 (9 pt)

Δ𝐺∘ =J mol−1

A4-5
English (Official)SYR-3 C-5 C-1

SYR-3 C-5 C
ANAS ABBAS

IChO
Problem 5
Cover sheet

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

Q5-1

IChO
SYR-3 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
SYR-3 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
SYR-3 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
SYR-3 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

‫‪Q5-1‬‬‫‪IChO‬‬‫‪SYR-3 C-5 Q-1‬‬‫)‪arabic (Syria‬‬‫ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﺍﻟﻐﺎﻣﺾ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪9‬‬‫‪A.2‬‬‫‪7‬‬‫‪total the of % 12‬‬‫‪A.3‬‬‫‪6‬‬‫‪A.4‬‬‫‪10‬‬‫‪B.1‬‬‫‪5‬‬‫‪B.2‬‬‫‪15‬‬‫‪B.3‬‬‫‪8‬‬‫‪Total‬‬‫‪60‬‬‫‪Score‬‬‫ﺃﻥ ﺧﻮﺍﺻﻬﻤﺎ ﻣﺨﺘﻠﻔﺔ ﺍﺧﺘﻼﻓﺎً ﻛﺒﻴﺮﺍً ‪.‬‬
‫ﺃﻥ ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﻋﻨﺼﺮ ﻣﻦ ﺍﻟﻤﺠﻤﻮﻋﺔ ‪ 14‬ﻣﺜﻞ ﺍﻟﻜﺮﺑﻮﻥ‪ ،‬ﺇﻻ ّ‬‫ﻋﻠﻰ ﺍﻟﺮﻏﻢ ﻣﻦ ّ‬‫ﺍﻟﺠﺰﺀ ‪A‬‬

‫ﻓﺈﻥ ﺍﻟﺮﺍﺑﻄﺔ ﺍﻟﺜﻼﺛﻴﺔ ﺳﻴﻠﻴﻜﻮﻥ‪-‬ﺳﻴﻠﻴﻜﻮﻥ ﻓﻲ ﻣﺮﻛﺐ ﺫﻱ ﺍﻟﺼﻴﻐﺔ‬‫ﻋﻠﻰ ﻧﺤﻮ ﻣﻐﺎﻳﺮ ﻟﻠﺮﺍﺑﻄﺔ ﺍﻟﺜﻼﺛﻴﺔ ﻛﺮﺑﻮﻥ‪-‬ﻛﺮﺑﻮﻥ‪،‬‬‫ّ‬‫‪) R1 –Si ≡ Si–R1‬ﺣﻴﺚ ‪ R‬ﻫﻮ ﻣﺴﺘﺒﺪﻝ ﻋﻀﻮﻱ( ﺗﻜﻮﻥ ﻓﻌﺎﻟﺔ ﻟﻠﻐﺎﻳﺔ‪ .‬ﻋﻠﻰ ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ‪ ،‬ﻓﻬﻮ ﻳﺘﻔﺎﻋﻞ ﻣﻊ ﺍﻹﺛﻴﻠﻴﻦ ﻟﻴﻌﻄﻲ‬‫ﻣﺮﻛﺒﺎً ﺣﻠﻘﻴﺎً‬‫ﻳﻀﻢ ﺣﻠﻘﺔ ﺭﺑﺎﻋﻴﺔ ﺍﻟﻌﻨﺎﺻﺮ‪.‬‬‫ّ‬‫ﻋﻨﺪﻣﺎ ُﻳﻌﺎﻟﺞ ‪ R1 –Si ≡ Si–R1‬ﻣﻊ ﺍﻷﻟﻜﻴﻦ) ‪ ،( R2 –C ≡ C–R2‬ﻳﺘﺸﻜﻞ ﺍﻟﻤﺮﻛﺐ ‪ A‬ﺫﻭ ﺍﻟﺤﻠﻘﺔ ﺍﻟﺮﺑﺎﻋﻴﺔ ﻛﻮﺳﻴﻂ ﺃﻭﻟﻲ‪ .‬ﻭﻳﻌﻄﻲ‬‫ﺗﻔﺎﻋﻞ ﺁﺧﺮ ﻟﺠﺰﻳﺌﺔ ‪ R2 –C ≡ C–R2‬ﻣﻊ ‪ A‬ﺍﻟﻤﺘﺼﺎﻭﻏﻴﻦ ‪ B‬ﻭ ‪ ،C‬ﺍﻟﻠﺬﻳﻦ ﻟﻬﻤﺎ ﺑﻨﻴﺘﺎﻥ ﺣﻠﻘﻴﺘﺎﻥ ﻣﺘﺮﺍﻓﻘﺘﺎﻥ ﺗﺸﺒﻬﺎﻥ ﺍﻟﺒﻨﺰﻥ‪،‬‬‫ﻭﻟﻬﺬﺍ ُﻳﺴﻤﻴﺎﻥ ﺛﻨﺎﺋﻲ ﺳﻴﻠﻴﻜﻮﻥ ﺍﻟﺒﻨﺰﻥ ‪ disilabenzene،‬ﻭﺑﺤﻴﺚ ﻳﻀﻢ ﻛﻞ ﻣﻨﻬﻤﺎ ﺣﻠﻘﺔ ﺳﺪﺍﺳﻴﺔ ﻭﻟﻬﻤﺎ ﺍﻟﺼﻴﻐﺔ‬‫‪Q5-2‬‬‫‪IChO‬‬‫‪SYR-3 C-5 Q-2‬‬‫)‪arabic (Syria‬‬‫‪.(R1 –Si)2 (R2 –C)4‬‬‫ﻳﻈﻬﺮ ﺗﺤﻠﻴﻞ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ‪ NMR 13 C‬ﻟﻠﻬﻴﻜﻠﻴﻦ ﺍﻟﺴﺪﺍﺳﻴﻲ ﺍﻟﺤﻠﻘﺔ ‪ Si2 C4‬ﺇﺷﺎﺭﺗﻴﻦ ﻟﻠﻤﺮﻛﺐ ‪ B‬ﻭﺇﺷﺎﺭﺓ ﻟﻠﻤﺮﻛﺐ ‪.C‬‬‫‪A.1‬‬‫ﺍﺭﺳﻢ ﺍﻟﺼﻴﻎ ﺍﻟﻤﻨﺸﻮﺭﺓ ﻟﻠﻤﺮﻛﺒﺎﺕ ‪ A‬ﻭ ‪ B‬ﻭ ‪ C‬ﺑﺎﺳﺘﻌﻤﺎﻝ ‪C. ، Si ،R2 ،R1‬‬‫‪9pt‬‬‫‪A.2‬‬‫ﺍﺣﺴﺐ ﻃﺎﻗﺔ ﺍﻻﺳﺘﻘﺮﺍﺭ ﺍﻟﻌﻄﺮﻱ )‪ (Aromatic Stabilization Energy‬ﻟﻜﻞ ﻣﻦ ﺍﻟﺒﻨﺰﻥ ﻭ ‪C‬‬‫‪7pt‬‬‫)ﻓﻲ ﺍﻟﺤﺎﻟﺔ ‪ ( R = R = H‬ﻛﻘﻴﻢ ﻣﻮﺟﺒﺔ ﻭﻣﻌﺘﺒﺮﺍً ﺗﻐﻴﺮﺍﺕ ﺍﻷﻧﺘﺎﻟﺒﻴﺔ ﻟﻌﺪﺩ ﻣﻦ ﺗﻔﺎﻋﻼﺕ‬‫ﺍﻟﻬﺪﺭﺟﺔ ﻟﺒﻌﺾ ﻣﻦ ﺍﻟﺠﻤﻞ ﻏﻴﺮ ﺍﻟﻤﺸﺒﻌﺔ ﺃﺩﻧﺎﻩ )ﺷﻜﻞ ‪.(1‬‬
‫‪2‬‬

‫‪1‬‬

‫ﺍﻟﺸﻜﻞ ‪1‬‬

‫‪Q5-3‬‬‫‪IChO‬‬‫‪SYR-3 C-5 Q-3‬‬‫)‪arabic (Syria‬‬‫ﻋﻨﺪﻣﺎ ﻳﺠﺮﻱ ﺗﺴﺨﻴﻦ ﻣﺤﻠﻮﻝ ﻟـ ‪ C‬ﻓﻲ ﺍﻟﻜﺰﻳﻠﻴﻦ‪ ،‬ﻓﺈﻧﻪ ﻳﺨﻀﻊ ﻟﺘﻔﺎﻋﻞ ﺗﺼﺎﻭﻍ ﻭﻳﻌﻄﻲ ﻣﺰﻳﺠﺎً ﺗﻮﺍﺯﻧﻴﺎً ﻣﻦ ﺍﻟﻤﺮﻛﺒﻴﻦ ‪ D‬ﻭ ‪.E‬‬‫ﻓﺈﺫﺍ ﻋﻠﻤﺖ ﺃﻥ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻤﺮﻛﺐ ‪ D‬ﺇﻟﻰ ‪ E‬ﻫﻲ ‪ 1‬ﺇﻟﻰ‪ 40.0‬ﺃﻱ ‪ D : E = 1 : 40.0‬ﻋﻨﺪ‪ 50.0‬ﺩﺭﺟﺔ ﻣﺌﻮﻳﺔ ﻭﺗﺼﺒﺢ ‪1‬‬‫ﺇﻟﻰ ‪ 20.0‬ﺃﻱ ‪ D : E = 1 : 20.0‬ﻋﻨﺪ ‪ 120.0‬ﺩﺭﺟﺔ ﻣﺌﻮﻳﺔ ‪.‬‬‫‪A.3‬‬‫ﺃﻥ 𝐻‪ Δ‬ﻻ ﻳﺘﻌﻠﻖ ﺑﺪﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‪.‬‬‫ﺍﺣﺴﺐ 𝐻‪ Δ‬ﻟﻠﺘﺤﻮﻝ ﻣﻦ ‪ D‬ﺇﻟﻰ ‪ . E‬ﺍﻓﺘﺮﺽ ّ‬‫‪6pt‬‬‫ﻳﺠﺮﻱ ﺗﻔﺎﻋﻞ ﺍﻟﺘﺼﺎﻭﻍ ﻣﻦ ‪ C‬ﺇﻟﻰ ‪ D‬ﻭ ﺇﻟﻰ ‪ E‬ﻣﻦ ﺧﻼﻝ ﺗﺤﻮﻻﺕ ﺍﻟﺮﻭﺍﺑﻂ 𝜋 ﺇﻟﻰ ﺭﻭﺍﺑﻂ 𝜎 ﻣﻦ ﺩﻭﻥ ﻛﺴﺮ ﺃﻱ ﻣﻦ ﺍﻟﺮﻭﺍﺑﻂ 𝜎‪.‬‬‫ﻳﻈﻬﺮ ﺗﺤﻠﻴﻞ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ‪ NMR 13 C‬ﻭﺟﻮﺩﺇﺷﺎﺭﺓ ﻭﺍﺣﺪﺓ ﻟﻠﻬﻴﻜﻞ ‪ Si2 C4‬ﻟﻠﻤﺮﻛﺐ ‪ D‬ﻭﺇﺷﺎﺭﺗﻴﻦ ﻟﻠﻤﺮﻛﺐ ‪ .E‬ﻻﻳﻀﻢ‬‫ﻫﻴﻜﻞ ‪ D‬ﺃﻱ ﺣﻠﻘﺎﺕ ﺛﻼﺛﻴﺔ ﺍﻟﻌﻨﺎﺻﺮ‪ ،‬ﺑﻴﻨﻤﺎ ﻳﻀﻢ ‪ E‬ﺣﻠﻘﺘﺎﻥ ﺛﻼﺛﻴﺘﺎ ﺍﻟﻌﻨﺎﺻﺮ ﺗﺘﺸﺎﺭﻛﺎﻥ ﺑﺤﺮﻑ‪.‬‬‫‪A.4‬‬‫ﺍﺭﺳﻢ ﺍﻟﺼﻴﻎ ﺍﻟﻤﻨﺸﻮﺭﺓ ﻟﻜﻞ ﻣﻦ ‪ D‬ﻭ ‪ E‬ﺑﺎﺳﺘﻌﻤﺎﻝ ‪C. ، Si ،R2 ،R1‬‬‫‪10pt‬‬‫ﺍﻟﺠﺰﺀ ‪B‬‬

‫ﺇﻥ ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﻗﺎﺩﺭ ﻋﻠﻰ ﺗﺸﻜﻴﻞ ﻣﺮﻛﺒﺎﺕ ﺗﺴﺎﻧﺪﻳﺔ ) ﺃﻱ ﺑﺄﻛﺜﺮ ﻣﻦ ‪ 4‬ﻣﺴﺘﺒﺪﻻﺕ( ﻣﻊ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﻜﻬﺮﺳﻠﺒﻴﺔ ﻣﺜﻞ ﺍﻟﻔﻠﻮﺭ‪.‬‬‫ﻓﺈﻥ ﻓﻠﻮﺭﻳﺪﺍﺕ ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﻋﺎﻟﻴﺔ ﺍﻟﺘﺴﺎﻧﺪ ﺗﻌﻤﻞ ﻛﻜﻮﺍﺷﻒ‬‫ﺃﻥ ﻓﻠﻮﺭﻳﺪﺍﺕ ﺍﻟﻤﻌﺎﺩﻥ ُﺗﺴﺘﻌﻤﻞ ﻏﺎﻟﺒﺎً ﻛﻜﻮﺍﺷﻒ ﻓﻠﻮﺭﻳﺔ‪،‬‬‫ّ‬‫ﻭﻛﻤﺎ ّ‬‫ﻓﻠﻮﺭﻳﺔ‪.‬‬‫ﺟﺮﻯ ﺗﻔﺎﻋﻞ ﻓﻠﻮﺭﺓ ‪ CCl4‬ﺑﺎﺳﺘﻌﻤﺎﻝ ‪ Na2 SiF6‬ﻛﻤﺎ ﻳﻠﻲ‪:‬‬‫• ﺗﻘﻴﻴﺲ ﻣﺤﻠﻮﻝ ‪:Na2 SiF6‬‬‫‪ .‬ﺍﻟﺘﺤﻀﻴﺮ‬‫ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ‪ : F‬ﺟﺮﻯ ﺣﻞ ‪0.855 g‬ﻣﻦ ‪ ( 188.053 g mol–1 ) Na2 SiF6‬ﻓﻲ ﺍﻟﻤﺎﺀ )ﺍﻟﺤﺠﻢ ﺍﻟﻜﻠﻲ ‪.( 200mL:‬‬‫ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ‪ : G‬ﺟﺮﻯ ﺣﻞ ‪ 6.86 g‬ﻣﻦ ‪ ( 568.424 g mol–1 ) Ce2 (SO4 )3‬ﻓﻲ ﺍﻟﻤﺎﺀ )ﺍﻟﺤﺠﻢ ﺍﻟﻜﻠﻲ‪.( : 200mL‬‬‫‪ .‬ﺍﻹﺟﺮﺍﺋﻴﺔ‪:‬‬‫ﺟﺮﻯ ﺗﻔﺎﻋﻞ ﻣﻌﺎﻳﺮﺓ ﺑﺎﻟﺘﺮﺳﻴﺐ ﻟﻠﻤﺤﻠﻮﻝ ‪ F‬ﺑﺤﺠﻢ ) ‪ (50.0 mL‬ﺑﺈﺿﺎﻓﺔ ﻗﻄﺮﺓ ﻗﻄﺮﺓ ﻣﻦ ﻣﺤﻠﻮﻝ ‪ G‬ﺑﻮﺟﻮﺩ ﺑﺮﺗﻘﺎﻟﻲ ﺍﻟﻜﺰﻳﻠﻨﻮﻝ‪،‬‬‫ﺍﻟﺬﻱ ﻳﺘﺴﺎﻧﺪ ﻣﻊ ‪ ،Ce3+‬ﺑﻮﺻﻔﻪ ﻣﺸﻌﺮﺍً ‪ .‬ﺑﻌﺪ ﺇﺿﺎﻓﺔ‪ 18.8 mL‬ﻣﻦ ﻣﺤﻠﻮﻝ ‪ G‬ﺗﺤﻮﻝ ﻟﻮﻥ ﺍﻟﻤﺤﻠﻮﻝ ﻣﻦ ﺍﻷﺻﻔﺮ ﺇﻟﻰ ﺍﻷﺭﺟﻮﺍﻧﻲ‪.‬‬‫ﺇﻥ ﺍﻟﺮﺍﺳﺐ ﺍﻟﻤﺘﺸﻜﻞ ﻫﻮ ﻣﺮﻛﺐ ﺛﻨﺎﺋﻲ ﻳﻀﻢ ‪ ، Ce3+‬ﻭﺍﻟﻤﺮﻛﺐ ﺍﻟﻮﺣﻴﺪ ﺍﻟﺬﻱ ﻳﻀﻢ ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﻫﻮ ‪.Si(OH)4‬‬‫ّ‬‫‪B.1‬‬‫‪5pt‬‬‫ﺍﻛﺘﺐ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻮﺯﻭﻧﺔ ﻟﺘﻔﺎﻋﻞ ‪ Na2 SiF6‬ﻣﻊ ‪.Ce2 (SO4 )3‬‬‫• ﺗﻔﺎﻋﻞ ‪ CCl4‬ﻣﻊ ‪:Na2 SiF6‬‬‫ﻣﺜﻼ ﻣﻬﻤﻞ ﺃﺛﻨﺎﺀ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ(‬‫)ﺿﻴﺎﻉ ﺍﻟﻤﻮﺍﺩ ﺑﺎﻟﺘﺒﺨﺮ‬‫ً‬‫ﻭﺳ ّﺨﻨﺖ ﺇﻟﻰ ‪ 300‬ﺩﺭﺟﺔ ﻣﺌﻮﻳﺔ ﻓﻲ ﻭﻋﺎﺀ ﺗﻔﺎﻋﻠﻲ ﻣﻐﻠﻖ‬‫ﺟﺮﺕ ﺇﺿﺎﻓﺔ )]‪ (𝑥 [g‬ﻣﻦ ‪ Na2 SiF6‬ﺇﻟﻰ )‪ (500.0 g‬ﻣﻦ ‪ُ CCl4‬‬‫ﻭﻣﻘﺎﻭﻡ ﻟﻠﻀﻐﻂ‪ .‬ﺟﺮﻯ ﺍﻟﺘﺨﻠﺺ ﺑﺎﻟﺘﺮﺷﻴﺢ ﻣﻦ ‪ Na2 SiF6‬ﻏﻴﺮ ﺍﻟﻤﺘﻔﺎﻋﻞ ﻭ ‪ NaCl‬ﺍﻟﻤﺘﻮﻟﺪ‪ .‬ﺟﺮﻯ ﺗﻤﺪﻳﺪ ﺍﻟﺮﺷﺎﺣﺔ ﺇﻟﻰ ﺣﺠﻢ‬‫‪ 1.00L‬ﺑﺎﺳﺘﻌﻤﺎﻝ ‪) CCl4‬ﻣﺤﻠﻮﻝ ‪ .(H‬ﺃﻇﻬﺮ ﻃﻴﻔﺎ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ‪ NMR 19 F‬ﻭ ‪ 29 Si‬ﻟﻤﺤﻠﻮﻝ ‪ H‬ﺃﻥ ‪ SiF4‬ﻫﻮ ﺍﻟﻤﺮﻛﺐ‬‫ﺍﻟﺴﻴﻠﻴﻜﻮﻧﻲ ﺍﻟﻮﺣﻴﺪ‪ .‬ﻓﻲ ﻃﻴﻒ ‪ NMR، 19 F‬ﺟﺮﻯ ﻣﻼﺣﻈﺔ ﺃﻧﻪ ﺑﺎﻹﺿﺎﻓﺔ ﺇﻟﻰ ‪ ،SiF4‬ﻫﻨﺎﻟﻚ ﺇﺷﺎﺭﺍﺕ ﺗﻮﺍﻓﻖ ‪CF2 Cl2 ، CFCl3‬‬‫‪ CF3 Cl ،‬ﻭ ‪) CF4‬ﺍﻧﻈﺮ ﺍﻟﺠﺪﻭﻝ ‪ .(1‬ﺗﺘﻨﺎﺳﺐ ﻧﺴﺐ ﺍﻟﺘﻜﺎﻣﻼﺕ ﻓﻲ ﻃﻴﻒ ‪ NMR 19 F‬ﻣﻊ ﻋﺪﺩ ﻣﻮﻻﺕ ﻧﻮﻯ ﺍﻟﻔﻠﻮﺭ‪.‬‬‫ﻣﻌﻄﻴﺎﺕ ‪NMR 19 F‬‬‫ﻧﺴﺐ ﺍﻟﺘﻜﺎﻣﻞ‬‫‪CFCl3‬‬‫‪45.0‬‬‫ﺍﻟﺠﺪﻭﻝ ‪1‬‬‫‪CF2 Cl2‬‬‫‪65.0‬‬‫‪CF3 Cl‬‬‫‪18.0‬‬‫‪CF4‬‬‫‪2.0‬‬

‫‪Q5-4‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-5 Q-4‬‬‫ﺟﺮﺕ ﺣﻠﻤﻬﺔ ‪ SiF4‬ﻟﺘﺸﻜﻴﻞ ‪ H2 SiF6‬ﻭﻓﻖ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪ 8‬ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬‫ﺃﺿﻴﻒ ﺍﻟﻤﺤﻠﻮﻝ ‪ (10 mL) H‬ﺇﻟﻰ ﻛﻤﻴﺔ ﻓﺎﺋﻀﺔ ﻣﻦ ﺍﻟﻤﺎﺀ‪ ،‬ﻣﻤﺎ ﻧﺘﺞ ﻋﻨﻪ ﺣﻠﻤﻬﺔ ﻛﺎﻣﻠﺔ ﻟـ ‪ .SiF4‬ﺑﻌﺪ ﺍﻟﻔﺼﻞ‪ ،‬ﺟﺮﻯ ﺗﻌﺪﻳﻞ‬‫‪ H2 SiF6‬ﺍﻟﻨﺎﺗﺞ ﻋﻦ ﺍﻟﺤﻠﻤﻬﺔ ﻓﻲ ﻭﺳﻂ ﻣﺎﺋﻲ‪ ،‬ﻭﺗﺤﻮﻳﻠﻪ ﻛﻠﻴﺎً ﺇﻟﻰ ‪) Na2 SiF6‬ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ‪.(J‬‬‫ﺟﺮﻯ ﺣﻞ ﺍﻟﺮﺍﺳﺐ ﻣﻦ ‪ Na2 SiF6‬ﻏﻴﺮ ﺍﻟﻤﺘﻔﺎﻋﻞ ﻭ ‪ NaCl‬ﺍﻟﺬﻱ ﺃﺯﻳﻞ ﺑﺎﻟﺘﺮﺷﻴﺢ ﻓﻲ ﺍﻟﻤﺮﺣﻠﺔ ﺍﻷﻭﻟﻰ )ﺍﻟﺠﺰﺀ ﺍﻟﻤﺴﻄﺮ ﺗﺤﺘﻪ ﺧﻂ(‬‫ﻣﺤﻠﻮﻻ ﻣﺎﺋﻴﺎً )ﻣﺤﻠﻮﻝ ‪.(10.0L ، K‬‬‫ﺣﻼ ﺗﺎﻣﺎً ﻓﻲ ﺍﻟﻤﺎﺀ ﻟﻴﻌﻄﻲ‬‫ً‬‫ً‬‫ﺑﻌﺪ ﺫﻟﻚ‪ ،‬ﺟﺮﺕ ﻣﻌﺎﻳﺮﺗﺎﻥ ﺑﺎﻟﺘﺮﺳﻴﺐ ﺑﺎﺳﺘﻌﻤﺎﻝ ﺍﻟﻤﺤﻠﻮﻝ ‪ ،G‬ﻭﻛﺎﻧﺖ ﻧﻘﻂ ﺍﻧﺘﻬﺎﺀ ﺍﻟﻤﻌﺎﻳﺮﺓ ﻟـ ‪ G‬ﻛﻤﺎ ﻳﻠﻲ‪:‬‬‫ﻣﻦ ﺃﺟﻞ ﺍﻟﻤﺤﻠﻮﻝ ‪) J‬ﻛﺎﻣﻞ ﺍﻟﻜﻤﻴﺔ( ﻟﺰﻡ ‪.61.6 mL‬‬‫ﻣﻦ ﺃﺟﻞ ‪ 100mL‬ﻣﻦ ﺍﻟﻤﺤﻠﻮﻝ ‪ K‬ﻟﺰﻡ ‪.44.4 mL :‬‬‫ﺃﻥ ﻭﺟﻮﺩ ‪ NaCl‬ﻭ ‪ SiO2‬ﻟﻦ ﻳﺆﺛﺮ ﻋﻠﻰ ﻣﻌﺎﻳﺮﺓ ﺍﻟﺘﺮﺳﻴﺐ‪.‬‬‫ﻳﺠﺐ ﺃﻥ ﻧﺸﻴﺮ ﻫﻨﺎ ﺇﻟﻰ ّ‬‫‪B.2‬‬‫ﺍﺣﺴﺐ ﻛﺘﻠﺔ ‪ NaCl‬ﺍﻟﻨﺎﺗﺞ ﻓﻲ ﻭﻋﺎﺀ ﺍﻟﺘﻔﺎﻋﻞ )ﺍﻟﻤﻌﻠﻮﻣﺔ ﺍﻟﻤﺴﻄﺮ ﺗﺤﺘﻬﺎ(‪ ،‬ﻭﺍﺣﺴﺐ ﺍﻟﻜﺘﻠﺔ‬‫)]‪ (𝑥 [g‬ﺍﻟﺨﺎﺻﺔ ﺑـ ‪ Na2 SiF6‬ﺍﻟﻤﺴﺘﻌﻤﻞ ﻛﻤﺎﺩﺓ ﺑﺪﺀ‪.‬‬‫‪15pt‬‬‫‪B.3‬‬‫ﺇﻥ ‪ % 77.8‬ﻣﻦ ‪ CCl4‬ﺍﻟﻤﺴﺘﻌﻤﻞ ﻛﻤﺎﺩﺓ ﺃﻭﻟﻴﺔ ﻟﻢ ﻳﺘﻔﺎﻋﻞ‪ .‬ﺍﺣﺴﺐ ﻛﺘﻠﺔ ‪ CF3 Cl‬ﺍﻟﻨﺎﺗﺞ‪.‬‬‫ّ‬‫‪8pt‬‬

IChO

A5-1

SYR-3 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
SYR-3 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
SYR-3 C-5 A-3

Part B
B.1 (5 pt)

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

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

B.2 (cont.)

NaCl ∶

g, Na2 SiF6 ∶g

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

B.3 (8 pt)

CF3 Cl ∶g

A5-5English (Official)SYR-3 C-6 C-1

SYR-3 C-6 C
ANAS ABBAS

IChO
Problem 6
Cover sheet

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

Q6-1

IChO
SYR-3 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
SYR-3 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
SYR-3 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
SYR-3 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
SYR-3 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
SYR-3 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

‫‪Q6-1‬‬‫‪IChO‬‬‫‪SYR-3 C-6 Q-1‬‬‫)‪arabic (Syria‬‬‫ﻛﻴﻤﻴﺎﺀ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺼﻠﺒﺔ ﻟﻠﻤﻌﺎﺩﻥ ﺍﻻﻧﺘﻘﺎﻟﻴﺔ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪6‬‬‫‪A.2‬‬‫‪3‬‬‫‪A.3‬‬‫‪3‬‬‫‪B.1‬‬‫‪6‬‬‫‪total the of % 13‬‬‫‪B.2‬‬‫‪4‬‬‫‪B.3‬‬‫‪4‬‬‫‪B.4‬‬‫‪4‬‬‫‪C.1‬‬‫‪5‬‬‫‪C.2‬‬‫‪5‬‬‫‪C.3‬‬‫‪5‬‬‫‪Total‬‬‫‪45‬‬‫‪Score‬‬‫ﺍﻟﺒﺮﻛﺎﻥ ﻓﻲ ﺟﺰﻳﺮﺓ ‪Sakurajima‬‬‫ﺍﻟﺠﺰﺀ ‪A:‬‬‫ﺇﻥ ﺍﻟﻴﺎﺑﺎﻥ ﻫﻲ ﻭﺍﺣﺪﺓ ﻣﻦ ﺍﻟﺪﻭﻝ ﺫﺍﺕ ﺍﻷﻋﺪﺍﺩ ﺍﻷﻋﻠﻰ ﻣﻦ ﺍﻟﺒﺮﺍﻛﻴﻦ ﻋﻠﻰ ﻣﺴﺘﻮﻯ ﺍﻟﻌﺎﻟﻢ‪ .‬ﻋﻨﺪﻣﺎ ﺗﺘﺒﻠﻮﺭ ﺍﻟﻔﻠﺰﺍﺕ ﺍﻟﻤﻌﺪﻧﻴﺔ‬‫ﺍﻟﺤﺎﻭﻳﺔ ﻋﻠﻰ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ ﺍﻧﻄﻼﻗﺎً ﻣﻦ ﺍﻟﻤﺼﻬﻮﺭ )ﺍﻟﺼﻬﺎﺭﺓ( ‪ magma،‬ﻓﺈﻥ ﺟﺰﺀﺍً ﻣﻦ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻤﻌﺪﻥ ﺍﻻﻧﺘﻘﺎﻟﻲ ) ‪(Mn+‬‬‫ﻓﻲ ﺍﻟﻤﺼﻬﻮﺭ ﺳﻮﻑ ﻳﺪﺧﻞ ﺇﻟﻰ ﺩﺍﺧﻞ ﺍﻟﻔﻠﺰﺍﺕ ﺍﻟﻤﻌﺪﻧﻴﺔ ﺍﻟﺤﺎﻭﻳﺔ ﻋﻠﻰ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ‪ .‬ﺗﺮﺗﺒﻂ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻤﻌﺪﻥ ﺍﻻﻧﺘﻘﺎﻟﻲ ‪Mn+‬‬‫‪،‬ﺍﻟﻤﺪﺭﻭﺳﺔ ﻓﻲ ﻫﺬﻩ ﺍﻟﻤﺴﺄﻟﺔ‪ ،‬ﺑﺮﻭﺍﺑﻂ ﺗﺴﺎﻧﺪﻳﺔ ﻣﻊ ﺃﻳﻮﻧﺎﺕ ﺍﻷﻭﻛﺴﻴﺪ ) ‪ (O2−‬ﻭﺗﺘﺒﻨﻰ ﻫﻨﺪﺳﺔ ﺭﺑﺎﻋﻲ ﺍﻟﻮﺟﻮﻩ ﺭﺑﺎﻋﻲ ﺍﻟﺘﺴﺎﻧﺪ‬‫) ‪ (𝑇d‬ﻓﻲ ﺍﻟﻤﺼﻬﻮﺭ‪ ،‬ﻭﻫﻨﺪﺳﺔ ﺛﻤﺎﻧﻲ ﻭﺟﻮﻩ ﺳﺪﺍﺳﻲ ﺍﻟﺘﺴﺎﻧﺪ ) ‪ (𝑂h‬ﻓﻲ ﺍﻟﻔﺰﺍﺕ ﺍﻟﻤﻌﺪﻧﻴﺔ ﺍﻟﺤﺎﻭﻳﺔ ﻋﻠﻰ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ‪ ،‬ﻭﻛﻼ‬‫ﺍﻟﻬﻨﺪﺳﺘﻴﻦ ﺗﺘﻤﺘﻌﺎﻥ ﺑﺘﺸﻜﻴﻞ ﺇﻟﻜﺘﺮﻭﻧﻲ ﻋﺎﻟﻲ ﺍﻟﺴﺒﻴﻦ‪ .‬ﻭﻳﻤﻜﻦ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻦ ﻣﻌﺎﻣﻞ ﺍﻟﺘﻮﺯﻳﻊ ‪ D،‬ﻷﻳﻮﻧﺎﺕ ﺍﻟﻤﻌﺪﻥ ﺍﻻﻧﺘﻘﺎﻟﻲ‬‫‪ Mn+‬ﺑﻴﻦ ﺍﻟﻔﻠﺰﺍﺕ ﺍﻟﻤﻌﺪﻧﻴﺔ ﺍﻟﺤﺎﻭﻳﺔ ﻋﻠﻰ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ ﻭﺍﻟﻤﺼﻬﻮﺭ‪ ،‬ﻋﺒﺮ ﺍﻟﻌﻼﻗﺔ ‪:‬‬

‫‪Q6-2‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-6 Q-2‬‬‫ﺣﻴﺚ ‪ [M]s‬ﻭ ‪ [M]l‬ﻫﻤﺎ ﺗﺮﺍﻛﻴﺰ ‪ Mn+‬ﻓﻲ ﺍﻟﻔﻠﺰﺍﺕ ﺍﻟﻤﻌﺪﻧﻴﺔ ﺍﻟﺤﺎﻭﻳﺔ ﻋﻠﻰ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ ﻭﻓﻲ ﺍﻟﻤﺼﻬﻮﺭ ﻋﻠﻰ ﺍﻟﺘﺮﺗﻴﺐ‪ .‬ﻭﻳﻈﻬﺮ‬‫ﺍﻟﺠﺪﻭﻝ ﺍﻟﺘﺎﻟﻲ ﻗﻴﻢ 𝐷 ﻷﻳﻮﻧﺎﺕ ﺍﻟــﻜﺮﻭﻡ ‪ Cr2+‬ﻭ ﺍﻟـــــﻤﻨﻐﻴﺰ ‪ Mn2+‬ﻛﺄﻣﺜﻠﺔ‪.‬‬‫ﻟﻨﺮﻣﺰ ﺑـــ ‪ ΔO‬ﻭ ‪ CFSEO‬ﻟﻠﻔﺎﺻﻞ ﺍﻟﻄﺎﻗﻲ ﻟﻠﻤﺪﺍﺭﺍﺕ ‪ d‬ﻟﻸﻳﻮﻧﺎﺕ ‪ Mn+‬ﻭﻃﺎﻗﺔ ﺍﺳﺘﻘﺮﺍﺭ ﺍﻟﺤﻘﻞ ﺍﻟﺒﻠﻮﺭﻱ ﻓﻲ ﺣﻘﻞ ‪ ،𝑂h‬ﻋﻠﻰ‬‫ﺍﻟﺘﺮﺗﻴﺐ‪ .‬ﻭﻟﺘﻜﻦ ‪ ΔT‬ﻭ ‪ CFSET‬ﺗﻠﻚ ﺍﻟﻤﻮﺍﻓﻘﺔ ﻓﻲ ﺣﻘﻞ ‪. 𝑇d‬‬‫‪A.1‬‬‫‪ A.1.‬ﺍﺣﺴﺐ‬‫‪|CFSEO −CFSET |=ΔCFSE‬‬‫ﺑﺪﻻﻟﺔ ‪ ΔO‬ﻣﻦ ﺃﺟﻞ ﻛﻞ ﻣﻦ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻜﺮﻭﻡ ‪ ،Cr2+‬ﺍﻟﻤﻨﻐﻴﺰ ‪ ،Mn2+‬ﻭ ‪.Co2+‬‬‫ﺍﻓﺘﺮﺽ ﺃﻥ‪:‬‬‫‪ΔT = 4/9 ΔO‬‬‫‪6pt‬‬‫‪A.2‬‬‫ﺗﻢ ﻣﻼﺣﻈﺔ ﻭﺟﻮﺩ ﻋﻼﻗﺔ ﺧﻄﻴﺔ ﻋﻨﺪ ﺭﺳﻢ 𝐷 ‪ ln‬ﺑﺪﻻﻟﺔ‬‫‪ّ A.2.‬‬‫‪ ΔCFSE/ΔO‬ﻓﻲ ﺟﻤﻠﺔ ﺍﻹﺣﺪﺍﺛﻴﺎﺕ ﺍﻟﺪﻳﻜﺎﺭﺗﻴﺔ ﺍﻟﻤﻌﺮﻭﺿﺔ ﻓﻴﻤﺎ ﻳﻠﻲ‪.‬‬‫ﺃﻭﺟﺪ ﻗﻴﻤﺔ ‪ D‬ﻣﻦ ﺃﺟﻞ ﺃﻳﻮﻥ ﺍﻟﻜﻮﺑﺎﻟﺖ ‪.Co2+‬‬‫‪3pt‬‬‫ﺗﺘﺒﻠﻮﺭ ﺃﻛﺎﺳﻴﺪ ﺍﻟﻤﻌﺎﺩﻥ ‪ Oxides Metal MO‬ﺣﻴﺚ ‪ : (M‬ﺍﻟﻜﺎﻟﺴﻴﻮﻡ ‪ Ca،‬ﺍﻟﺘﻴﺘﺎﻧﻴﻮﻡ ‪ ، Ti‬ﺍﻟﻔﺎﻧﺎﺩﻳﻮﻡ ‪ ، V‬ﺍﻟﻤﻨﻐﻨﻴﺰ ‪ Mn،‬ﺃﻭ‬‫‪Q6-3‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-6 Q-3‬‬‫َ‬‫ﻫﻨﺪﺳﺔ ﺛﻤﺎﻧﻲ ﻭﺟﻮﻩ ‪ Oh‬ﻣﻊ ﺗﺸﻜﻴﻞ ﺇﻟﻜﺘﺮﻭﻧﻲ‬‫ﺍﻟﻜﻮﺑﺎﻟﺖ )‪ Co‬ﻭﻓﻖ ﺑﻨﻴﺔ ﻣﻠﺢ ﺻﺨﺮﻱ‪ ،‬ﻭﺍﻟﺬﻱ ﺗﺘﺒﻨﻰ ﻓﻴﻪ ﺍﻷﻳﻮﻧﺎﺕ ‪Mn+‬‬‫ﻋﺎﻟﻲ ﺍﻟﺴﺒﻴﻦ‪ .‬ﻭﺗﻜﻮﻥ ﺃﻧﻄﺎﻟﺒﻴﺔ ﺍﻟﺸﺒﻜﺔ ﻟﻬﺬﻩ ﺍﻷﻛﺎﺳﻴﺪ ﻣﺘﺤﻜﻢ ﺑﻬﺎ ﻋﺒﺮ ﺗﺄﺛﻴﺮﺍﺕ ﻛﻮﻟﻮﻥ ﺍﻟﻤﺘﺒﺎﺩﻟﺔ ﺍﻋﺘﻤﺎﺩﺍً ﻋﻠﻰ ﻧﺼﻒ‬‫ﻗﻄﺮ ﻭﺷﺤﻨﺔ ﺍﻷﻳﻮﻧﺎﺕ‪ ،‬ﻣﻊ ﺑﻌﺾ ﺍﻟﻤﺴﺎﻫﻤﺔ ﻟﻄﺎﻗﺔ ﺍﺳﺘﻘﺮﺍﺭ ﺍﻟﺤﻘﻞ ﺍﻟﺒﻠﻮﺭﻱ )‪ (CFSE‬ﻟﻸﻳﻮﻧﺎﺕ ‪ Mn+‬ﻓﻲ ﺍﻟﺤﻘﻞ ‪Oh.‬‬‫‪A.3‬‬‫‪ A.3.‬ﺍﺧﺘﺮ ﺍﻟﻤﺠﻤﻮﻋﺔ ﺍﻟﻤﻨﺎﺳﺒﺔ ﻣﻦ ﺃﻧﻄﺎﻟﺒﻴﺎﺕ ﺍﻟﺸﺒﻜﺔ )ﻣﻘﺪﺭﺓ ﺑﺎﻟـــ ] ‪ [kJ mol−1‬ﻣﻦ ﺑﻴﻦ ﻭﺍﺣﺪ‬‫ﻣﻦ ﺍﻟﺨﻴﺎﺭﺍﺕ ﻣﻦ )‪ (a‬ﺇﻟﻰ ‪(f).‬‬‫‪3pt‬‬

‫‪Q6-4‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-6 Q-4‬‬‫ﺍﻟﺠﺰﺀ ‪B‬‬‫ﻣﺰﻳﺞ ‪ A،‬ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺍﻟﻼﻧﺘﺎﻧﻮﻡ ‪ La3+‬ﻭﺍﻟﻨﺤﺎﺱ ‪ ،Cu2+‬ﻓﻲ ﺧﻠﻴﺔ ﻭﺍﺣﺪﻳﺔ ﺭﺑﺎﻋﻴﺔ ﺍﻟﺰﻭﺍﻳﺎ ‪ tetragonal‬ﺗﻤﺎﻣﺎً‬‫ﺃﻭﻛﺴﻴﺪ‬‫ﻳﺘﺒﻠﻮﺭ‬‫ٌ‬‫ٌ‬‫ﻛﻤﺎ ﻳﻈﻬﺮ ﻓﻲ ﺍﻟﺸﻜﻞ )‪ .(1‬ﻓﻲ ﺛﻤﺎﻧﻲ ﺍﻟﻮﺟﻮﻩ ]] ‪ ،[[CuO6‬ﺗﻜﻮﻥ ﺍﻟﻤﺴﺎﻓﺔ ‪ Cu–O‬ﻋﺒﺮ ﺍﻟﻤﺤﻮﺭ ‪ (𝑙𝑧 ) z‬ﺃﻃﻮﻝ ﻣﻨﻬﺎ ﻋﺒﺮ‬‫ﺍﻟﻤﺤﻮﺭ ‪ ،(𝑙𝑥 ) x‬ﻭﻳﻜﻮﻥ ﺛﻤﺎﻧﻲ ﺍﻟﻮﺟﻮﻩ ]] ‪ [[CuO6‬ﻣﺸﻮﻫﺎً ﻋﻦ ﻫﻨﺪﺳﺔ ﺛﻤﺎﻧﻲ ﺍﻟﻤﻮﺟﻮﻩ ﺍﻟﻨﻈﺎﻣﻴﺔ ‪ .𝑂h‬ﻭﻫﺬﺍ ﺍﻟﺘﺸﻮﻩ ﻳﺰﻳﻞ‬‫ﺍﻻﻧﺤﻄﺎﻁ )ﺃﻱ ﺗﺴﺎﻭﻱ ﻃﺎﻗﺘﻬﺎ( ‪ degeneracy‬ﻟﻠﻤﺪﺍﺭﺍﺕ ‪ d𝑥2 −𝑦2 ) eg‬ﻭ ‪.( d𝑧2‬‬‫‪1 Fig.‬‬‫ﻟﻤﻌﻘﺪ ‪ B،‬ﺍﻟﺬﻱ ﻳﺘﺸﻜﻞ ﻋﺒﺮ ﻣﺰﺝ ﻛﻠﻮﺭﻳﺪﺍﺕ ﺍﻟﻤﻌﺎﺩﻥ ‪metal‬‬‫ﻳﻤﻜﻦ ﺍﺻﻄﻨﺎﻉ ﺍﻷﻭﻛﺴﻴﺪ ﺍﻟﻤﺰﻳﺞ ‪ A‬ﻋﺒﺮ ﺍﻟﺘﻔﻜﻚ ﺍﻟﺤﺮﺍﺭﻱ‬‫ﱠ‬‫ﻣﻤﺪﺩ ﻣﻦ ﺍﻷﻣﻮﻧﻴﺎ ﺍﻟﺤﺎﻭﻱ ﻋﻠﻰ ﺣﻤﺾ ﺍﻟﺴﻜﻮﺍﺭﻳﻚ ‪C4 H2 O4‬؛ ﺃﻱ ﺣﻤﺾ ﺛﻨﺎﺋﻲ ‪diacid.‬‬‫‪ chlorides‬ﻓﻲ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ‬‫ّ‬‫ﻓﻘﺪﺍﻥ ﻭﺯﻥ ﻗﺪﺭﻩ ‪ % 29.1‬ﺣﺘﻰ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ‪ 200‬ﺩﺭﺟﺔ‬‫ُﻳﻈﻬﺮ ﺳﻠﻮﻙ ﺍﻟﺘﻔﻜﻚ ﺑﺎﻟﺤﺮﺍﺭﺓ ‪ pyrolysis‬ﻟــ ‪ B‬ﻓﻲ ﺍﻟﻬﻮﺍﺀ ﺍﻟﺠﺎﻑ‬‫َ‬‫ﻣﺌﻮﻳﺔ ﻳﻌﺰﺍ ﺇﻟﻰ ﻓﻘﺪﺍﻥ ﻣﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ‪ ،‬ﻣﺘﺒﻮﻋﺎً ﺑﻔﻘﺪﺍﻥ ﻭﺯﻥ ﺁﺧﺮ ﺣﺘﻰ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ‪ 700‬ﺩﺭﺟﺔ ﻣﺌﻮﻳﺔ ﻣﻌﺰﻭ ﺇﻟﻰ ﺗﺤﺮﺭ ‪.CO2‬‬‫ﻭﻳﻜﻮﻥ ﻓﻘﺪﺍﻥ ﺍﻟﻮﺯﻥ ﺍﻟﻜﻠﻲ ﺧﻼﻝ ﺗﺸﻜﻞ ﺍﻟﻤﺮﻛﺐ ‪ A‬ﺍﻧﻄﻼﻗﺎً ﻣﻦ ﺍﻟﻤﺮﻛﺐ ‪ B‬ﻫﻮ‪ . 63.6%‬ﻭﻳﺠﺐ ﻣﻼﺣﻈﺔ ﺃﻧﻪ ﻳﺘﻢ ﺗﺤﺮﻳﺮ‬‫ﻓﻘﻂ ﺍﻟﻤﺎﺀ ﻭ ‪ CO2‬ﺃﺛﻨﺎﺀ ﺗﻔﺎﻋﻞ ﺍﻟﺘﻔﻜﻚ ﺑﺎﻟﺤﺮﺍﺭﺓ‪.‬‬‫‪B.1‬‬‫‪ B.1.‬ﺍﻛﺘﺐ ﺍﻟﺼﻴﻎ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﻤﺮﻛﺒﻴﻦ ‪ A‬ﻭ‪.B‬‬‫‪6pt‬‬‫‪B.2‬‬‫‪ B.2.‬ﺍﺣﺴﺐ 𝑥𝑙 ﻭ 𝑧𝑙 ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﺸﻜﻞ )‪.(1‬‬‫‪4pt‬‬‫‪B.3‬‬‫ﺑﺎﻟﻨﺴﺒﺔ ﻷﻳﻮﻥ ‪ Cu2+‬ﻓﻲ ﺛﻤﺎﻧﻲ ﺍﻟﻮﺟﻮﻩ ﺍﻟﻤﺸﻮﻩ ] ‪ [CuO6‬ﻓﻲ ﺍﻟﻤﺮﻛﺐ ‪ A‬ﻓﻲ ﺍﻟﺸﻜﻞ‬‫‪ .1‬ﺍﻛﺘﺐ ﺃﺳﻤﺎﺀ ﺍﻟﻤﺪﺍﺭﺍﺕ ‪ eg‬ﺍﻟﻤﻨﻘﺴﻤﺔ ) ‪ d𝑥2 −𝑦2‬ﻭ ‪ (d𝑧2‬ﻓﻲ )‪ (i‬ﻭ ‪ (ii)،‬ﻭﺍﺭﺳﻢ ﺍﻟﺘﺸﻜﻴﻞ‬‫ﺍﻹﻟﻜﺘﺮﻭﻧﻲ ﻓﻲ ﺍﻟﺼﻨﺪﻭﻕ ﺍﻟﻤﻨﻘﻂ ﻓﻲ ﻭﺭﻗﺔ ﺇﺟﺎﺑﺘﻚ‪.‬‬‫‪4pt‬‬

‫‪Q6-5‬‬‫‪IChO‬‬‫‪SYR-3 C-6 Q-5‬‬‫)‪arabic (Syria‬‬‫ﻋﺎﺯﻻ‪ .‬ﻋﻨﺪﻣﺎ ﻧﺴﺘﺒﺪﻝ ﺃﻳﻮﻥ ‪ La3+‬ﻭﺍﺣﺪ ﻋﺒﺮ ﺃﻳﻮﻥ ‪ Sr2+‬ﻭﺍﺣﺪ‪ ،‬ﻳﺠﺮﻱ ﺗﻮﻟﻴﺪ ﺛﻘﺐ ﻭﺍﺣﺪ ﻓﻲ ﺍﻟﺸﺒﻜﺔ ﺍﻟﺒﻠﻮﺭﻳﺔ‬‫ﻳﻜﻮﻥ ﺍﻟﻤﺮﻛﺐ ‪A‬‬
‫ً‬‫ﺍﻟﺘﻲ ﻳﻤﻜﻨﻬﺎ ﻧﻘﻞ ﺍﻟﻜﻬﺮﺑﺎﺀ‪ .‬ﻭﻛﻨﺘﻴﺠﺔ‪ ،‬ﻳﻈﻬﺮ ﺍﻟﻤﺮﻛﺐ ‪ A‬ﺍﻟﻤﺸﺎﺏ ﺑﺎﻟـــ ‪ Sr2+‬ﻧﺎﻗﻠﻴﺔ ﻓﺎﺋﻘﺔ ﺗﺤﺖ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ‪ 38‬ﻛﻠﻔﻦ‪.‬‬‫ﻋﻨﺪﻣﺎ ﺣﺼﻞ ﺗﻔﺎﻋﻞ ﺍﻻﺳﺘﺒﺪﺍﻝ ﻟﻠﻤﺮﻛﺐ ‪ A،‬ﺟﺮﻯ ﺗﻮﻟﻴﺪ ‪ 2.05 × 1027‬ﺛﻘﺒﺎً ﻓﻲ ﺍﻟﻤﺘﺮ ﺍﻟﻤﻜﻌﺐ )‪.( .holesm−3 .‬‬‫‪B.4‬‬‫ﺍﺣﺴﺐ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﻷﻳﻮﻧﺎﺕ ﺍﻟﺴﺘﺮﻭﻧﺴﻴﻮﻡ ‪ Sr2+‬ﺍﻟﻤﺴﺘﺒﺪﻟﺔ ﻣﻜﺎﻥ ﺃﻳﻮﻧﺎﺕ ‪ Sr2+‬ﺍﻋﺘﻤﺎﺩﺍً‬‫ﻋﻠﻰ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻓﻲ ﺗﻔﺎﻋﻞ ﺍﻻﺳﺘﺒﺪﺍﻝ‪ .‬ﻻﺣﻆ ﺃﻥ ﺗﻜﺎﻓﺆﺍﺕ ﺍﻷﻳﻮﻧﺎﺕ ﺍﻟﻤﻜﻮﻧﺔ ﻭﺍﻟﺒﻨﻴﺔ‬‫ﺍﻟﺒﻠﻮﺭﻳﺔ ﻻ ﺗﺘﻐﻴﺮ ﻋﺒﺮ ﺗﻔﺎﻋﻞ ﺍﻻﺳﺘﺒﺪﺍﻝ‪.‬‬‫‪4pt‬‬‫ﺍﻟﺠﺰﺀ ‪C‬‬‫‪) Cu‬ﺍﻟﺸﻜﻞ ‪ .(2A‬ﻳﺘﻤﺘﻊ‬‫ﻳﺘﻜﻮﻥ ‪ Cu2 (CH3 CO2 )4‬ﻣﻦ ﺃﺭﺑﻊ ‪ CH3 CO2‬ﻣﺮﺗﺒﻄﺔ ﺑﺮﻭﺍﺑﻂ ﺗﺴﺎﻧﺪﻳﺔ ﻣﻊ ﺃﻳﻮﻧﻴﻦ‬‫‪−‬‬‫‪ Cu2 (CH3 CO2 )4‬ﺑﻤﺴﺘﻮﻳﺎﺕ ﻋﺎﻟﻴﺔ ﻣﻦ ﺍﻟﺘﻨﺎﻇﺮ ﺍﻟﺒﻨﻴﻮﻱ‪ ،‬ﻣﻊ ﻣﺤﻮﺭﻳﻦ ﻣﺎﺭﻳﻦ ﻋﺒﺮ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ﻷﻳﻮﻧﺎﺕ ‪ CH3 CO2‬ﺍﻷﺭﺑﻊ‪ ،‬ﻭ‬‫‪−‬‬‫ﻭﻣﺤﻮﺭ ﻣﺎﺭ ﻋﺒﺮ ﺍﻷﻳﻮﻧﻴﻦ‬‫‪2+‬‬‫‪ ، Cu‬ﻭﺍﻟﺘﻲ ﺗﻜﻮﻥ ﻛﻞ ﻣﻨﻬﺎ ﻣﺘﻮﺟﻬﺔ ﺑﺸﻜﻞ ﻣﺘﻌﺎﻣﺪ ﺑﺎﻟﻨﺴﺒﺔ ﻟﺒﻌﻀﻬﺎ ﺍﻷﺧﺮ‪ .‬ﻋﻨﺪﻣﺎ ﻧﺴﺘﺨﺪﻡ‬‫‪2+‬‬‫ﺑﺪﻻ ﻣﻦ‬‫ﺭﺑﻴﻄﺔ ﺛﻨﺎﺋﻲ ﻛﺮﺑﻮﻛﺴﻴﻼﺕ‬
‫ً‬

‫‪−‬‬

‫‪ ،CH3 CO2‬ﻳﺘﺸﻜﻞ ”ﻣﻌﻘﺪ ﻗﻔﺺ ‪ .” cage complex‬ﻳﺘﻜﻮﻥ ﺍﻟﻤﻌﻘﺪ ﺍﻟﻘﻔﺺ‬‫‪L1‬‬‫‪ Cu4 (L1‬ﻣﻦ ﺛﻨﺎﺋﻲ ﻛﺮﺑﻮﻛﺴﻴﻼﺕ ‪ L1‬ﻣﺴﺘﻮﻱ )ﺍﻟﺸﻜﻞ ‪ (2B‬ﻭﺃﻳﻮﻥ ‪) Cu‬ﺍﻟﺸﻜﻞ ‪ .(2C‬ﺗﺤﺪﺩ ﺍﻟﺰﺍﻭﻳﺔ ‪ θ‬ﺑﻴﻦ ﺍﺗﺠﺎﻫﺎﺕ‬‫‪L1)4‬‬‫َ‬‫ﺑﻨﻴﺔ ﺍﻟﻤﻌﻘﺪ ﺍﻟﻘﻔﺺ‪ .‬ﺗﻜﻮﻥ ﺍﻟﺰﺍﻭﻳﺔ ‪θ‬‬‫ﺍﻟﺘﺴﺎﻧﺪ ﻟﻠﻜﺮﺑﻮﻛﺴﻴﻼﺕ ﺍﻻﺛﻨﺘﻴﻦ‪ ،‬ﻭﺍﻟﻤﺸﺎﺭ ﺇﻟﻴﻬﺎ ﻋﺒﺮ ﺍﻷﺳﻬﻢ ﻓﻲ ﺍﻟﺸﻜﻞ ‪2B،‬‬‫‪2+‬‬‫ﻫﻲ ‪ °0‬ﻣﻦ ﺃﺟﻞ ‪ L1.‬ﻻﺣﻆ ﺃﻧﻪ ﻟﻢ ﻳﺘﻢ ﺗﻤﺜﻴﻞ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻓﻲ ﺍﻟﺸﻜﻞ ‪.2‬‬‫‪2 Fig.‬‬‫‪Q6-6‬‬‫)‪arabic (Syria‬‬‫‪C.1‬‬‫‪IChO‬‬‫‪SYR-3 C-6 Q-6‬‬‫ﺗﺜﺒﺖ ﺍﻟﺰﺍﻭﻳﺔ ‪ θ‬ﻟﺜﻨﺎﺋﻲ ﺍﻟﻜﺮﺑﻮﻛﺴﻴﻼﺕ ‪ L2‬ﻓﻴﻤﺎ ﻳﻠﻲ ﻋﻠﻰ ﺍﻟﻘﻴﻤﺔ ˚‪ .90‬ﺇﺫﺍ ﻛﺎﻥ ﺗﺮﻛﻴﺐ‬‫‪L2‬‬‫‪ ، Cu𝑛 (L2‬ﺃﻋﻂ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻷﺻﻐﺮ ﻟﻸﻋﺪﺍﺩ‬‫ﺍﻟﻤﻌﻘﺪ ﺍﻟﻘﻔﺺ ﺍﻟﻤﺸﻜﻞ ﻣﻦ ‪ L2‬ﻭ ‪ Cu‬ﻫﻮ 𝑚)‪L2‬‬‫‪−‬‬‫ﺍﻟﺼﺤﻴﺤﺔ ‪ n‬ﻭ ‪ m.‬ﺍﻓﺘﺮﺽ ﺃﻧﻪ ﻓﻘﻂ ﺯﻣﺮ ‪ CO2‬ﻟـــ ‪ L2‬ﺗﺸﻜﻞ ﺭﺍﺑﻄﺔ ﺗﺴﺎﻧﺪﻳﺔ ﻣﻊ ﺃﻳﻮﻧﺎﺕ‬‫‪2+‬‬‫‪Cu .‬‬‫‪2+‬‬‫‪5pt‬‬

‫‪Q6-7‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-6 Q-7‬‬‫ﻳﺤﺘﻮﻱ ﻣﻌﻘﺪ ﺗﻮﺗﻴﺎﺀ‪ ،Zn4 O(CH3 CO2 )6 ،‬ﻋﻠﻰ ﺃﺭﺑﻊ‬‫‪.(3A‬‬‫‪2+‬‬‫‪ Zn‬ﺭﺑﺎﻋﻴﺔ ﺍﻟﻮﺟﻮﻩ ‪ ،‬ﺳﺘﺔ‬‫‪−‬‬‫‪ ،CH3 CO2‬ﻭﺃﻳﻮﻥ ‪ O2−‬ﻭﺍﺣﺪ )ﺍﻟﺸﻜﻞ‬‫ﻓﻲ ﺍﻟﻤﻌﻘﺪ ‪ ،Zn4 O(CH3 CO2 )6‬ﻳﻜﻮﻥ ‪ O2−‬ﻣﺘﻮﺿﻌﺎً ﻓﻲ ﺍﻟﻤﺮﻛﺰ‪ ،‬ﻭﺗﺘﻮﺟﻪ ﺍﻟﻤﺤﺎﻭﺭ ﺍﻟﺜﻼﺛﺔ ﺍﻟﻤﺎﺭﺓ ﻣﻦ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ﻟــ‬‫‪−‬‬‫‪ CH3 CO2‬ﺑﺸﻜﻞ ﻣﺘﻌﺎﻣﺪ ﺑﺎﻟﻨﺴﺒﺔ ﻟﺒﻌﻀﻬﺎ ﺍﻟﺒﻌﺾ‪.‬‬‫ﺑﺪﻻ ﻣﻦ ‪ ،CH3 CO2‬ﺗﻜﻮﻥ ﻋﻨﺎﻗﻴﺪ ‪Zn‬‬‫ﻋﻨﺪﻣﺎ ﻧﺴﺘﺨﺪﻡ ‪) p-benzenedicarboxylate‬ﺍﻟﺸﻜﻞ ‪(L3, θ = 180° ، 3B‬‬
‫ً‬‫ﻣﺮﺗﺒﻄﺔ ﻣﻊ ﺑﻌﻀﻬﺎ ﺍﻟﺒﻌﺾ ﻟﺘﺸﻜﻞ ﺟﺴﻤﺎً ﺻﻠﺒﺎً ﺑﻠﻮﺭﻳﺎً )‪ (X‬ﺍﻟﺬﻱ ﻳﺴﻤﻰ ”ﺑﻮﻟﻴﻤﻴﺮ ﺗﺴﺎﻧﺪ ﻣﺴﺎﻣﻲ ‪coordination porous‬‬‫‪L3‬‬‫‪ ،[Zn4 O(L3‬ﻭﻳﺘﻤﺘﻊ ﺑﺒﻨﻴﺔ ﺑﻠﻮﺭﻳﺔ ﻣﻜﻌﺒﺔ ﻣﻊ ﻣﺴﺎﻣﺎﺕ ﺑﺄﺑﻌﺎﺩ ﻧﺎﻧﻮﻳﺔ‪ .‬ﺟﺮﻯ‬‫”‪ polymer‬ﺍﻟﺸﻜﻞ ‪ (3C).‬ﻳﻜﻮﻥ ﺗﺮﻛﻴﺐ ‪ X‬ﻫﻮ 𝑛] ‪L3)3‬‬‫‪2+‬‬‫ﻣﻤﺜﻼ ﻋﻠﻰ ﺷﻜﻞ ﻣﺠﺴﻢ‬‫ﺗﻤﺜﻴﻞ ﻣﺴﺎﻡ ﻭﺍﺣﺪ ﻋﻠﻰ ﺷﻜﻞ ﻛﺮﺓ ﻓﻲ ﺍﻟﺸﻜﻞ ‪ ، 3D‬ﻭﻳﻜﻮﻥ ﻛﻞ ﻋﻨﻘﻮﺩ ‪ Zn‬ﺭﺑﺎﻋﻲ ﺍﻟﻮﺟﻮﻩ‬‫ً‬‫ﻏﺎﻣﻖ ﺭﻣﺎﺩﻱ ﻓﻲ ﺍﻟﺸﻜﻞ ‪ 3C‬ﻭ ‪ 3D.‬ﻻﺣﻆ ّﺃﻧﻪ ﻟﻢ ﻳﺘﻢ ﺗﻤﺜﻴﻞ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻓﻲ ﺍﻟﺸﻜﻞ ‪.3‬‬‫‪−‬‬‫‪2+‬‬‫‪3 Fig.‬‬‫‪C.2‬‬‫ﻳﺘﻤﺘﻊ ‪ X‬ﺑﺨﻠﻴﺔ ﻭﺍﺣﺪﻳﺔ ﻣﻜﻌﺒﺔ ﻣﻊ ﻃﻮﻝ ﺣﺮﻑ ﺃﻭ ﺟﺎﻧﺐ 𝑎 ‪ ،‬ﺍﻟﺸﻜﻞ ‪ 3C،‬ﻭﻛﺜﺎﻓﺔ ﻗﺪﺭﻫﺎ‬‫ّ‬‫‪ . g cm−3 0.592‬ﺍﺣﺴﺐ 𝑎 ﻣﻘﺪﺭﺍً ﺑﺎﻟـــ ‪[cm].‬‬‫‪5pt‬‬‫‪C.3‬‬‫ﻳﺤﺘﻮﻱ ‪ X‬ﻋﻠﻰ ﻋﺪﺩ ﻣﻌﺘﺒﺮ ﻣﻦ ﺍﻟﻤﺴﺎﻣﺎﺕ‪ ،‬ﻭﻳﻤﻜﻦ ﻟـــ ‪ 1g‬ﻣﻦ ‪ X‬ﺃﻥ ﻳﺴﺘﻮﻋﺐ ‪3.0 × 102 mL‬‬‫ﻣﻦ ﻏﺎﺯ ‪ CO2‬ﻓﻲ ﺍﻟﻤﺴﺎﻣﺎﺕ ﻋﻨﺪ ﺿﻐﻂ ‪ bar 1‬ﻭﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ‪ 25‬ﺩﺭﺟﺔ ﻣﺌﻮﻳﺔ‪ .‬ﺍﺣﺴﺐ ﺍﻟﻌﺪﺩ‬‫ﺍﻟﻮﺳﻄﻲ ﻟﺠﺰﻳﺌﺎﺕ ‪ CO2‬ﻓﻲ ﻭﺍﺣﺪﺓ ﺍﻟﻤﺴﺎﻡ‪.‬‬‫‪5pt‬‬IChO
SYR-3 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
SYR-3 C-6 A-2

A.2 (3 pt)

𝐷∶

A.3 (3 pt)

A6-2
English (Official)

IChO

A6-3

SYR-3 C-6 A-3

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

A∶

,B∶

B.2 (4 pt)

𝑙𝑥 =nm, 𝑙𝑧 =nm

IChO
SYR-3 C-6 A-4

B.3 (4 pt), (ii) ∶(i) ∶B.4 (4 pt)%

A6-4
English (Official)IChO
SYR-3 C-6 A-5Part C
C.1 (5 pt)𝑛=,𝑚=

C.2 (5 pt)

𝑎=

cm

A6-5
English (Official)IChO
SYR-3 C-6 A-6

C.3 (5 pt)

A6-6
English (Official)SYR-3 C-7 C-1

SYR-3 C-7 C
ANAS ABBAS

IChO
Problem 7
Cover sheet

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

Q7-1

IChO
SYR-3 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
SYR-3 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
SYR-3 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
SYR-3 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
SYR-3 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‫‪Q7-1‬‬‫‪IChO‬‬‫‪SYR-3 C-7 Q-1‬‬‫)‪arabic (Syria‬‬‫ﺍﻟﻠﻌﺐ ﻣﻊ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﻌﻄﺮﻳﺔ ﺍﻟﻼﺑﻨﺰﻭﺋﻴﺔ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪5‬‬‫‪total the of % 13‬‬‫‪A.2‬‬‫‪2‬‬‫‪A.3‬‬‫‪19‬‬‫‪B.1‬‬‫‪10‬‬‫‪Total‬‬‫‪36‬‬‫‪Score‬‬‫ﺍﻟﺒﺮﻭﻓﻴﺴﻮﺭ ﻧﻮﺯﻭﻱ )‪ (1996-1902‬ﺍﻓﺘﺘﺢ ﻣﺠﺎﻝ ﺍﻟﺒﺤﺚ ﻋﻦ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﻌﻄﺮﻳﺔ ﺍﻟﻼﺑﻨﺰﻭﺋﻴﺔ ‪ ،‬ﻭﺍﻟﺘﻲ ﺃﺻﺒﺤﺖ ﺗﻄﺒﻴﻘﺎﺗﻬﺎ ﻣﻨﺘﺸﺮﺓ‬‫ﺑﺸﻜﻞ ﻛﺒﻴﺮ ﻓﻲ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻌﻀﻮﻳﺔ‪.‬‬‫‪Univ. Tohoku courtesy: Photo‬‬‫ﺍﻟﺠﺰﺀ ‪A‬‬‫‪ Lineariifolianone‬ﻫﻮ ﻣﻨﺘﺞ ﻃﺒﻴﻌﻲ ﺑﻬﻴﻜﻞ ﻓﺮﻳﺪ ﺗﻢ ﻋﺰﻟﻪ ﻣﻦ ‪ .linariifolia Inula‬ﺍﻧﻄﻼﻗﺎ ﻣﻦ ‪(1) valencene‬‬‫‪،‬ﻭﻋﺒﺮ ﺧﻄﻮﺓ ﻭﺍﺣﺪﺓ ﻳﻨﺘﺞ ﺍﻟﻤﺮﻛﺐ ‪ ، 2‬ﺍﻟﻜﻴﺘﻮﻥ ‪ 4‬ﻧﺤﺼﻞ ﻋﻠﻴﻪ ﻣﻦ ﺧﻼﻝ ﺛﻼﺙ ﺧﻄﻮﺍﺕ ﻣﺮﻭﺭﺍً ﺑﺎﻟﻤﺮﻛﺐ ‪. 3‬ﻳﺘﻢ ﺗﺤﻮﻳﻞ‬‫‪ (5) Eremophilene‬ﺇﻟﻰ ﺍﻟﻤﺮﻛﺐ ‪ 6‬ﻋﻦ ﻃﺮﻳﻖ ﺇﺟﺮﺍﺀ ﻧﻔﺲ ﺍﻟﺨﻄﻮﺍﺕ ﺍﻷﺭﺑﻌﺔ ﺍﻟﺴﺎﺑﻘﺔ‪.‬‬‫‪Q7-2‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-7 Q-2‬‬‫‪linariifolia Inula‬‬‫‪A.1‬‬‫ﺍﺭﺳﻢ ﺍﻟﺒﻨﻴﺔ ﺍﻟﻬﻴﻜﻠﻴﺔ ﻟﻠﻤﺮﻛﺒﺎﺕ ‪ 2‬ﻭ ‪ 6‬ﻭﺣﺪﺩ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺑﻮﺿﻮﺡ ﻋﻨﺪ ﺍﻟﻀﺮﻭﺭﺓ‬‫‪5pt‬‬‫ﺑﻌﺪ ﺫﻟﻚ ‪ ،‬ﻳﺘﻢ ﺗﺤﻮﻳﻞ ﺍﻟﻜﻴﺘﻮﻥ ‪ 4‬ﺇﻟﻰ ﺍﺳﺘﺮ ‪ .15‬ﺍﻟﻤﺮﻛﺐ ‪) 8‬ﺍﻟﻮﺯﻥ ﺍﻟﺠﺰﻳﺌﻲ‪ (188 :‬ﻳﺤﺘﻔﻆ ﺑﺠﻤﻴﻊ ﻣﺮﺍﻛﺰ ﺍﻟﻔﻌﺎﻟﻴﺔ ﺍﻟﻀﻮﺋﻴﺔ‬‫ﺍﻟﻤﻮﺟﻮﺩﺓ ﻓﻲ ﺍﻟﻤﺮﻛﺐ ‪ .7‬ﺍﻟﻤﺮﻛﺒﺎﺕ ‪ 9‬ﻭ ‪ 10‬ﺗﺤﺘﻮﻱ ﻋﻠﻰ ﺧﻤﺴﺔ ﻣﺮﺍﻛﺰ ﻓﻌﺎﻟﻴﺔ ﺿﻮﺋﻴﺔ ﻭﻻ ﺗﻮﺟﺪ ﺭﻭﺍﺑﻂ ﻛﺮﺑﻮﻥ‪-‬ﻛﺮﺑﻮﻥ‬‫‪Q7-3‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-7 Q-3‬‬‫ﺑﺪﻻ ‪ H2 16 O‬ﻻﺻﻄﻨﺎﻉ ﺍﻟﻤﺮﻛﺒﺎﺕ )‪ ( 13 lineariifolianones‬ﺍﻟﻤﻮﺳﻮﻣﺔ ﺑﻨﻈﻴﺮ‬‫ﻣﺰﺩﻭﺟﺔ‪ .‬ﺍﻓﺘﺮﺽ ﺃﻧﻪ ﺗﻢ ﺍﺳﺘﺨﺪﺍﻡ ‪H2 18 O‬‬‫ً‬‫ﺍﻷﻭﻛﺴﺠﻴﻦ ‪ 18 O-‬ﻭ ﺍﻟﻤﺮﻛﺐ ‪ 14‬ﻣﻦ ﺍﻟﻤﺮﻛﺒﻴﻦ ‪ 11‬ﻭ ‪ 12‬ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‪ .‬ﺍﻟﻤﺮﻛﺒﺎﺕ ‪ 13‬ﻭ ‪ 14‬ﺍﻟﻤﻮﺳﻮﻣﺔ ﺑﻨﻈﻴﺮ ﺍﻷﻭﻛﺴﺠﻴﻦ‬‫‪ 18 O‬ﻭﻫﻲ ﻣﺼﺎﻭﻏﺎﺕ ﺍﻳﺰﻭﺗﻮﺑﻴﺔ )ﺍﻟﺘﻲ ﺗﺨﺘﻠﻒ ﺑﻤﻮﺿﻮﻉ ﺍﻻﻳﺰﻭﺗﻮﺏ(‪ ،‬ﺑﺄﻫﻤﺎﻝ ﻭﺟﻮﺩ ﺍﻟﻌﻨﺎﺻﺮ ﺍﻟﻤﻮﺳﻮﻣﺔ ﻳﻤﺘﻠﻚ ﺍﻟﻤﺮﻛﺒﺎﻥ‬‫‪ 13‬ﻭ ‪ 14‬ﻧﻔﺲ ﻓﺮﺍﻏﻴﺔ ﺍﻟﻤﺮﻛﺐ ‪. 15‬‬

IChO
SYR-3 C-7 Q-4

Q7-4
arabic (Syria)

‫‪Q7-5‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-7 Q-5‬‬‫‪A.2‬‬‫ﺍﺧﺘﺮ ﺍﻟﺒﻨﻴﺔ ﺍﻟﻬﻴﻜﻠﻴﺔ ﺍﻟﻤﻨﺎﺳﺒﺔ ﻟـ ‪.A‬‬‫‪2pt‬‬‫‪A.3‬‬‫ﺃﻳﻀﺎ‬‫ﺍﺭﺳﻢ ﺍﻟﺒﻨﻴﺔ ﺍﻟﻬﻴﻜﻠﻴﺔ ﻟﻠﻤﺮﻛﺒﺎﺕ ‪ 14-8‬ﻭ ﻣﻮﺿﺤﺎً ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ ﻋﻨﺪ ﺍﻟﻀﺮﻭﺭﺓ‪ .‬ﺃﺷﺮ ً‬‫‪19pt‬‬‫ﺇﻟﻰ ﺫﺭﺍﺕ ﺍﻷﻭﻛﺴﺠﻴﻦ ﺍﻟﻤﻮﺳﻮﻣﺔ ‪O‬‬
‫ﻓﻲ ﺍﻟﻤﺜﺎﻝ ﺃﺩﻧﺎﻩ‪.‬‬‫‪18‬‬‫ﻭﺍﻟﺘﻲ ﺗﻢ ﺇﺩﺧﺎﻟﻬﺎ ﻟـﻠﻤﺮﻛﺒﺎﺕ ‪ 13‬ﻭ ‪ 14‬ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ‬‫‪Q7-6‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-7 Q-6‬‬‫‪B Part‬‬‫ٌﻳﺼﻨﻊ ﺍﻟﻤﺮﻛﺐ ‪ 19‬ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺃﺩﻧﺎﻩ‪ .‬ﻓﻴﻤﺎ ﻳﺘﻌﻠﻖ ﺑﺎﻟﻤﺮﻛﺐ ﺍﻟﻌﻄﺮﻱ ﺍﻟﻼﺑﻨﺰﻭﺋﻲ ‪ ،‬ﻳﻤﻜﻦ ﺍﺳﺘﺨﺪﺍﻡ ﺍﻟﻤﺮﻛﺐ ‪ 19‬ﻛﻤﻨﺸﻂ‬‫ﻟﻠﻜﺤﻮﻝ ‪ ،‬ﻭﺗﻢ ﺗﺤﻮﻳﻞ ﺍﻟﻤﺮﻛﺐ ‪ 20‬ﺇﻟﻰ ‪ 22‬ﻋﺒﺮ ﻣﺮﻛﺐ ﻭﺳﻄﻲ )ﺯﻭﺝ ﺃﻳﻮﻧﻲ( ﺍﻟﻤﺮﻛﺐ‪ . 21‬ﻋﻠﻰ ﺍﻟﺮﻏﻢ ﻣﻦ ﺃﻥ ﺗﺸﻜﻞ ﺍﻟﻤﺮﻛﺐ‬‫ﺗﺪﺭﻳﺠﻴﺎ ﻟﻴﻌﻄﻲ ﺍﻟﻤﺮﻛﺒﻴﻦ ‪ 18‬ﻭ ‪.22‬‬‫‪ 21‬ﻗﺪ ﻟﻮﺣﻆ ﺑﻮﺍﺳﻄﺔ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﺍﻟﻨﻮﻭﻱ‪ ،‬ﻓﺈﻥ ﺍﻟﻤﺮﻛﺐ ‪ 21‬ﻳﺘﺤﻠﻞ‬‫ً‬‫‪B.1‬‬‫ﺍﺭﺳﻢ ﺍﻟﺒﻨﻰ ﺍﻟﻬﻴﻜﻠﻴﺔ ﻟﻠﻤﺮﻛﺒﺎﺕ ‪ 19-17‬ﻭ ‪ .21‬ﺗﺤﺪﻳﺪ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ ﻟﻠﻤﺮﻛﺒﺎﺕ ﻏﻴﺮ‬‫ﺿﺮﻭﺭﻱ‪.‬‬‫‪10pt‬‬IChO
SYR-3 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
SYR-3 C-7 A-2

A.3 (19 pt)

A7-2
English (Official)IChO
SYR-3 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)SYR-3 C-8 C-1

SYR-3 C-8 C
ANAS ABBAS

IChO
Problem 8
Cover sheet

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

Q8-1

IChO
SYR-3 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
SYR-3 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
SYR-3 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
SYR-3 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
SYR-3 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

‫‪Q8-1‬‬‫‪IChO‬‬‫‪SYR-3 C-8 Q-1‬‬‫)‪arabic (Syria‬‬‫ﺍﻟﺠﺰﻳﺌﺎﺕ ﺍﻟﻌﻀﻮﻳﺔ ﺍﻟﺪﻳﻨﺎﻣﻴﻜﻴﺔ ﻭﻛﻴﺮﺍﻟﻴﺘﻬﺎ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪9‬‬‫‪total the of % 11‬‬‫‪A.2‬‬‫‪3‬‬‫‪A.3‬‬‫‪7‬‬‫‪B.1‬‬‫‪3‬‬‫‪B.2‬‬‫‪4‬‬‫‪Total‬‬‫‪26‬‬‫‪Score‬‬‫‪A Part‬‬‫ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﻬﻴﺪﺭﻭﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﻌﻄﺮﻳﺔ ﻣﺘﻌﺪﺩﺓ ﺍﻟﺤﻠﻘﺎﺕ ﺍﻟﺘﻲ ﺗﻨﺪﻣﺞ ﻓﻴﻬﺎ ﺍﻟﺤﻠﻘﺎﺕ ﺑﺸﻜﻞ ﻣﺘﻌﺎﻗﺐ ﻓﻲ ﺍﻟﻤﻮﻗﻊ ﺍﻭﺭﺗﻮ ﺗﺪﻋﻰ‬‫]‪ [n‬ﻛﺮﺑﻮ ﻫﻴﻠﺴﻴﻨﺎﺕ )ﻳﻤﺜﻞ ‪ n‬ﻋﺪﺩ ﺍﻟﺤﻠﻘﺎﺕ ﺍﻟﺴﺪﺍﺳﻴﺔ ( )ﺍﻧﻈﺮ ﺃﺩﻧﺎﻩ(‪[4] .‬ﻛﺮﺑﻮ ﻫﻴﻠﺴﻴﻦ )‪ُ ([4]C‬ﻳﺤﻀﺮ ﺑﻜﻔﺎﺀﺓ ﻋﻦ ﻃﺮﻳﻖ‬‫ﺍﺳﺘﺨﺪﺍﻡ ﺗﻔﺎﻋﻞ ﺿﻮﺋﻲ ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺃﺩﻧﺎﻩ ‪ ،‬ﻣﺮﻭﺭﺍً ﺑﻤﺮﻛﺐ ﻭﺳﻴﻄﻲ )‪ (.Int‬ﻳﺘﺄﻛﺴﺪ ﺑﺴﻬﻮﻟﺔ ﺑﻮﺍﺳﻄﺔ ﺍﻟﻴﻮﺩ‪.‬‬‫ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻀﻮﺋﻲ ﻳﺠﺮﻱ ﺑﻄﺮﻳﻘﺔ ﻣﺸﺎﺑﻬﺔ ﻟﻠﻤﺜﺎﻝ ﺍﻟﺘﺎﻟﻲ‪.‬‬‫‪Q8-2‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-8 Q-2‬‬‫ﻣﻼﺣﻈﺔ‪ :‬ﻟﺠﻤﻴﻊ ﺍﻷﺳﺌﻠﺔ ﻓﻲ ﺍﻟﻤﺴﺄﻟﺔ ‪ ، 8‬ﻳﺮﺟﻰ ﺭﺳﻢ ﺭﻭﺍﺑﻂ ﻣﻔﺮﺩﺓ ﻭﻣﺰﺩﻭﺟﺔ ﻣﺘﻨﺎﻭﺑﺔ ﻓﻲ ﺇﺟﺎﺑﺎﺗﻚ ﻋﻠﻰ‬‫ﺍﻷﺳﺌﻠﺔ ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﻓﻲ ﺃﻣﺜﻠﺔ ﺍﻟﻜﺮﺑﻮ ﻫﻴﻠﺴﻴﻦ ‪ (carbohelicene).‬ﻻ ﺗﺴﺘﺨﺪﻡ ﺍﻟﺪﻭﺍﺋﺮ ﻟﻸﻧﻈﻤﺔ ﺍﻟﻌﻄﺮﻳﺔ‬‫ﺍﻟﻤﺘﺮﺍﻓﻘﺔ‪.‬‬‫‪A.1‬‬‫ﺍﺭﺳﻢ ﺍﻟﺒﻨﻰ ﺍﻟﻬﻴﻜﻠﻴﺔ ‪ .C – A‬ﻳﺠﺐ ﺗﻤﻴﻴﺰ ﺍﻟﻤﺘﺼﺎﻭﻏﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ‪.‬‬‫‪9pt‬‬‫‪A.2‬‬‫ﺃﺩﺕ ﻣﺤﺎﻭﻻﺕ ﺗﺼﻨﻴﻊ ﺍﻟﻤﺮﻛﺐ ]‪ [5‬ﻛﺮﺑﻮ ﻫﻴﻠﺴﻴﻦ ﻣﻦ ﻣﻠﺢ ﺍﻟﻔﻮﺳﻔﻮﻧﻴﻮﻡ ﻭﻣﺮﻛﺐ ﺑﺪﺀ ﻣﻨﺎﺳﺐ‬‫ﺇﻟﻰ ﺗﻜﻮﻳﻦ ﻛﻤﻴﺔ ﺿﺌﻴﻠﺔ ﻣﻦ ﺍﻟﻤﺮﻛﺐ ]‪[5‬ﻛﺮﺑﻮﻫﻴﻠﺴﻴﻦ ‪،‬ﻭﺑﺎﻟﻤﻘﺎﺑﻞ ﻧﺘﺞ ﻟﺪﻳﻨﺎ ﺍﻟﻤﺮﻛﺐ ‪ D‬ﺍﻟﺬﻱ‬‫ﻭﺯﻧﻪ ﺍﻟﺠﺰﻳﺌﻲ ﺃﻗﻞ ﺑﻤﻘﺪﺍﺭ ‪ 2‬ﺩﺍﻟﺘﻮﻥ ﻣﻦ ﺍﻟﻤﺮﻛﺐ ]‪ [5‬ﻛﺮﺑﻮﻫﻴﻠﺴﻴﻦ‪ .‬ﺍﻟﻤﻌﻄﻴﺎﺕ ﺍﻟﻄﻴﻔﻴﺔ ‪NMR‬‬‫ﻟﻠﻤﺮﻛﺐ ‪ D‬ﻣﻮﺟﻮﺩﺓ ﺃﺩﻧﺎﻩ‪ .‬ﺍﺭﺳﻢ ﺍﻟﺒﻨﻴﺔ ﺍﻟﻬﻴﻜﻠﻴﺔ ﻟﻠﻤﺮﻛﺐ ‪.D‬‬‫‪[D (δδ, ppm in CS2CS2, r.t.), 8.85 (2H), 8.23 (2H), 8.07 (2H), 8.01 (2H), 7.97 (2H),‬‬‫])‪7.91 (2H‬‬‫‪3pt‬‬‫]‪ - [5‬ﻭﺍﻟﻜﺮﺑﻮﻫﻴﻠﺴﻴﻨﺎﺕ ﺍﻷﻛﺒﺮ ﻣﻦ ﺣﻴﺚ ﻋﺪﺩ ﺍﻟﺤﻠﻘﺎﺕ ]‪ [n‬ﻟﻬﺎ ﺑﻨﻴﺔ ﺣﻠﺰﻭﻧﻴﺔ ﻭﺗﻤﺘﺎﺯ ﺑﺎﻟﻜﻴﺮﺍﻟﻴﺔ ﻭﺍﻟﺘﺤﻮﻝ ﺍﻟﺒﻴﻨﻲ ﺑﻴﻦ‬‫ﺍﻟﻤﺼﺎﻭﻏﺎﺕ ﺍﻻﻳﻨﺎﺗﻮﻣﻴﺮﻳﺔ ﻓﻲ ﻫﺬﻩ ﺍﻟﻬﻴﻠﺴﻴﻨﺎﺕ ﻳﻜﻮﻥ ﺑﻄﻴﺌً ﺎ ﺑﺸﻜﻞ ﻣﻠﺤﻮﻅ ﻓﻲ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﺍﻟﻐﺮﻓﺔ‪ .‬ﻳﺘﻢ ﺗﻌﺮﻳﻒ‬‫ﺍﻟﻜﻴﺮﺍﻟﻴﺔ ﻟﻤﺮﻛﺒﺎﺕ ]‪ [n‬ﻛﺮﺑﻮ ﻫﻴﻠﺴﻴﻨﺎﺕ ]‪ ( carbohelicenes ([n‬ﻋﻠﻰ ﺃﻧﻬﺎ )‪ (M‬ﺃﻭ )‪ (P‬ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺃﺩﻧﺎﻩ‪.‬‬‫ﻳﻤﻜﻦ ﻓﺼﻞ ﺍﻟﻤﺘﺼﺎﻭﻏﺎﺕ ﺍﻻﻳﻨﺎﺗﻮﻣﻴﺮﻳﺔ ﻟﻠﻜﺮﺑﻮﻫﻴﻠﻴﺴﻴﻨﺎﺕ ﺣﻴﺚ ‪ n‬ﺃﻛﺒﺮ ﻣﻦ ‪ 4‬ﺑﺎﺳﺘﺨﺪﺍﻡ ﻋﻤﻮﺩ ﻛﺮﻭﻣﺎﺗﻮﻏﺮﺍﻓﻴﺎ ﻛﻴﺮﺍﻟﻲ‬‫‪،‬ﻭﺍﻟﺬﻱ ﻃﻮﺭﻩ ﺍﻟﺒﺮﻭﻓﻴﺴﻮﺭ ﻳﻮﺷﻴﻮ ﺃﻭﻛﺎﻣﻮﺗﻮ‪.‬‬‫‪Foundation Prize Japan The courtesy: Photo‬‬‫‪Q8-3‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-8 Q-3‬‬‫ﺍﻟﻬﻴﻠﻴﺴﻴﻨﺎﺕ ﺍﻟﻤﺘﻌﺪﺩﺓ ﻋﺒﺎﺭﺓ ﻋﻦ ﺟﺰﻳﺌﺎﺕ ﺗﺤﺘﻮﻱ ﻋﻠﻰ ﺍﺛﻨﻴﻦ ﺃﻭ ﺃﻛﺜﺮ ﻣﻦ ﺍﻟﻬﻴﺎﻛﻞ ﺍﻟﺸﺒﻴﻬﺔ ﺑﺎﻟﻬﻴﻠﻴﺴﻴﻦ‪.‬ﺍﺫﺍ ُﺃﺧﺬﺕ ﺍﻟﻜﻴﺮﺍﻟﻴﺔ‬‫ﺍﻟﺤﻠﺰﻭﻧﻴﺔ ﺑﻌﻴﻦ ﺍﻻﻋﺘﺒﺎﺭ ‪ ،‬ﻓﺈﻥ ﺍﻟﻌﺪﻳﺪ ﻣﻦ ﺍﻟﻤﺼﺎﻭﻏﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ ﻣﻮﺟﻮﺩﺓ ﻓﻲ ﺍﻟﻬﻴﻠﻴﺴﻴﻨﺎﺕ ﺍﻟﻤﺘﻌﺪﺩﺓ‪ .‬ﻋﻠﻰ ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ‬‫‪ ،‬ﻳﺤﺘﻮﻱ ﺍﻟﻤﺮﻛﺐ ‪ E‬ﻋﻠﻰ ﺛﻼﺛﺔ ﻗﻄﻊ ﻣﻦ ﻣﺸﺎﺑﻬﺎﺕ ﺍﻟـ ]‪[5‬ﻛﺮﺑﻮﻫﻴﻠﻴﺴﻴﻦ ﻓﻲ ﺟﺰﻱﺀ ﻭﺍﺣﺪ‪.‬ﻳﺘﻢ ﺗﻮﺻﻒ ﺃﺣﺪ ﺍﻷﻳﺰﻭﻣﺮﺍﺕ‬‫ﺍﻟﻔﺮﺍﻏﻴﺔ ﺑﺄﻧﻪ )‪ (P ، P ، P‬ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺃﺩﻧﺎﻩ‪.‬‬‫‪A.3‬‬‫ﺗﺸﻜﻴﻞ ﺗﺮﻳﻤﻴﺮ ﻣﻦ ﺧﻼﻝ ﺍﻟﻨﻴﻜﻞ ﻟـﻠﻤﺮﻛﺐ ‪-2،1‬ﺛﻨﺎﺋﻲ ﺑﺮﻭﻣﻮﺑﻴﻨﺰﻳﻦ ﻳﻌﻄﻲ ﺍﻟﻤﺮﻛﺐ ﺛﻼﺛﻲ‬‫ﺍﻟﻔﻴﻨﻠﻴﻦ‪ .‬ﻋﻨﺪﻣﺎ ﻳﺘﻢ ﺗﻄﺒﻴﻖ ﻧﻔﺲ ﺍﻟﺘﻔﺎﻋﻞ ﻋﻠﻰ ﺍﻟﻤﺼﺎﻭﻍ ‪ F‬ﻣﻦ ‪ ، (P) -F‬ﻳﺘﻢ ﺍﻟﺤﺼﻮﻝ‬‫ﻋﻠﻰ ﺍﻟﻬﻴﻠﻴﺴﻴﻦ ﺍﻟﻤﺘﻌﺪﺩ ‪ .(C66 H36 ) G‬ﻓﺈﺫﺍ ﻋﻠﻤﺖ ﺃﻥ ﺍﻟﺘﺤﻮﻝ ﺍﻟﺒﻴﻨﻲ ﺑﻴﻦ ﺍﻟﻤﺘﺼﺎﻭﻏﺎﺕ‬‫ﺍﻟﻔﺮﺍﻏﻴﺔ ﻻ ﻳﺤﺪﺙ ﺃﺛﻨﺎﺀ ﺍﻟﺘﻔﺎﻋﻞ ‪ ،‬ﺣﺪﺩ ﺟﻤﻴﻊ ﺍﻟﻤﺘﺼﺎﻭﻏﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺍﻟﻤﻤﻜﻨﺔ ﻟـ ‪.(C66 H36 ) G‬‬‫ﺍﻟﻤﺘﻜﻮﻧﺔ ﻓﻲ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ‪ ،‬ﺑﺪﻭﻥ ﺗﻜﺮﺍﺭﻳﺔ ‪ .‬ﻛﻤﺮﺟﻊ ﻟﻠﺤﻞ ‪ ،‬ﻳﺠﺐ ﺭﺳﻢ ﺍﻟﻤﺘﺼﺎﻭﻏﺎﺕ ﺑﺎﻟﻜﺎﻣﻞ‬‫ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻜﻴﺮﺍﻟﻴﺔ ﺍﻟﻤﺤﺪﺩﺓ ﻛﻤﺎ ﻓﻲ ﺍﻟﻤﺜﺎﻝ ﺃﻋﻼﻩ ‪ ،‬ﻣﻊ ﺗﺴﻤﻴﺎﺕ ﻋﺪﺩﻳﺔ ؛ ﻳﺠﺐ ﺃﻥ ﻳﺘﻢ ﺳﺮﺩ‬‫ﻭﻓﻘﺎ ﻟﻨﻔﺲ ﺍﻟﺘﺮﻗﻴﻢ‪ .‬ﻋﻠﻰ‬‫ﺍﻟﻤﺘﺼﺎﻭﻏﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺍﻷﺧﺮﻯ ﺑﺄﺭﻗﺎﻡ ﺍﻟﻤﻮﻗﻊ ﻭﺗﺴﻤﻴﺎﺕ ‪ M‬ﻭ ‪P‬‬
‫ً‬‫ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ ‪ ،‬ﻳﺠﺐ ﺇﺩﺭﺍﺝ ﺍﻟﻤﺘﺼﺎﻭﻏﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺍﻷﺧﺮﻯ ﻟـﻠﻤﺮﻛﺐ ‪ E‬ﻋﻠﻰ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﻟﻲ‪:‬‬‫)‪(1, 2, 3) = (P, M, P), (P, M, M), (P, P, M), (M, M, M), (M, M, P), (M, P, P), and (M, P, M‬‬‫‪7pt‬‬

‫‪Q8-4‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-8 Q-4‬‬‫‪B Part‬‬‫ﺍﻟﺴﻮﻣﺎﻧﻴﻦ ﻋﺒﺎﺭﺓ ﻋﻦ ﻫﻴﺪﺭﻭﻛﺮﺑﻮﻥ ﻋﻠﻰ ﺷﻜﻞ ﻭﻋﺎﺀ ﺗﻢ ﺫﻛﺮﻩ ﻷﻭﻝ ﻣﺮﺓ ﻓﻲ ﺍﻟﻴﺎﺑﺎﻥ ﻓﻲ ﻋﺎﻡ ‪ .2003‬ﺍﺳﻢ ”ﺳﻮﻣﺎﻧﻴﻦ”‬‫ﻣﺸﺘﻖ ﻣﻦ ﺍﻟﻜﻠﻤﺔ ﺍﻟﺴﻨﺴﻜﺮﻳﺘﻴﺔ ﺍﻟﻬﻨﺪﻳﺔ ”ﺳﻮﻣﺎﻥ” ﻭﺍﻟﺘﻲ ﺗﻌﻨﻲ ﻋﺒﺎﺩ ﺍﻟﺸﻤﺲ‪ .‬ﺗﻢ ﺍﺻﻄﻨﺎﻉ ﺍﻟﺴﻮﻣﺎﻧﻴﻦ ﻣﻦ ﺧﻼﻝ ﺳﻠﺴﻠﺔ‬‫ﻣﻦ ﺗﻔﺎﻋﻼﺕ ﺍﻟﺘﺒﺎﺩﻝ ﺍﻟﺘﻤﺎﺛﻠﻲ )‪ (metathesis‬ﺍﻟﻤﺘﻀﻤﻨﺔ ﻓﺘﺢ ﻭﺇﻏﻼﻕ ﺍﻟﺤﻠﻘﺔ‪.‬‬‫ﺗﻔﺎﻋﻼﺕ ﺍﻟﺘﺒﺎﺩﻝ ﺍﻟﺘﻤﺎﺛﻠﻲ ﺍﻟﻤﺤﻔﺰﺓ ﺑﻮﺍﺳﻄﺔ ﺣﻔﺎﺯ ﺍﻟﺮﻭﺛﻴﻨﻴﻮﻡ )* ‪ (Ru‬ﻣﻮﺿﺤﺔ ﺃﺩﻧﺎﻩ‪.‬‬‫‪B.1‬‬‫ﺍﺭﺳﻢ ﺍﻟﺒﻨﻴﺔ ﺍﻟﻬﻴﻜﻠﻴﺔ ﻟﻠﻤﺮﻛﺐ ﺍﻟﻮﺳﻴﻄﻲ ‪) I‬ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺍﻟﺨﺎﺻﺔ ﺑﻪ ﻏﻴﺮ ﻣﻄﻠﻮﺑﺔ(‪.‬‬‫‪3pt‬‬

‫‪Q8-5‬‬‫)‪arabic (Syria‬‬‫‪B.2‬‬‫‪IChO‬‬‫‪SYR-3 C-8 Q-5‬‬‫ﺑﺪﺀﺍ ﻣﻦ ﺍﻟﻄﻠﻴﻌﺔ ﺍﻟﻔﻌﺎﻟﺔ ﺿﻮﺋﻴﺎً ‪ ، J‬ﺍﺳﺘﺨﺪﻣﺖ ﻧﻔﺲ ﺳﻠﺴﻠﺔ ﺍﻟﺘﻔﺎﻋﻼﺕ ﻟﺘﺤﻀﻴﺮ ﻣﺸﺘﻖ‬‫ً‬‫ﺍﻟﺴﻮﻣﺎﻧﻴﻦ ﺍﻟﻔﻌﺎﻝ ﺿﻮﺋﻴﺎً ‪ .K‬ﻣﺮﺍﻛﺰ ﺍﻟﻔﻌﺎﻟﻴﺔ ﺍﻟﻔﺮﺍﻏﻴﺔ ﻓﻲ ‪ J‬ﻻ ﺗﻌﺎﻧﻲ ﻣﻦ ﺃﻱ ﺍﻧﻌﻜﺎﺱ‬‫ﺃﺛﻨﺎﺀ ﺗﻔﺎﻋﻞ ﺍﻟﺘﺒﺎﺩﻝ ﺍﻟﺘﻤﺎﺛﻠﻲ )‪ . (metathesis‬ﺍﺭﺳﻢ ﻫﻴﻜﻞ ‪ K‬ﻣﺴﺘﻌﻴﻨﺎً ﺑﺎﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ‬‫ﺍﻟﻤﻨﺎﺳﺒﺔ‪.‬‬‫‪4pt‬‬

IChO

A8-1

SYR-3 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
SYR-3 C-8 A-2

A.3 (7 pt)

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

Part B
B.1 (3 pt)

B.2 (4 pt)

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

SYR-3 C-9 C
ANAS ABBAS

IChO
Problem 9
Cover sheet

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

Q9-1

IChO
SYR-3 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
SYR-3 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
SYR-3 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
SYR-3 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
SYR-3 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

‫‪Q9-1‬‬‫‪IChO‬‬‫‪SYR-3 C-9 Q-1‬‬‫)‪arabic (Syria‬‬‫ﻣﺎ ﺗﺤﺒﻪ ﻭﻣﺎ ﺗﻜﺮﻫﻪ ﺍﻟﻜﺒﺴﻮﻟﺔ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪13‬‬‫‪total the of % 10‬‬‫‪A.2‬‬‫‪2‬‬‫‪A.3‬‬‫‪2‬‬‫‪A.4‬‬‫‪3‬‬‫‪A.5‬‬‫‪3‬‬‫‪Total‬‬‫‪23‬‬‫‪Score‬‬‫ﻓﺈﻧﻪ ﻳﻤﻜﻨﻚ ﺃﻥ ﺗﻔﻜﻜﻬﺎ ﺇﻟﻰ ﻗﻄﻌﺘﻴﻦ ﺑﺸﻜﻞ‬‫ﺍﻟﺼﻐﺎﺭ ﺍﻟﺠﻴﺪﻭﻥ ﻻ ﻳﻔﻌﻠﻮﻥ ﻫﺬﺍ‪ ،‬ﻭﻟﻜﻦ ﺇﺫﺍ ﻓﺘﻘﺖ ﺃﻭ ﻗﻄﻌﺖ ﻛﺮﺓ ﺗﻨﺲ ‪ّ ،‬‬‫ﺣﺮﻑ ‪U.‬‬‫ﺗﻢ ﺍﺻﻄﻨﺎﻉ ﺍﻟﻤﺮﻛﺒﻴﻦ ‪ 1‬ﻭ‪ 2‬ﻛﺠﺰﻳﺌﺎﺕ ﻋﻠﻰ ﺷﻜﻞ ﺣﺮﻑ ‪ U‬ﺑﺄﺑﻌﺎﺩ ﻣﺨﺘﻠﻔﺔ‪ .‬ﻭﺟﺮﻯ ﺍﺻﻄﻨﺎﻉ ﺍﻟﻤﺮﻛﺐ‬‫ﺑﻨﺎﺀ ﻋﻠﻰ ﻫﺬﻩ ﺍﻟﻔﻜﺮﺓ‪ّ ،‬‬‫‪ 3‬ﻛﺸﺒﻴﻪ ﻟﻠﻤﺮﻛﺐ ‪ ، 1‬ﻛﻤﺎ ﺟﺮﻯ ﺗﻘﺼﻲ ﺳﻠﻮﻙ ﺍﻟﻜﺒﺴﻠﺔ ﻟﻬﺬﻩ ﺍﻟﻤﺮﻛﺒﺎﺕ‪.‬‬‫‪Q9-2‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-9 Q-2‬‬‫ﻧﻌﺮﺽ ﻓﻴﻤﺎ ﻳﻠﻲ ﻃﺮﻳﻖ ﺍﻻﺻﻄﻨﺎﻉ ﻟﻠﻤﺮﻛﺐ ‪ .2‬ﺃﻋﻄﻰ ﺍﻟﺘﺤﻠﻴﻞ ﺍﻟﻌﻨﺼﺮﻱ ﻟﻠﻤﺮﻛﺐ ‪ 9‬ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻜﺘﻠﻲ ﺍﻟﺘﺎﻟﻲ‪:‬‬‫)‪(C; 40.49%, H; 1.70%, and O; 17.98%‬‬

‫‪Q9-3‬‬‫)‪arabic (Syria‬‬‫‪A.1‬‬‫‪IChO‬‬‫‪SYR-3 C-9 Q-3‬‬‫ﺍﺭﺳﻢ ﺑﻨﻴﺔ ﺍﻟﻤﺮﻛﺒﺎﺕ ﻣﻦ ‪ 4‬ﺇﻟﻰ ‪9‬؛ ﻳﻤﻜﻦ ﺇﻫﻤﺎﻝ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ‪.‬‬‫ﺍﺳﺘﺨﺪﻡ ”‪”PMB‬‬‫ِ‬‫ﺑﺪﻻ ﻣﻦ ﺭﺳﻢ ﺍﻟﺒﻨﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ ﻟــﺰﻣﺮﺓ‪ p-methoxybenzyl‬ﻓﻲ ﺍﻟﻤﺨﻄﻂ ﺃﻋﻼﻩ‪.‬‬‫ﻛﻤﺴﺘﺒﺪﻝ‬‫ً‬‫‪13pt‬‬‫‪Q9-4‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-9 Q-4‬‬‫ﻓﻲ ﻃﻴﻒ ﺍﻟﻜﺘﻠﺔ ﻟﻠﻤﺮﻛﺐ ‪ ،1‬ﺗﻢ ﺑﻮﺿﻮﺡ ﻣﻼﺣﻈﺔ ﻗﻤﺔ ﺍﻷﻳﻮﻥ ﺍﻟﺘﻲ ﺗﺘﻔﻖ ﻣﻊ ﺩﻳﻤﻴﺮﻩ ﺃﻱ ) ‪ ،(112‬ﻓﻲ ﺣﻴﻦ ّﺃﻧﻪ ﻟﻢ ﺗﺘﻢ‬‫ﻣﻼﺣﻈﺔ ﻗﻤﺔ ﺃﻳﻮﻥ ﻟــ ) ‪ (332‬ﻓﻲ ﻃﻴﻒ ﺍﻟﻜﺘﻠﺔ ﻟﻠﻤﺮﻛﺐ ‪ .3‬ﻓﻲ ﺃﻃﻴﺎﻑ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﺍﻟﻨﻮﻭﻱ ‪ 1 H‬ﻟﻤﺤﻠﻮﻝ ﻣﻦ ) ‪،(112‬‬‫ﻭﺃﻥ ﺍﻧﺰﻳﺎﺣﻬﺎ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻛﺎﻥ‬‫ﺃﻥ ﻛﻞ ﺍﻟﺒﺮﻭﺗﻮﻧﺎﺕ ‪ NH‬ﺍﻵﺗﻴﺔ ﻣﻦ ﺍﻟﻤﺮﻛﺐ ‪ 1‬ﻳﺠﺐ ﺃﻥ ﺗﻜﻮﻥ ﻣﺘﻜﺎﻓﺌﺔ ﻛﻴﻤﻴﺎﺋﻴﺎً ‪،‬‬‫ّ‬‫ﺗﻢ ﻣﻼﺣﻈﺔ ّ‬‫ّ‬‫ﻣﺨﺘﻠﻔﺎً ﺑﺸﻜﻞ ﻣﻌﺘﺒﺮ ﻋﻦ ﺗﻠﻚ ﺍﻟﻌﺎﺋﺪﺓ ﻟـﺒﺮﻭﺗﻮﻧﺎﺕ ‪ NH‬ﻣﻦ ﺍﻟﻤﺮﻛﺐ ‪ .3‬ﺗﺸﻴﺮ ﻫﺬﻩ ﺍﻟﻤﻌﻄﻴﺎﺕ ﺇﻟﻰ ﺗﺸﻜﻞ ﺭﻭﺍﺑﻂ ﻫﻴﺪﺭﻭﺟﻴﻨﻴﺔ‬‫ﺑﻴﻦ ﺍﻷﺟﺰﺍﺀ ‪ NH‬ﻓﻲ ﺍﻟﻤﺮﻛﺐ ‪ 1‬ﻭﺍﻟﺬﺭﺍﺕ ‪ X‬ﻣﻦ ﺟﺰﻳﺌﺔ ﺃﺧﺮﻯ ﻣﻦ ﺍﻟﻤﺮﻛﺐ ‪ 1‬ﻟﺘﺸﻜﻴﻞ ﻛﺒﺴﻮﻟﺔ ﺍﻟﺪﻳﻤﻴﺮ‪.‬‬‫‪A.2‬‬‫ﺃﺣﻂ ﻛﻞ ﺍﻟﺬﺭﺓ )ﺍﻟﺬﺭﺍﺕ( ‪ X‬ﺍﻟﻤﻨﺎﺳﺒﺔ ﻓﻲ ﺍﻟﻤﺮﻛﺐ ‪.1‬‬‫‪2pt‬‬‫‪A.3‬‬‫ﺃﻋﻂ ﻋﺪﺩ ﺍﻟﺮﻭﺍﺑﻂ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻨﻴﺔ ﻓﻲ ﻛﺒﺴﻮﻟﺔ ﺍﻟﺪﻳﻤﻴﺮ ) ‪.(112‬‬‫‪2pt‬‬

‫‪Q9-5‬‬‫)‪arabic (Syria‬‬‫‪IChO‬‬‫‪SYR-3 C-9 Q-5‬‬‫ﺗﺘﻤﺘﻊ ﺍﻟﻜﺒﺴﻮﻟﺔ ﺍﻟﺪﻳﻤﻴﺮﻳﺔ ) ‪ (112‬ﻟﻠﻤﺮﻛﺐ ‪ 1‬ﺑﻔﺮﺍﻍ ﺩﺍﺧﻠﻲ ‪ ، space internal‬ﻳﻤﻜﻦ ﺃﻥ ﺗﺘﻢ ﻓﻴﻪ ﻛﺒﺴﻠﺔ ﺟﺰﻳﺌﺔ ﺻﻐﻴﺮﺓ‬‫ﻣﻨﺎﺳﺒﺔ ‪ . Z‬ﻭﻳﻤﻜﻦ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻦ ﻫﺬﻩ ﺍﻟﻈﺎﻫﺮﺓ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬‫ﻭﻳﻌﻄﻰ ﺛﺎﺑﺖ ﺍﻟﺘﻮﺍﺯﻥ ﻟﻜﺒﺴﻠﺔ ‪ Z‬ﻓﻲ ‪ 1 2‬ﻛﻤﺎ ﻳﻠﻲ‪:‬‬‫ﻳﻤﻜﻦ ﺗﺘﺒﻊ ﻛﺒﺴﻠﺔ ﺟﺰﻳﺌﺔ ﺿﻤﻦ ﻛﺒﺴﻮﻟﺔ ﺑﺎﺳﺘﺨﺪﺍﻡ ﻣﻄﻴﺎﻓﻴﺔ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﺍﻟﻨﻮﻭﻱ ‪ NMR.‬ﻓﻌﻠﻰ ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ‬‫ﺃﻋﻄﻰ ‪ 1 2‬ﻓﻲ ‪ C6 D6‬ﺇﺷﺎﺭﺍﺕ ﻣﺨﺘﻠﻔﺔ ﻓﻲ ﺃﻃﻴﺎﻑ ‪ 1 H NMR‬ﻗﺒﻞ ﻭﺑﻌﺪ ﺇﺿﺎﻓﺔ ‪.CH4‬‬‫ﺗﻢ ﻗﻴﺎﺱ ﻃﻴﻒ ‪ 1 H NMR‬ﻟﻠـﻜﺒﺴﻮﻟﺔ ﺍﻟﺪﻳﻤﻴﺮﻳﺔ ‪ 2 2‬ﻓﻲ‬‫ﻳﺸﻜﻞ ﺍﻟﻤﺮﻛﺐ ‪ 2‬ﺃﻳﻀﺎً ﻛﺒﺴﻮﻟﺔ ﺩﻳﻤﻴﺮﻳﺔ ﻗﺎﺳﻴﺔ ﻭﻛﺒﻴﺮﺓ ) ‪ّ .(222‬‬‫‪ ، C6 D5 F ،C6 D6‬ﻭﻓﻲ ﻣﺰﻳﺞ ﻣﺬﻳﺒﺎﺕ ‪ ،C6 D6 / C6 D5 F‬ﻣﻊ ﺍﻟﻤﺤﺎﻓﻈﺔ ﻋﻠﻰ ﻛﻞ ﺍﻟﺸﺮﻭﻁ ﺍﻷﺧﺮﻯ ﺛﺎﺑﺘﺔ‪ .‬ﻭﻧﻠﺨﺺ ﻓﻴﻤﺎ ﻳﻠﻲ‬‫ﺍﻻﻧﺰﻳﺎﺣﺎﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﺒﺮﻭﺗﻮﻥ ‪ Ha‬ﻟﻠﻤﺮﻛﺐ ‪ 2‬ﻓﻲ ﺍﻟﻤﺬﻳﺒﺎﺕ ﺍﻟﻤﺬﻛﻮﺭﺓ ﺃﻋﻼﻩ‪ ،‬ﻭﻟﻢ ﺗﺘﻢ ﻣﻼﺣﻈﺔ ﺇﺷﺎﺭﺍﺕ ﺃﺧﺮﻯ ﺍﻧﻄﻼﻗﺎً ﻣﻦ‬‫‪a‬‬‫ﺃﻥ ﺩﺍﺧﻞ ﺍﻟﻜﺒﺴﻮﻟﺔ ﻣﻤﺘﻠﻰﺀ ﺩﺍﺋﻤﺎً ﻣﻊ ﺃﻛﺒﺮ ﻋﺪﺩ ﻣﻤﻜﻦ ﻣﻦ ﺟﺰﻳﺌﺎﺕ‬‫‪ H‬ﻓﻲ ‪ ،2‬ﺑﺎﺳﺘﺜﻨﺎﺀ ﺍﻟﻤﺬﻛﻮﺭﺓ ﻓﻲ ﺍﻟﻘﺎﺋﻤﺔ‪ .‬ﺍﻓﺘﺮﺽ ّ‬‫ﻭﺃﻥ ﻛﻞ ﺇﺷﺎﺭﺓ ﺗﻌﻮﺩ ﻟﻨﻮﻉ ﺃﻭ ﺻﻨﻒ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﻜﺒﺴﻮﻟﺔ ﺍﻟﻤﻤﺘﻠﺌﺔ‪.‬‬‫ﺍﻟﻤﺬﻳﺐ‪،‬‬‫ّ‬‫ﻣﺬﻳﺐ‬‫𝛿 )‪Ha of (ppm‬‬‫‪C6 D5 F‬‬‫‪4.71‬‬‫‪C6 D6‬‬‫‪C6 D5 F / C6 D6‬‬‫‪A.4‬‬‫‪4.60‬‬‫‪4.60, 4.71, 4.82‬‬‫ﺣﺪﺩ ﻋﺪﺩ ﺟﺰﻳﺌﺎﺕ ‪ C6 D6‬ﻭ ‪ C6 D5 F‬ﺍﻟﻤﺘﻜﺒﺴﻠﺔ ﻓﻲ ‪ 2 2‬ﻣﻌﻄﻴﺎً ﻛﻞ ﺇﺷﺎﺭﺓ ‪.Ha‬‬‫‪3pt‬‬

‫‪Q9-6‬‬‫‪IChO‬‬‫‪SYR-3 C-9 Q-6‬‬‫)‪arabic (Syria‬‬‫‪1‬‬‫ﺃﻥ ‪ 2 2‬ﻳﻤﻜﻦ ﺃﻥ ﻳﺘﻀﻤﻦ ﺟﺰﻳﺌﺔ ﻭﺍﺣﺪﺓ ﻣﻦ ﺍﻟﺤﻤﺾ )‪1-adamantanecarboxylic (AdA‬‬‫ﻛﺸﻔﺖ ﻗﻴﺎﺳﺎﺕ ‪ H NMR‬ﻓﻲ ‪ّ C6 D6‬‬
‫ﺗﻢ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻨﻬﺎ ﻓﻴﻤﺎ ﻳﻠﻲ‪ ،‬ﻋﻨﺪ ﺩﺭﺟﺎﺕ ﺣﺮﺍﺭﺓ ﻣﺨﺘﻠﻔﺔ‪ .‬ﺣﻴﺚ ﺗﺸﻴﺮ‬‫‪ ، acid‬ﻭﺟﺮﻯ ﺗﺤﺪﻳﺪ ﺍﻟﺜﻮﺍﺑﺖ ﺍﻟﻤﻮﺍﻓﻘﺔ ) ‪ ،(𝐾a‬ﺍﻟﺘﻲ ّ‬‫] ‪[solvent@222‬ﺇﻟﻰ ﺍﻧﻮﺍﻉ ﺣﺎﻭﻳﺔ ﻋﻠﻰ ﺟﺰﻳﺌﺔ ﻣﺬﻳﺐ ﻭﺍﺣﺪﺓ ﺃﻭ ﺃﻛﺜﺮ‬‫ﺗﻢ ﺗﺤﺪﻳﺪ ﻗﻴﻢ ‪ 𝐾a‬ﻟﻠــ ‪CH4‬ﻭﺍﻟـــ ‪ ،112‬ﺍﻟﻤﻌﻄﺎﺓ ﻛﻤﺎ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ )‪ (2‬ﻋﻨﺪ ﺩﺭﺟﺎﺕ ﺣﺮﺍﺭﺓ ﻣﺤﺘﻠﻔﺔ ﻓﻲ ‪C6 D6‬‬
‫ﻭﺑﺸﻜﻞ ﻣﺸﺎﺑﻪ‪ّ ،‬‬‫‪1‬‬‫‪ ،‬ﻋﺒﺮ ﻗﻴﺎﺳﺎﺕ ‪ . H NMR‬ﻭﻧﻌﺮﺽ ﻓﻴﻤﺎ ﻳﻠﻲ ﺭﺳﻢ ﺍﻟﺨﻂ ﺍﻟﺒﻴﺎﻧﻲ ﻟﻠﺜﺎﺑﺘﻴﻦ ﺍﻟﻤﻮﺍﻓﻘﻴﻦ‪(ln 𝐾a vs 1/T ).‬‬

‫ﻟﻢ ﺗﺘﻜﺒﺴﻞ ﺃﻱ ﺟﺰﻳﺌﺔ ﻣﻦ ‪ C6 D6‬ﻓﻲ ) ‪ .(112‬ﻓﻲ ﺍﻟﺨﻂ ‪ ، II‬ﻳﻜﻮﻥ ﺗﻐﻴﺮ ﺍﻹﻧﺘﺮﻭﺑﻲ )𝑆‪ (Δ‬ﻫﻮ )‪ (1‬ﻭﺗﻐﻴﺮ ﺍﻹﻧﺘﺎﻟﺒﻲ )𝐻‪ (Δ‬ﻫﻮ‬‫ﺃﻥ ﺍﻟﻘﻮﺓ ﺍﻟﻤﺤﺮﻛﺔ ﻟﻠﻜﺒﺴﻠﺔ ﻓﻲ ﺍﻟﺨﻂ )‪ (II‬ﻫﻲ )‪ .(3‬ﻭﻫﻜﺬﺍ ﻳﺘﻮﺍﻓﻖ ﺍﻟﺨﻂ ‪ I‬ﻣﻊ )‪ ،(4‬ﻓﻲ ﺣﻴﻦ ﻳﺘﻮﺍﻓﻖ‬‫)‪ ،(2‬ﻣﺸﻴﺮﺍً ﺇﻟﻰ ّ‬‫ﺍﻟﺨﻂ ‪ II‬ﻣﻊ )‪.(5‬‬‫‪A.5‬‬‫ﺍﺧﺘﺮ ﺍﻟﺨﻴﺎﺭﺍﺕ ﺍﻟﺼﺤﻴﺤﺔ ﻓﻲ ﺍﻟﻔﺮﺍﻏﺎﺕ ﻣﻦ )‪ (1‬ﺇﻟﻰ )‪ (5‬ﻓﻲ ﺍﻟﻔﻘﺮﺓ ﺍﻟﺘﺎﻟﻴﺔ ﻣﻦ ‪ A‬ﻭ ‪B.‬‬
‫‪A‬‬

‫‪B‬‬

‫)‪(1‬‬‫‪positive‬‬‫‪negative‬‬‫)‪(3‬‬‫𝑆‪Δ‬‬‫𝐻‪Δ‬‬‫‪CH4 and 1 2‬‬‫‪AdA and 2 2‬‬‫)‪(2‬‬‫)‪(4‬‬‫)‪(5‬‬‫‪positive‬‬‫‪CH4 and 1 2‬‬‫‪negative‬‬‫‪AdA and 2 2‬‬‫‪3pt‬‬IChO
SYR-3 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
SYR-3 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) ∶