IRN-1 C-0 C
Seyed
IRN-1 C-0
C-1 Mohammad Hossein Barakati

IChO
General instructions
Cover sheet

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

IChO
IRN-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

IRN-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
IRN-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
IRN-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

IRN-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
IRN-1 C-0 G-6

Periodic Table

G0-6
English (Official)

IChO
IRN-1 C-0 G-7

1

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

G0-7
English (Official)‫‪G0-1‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-0 G-1‬‬‫ﺍﻟﻤﭙﯿﺎﺩ ﺑﯿﻦﺍﻟﻤﻠﻠﯽ ﺷﯿﻤﯽ ‪ 2021‬ﮊﺍﭘﻦ‬‫ﺩﻭﺭﻩ ﭘﻨﺠﺎﻩﻭﺳﻮﻡ‪ IChO2021 ،‬ﮊﺍﭘﻦ‬‫‪ 25‬ﺟﻮﻻﯼ ﺗﺎ ‪ 2‬ﺁﮔﻮﺳﺖ ‪2021‬‬‫‪https://www.icho2021.org‬‬‫‪IChO‬‬‫‪G0-2‬‬‫‪IRN-1 C-0 G-2‬‬‫)‪persian (Iran‬‬‫ﺩﺳﺘﻮﺭﺍﻟﻌﻤﻞ ﻋﻤﻮﻣﯽ‬‫• ﺑﺮﺍﯼ ﻧﻮﺷﺘﻦ ﭘﺎﺳﺦﻫﺎ ﻓﻘﻂ ﻣﺠﺎﺯﯾﺪ ﺍﺯ ﺧﻮﺩﮐﺎﺭ ﺍﺳﺘﻔﺎﺩﻩ ﮐﻨﯿﺪ‪.‬‬‫• ﻣﺎﺷﯿﻦﺣﺴﺎﺏ ﺷﻤﺎ ﺑﺎﯾﺪ ﺍﺯ ﻧﻮﻉ ﻏﯿﺮ ﻗﺎﺑﻞ ﺑﺮﻧﺎﻣﻪﺭﯾﺰﯼ ﺑﺎﺷﺪ‪.‬‬‫• ﺍﯾﻦ ﺁﺯﻣﻮﻥ ﺷﺎﻣﻞ ‪ ۹‬ﻣﺴﺎﻟﻪ ﺍﺳﺖ‪.‬‬‫• ﻣﯽﺗﻮﺍﻧﯿﺪ ﻣﺴﺎﺋﻞ ﺭﺍ ﺑﺎ ﻫﺮ ﺗﺮﺗﯿﺒﯽ ﺣﻞ ﮐﻨﯿﺪ‪.‬‬‫• ﺑﺮﺍﯼ ﺣﻞ ﻣﺴﺎﺋﻞ ‪ ۵‬ﺳﺎﻋﺖ ﺯﻣﺎﻥ ﺩﺍﺭﯾﺪ‪.‬‬‫• ﭘﺲ ﺍﺯ ﺩﺍﺩﻩ ﺷﺪﻥ ﺩﺳﺘﻮﺭ ﺷﺮﻭﻉ ﻣﯽﺗﻮﺍﻧﯿﺪ ﺷﺮﻭﻉ ﺑﻪ ﮐﺎﺭ ﮐﻨﯿﺪ‪.‬‬‫• ﺗﻤﺎﻡ ﻧﺘﺎﯾﺞ ﺑﺎﯾﺪ ﺑﺎ ﺧﻮﺩﮐﺎﺭ ﺩﺭ ﮐﺎﺩﺭﻫﺎﯼ ﻣﺸﺨﺺﺷﺪﻩ ﺩﺭ ﺑﺮﮔﻪﻫﺎﯼ ﭘﺎﺳﺦﻧﺎﻣﻪ ﻧﻮﺷﺘﻪ ﺷﻮﺩ‪ .‬ﻣﯽﺗﻮﺍﻧﯿﺪ ﺍﺯ ﭘﺸﺖ ﺑﺮﮔﻪﻫﺎﯼ ﺳﺆﺍﻝ ﺑﻪﻋﻨﻮﺍﻥ‬‫ﭼﺮﮐﻨﻮﯾﺲ ﺍﺳﺘﻔﺎﺩﻩ ﮐﻨﯿﺪ‪ .‬ﺗﻮﺟﻪ ﮐﻨﯿﺪ ﮐﻪ ﺑﻪ ﺟﻮﺍﺏﻫﺎﯼ ﻧﻮﺷﺘﻪﺷﺪﻩ ﺩﺭ ﺧﺎﺭﺝ ﺍﺯ ﮐﺎﺩﺭﻫﺎﯼ ﻣﺸﺨﺺ‪ ،‬ﻧﻤﺮﻩﺍﯼ ﺗﻌﻠﻖ ﻧﻤﯽﮔﯿﺮﺩ‪.‬‬‫ﭘﺎﺳﺦ ﺩﺭﺳﺖ ﺑﺎﯾﺪ‬‫• ﻫﺮ ﺟﺎ ﮐﻪ ﻻﺯﻡ ﺍﺳﺖ ﻣﺤﺎﺳﺒﺎﺕ ﻣﺮﺑﻮﻁ ﺑﻪ ﭘﺎﺳﺦﻫﺎ ﺭﺍ ﻫﻢ ﺩﺭ ﮐﺎﺩﺭﻫﺎ ﺑﻨﻮﯾﺴﯿﺪ‪ .‬ﺑﺮﺍﯼ ﮔﺮﻓﺘﻦ ﻧﻤﺮﻩ ﮐﺎﻣﻞ ﻋﻼﻭﻩ ﺑﺮ ﻧﻮﺷﺘﻦ‬‫ِ‬‫ﻧﺸﺎﻥ ﺩﻫﯿﺪ ﮐﻪ ﺭﻭﯼ ﺳﺆﺍﻝ ﮐﺎﺭ ﮐﺮﺩﻩﺍﯾﺪ‪.‬‬‫ﺩﺳﺘﻮﺭ ﺗﻮﻗﻒ‪ ،‬ﺍﺧﻄﺎﺭ ﺧﻮﺍﻫﺪ ﺩﺍﺩ‪.‬‬‫• ﻣﺮﺍﻗﺐ ﺁﺯﻣﻮﻥ ‪ ۳۰‬ﺩﻗﯿﻘﻪ ﻗﺒﻞ ﺍﺯ‬‫ِ‬‫• ﻭﻗﺘﯽ ﺩﺳﺘﻮﺭ ﺗﻮﻗﻒ ﺩﺍﺩﻩ ﺷﺪﻩ ﺑﺎﯾﺪ ﮐﺎﺭ ﺭﺍ ﻣﺘﻮﻗﻒ ﮐﻨﯿﺪ‪ .‬ﺍﮔﺮ ﺍﺯ ﻧﻮﺷﺘﻦ ﺩﺳﺖ ﺑﺮﻧﺪﺍﺭﯾﺪ ﺑﺮﺍﯼ ﺁﺯﻣﻮﻥ ﺷﻤﺎ ﻧﻤﺮﻩ ﺻﻔﺮ ﻣﻨﻈﻮﺭ ﺧﻮﺍﻫﺪ ﺷﺪ‪.‬‬‫• ﻧﺴﺨﻪ ﺭﺳﻤﯽ ﺩﻓﺘﺮﭼﻪ ﺁﺯﻣﻮﻥ ﺑﻪ ﺯﺑﺎﻥ ﺍﻧﮕﻠﯿﺲ ﺑﻨﺎ ﺑﻪ ﺩﺭﺧﻮﺍﺳﺖ ﺷﻤﺎ ﻭ ﻓﻘﻂ ﺑﺮﺍﯼ ﺭﻓﻊ ﺍﺑﻬﺎﻡ ﺩﺭ ﺍﺧﺘﯿﺎﺭﺗﺎﻥ ﻗﺮﺍﺭ ﺧﻮﺍﻫﺪ ﮔﺮﻓﺖ‪.‬‬‫• ﺷﻤﺎ ﻣﺠﺎﺯ ﻧﯿﺴﺘﯿﺪ ﺑﺪﻭﻥ ﮔﺮﻓﺘﻦ ﺍﺟﺎﺯﻩ ﺻﻨﺪﻟﯽ ﺧﻮﺩ ﺭﺍ ﺗﺮﮎ ﮐﻨﯿﺪ‪ .‬ﺍﮔﺮ ﺑﻪ ﮐﻤﮏ ﻧﯿﺎﺯ ﺩﺍﺷﺘﯿﺪ )ﻣﺸﮑﻞ ﻣﺎﺷﯿﻦ ﺣﺴﺎﺏ‪ ،‬ﻧﯿﺎﺯ ﺑﻪ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺳﺮﻭﯾﺲ‬‫ﺑﻬﺪﺍﺷﺘﯽ‪ ،(... ،‬ﺩﺳﺖ ﺧﻮﺩ ﺭﺍ ﺑﺎﻻ ﺑﺒﺮﯾﺪ ﻭ ﻣﻨﺘﻈﺮ ﻣﺮﺍﻗﺐ ﺑﻤﺎﻧﯿﺪ‪.‬‬‫ﻣﻮﻓﻖ ﺑﺎﺷﯿﺪ!‬‫ﻣﺴﺎﺋﻞ ﻭ ﺑﺎﺭﻡﻫﺎ‬‫ﺷﻤﺎﺭﻩ ﻣﺴﺄﻟﻪ‬
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‫‪9‬‬

‫ﺟﻤﻊ ﮐﻞ‬‫ﻋﻨﻮﺍﻥ‬‫ﻧﻤﺮﻩ ﮐﻞ ﺳﺆﺍﻝ‬‫ﺩﺭﺻﺪ ﺩﺭ ﻧﻤﺮﻩ ﮐﻞ‬‫‪35‬‬‫‪11‬‬‫ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ﺳﻄﺢ ﯾﮏ ﻓﻠﺰ‬‫‪24‬‬‫ﻗﺎﻧﻮﻥ ﻻﻣﺒﺮﺕ‪ -‬ﺑﯿﺮ؟‬‫‪22‬‬‫ﮐﭙﺴﻮﻝ ﺯﻣﺎﻥ ﺍﯾﺰﻭﺗﻮﭘﯽ‬‫ﺷﯿﻤﯽ ﺍﮐﺴﺎﯾﺶ‪ -‬ﮐﺎﻫﺶ ﺭﻭﯼ‬‫ﺳﯿﻠﯿﺴﯿﻢ ﻣﺮﻣﻮﺯ‬‫ﺷﯿﻤﯽ ﺣﺎﻟﺖ ﺟﺎﻣﺪ ﻓﻠﺰﺍﺕ ﻭﺍﺳﻄﻪ‬‫ﺑﺎﺯﯼ ﺑﺎ ﺁﺭﻭﻣﺎﺗﯿﺴﯿﺘﻪ ﻏﯿﺮ ﺑﻨﺰﻭﺋﯿﺪﯼ‬‫ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ﺁﻟﯽ ﺩﯾﻨﺎﻣﯿﮏ ﻭ ﮐﺎﯾﺮﺍﻟﯿﻪ ﺁﻥﻫﺎ‬‫ﭼﯿﺰﻫﺎﯾﯽ ﮐﻪ ﮐﭙﺴﻮﻝ ﺩﻭﺳﺖ ﺩﺍﺭﺩ ﻭ ﭼﯿﺰﻫﺎﯾﯽ ﮐﻪ ﺩﻭﺳﺖ ﻧﺪﺍﺭﺩ‬‫‪32‬‬‫‪60‬‬‫‪45‬‬‫‪36‬‬‫‪26‬‬‫‪23‬‬‫‪11‬‬‫‪8‬‬‫‪11‬‬‫‪12‬‬‫‪13‬‬‫‪13‬‬‫‪11‬‬‫‪10‬‬‫‪100‬‬

‫‪G0-3‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-0 G-3‬‬‫ﺛﺎﺑﺖﻫﺎﯼ ﻓﯿﺰﯾﮑﯽ ﻭ ﻣﻌﺎﺩﻟﻪﻫﺎ‬‫ﺙﺍﺏﺕﻩﺍ‬‫ﺳﺮﻋﺖ ﻧﻮﺭ ﺩﺭ ﺧﻸ‬‫‪𝑐 = 2.99792458 × 108 m s−1‬‬‫ﺑﺎﺭ ﺑﻨﯿﺎﺩﯼ‬‫‪𝑒 = 1.602176634 × 10−19 C‬‬‫ﺛﺎﺑﺖ ﭘﻼﻧﮏ‬‫ﺟﺮﻡ ﺍﻟﮑﺘﺮﻭﻥ‬‫ﺛﺎﺑﺖ ﺍﻟﮑﺘﺮﯾﮏ )ﮔﺬﺭﺩﻫﯽ ﺧﻸ(‬‫ﺛﺎﺑﺖ ﺁﻭﻭﮔﺎﺩﺭﻭ‬‫ﺛﺎﺑﺖ ﺑﻮﻟﺘﺴﻤﺎﻥ‬‫ﺛﺎﺑﺖ ﻓﺎﺭﺍﺩﯼ‬‫ﺛﺎﺑﺖ ﮔﺎﺯﻫﺎ‬‫ﻭﺍﺣﺪ ﺟﺮﻡ ﺍﺗﻤﯽ‬‫ﻓﺸﺎﺭ ﺍﺳﺘﺎﻧﺪﺍﺭﺩ‬‫ﻓﺸﺎﺭ ﺍﺗﻤﺴﻔﺮﯼ‬‫ﺻﻔﺮ ﺩﺭﺟﻪ ﺳﻠﺴﯿﻮﺱ‬‫ﺁﻧﮕﺴﺘﺮﻭﻡ‬‫ﭘﯿﮑﻮﻣﺘﺮ‬‫ﺍﻟﮑﺘﺮﻭﻥ ﻭﻟﺖ‬‫ﺟﺰﺀ ﺩﺭ ﻣﯿﻠﯿﻮﻥ‬‫ﺟﺰﺀ ﺩﺭ ﺑﯿﻠﯿﻮﻥ‬‫ﺟﺰﺀ ﺩﺭ ﺗﺮﯾﻠﯿﻮﻥ‬‫ﻋﺪﺩ ﭘﯽ‬‫ﭘﺎﯾﻪ ﻟﮕﺎﺭﯾﺘﻢ ﻃﺒﯿﻌﯽ )ﻋﺪﺩ ﺍﻭﯾﻠﺮ(‬‫‪ℎ = 6.62607015 × 10−34 J s‬‬‫‪𝑚e = 9.10938370 × 10−31 kg‬‬‫‪𝜀0 = 8.85418781 × 10−12 F m−1‬‬‫‪mol‬‬‫‪−1‬‬‫‪−1‬‬‫‪−1‬‬‫‪JK‬‬‫‪−23‬‬‫‪𝑁A = 6.02214076 × 10‬‬‫‪𝑘B = 1.380649 × 10‬‬‫‪𝐹 = 𝑁A × 𝑒 = 9.64853321233100184 × 10 C mol‬‬‫‪4‬‬‫‪mol‬‬‫‪−1‬‬‫‪23‬‬‫‪mol−1‬‬‫‪−1‬‬‫‪−1‬‬‫‪𝑅 = 𝑁A × 𝑘B = 8.31446261815324 J K‬‬‫‪L atm K‬‬‫‪−2‬‬‫‪= 8.2057366081 × 10‬‬‫‪𝑢 = 1 Da = 1.66053907 × 10−27 kg‬‬‫‪𝑝 = 1 bar = 10 Pa‬‬‫‪5‬‬‫‪𝑝atm = 1.01325 × 10 Pa‬‬‫‪5‬‬‫‪0 C = 273.15 K‬‬
‫∘‬

‫‪m‬‬
‫‪m‬‬

‫‪J‬‬

‫‪−10‬‬‫‪−12‬‬‫‪−19‬‬‫‪−6‬‬‫‪−9‬‬‫‪1 Å = 10‬‬‫‪1 pm = 10‬‬‫‪1 eV = 1.602176634 × 10‬‬‫‪1 ppm = 10‬‬‫‪1 ppb = 10‬‬‫‪−12‬‬‫‪1 ppt = 10‬‬‫‪𝜋 = 3.141592653589793‬‬‫‪𝑒 = 2.718281828459045‬‬

‫‪IChO‬‬‫‪G0-4‬‬‫‪IRN-1 C-0 G-4‬‬‫)‪persian (Iran‬‬‫ﻡﻉﺍﺩﻝﻩﻩﺍ‬‫ﻗﺎﻧﻮﻥ ﮔﺎﺯ ﺍﯾﺪﻩﺍﻝ‬‫ﻗﺎﻧﻮﻥ ﮐﻮﻟﻦ‬‫ﻗﺎﻧﻮﻥ ﺍﻭﻝ ﺗﺮﻣﻮﺩﯾﻨﺎﻣﯿﮏ‬‫ﺁﻧﺘﺎﻟﭙﯽ ‪H‬‬‫ﺁﻧﺘﺮﻭﭘﯽ ﺑﺮ ﻣﺒﻨﺎﯼ ﺍﺻﻞ ﺑﻮﻟﺘﺴﻤﺎﻥ‬‫‪S‬‬‫ﺗﻐﯿﯿﺮ ﺁﻧﺘﺮﻭﭘﯽ 𝑆‪Δ‬‬‫ﺍﻧﺮﮊﯼ ﺁﺯﺍﺩ ﮔﯿﺒﺲ ‪G‬‬
‫ﺧﺎﺭﺝ ﻗﺴﻤﺖ ﻭﺍﮐﻨﺶ ‪Q‬‬

‫𝑇𝑅𝑛 = 𝑉 𝑃‬‫ﮐﻪ ‪ P‬ﻓﺸﺎﺭ‪ V ،‬ﺣﺠﻢ‪ n ،‬ﻣﻘﺪﺍﺭ ﻣﺎﺩﻩ ﻭ ‪ T‬ﺩﻣﺎﯼ ﻣﻄﻠﻖ ﮔﺎﺯ ﺍﯾﺪﻩﺍﻝ ﺍﺳﺖ‪.‬‬‫𝑞 𝑞‬‫‪𝐹 = 𝑘e 1 2 2‬‬‫𝑟‬‫ﮐﻪ ‪ F‬ﻧﯿﺮﻭﯼ ﺍﻟﮑﺘﺮﻭﺍﺳﺘﺎﺗﯿﮑﯽ‪ 𝑘e (≃ 9.0 × 109 N m2 C−2 ) ،‬ﺛﺎﺑﺖ ﮐﻮﻟﻦ‪ 𝑞1 ،‬ﻭ ‪ 𝑞2‬ﻣﻘﺎﺩﯾﺮ ﺑﺎﺭﻫﺎ ﻭ ‪r‬‬‫ﻓﺎﺻﻠﻪ ﺑﯿﻦ ﺑﺎﺭﻫﺎﺳﺖ‪.‬‬‫𝑤 ‪Δ𝑈 = 𝑞 +‬‬‫ﮐﻪ 𝑈‪ Δ‬ﺗﻐﯿﯿﺮ ﺍﻧﺮﮊﯼ ﺩﺭﻭﻧﯽ‪ 𝑞 ،‬ﮔﺮﻣﺎﯼ ﺩﺍﺩﻩﺷﺪﻩ ﻭ 𝑤 ﮐﺎﺭ ﺍﻧﺠﺎﻡﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫𝑉𝑃 ‪𝐻 = 𝑈 +‬‬‫𝑊 ‪𝑆 = 𝑘B ln‬‬‫ﮐﻪ ‪ W‬ﺗﻌﺪﺍﺩ ﺭﯾﺰﺣﺎﻟﺖﻫﺎﺳﺖ‪.‬‬‫𝑞‬‫‪Δ𝑆 = rev‬‬‫𝑇‬‫ﮐﻪ ‪ 𝑞rev‬ﮔﺮﻣﺎﯼ ﻣﺮﺑﻮﻁ ﺑﻪ ﻓﺮﺍﯾﻨﺪ ﺑﺮﮔﺸﺖﭘﺬﯾﺮ ﺍﺳﺖ‪.‬‬‫𝑆𝑇 ‪𝐺 = 𝐻 −‬‬‫∘ 𝐸 𝐹𝑧‪Δr 𝐺∘ = −𝑅𝑇 ln 𝐾 = −‬‬‫ﮐﻪ ‪ K‬ﺛﺎﺑﺖ ﺗﻌﺎﺩﻝ‪ 𝑧 ،‬ﺗﻌﺪﺍﺩ ﺍﻟﮑﺘﺮﻭﻥﻫﺎ ﻭ ∘ 𝐸 ﭘﺘﺎﻧﺴﯿﻞ ﺍﺳﺘﺎﻧﺪﺍﺭﺩ ﺍﻟﮑﺘﺮﻭﺩ ﺍﺳﺖ‪.‬‬‫𝑄 ‪Δr 𝐺 = Δr 𝐺∘ + 𝑅𝑇 ln‬‬‫ﻭﺍﮐﻨﺶ‬‫ﺑﺮﺍﯼ‬‫ِ‬‫‪𝑎A + 𝑏B ⇌ 𝑐C + 𝑑D‬‬‫𝑐‬‫𝑑‬‫]‪[C] [D‬‬
‫=𝑄‬
‫𝑎‬
‫𝑏‬
‫]‪[A] [B‬‬‫ﮐﻪ ]‪ [A‬ﻏﻠﻈﺖ ‪ A‬ﺍﺳﺖ‪.‬‬‫‪IChO‬‬‫‪G0-5‬‬‫‪IRN-1 C-0 G-5‬‬‫)‪persian (Iran‬‬‫ﺗﻐﯿﯿﺮ ﮔﺮﻣﺎ 𝑞‪Δ‬‬‫ﻣﻌﺎﺩﻟﻪ ﻧﺮﻧﺴﺖ ﺑﺮﺍﯼ ﻭﺍﮐﻨﺶﻫﺎﯼ‬‫ﺍﮐﺴﺎﯾﺶ‪ -‬ﮐﺎﻫﺶ‬‫ﻣﻌﺎﺩﻟﻪ ﺁﺭﻧﯿﻮﺱ‬‫ﻗﺎﻧﻮﻥ ﻻﻣﺒﺮﺕ‪ -‬ﺑﯿﺮ‬‫ﻣﻌﺎﺩﻟﻪ ﻫﻨﺪﺭﺳﻮﻥ‪ -‬ﻫﺴﻞﺑﺎﺥ‬‫ﺍﻧﺮﮊﯼ ﯾﮏ ﻓﻮﺗﻮﻥ‬‫𝑇‪Δ𝑞 = 𝑛𝑐m Δ‬‬‫ﻣﻮﻟﯽ ﻣﺴﺘﻘﻞ ﺍﺯ ﺩﻣﺎﺳﺖ‪.‬‬‫ﮔﺮﻣﺎﯾﯽ‬‫ﮐﻪ ‪ 𝑐m‬ﻇﺮﻓﯿﺖ‬
‫ِ‬
‫𝐶‬
‫𝑇𝑅‬
‫‪ln ox‬‬‫‪𝐸 = 𝐸∘ +‬‬‫𝐹𝑧‬‫‪𝐶red‬‬‫ﮐﻪ ‪ 𝐶ox‬ﻏﻠﻈﺖ ﻣﺎﺩﻩﯼ ﺍﮐﺴﯿﺪﺷﺪﻩ ﻭ ‪ 𝐶red‬ﻏﻠﻈﺖ ﻣﺎﺩﻩﯼ ﺍﺣﯿﺎﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫𝐸‬‫) 𝑎 ‪𝑘 = 𝐴 exp (−‬‬‫𝑇𝑅‬‫ﮐﻪ ‪ k‬ﺛﺎﺑﺖ ﺳﺮﻋﺖ‪ A ،‬ﺿﺮﯾﺐ ﭘﯿﺶﻧﻤﺎﯾﯽ ﻭ 𝑎𝐸 ﺍﻧﺮﮊﯼ ﻓﻌﺎﻟﺴﺎﺯﯼ ﺍﺳﺖ‪.‬‬‫𝑥𝑒 = )𝑥(‪exp‬‬‫𝑐𝑙𝜀 = 𝐴‬‫ﮐﻪ ‪ A‬ﻣﯿﺰﺍﻥ ﺟﺬﺏ‪ 𝜀 ،‬ﺿﺮﯾﺐ ﺟﺬﺏ ﻣﻮﻟﯽ‪ l ،‬ﻃﻮﻝ ﻣﺴﯿﺮ ﻧﻮﺭ ﻭ ‪ c‬ﻏﻠﻈﺖ ﻣﺤﻠﻮﻝ ﺍﺳﺖ‪.‬‬‫ﺗﻌﺎﺩﻝ‬‫ﺑﺮﺍﯼ ﯾﮏ‬‫ِ‬‫‪HA ⇌ H+ + A−‬‬‫ﮐﻪ ﺛﺎﺑﺖ ﺗﻌﺎﺩﻝ ﺑﺮﺍﺑﺮ ‪ 𝐾a‬ﺑﺎﺷﺪ‪،‬‬‫] ‪[A−‬‬‫)‬‫( ‪pH = p𝐾a + log‬‬‫]‪[HA‬‬‫𝑐‬‫‪𝐸 = ℎ𝜈 = ℎ‬‬‫𝜆‬‫ﮐﻪ 𝜈 ﻓﺮﮐﺎﻧﺲ ﻭ 𝜆 ﻃﻮﻝ ﻣﻮﺝ ﻧﻮﺭ ﺍﺳﺖ‪.‬‬‫ﻣﺠﻤﻮﻉ ﯾﮏ ﺳﺮﯼ ﻫﻨﺪﺳﯽ‬‫ﻭﻗﺘﯽ ‪،𝑥 ≠ 1‬‬‫ﻣﻌﺎﺩﻟﻪ ﺗﻘﺮﯾﺒﯽ ﮐﻪ ﻣﯽﺗﻮﺍﻥ ﺑﺮﺍﯼ‬‫ﺣﻞ ﻣﺴﺎﺋﻞ ﺑﻪ ﮐﺎﺭ ﺑﺮﺩ‬‫ﻭﻗﺘﯽ ﮐﻪ ‪،𝑥 ≪ 1‬‬‫‪1‬‬‫𝑥‪≃1+‬‬‫𝑥‪1−‬‬‫‪1 − 𝑥𝑛+1‬‬‫𝑛‬‫= 𝑖𝑥 ‪1 + 𝑥 + 𝑥2 + ⋯ + 𝑥𝑛 = ∑𝑖=0‬‬‫𝑥‪1−‬‬IChO
IRN-1 C-0 G-6

G0-6
persian (Iran)‫ﺟﺪﻭﻝ ﺗﻨﺎﻭﺑﯽ‬IChO
IRN-1 C-0 G-7

G0-7
persian (Iran)

NMR ‫ﺟﺎﺑﻪﺟﺎﯾﯽﻫﺎﯼ ﺷﯿﻤﯿﺎﯾﯽ ﭘﺮﻭﺗﻮﻥ‬‫‪G0-8‬‬‫)‪persian (Iran‬‬‫ً‬‫ﺗﻘﺮﯾﺒﺎ 𝑚𝑝𝑝 ‪+0.4‬‬‫𝛿‪ Δ‬ﺑﺮﺍﯼ ﯾﮏ ﺍﺳﺘﺨﻼﻑ ﮔﺮﻭﻩ ﺁﻟﮑﯿﻞ‪:‬‬‫‪IChO‬‬‫‪IRN-1 C-0 G-8‬‬

IRN-1 C-1 C
Seyed
IRN-1 C-1
C-1 Mohammad Hossein Barakati

IChO
Problem 1
Cover sheet

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

Q1-1

IChO
IRN-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
IRN-1 C-1 Q-2

Q1-2

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

Fig.2

English (Official)

Q1-3

IChO
IRN-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
IRN-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
IRN-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‫‪Q1-1‬‬‫‪IChO‬‬‫‪IRN-1 C-1 Q-1‬‬‫)‪persian (Iran‬‬‫ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ﺳﻄﺢ ﯾﮏ ﻓﻠﺰ‬‫‪11% of the total‬‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪A.2‬‬‫‪B.1‬‬‫‪B.2‬‬‫‪B.3‬‬‫‪B.4‬‬

‫‪6‬‬

‫‪4‬‬

‫‪5‬‬

‫‪3‬‬

‫‪3‬‬

‫‪3‬‬

‫‪Total‬‬‫‪24‬‬‫‪Score‬‬‫ﺍﻧﺘﻈﺎﺭ ﻣﯽﺭﻭﺩ ﮐﻪ ﺩﺭ ﺁﯾﻨﺪﻩ ﺑﺘﻮﺍﻥ ﻫﯿﺪﺭﻭﮊﻥ ﺭﺍ ﺑﻪﻋﻨﻮﺍﻥ ﯾﮏ ﻣﻨﺒﻊ ﺍﻧﺮﮊﯼ ﺍﺳﺘﻔﺎﺩﻩ ﮐﺮﺩ ﮐﻪ ﺑﻪ ﺳﻮﺧﺖﻫﺎﯼ ﻓﺴﯿﻠﯽ ﻭﺍﺑﺴﺘﮕﯽ ﻧﺪﺍﺭﺩ‪ .‬ﺩﺭ ﺍﯾﻦ ﻣﺴﺄﻟﻪ ﻓﺮﺍﯾﻨﺪ‬‫ﺫﺧﯿﺮﻩﺳﺎﺯﯼ ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ﯾﮏ ﻓﻠﺰ ﺭﺍ ﺑﺮﺭﺳﯽ ﻣﯽﮐﻨﯿﻢ ﮐﻪ ﺑﻪ ﺗﮑﻨﻮﻟﻮﮊﯼ ﺩﺧﯿﺮﻩﺳﺎﺯﯼ ﺍﻧﺮﮊﯼ ﻭ ﺣﻤﻞ ﻭ ﻧﻘﻞ ﺑﺮ ﭘﺎﯾﻪ ﻫﯿﺪﺭﻭﮊﻥ ﻣﺮﺑﻮﻁ ﺍﺳﺖ‪.‬‬‫ﻗﺴﻤﺖ ‪A‬‬
‫ﺳﻄﺢ ﻓﻠﺰﺍﺕ ﺑﻪ ﺩﺭﻭﻥ ﺁﻧﻬﺎ ﺟﺬﺏ ﻣﯽﺷﻮﺩ‪ ،‬ﺩﺭ ﺍﺑﺘﺪﺍ ﻓﺮﺍﯾﻨﺪ ﺟﺬﺏ ﺳﻄﺤﯽ ﻫﯿﺪﺭﻭﮊﻥ ﺭﻭﯼ ﺳﻄﺢ ﻓﻠﺰ‪H2 (g) → ،‬‬‫ﺍﺯ ﺁﻧﺠﺎ ﮐﻪ ﻫﯿﺪﺭﻭﮊﻥ ﺍﺯ ﻃﺮﯾﻖ‬‫ِ‬‫)‪ ،2H(ad‬ﺭﺍ ﺑﺮﺭﺳﯽ ﻣﯽﮐﻨﯿﻢ‪ .‬ﺩﺭ ﺍﯾﻦ ﺭﺍﺑﻄﻪ ﺣﺎﻟﺖﻫﺎﯼ ﮔﺎﺯﯼ ﻭ ﺟﺬﺏﺷﺪﻩﯼ ﻫﯿﺪﺭﻭﮊﻥ ﺑﻪﺗﺮﺗﯿﺐ ﺑﺎ )‪ (g‬ﻭ )‪ (ad‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﻣﻮﻟﮑﻮﻝﻫﺎﯼ‬‫ﺟﺬﺏ ﺳﻄﺢ ﻣﯽﺷﻮﻧﺪ )ﺷﮑﻞ ‪ .(1‬ﺩﺭ ﺍﯾﻨﺠﺎ ﺍﻧﺮﮊﯼ ﭘﺘﺎﻧﺴﯿﻞ‬‫ﺻﻮﺭﺕ ﺍﺗﻢﻫﺎﯼ ‪H‬‬‫ﻫﯿﺪﺭﻭﮊﻥ ) ‪ (H2‬ﮐﻪ ﺑﻪ ﺳﻄﺢ ﻓﻠﺰ )‪ (M‬ﻣﯽﺭﺳﻨﺪ‪ ،‬ﺭﻭﯼ ﺳﻄﺢ ﺗﻔﮑﯿﮏ ﻭ ﺑﻪ‬‫ِ‬‫ِ‬‫ﺳﻄﺤﯽ ﻓﻠﺰ )‪ .(z‬ﻓﺮﺽ ﻣﯽﺷﻮﺩ ﻣﺤﻮﺭﯼ ﮐﻪ ﺩﻭ ﺍﺗﻢ ‪ H‬ﺭﺍ ﺑﻪ ﻫﻢ‬‫‪ H2‬ﺑﺎ ﺩﻭ ﻣﺘﻐﯿﺮ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﻣﯽﺷﻮﺩ‪ :‬ﻓﺎﺻﻠﻪ ﺑﯿﻦﺍﺗﻤﯽ )‪ (d‬ﻭ ﺍﺭﺗﻔﺎﻉ ﻧﺴﺒﺖ ﺑﻪ ﺍﺗﻢ‬‫ِ‬‫ﻋﻤﻮﺩﯼ ﻧﻘﻄﻪﭼﯿﻦ ﺩﺭ ﺷﮑﻞ ‪ 1‬ﻗﺮﺍﺭ ﺩﺍﺭﺩ‪ .‬ﺷﮑﻞ ‪ 2‬ﻧﻤﻮﺩﺍﺭ ﮐﺎﻧﺘﻮﺭﯼ ﺍﻧﺮﮊﯼ‬‫ﻣﺘﺼﻞ ﻣﯽﮐﻨﺪ ﺑﺎ ﺳﻄﺢ ﻓﻠﺰ ﻣﻮﺍﺯﯼ ﺍﺳﺖ ﻭ ﻣﺮﮐﺰ ﮔﺮﺍﻧﺶ ﻫﻤﯿﺸﻪ ﺭﻭﯼ ﺧﻂ‬‫ِ‬‫ﻣﻮﻝ ‪ H2‬ﻧﺸﺎﻥ ﻣﯽﺩﻫﻨﺪ‪ .‬ﻓﺎﺻﻠﻪ ﺑﯿﻦ ﺧﻂﻫﺎﯼ‬‫ﭘﺘﺎﻧﺴﯿﻞ ﺭﺍ ﺑﺮﺍﯼ ﺗﻔﮑﯿﮏ ﺭﻭﯼ ﺳﻄﺢ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ‪ .‬ﻣﻘﺎﺩﯾﺮ ﻋﺪﺩﯼ ﺍﻧﺮﮊﯼ ﭘﺘﺎﻧﺴﯿﻞ ﺭﺍ ﺑﺮﺣﺴﺐ ‪ kJ‬ﺑﺮ‬‫ِ‬‫ﺗﻮﭘﺮ ‪ ،20 kJ mol−1‬ﻓﺎﺻﻠﻪ ﺑﯿﻦ ﺧﻂﻫﺎﯼ ﺑﺮﯾﺪﻩ ‪ 100 kJ mol−1‬ﻭ ﻓﺎﺻﻠﻪ ﺑﯿﻦ ﺧﻂﻫﺎﯼ ﺗﻮﭘﺮ ﻭ ﺑﺮﯾﺪﻩ ‪ 80 kJ mol−1‬ﺍﺳﺖ‪ .‬ﺍﺯ ﺍﻧﺮﮊﯼ ﻧﻘﻄﻪ ﺻﻔﺮ‬‫ﺍﺭﺗﻌﺎﺷﯽ ﺻﺮﻓﻨﻈﺮ ﻣﯽﺷﻮﺩ‪.‬‬‫‪Q1-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-1 Q-2‬‬‫ﺷﮑﻞ ‪ .1‬ﺗﻌﺮﯾﻒ ﻣﺘﻐﯿﺮﻫﺎ‪ .‬ﺷﮑﻞ ﺑﺎ ﻣﻘﯿﺎﺱﻫﺎﯼ ﻭﺍﻗﻌﯽ ﺭﺳﻢ ﻧﺸﺪﻩ ﺍﺳﺖ‪.‬‬‫ﺷﮑﻞ ‪2‬‬

‫‪Q1-3‬‬‫‪IChO‬‬‫‪IRN-1 C-1 Q-3‬‬‫)‪persian (Iran‬‬‫‪A.1‬‬‫ﺑﺮﺍﯼ ﻫﺮ ﯾﮏ ﺍﺯ ﻣﻮﺍﺭﺩ )‪ (i‬ﺗﺎ )‪ (iii‬ﺯﯾﺮ‪ ،‬ﻧﺰﺩﯾﮏﺗﺮﯾﻦ ﻣﻘﺪﺍﺭ ﺭﺍ ﺍﺯ ﺑﯿﻦ ﻣﻘﺎﺩﯾﺮ ‪ A‬ﺗﺎ ‪ G‬ﺍﻧﺘﺨﺎﺏ ﮐﻨﯿﺪ‪.‬‬‫)‪ (i‬ﻓﺎﺻﻠﻪ ﺑﯿﻦﺍﺗﻤﯽ ﺑﺮﺍﯼ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ‪ H2‬ﮔﺎﺯﯼ ﺷﮑﻞ‬‫)‪ (ii‬ﻓﺎﺻﻠﻪ ﺑﯿﻦﺍﺗﻤﯽ ﺑﯿﻦ ﺍﺗﻢﻫﺎﯼ ﻓﻠﺰ ) ‪ 𝑑M‬ﺩﺭ ﺷﮑﻞ ‪(1‬‬‫)‪ (iii‬ﻓﺎﺻﻠﻪ ﺍﺗﻢﻫﺎﯼ ‪ H‬ﺟﺬﺏﺷﺪﻩ ﺍﺯ ﺳﻄﺢ ) ‪ ℎad‬ﺩﺭ ﺷﮑﻞ ‪(1‬‬‫‪D. 0.15 nm‬‬‫‪A.2‬‬‫‪C. 0.11 nm‬‬‫‪G. 0.27 nm‬‬‫‪B. 0.07 nm‬‬‫‪F. 0.23 nm‬‬‫‪A. 0.03 nm‬‬‫‪E. 0.19 nm‬‬‫ﺑﺮﺍﯼ ﻫﺮ ﯾﮏ ﺍﺯ ﻣﻮﺍﺭﺩ )‪ (i‬ﻭ )‪ (ii‬ﺯﯾﺮ‪ ،‬ﻧﺰﺩﯾﮏﺗﺮﯾﻦ ﻣﻘﺪﺍﺭ ﺭﺍ ﺍﺯ ﺑﯿﻦ ﻣﻘﺎﺩﯾﺮ ‪ A‬ﺗﺎ ‪ H‬ﺍﻧﺘﺨﺎﺏ ﮐﻨﯿﺪ‪.‬‬‫)‪ (i‬ﺍﻧﺮﮊﯼ ﻣﻮﺭﺩ ﻧﯿﺎﺯ ﺑﺮﺍﯼ ﺗﻔﮑﯿﮏ ‪ H2‬ﮔﺎﺯﯼ ﺑﻪ ‪ H‬ﮔﺎﺯﯼ‬‫)‪H2 (g) → 2H(g‬‬‫)‪ (ii‬ﺍﻧﺮﮊﯼ ﺁﺯﺍﺩﺷﺪﻩ ﻃﯽ ﺟﺬﺏ ﺳﻄﺤﯽ ‪ H2‬ﮔﺎﺯﯼ‬‫)‪H2 (g) → 2H(ad‬‬‫‪D. 100 kJ mol−1‬‬‫‪H. 400 kJ mol−1‬‬‫‪C. 60 kJ mol−1‬‬‫‪G. 300 kJ mol−1‬‬‫‪6pt‬‬‫‪B. 40 kJ mol−1‬‬‫‪F. 200 kJ mol−1‬‬‫‪A. 20 kJ mol−1‬‬‫‪E. 150 kJ mol−1‬‬‫‪4pt‬‬

‫‪Q1-4‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-1 Q-4‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫ﻫﻤﺎﻧﻄﻮﺭ ﮐﻪ ﺩﺭ ﻭﺍﮐﻨﺶﻫﺎﯼ )‪ (1a‬ﻭ )‪ (1b‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﺍﺗﻢﻫﺎﯼ ﻫﯿﺪﺭﻭﮊﻧﯽ ﮐﻪ ﺟﺬﺏ ﺳﻄﺢ ﺷﺪﻩﺍﻧﺪ‪ ،‬ﺩﺭ ﻣﺮﺣﻠﻪ ﺑﻌﺪﯼ ﯾﺎ ﺑﻪ ﺩﺭﻭﻥ ﻓﻠﺰ ﺟﺬﺏ‬‫ﻣﯽﺷﻮﻧﺪ ﯾﺎ ﺩﻭﺑﺎﺭﻩ ﺑﺎ ﻫﻢ ﺗﺮﮐﯿﺐ ﻣﯽﺷﻮﻧﺪ ﻭ ﺑﺎ ﻭﺍﺟﺬﺏ‪ ،‬ﻭﺍﺭﺩ ﻓﺎﺯ ﮔﺎﺯﯼ ﻣﯽﺷﻮﻧﺪ‪ .‬ﺩﺭ ﺍﯾﻦ ﻭﺍﮐﻨﺶ‪ H(ab) ،‬ﺍﺗﻢ ﻫﯿﺪﺭﻭﮊﻧﯽ ﺭﺍ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ ﮐﻪ ﺑﻪ‬‫ﺩﺭﻭﻥ ﻓﻠﺰ ﺟﺬﺏ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫ﺳﺮﻋﺖ ﻭﺍﮐﻨﺶﻫﺎﯼ ﺟﺬﺏ ﺳﻄﺤﯽ )‪ ، (adsorption‬ﻭﺍﺟﺬﺏ )‪ (desorption‬ﻭ ﺟﺬﺏ )‪ (absorption‬ﺑﻪﺗﺮﺗﯿﺐ ﺑﺎ ] ‪ 𝑟2 [s−1 ] ،𝑟1 [s−1‬ﻭ ] ‪𝑟3 [s−1‬‬‫ﺑﺮﺍﺑﺮ ﺍﺳﺖ‪ .‬ﺍﯾﻦ ﺳﺮﻋﺖﻫﺎ ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﭼﻨﯿﻦ ﻧﻮﺷﺖ‪:‬‬‫ﺩﺭ ﺍﯾﻦ ﺭﻭﺍﺑﻂ ] ‪ 𝑘2 [s−1 ] ، 𝑘1 [s−1 Pa−1‬ﻭ ] ‪ 𝑘3 [s−1‬ﺛﺎﺑﺖﻫﺎﯼ ﺳﺮﻋﺖ ﻭﺍﮐﻨﺶ ﻫﺴﺘﻨﺪ ﻭ ‪ 𝑃H2‬ﻓﺸﺎﺭ ‪ H2‬ﺭﺍ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ‪ 𝜃 .‬ﮐﺴﺮﯼ ﺍﺯ ﺟﺎﯾﮕﺎﻩﻫﺎﯼ‬‫ﺩﺭ ﺩﺳﺘﺮﺱ ﺭﻭﯼ ﺳﻄﺢ ﺍﺳﺖ ﮐﻪ ﺑﺎ ﺍﺗﻢﻫﺎﯼ ‪ H‬ﺍﺷﻐﺎﻝ ﺷﺪﻩﺍﻧﺪ )‪ .(0 ≤ 𝜃 ≤ 1‬ﻓﺮﺽ ﻣﯽﺷﻮﺩ ﮐﻪ ﺟﺬﺏ ﺳﻄﺤﯽ )‪ (adsorption‬ﻭ ﻭﺍﺟﺬﺏ ﺩﺭ‬‫ﻣﻘﺎﯾﺴﻪ ﺑﺎ ﺟﺬﺏ )‪ (absorption‬ﺳﺮﯾﻊﺍﻧﺪ ) ‪ (𝑟1 , 𝑟2 ≫ 𝑟3‬ﻭ 𝜃 ﺛﺎﺑﺖ ﻣﯽﻣﺎﻧﺪ‪.‬‬‫‪B.1‬‬‫‪ 𝑟3‬ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﭼﻨﯿﻦ ﻧﻮﺷﺖ‪:‬‬‫𝐶 ﺭﺍ ﺑﺮﺣﺴﺐ ‪ 𝑘1‬ﻭ ‪ 𝑘2‬ﺑﻪﺩﺳﺖ ﺁﻭﺭﯾﺪ‪.‬‬‫‪5pt‬‬

‫‪Q1-5‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-1 Q-5‬‬‫ﻧﻤﻮﻧﻪﺍﯼ ﺍﺯ ﻓﻠﺰ ﺑﺎ ﻣﺴﺎﺣﺖ ﺳﻄﺢ ‪ 𝑆 = 1.0×10−3 m2‬ﺩﺭ ﻇﺮﻓﯽ ﺑﻪ ﺣﺠﻢ ﯾﮏ ﻟﯿﺘﺮ ) ‪ (1.0×10−3 m3‬ﺷﺎﻣﻞ ‪ H2‬ﺑﺎ ﻓﺸﺎﺭ ‪𝑃H2 = 1.0×102 Pa‬‬‫ﺳﻄﺤﯽ ﺍﺗﻢ ﻫﯿﺪﺭﻭﮊﻥ ﺭﻭﯼ ﺳﻄﺢ ﻓﻠﺰ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ‪ .𝑁 = 1.3 × 1018 m−2‬ﺩﻣﺎﯼ ﺳﻄﺢ ﺩﺭ‬‫ﻗﺮﺍﺭ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﺩﺍﻧﺴﯿﺘﻪﯼ ﺟﺎﯾﮕﺎﻩﻫﺎﯼ ﺟﺬﺏ‬‫ِ‬‫‪−4‬‬‫‪−1‬‬‫ﺛﺎﺑﺖ ‪ 𝑣 = 4.0 × 10 Pa s‬ﮐﺎﻫﺶ ﻣﯽﯾﺎﺑﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ‬‫‪ 𝑇 = 400 K‬ﺛﺎﺑﺖ ﻧﮕﻪ ﺩﺍﺷﺘﻪ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﺑﺎ ﭘﯿﺸﺮﻓﺖ ﻭﺍﮐﻨﺶ )‪ 𝑃H2 ،(1‬ﺑﺎ ﺳﺮﻋﺖ‬‫ِ‬‫‪ H2‬ﮔﺎﺯﯼ ﺍﯾﺪﻩﺍﻝ ﺍﺳﺖ ﻭ ﻣﯽﺗﻮﺍﻥ ﺍﺯ ﺣﺠﻢ ﻧﻤﻮﻧﻪ ﻓﻠﺰ ﺻﺮﻓﻨﻈﺮ ﮐﺮﺩ‪.‬‬‫‪B.2‬‬‫ﻣﻘﺪﺍﺭ ﺍﺗﻢﻫﺎﯼ ‪ H‬ﺟﺬﺏﺷﺪﻩ ﺭﺍ ﺑﺮﺣﺴﺐ ﻣﻮﻝ‪ ،‬ﺑﻪﺍﺯﺍﯼ ﻭﺍﺣﺪ ﻣﺴﺎﺣﺖ ﺳﻄﺢ ﻭ ﺑﻪﺍﺯﺍﯼ ﻭﺍﺣﺪ ﺯﻣﺎﻥ‪،‬‬‫] ‪ ،𝐴 [mol s−1 m−2‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪3pt‬‬‫‪B.3‬‬‫ﺩﺭ ﺩﻣﺎﯼ ‪ 𝐶 ، 𝑇 = 400 K‬ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ‪ . 1.0 × 102 Pa−1‬ﻣﻘﺪﺍﺭ ‪ 𝑘3‬ﺭﺍ ﺩﺭ ‪ 400‬ﮐﻠﻮﯾﻦ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ .‬ﺍﮔﺮ‬‫ﺟﻮﺍﺏ ﻗﺴﻤﺖ ‪ B.2‬ﺭﺍ ﺑﻪﺩﺳﺖ ﻧﯿﺎﻭﺭﺩﻩﺍﯾﺪ‪ ،‬ﺍﺯ ﻣﻘﺪﺍﺭ ‪ 𝐴 = 3.6 × 10−7 mol s−1 m−2‬ﺍﺳﺘﻔﺎﺩﻩ ﮐﻨﯿﺪ‪.‬‬‫ِ‬‫‪3pt‬‬‫‪B.4‬‬‫ﺩﺭ ﯾﮏ ﺩﻣﺎﯼ ‪ T‬ﻣﺘﻔﺎﻭﺕ‪ ،‬ﻣﻘﺎﺩﯾﺮ ‪ 𝐶 = 2.5 × 103 Pa−1‬ﻭ ‪ 𝑘3 = 4.8 × 10−2 s−1‬ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﺍﺯ ﺑﯿﻦ‬‫ﻣﻨﺤﻨﯽﻫﺎﯼ )‪ (a‬ﺗﺎ )‪ (h‬ﺩﺭ ﺩﻭ ﻧﻤﻮﺩﺍﺭ ﺯﯾﺮ‪ ،‬ﻣﻨﺤﻨﯽ ﺩﺭﺳﺖ ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﮐﻨﯿﺪ ﮐﻪ ‪ 𝑟3‬ﺭﺍ ﺑﻪﺻﻮﺭﺕ ﺗﺎﺑﻌﯽ ﺍﺯ ‪ 𝑃H2‬ﺩﺭ ﺍﯾﻦ‬‫ﺩﻣﺎ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ‪.‬‬‫‪3pt‬‬

IChO

A1-1

IRN-1 C-1 A-1

persian (Iran)‫ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ﺳﻄﺢ ﯾﮏ ﻓﻠﺰ‬A ‫ﻗﺴﻤﺖ‬(6 pt) A.1
(iii)

(ii)

(i)

(4 pt) A.2
(ii)

(i)

IChO

A1-2

IRN-1 C-1 A-2

persian (Iran)B ‫ﻗﺴﻤﺖ‬(5 pt) B.1𝐶=(3 pt) B.2

𝐴=

mol s−1 m−2

IChO

A1-3

IRN-1 C-1 A-3

persian (Iran)(3 pt) B.3

𝑘3 =

s−1
(3 pt) B.4

IRN-1 C-2 C
Seyed
IRN-1 C-2
C-1 Mohammad Hossein Barakati

IChO
Problem 2
Cover sheet

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

Q2-1

IChO
IRN-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
IRN-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
IRN-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
IRN-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

‫‪Q2-1‬‬‫‪IChO‬‬‫‪IRN-1 C-2 Q-1‬‬‫)‪persian (Iran‬‬‫ﮐﭙﺴﻮﻝ ﺯﻣﺎﻥ ﺍﯾﺰﻭﺗﻮﭘﯽ‬‫‪11% of the total‬‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪8‬‬‫‪A.2‬‬‫‪8‬‬‫‪A.3‬‬‫‪10‬‬‫‪A.4‬‬‫‪9‬‬‫‪Total‬‬‫‪35‬‬‫‪Score‬‬‫ﻣﻮﻟﮑﻮﻝﻫﺎﯾﯽ ﻣﺎﻧﻨﺪ ‪ CH4‬ﻭ ‪ CH3 D‬ﺭﺍ ﮐﻪ ﻓﻘﻂ ﺩﺭ ﺗﺮﮐﯿﺐ ﺍﯾﺰﻭﺗﻮﭘﯽ ﺑﺎ ﻫﻢ ﺗﻔﺎﻭﺕ ﺩﺍﺭﻧﺪ‪ ،‬ﺍﯾﺰﻭﺗﻮﭘﻮﻟﻮﮒ ﻣﯽﻧﺎﻣﻨﺪ‪ .‬ﺧﻮﺍﺹ ﺷﯿﻤﯿﺎﯾﯽ ﺍﯾﺰﻭﺗﻮﭘﻮﻟﻮﮒﻫﺎ‬‫ﯾﮑﺴﺎﻥ ﺩﺭ ﻧﻈﺮﮔﺮﻓﺘﻪ ﻣﯽﺷﻮﺩ‪ .‬ﺑﺎ ﺍﯾﻦ ﺣﺎﻝ‪ ،‬ﺩﺭ ﻃﺒﯿﻌﺖ ﺗﻔﺎﻭﺕﻫﺎﯼ ﺟﺰﯾﯽ ﺩﺭ ﺧﻮﺍﺹ ﺷﯿﻤﯿﺎﯾﯽ ﺁﻧﻬﺎ ﻭﺟﻮﺩ ﺩﺍﺭﺩ‪.‬‬‫ﻓﺮﺽ ﮐﻨﯿﺪ ﺗﻤﺎﻡ ﻣﻮﺍﺩﯼ ﮐﻪ ﺩﺭ ﺍﯾﻦ ﻣﺴﺎﻟﻪ ﻣﻄﺮﺡ ﺷﺪﻩﺍﻧﺪ‪ ،‬ﺩﺭ ﻓﺎﺯ ﮔﺎﺯﯼ ﻫﺴﺘﻨﺪ‪.‬‬‫ﺗﻌﺎﺩﻝ ﺯﯾﺮ ﺭﺍ ﺩﺭ ﻧﻈﺮ ﻣﯽﮔﯿﺮﯾﻢ‪:‬‬‫ﺁﻧﺘﺮﻭﭘﯽ )‪ (S‬ﺑﺎ ﺍﻓﺰﺍﯾﺶ ﺗﻌﺪﺍﺩ ﺣﺎﻟﺖﻫﺎﯼ ﻣﯿﮑﺮﻭﺳﮑﻮﭘﯽ ﻣﻤﮑﻦ ﺑﺮﺍﯼ ﺳﯿﺴﺘﻢ )‪ (W‬ﺍﻓﺰﺍﯾﺶ ﻣﯽﯾﺎﺑﺪ‪:‬‬‫ﺑﺮﺍﯼ ‪ 12 C16 O2‬ﻭ ‪ 𝑊 = 1 ،12 C18 O2‬ﺍﺳﺖ‪ .‬ﺩﺭ ﻣﻘﺎﺑﻞ‪ ،‬ﺑﺮﺍﯼ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ‪ 𝑊 = 2 ،12 C16 O18 O‬ﺍﺳﺖ‪ ،‬ﺯﯾﺮﺍ ﺍﺗﻢﻫﺎﯼ ﺍﮐﺴﯿﮋﻥ ﺩﺭ ﺍﯾﻦ ﻣﻮﻟﮑﻮﻝ‬‫ﻣﻮﻟﮑﻮﻝ ‪ 12 C16 O18 O‬ﻭﺟﻮﺩ ﺩﺍﺭﺩ‪.𝑊 = 22 = 4 ،‬‬‫ﺗﻤﯿﺰﭘﺬﯾﺮﻧﺪ‪ .‬ﺍﺯ ﺁﻧﺠﺎ ﮐﻪ ﺩﺭ ﺳﻤﺖ ﺭﺍﺳﺖ ﺗﻌﺎﺩﻝ ﻧﺸﺎﻥ ﺩﺍﺩﻩﺷﺪﻩ ﺩﺭ ﻣﻌﺎﺩﻟﻪ )‪ ،(1‬ﺩﻭ‬‫ِ‬‫‪Q2-2‬‬‫)‪persian (Iran‬‬‫‪A.1‬‬‫‪IChO‬‬‫‪IRN-1 C-2 Q-2‬‬‫ﺗﻐﯿﯿﺮ ﺁﻧﺘﺎﻟﭙﯽ )𝐻‪ (Δ‬ﺑﺮﺍﯼ ﻣﻌﺎﺩﻟﻪ )‪ (3‬ﺑﺪﻭﻥ ﺗﻮﺟﻪ ﺑﻪ ﺩﻣﺎ ﻣﺜﺒﺖ ﺍﺳﺖ‪.‬‬‫‪8pt‬‬‫ﺛﺎﺑﺖ ﺗﻌﺎﺩﻝ )‪ (K‬ﺭﺍ ﺑﺮﺍﯼ ﻣﻌﺎﺩﻟﻪ )‪ (3‬ﺩﺭ ﺩﻣﺎﻫﺎﯼ ﺑﺴﯿﺎﺭ ﭘﺎﯾﯿﻦ )ﺩﺭ ﺣﺪ ‪ (𝑇 → 0‬ﻭ ﺩﻣﺎﻫﺎﯼ ﺑﺴﯿﺎﺭ ﺑﺎﻻ )ﺩﺭ ﺣﺪ‬‫∞‪ (𝑇 → +‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﻭﺍﮐﻨﺶ ﺩﺭ ﺍﯾﻦ ﺩﻣﺎﻫﺎ ﺗﻐﯿﯿﺮ ﻧﻤﯽﮐﻨﺪ ﻭ 𝐻‪ Δ‬ﺩﺭ ﺩﻣﺎﻫﺎﯼ ﺑﺎﻻ ﺑﻪ ﻣﻘﺪﺍﺭﯼ‬‫ﺛﺎﺑﺖ ﻫﻤﮕﺮﺍ ﻣﯽﺷﻮﺩ‪.‬‬‫ﺑﺮﺍﯼ ﻭﺍﮐﻨﺶ ﺯﯾﺮ‪ ،‬ﺗﻐﯿﯿﺮ ﺁﻧﺘﺎﻟﭙﯽ )𝐻‪ (Δ‬ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﺑﺎ ﺍﺭﺗﻌﺎﺷﺎﺕ ﻣﻮﻟﮑﻮﻟﯽ ﺗﻮﺻﯿﻒ ﮐﺮﺩ‪.‬‬‫ﺍﺭﺗﻌﺎﺷﯽ ] ‪ 𝜈 [s−1‬ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ‪:‬‬‫ﺩﺭ ﺩﻣﺎﯼ ﺻﻔﺮ ﮐﻠﻮﯾﻦ‪ ،‬ﺍﻧﺮﮊﯼ ﺍﺭﺗﻌﺎﺷﯽ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﺩﻭﺍﺗﻤﯽ ﺑﺎ ﻓﺮﮐﺎﻧﺲ‬‫ِ‬‫ﺩﺭ ﺍﯾﻦ ﺭﺍﺑﻄﻪ ‪ k‬ﺛﺎﺑﺖ ﻧﯿﺮﻭ ﻭ 𝜇 ﺟﺮﻡ ﮐﺎﻫﺶﯾﺎﻓﺘﻪ ﺍﺳﺖ‪ .‬ﺟﺮﻡ ﮐﺎﻫﺶﯾﺎﻓﺘﻪ ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﺑﺮ ﺍﺳﺎﺱ ﻣﻌﺎﺩﻟﻪ ﺯﯾﺮ ﺑﺮﺣﺴﺐ ﺟﺮﻡ ﺍﺗﻢﻫﺎ ) ‪ 𝑚1‬ﻭ ‪ (𝑚2‬ﺩﺭ‬‫ﻣﻮﻟﮑﻮﻝ ﺩﻭﺍﺗﻤﯽ ﺑﯿﺎﻥ ﮐﺮﺩ‪:‬‬‫‪A.2‬‬‫ﺍﺭﺗﻌﺎﺵ ‪ H2‬ﺑﺮﺣﺴﺐ ﻋﺪﺩ ﻣﻮﺟﯽ ﺩﺭ ‪ 4161.0 cm−1‬ﺭﺥ ﻣﯽﺩﻫﺪ‪ .‬ﺑﺮﺍﯼ ﻭﺍﮐﻨﺶ ﺯﯾﺮ‪ Δ𝐻 ،‬ﺭﺍ ﺩﺭ ﺩﻣﺎﯼ ﺻﻔﺮ‬‫ﮐﻠﻮﯾﻦ ﺑﺮﺣﺴﺐ ﻭﺍﺣﺪ ‪ J mol−1‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫ﻓﺮﺽ ﮐﻨﯿﺪ‪:‬‬‫• ﻓﻘﻂ ﺍﻧﺮﮊﯼ ﺍﺭﺗﻌﺎﺷﯽ ﺩﺭ 𝐻‪ Δ‬ﺳﻬﻢ ﺩﺍﺭﺩ‪.‬‬‫• ﻣﻘﺎﺩﯾﺮ ‪ k‬ﺑﺮﺍﯼ ‪ HD ،H2‬ﻭ ‪ D2‬ﯾﮑﺴﺎﻥ ﺍﺳﺖ‪.‬‬‫• ﺟﺮﻡ ‪ H‬ﺑﺮﺍﺑﺮ ‪ 1‬ﺩﺍﻟﺘﻮﻥ )‪ (Da‬ﻭ ﺟﺮﻡ ‪ D‬ﺑﺮﺍﺑﺮ ‪ 2‬ﺩﺍﻟﺘﻮﻥ ﺍﺳﺖ‪.‬‬‫‪8pt‬‬

‫‪Q2-3‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-2 Q-3‬‬‫ﻣﻮﻟﯽ ‪ D2‬ﺗﻌﺮﯾﻒ ﻣﯽﺷﻮﺩ‪.‬‬‫ﻧﺴﺒﺖ ﻣﻮﻟﯽ ‪ HD ،H2‬ﻭ ‪ D2‬ﺩﺭ ﺳﯿﺴﺘﻤﯽ ﮐﻪ ﺩﺭ ﺗﻌﺎﺩﻝ ﺑﺎﺷﺪ‪ ،‬ﺑﻪ ﺩﻣﺎ ﻭﺍﺑﺴﺘﻪ ﺍﺳﺖ‪ .‬ﺩﺭ ﺍﯾﻨﺠﺎ ‪ ΔD2‬ﺑﻪﺻﻮﺭﺕ ﺗﻐﯿﯿﺮ ﻧﺴﺒﺖ‬‫ِ‬‫] ‪[D2‬‬‫] ‪[D‬‬‫ﺩﺭ ﺍﯾﻦ ﺭﺍﺑﻄﻪ‪ 𝑅D2 ،‬ﻧﺸﺎﻥﺩﻫﻨﺪﻩﯼ ‪ 2‬ﺩﺭ ﻧﻤﻮﻧﻪ ﺍﺳﺖ ﻭ ‪ 𝑅D∗ 2‬ﺑﻪ‬‫] ‪[H2‬‬‫] ‪[H2‬‬‫∞‪ ،𝑇 → +‬ﺗﻮﺯﯾﻊ ﺍﯾﺰﻭﺗﻮﭖﻫﺎ ﺗﺼﺎﺩﻓﯽ ﻣﯽﺷﻮﺩ‪.‬‬‫‪A.3‬‬‫ﺩﺭ ∞‪ 𝑇 → +‬ﺍﺷﺎﺭﻩ ﻣﯽﮐﻨﺪ‪ .‬ﺩﺭ ﺍﯾﻨﺠﺎ ﻻﺯﻡ ﺑﻪ ﺫﮐﺮ ﺍﺳﺖ ﮐﻪ ﺩﺭ ﺣﺪ‬‫ﻃﺒﯿﻌﯽ ‪ ، D‬ﺩﺭ ﺣﺎﻟﺘﯽ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ ﮐﻪ ﺗﺒﺎﺩﻝ ﺍﯾﺰﻭﺗﻮﭘﯽ ﺩﺭ ﺗﻌﺎﺩﻝ ﺍﺳﺖ ﻭ ﺩﺭ ﺩﻣﺎﯼ‬‫‪ ΔD2‬ﺭﺍ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻓﺮﺍﻭﺍﻧﯽ‬‫ِ‬‫ﺗﻌﺎﺩﻝ‪ ،‬ﺛﺎﺑﺖ ﺗﻌﺎﺩﻝ )‪ (K‬ﺩﺭ ﻣﻌﺎﺩﻟﻪ )‪ (4‬ﺑﺎ ‪ 0.300‬ﺑﺮﺍﺑﺮ ﺍﺳﺖ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﮐﻪ ﻓﺮﺍﻭﺍﻧﯽ ﻃﺒﯿﻌﯽ ‪ D‬ﻭ ‪ H‬ﺑﻪﺗﺮﺗﯿﺐ‬‫ﺑﺮﺍﺑﺮ ‪ 1.5576 × 10–4‬ﻭ ‪ 1 − 1.5576 × 10–4‬ﺍﺳﺖ‪.‬‬‫‪10pt‬‬‫‪Q2-4‬‬‫‪IChO‬‬‫‪IRN-1 C-2 Q-4‬‬‫)‪persian (Iran‬‬‫ﺩﺭ ﺣﺎﻟﺖ ﮐﻠﯽ‪ ،‬ﺑﺎ ﮐﺎﻫﺶ ﺩﻣﺎ ﻧﺴﺒﺖ ﻣﻮﻟﯽ ﺑﺮﺍﯼ ﺍﯾﺰﻭﺗﻮﭘﻮﻟﻮﮒﻫﺎﯼ ﺩﻭﺍﺳﺘﺨﻼﻓﯽ )ﮐﻪ ﺩﻭ ﺍﺗﻢ ﺍﯾﺰﻭﺗﻮﭖ ﺳﻨﮕﯿﻦ ﺩﺭ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﺩﺍﺭﻧﺪ(‪ ،‬ﺍﻓﺰﺍﯾﺶ ﻣﯽﯾﺎﺑﺪ‪.‬‬‫ﻧﺴﺒﺖ ﻣﻮﻟﯽ ﺩﻭ ﻣﻮﻟﮑﻮﻝ ‪ CO2‬ﺑﺎ ﻭﺯﻥﻫﺎﯼ ﻣﻮﻟﮑﻮﻟﯽ ‪ 44‬ﻭ ‪ 47‬ﺭﺍ ﺩﺭ ﻧﻈﺮ ﻣﯽﮔﯿﺮﯾﻢ‪ .‬ﺍﯾﻦ ﺩﻭ ﻣﻮﻟﮑﻮﻝ ﺭﺍ ﺩﺭ ﺍﺩﺍﻣﻪ ﺑﺎ ]‪ CO2 [44‬ﻭ ]‪ CO2 [47‬ﻧﺸﺎﻥ‬‫ﻣﯽﺩﻫﯿﻢ‪ .‬ﮐﻤﯿﺖ ‪ Δ47‬ﺑﻪﺻﻮﺭﺕ ﺯﯾﺮ ﺗﻌﺮﯾﻒ ﻣﯽﺷﻮﺩ‪:‬‬‫]]‪[CO2 [47‬‬‫]]‪[CO2 [47‬‬‫∗‬‫‪ 𝑅47‬ﺑﻪ‬‫ﺩﺭ ﻧﻤﻮﻧﻪ ﺍﺳﺖ ﻭ‬‫‪ 𝑅47‬ﻧﺸﺎﻥﺩﻫﻨﺪﻩﯼ‬‫]]‪[CO2 [44‬‬‫]]‪[CO2 [44‬‬‫ﺩﺭ ﺟﺪﻭﻝ ﺯﯾﺮ ﺁﻣﺪﻩ ﺍﺳﺖ‪ .‬ﺍﺯ ﺍﯾﺰﻭﺗﻮﭖﻫﺎﯾﯽ ﮐﻪ ﺩﺭ ﺍﯾﻨﺠﺎ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﻧﺸﺪﻩﺍﻧﺪ‪ ،‬ﺻﺮﻓﻨﻈﺮ ﮐﻨﯿﺪ‪.‬‬‫ﺩﺭ ∞‪ 𝑇 → +‬ﺍﺷﺎﺭﻩ ﻣﯽﮐﻨﺪ‪ .‬ﻓﺮﺍﻭﺍﻧﯽ ﻃﺒﯿﻌﯽ ﺍﺗﻢﻫﺎﯼ ﮐﺮﺑﻦ ﻭ ﺍﮐﺴﯿﮋﻥ‬‫‪C‬‬‫ﻓﺮﺍﻭﺍﻧﯽ ﻃﺒﯿﻌﯽ‬‫ﻓﺮﺍﻭﺍﻧﯽ ﻃﺒﯿﻌﯽ‬‫‪O‬‬‫‪12‬‬‫‪C‬‬‫‪0.988888‬‬‫‪16‬‬‫‪0.997621‬‬‫‪O‬‬‫‪13‬‬‫‪0.011112‬‬‫‪17‬‬‫‪0.0003790‬‬‫‪O‬‬‫‪18‬‬‫‪0.0020000‬‬‫ﻭﺍﺑﺴﺘﮕﯽ ﺩﻣﺎﯾﯽ ‪ Δ47‬ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﺯﯾﺮ ﺗﻌﯿﯿﻦ ﻣﯽﺷﻮﺩ ﮐﻪ ‪ T‬ﺩﻣﺎﯼ ﻣﻄﻠﻖ ﺑﺮﺣﺴﺐ ﻭﺍﺣﺪ ﮐﻠﻮﯾﻦ ﺍﺳﺖ‪:‬‬‫‪A.4‬‬‫ﺑﺮﺍﯼ ﯾﮏ ﭘﻼﻧﮑﺘﻮﻥ ﻓﺴﯿﻞﺷﺪﻩ ﮐﻪ ﺍﺯ ﺑﺴﺘﺮ ﺩﺭﯾﺎﯼ ﻗﻄﺐ ﺟﻨﻮﺏ ﺑﻪﺩﺳﺖ ﺁﻣﺪﻩ ‪ 𝑅47‬ﺑﺮﺍﺑﺮ ﺑﺎ ‪4.50865 × 10–5‬‬‫ﺍﺳﺖ‪ .‬ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺍﯾﻦ ‪ ،𝑅47‬ﺩﻣﺎ ﺭﺍ ﺗﺨﻤﯿﻦ ﺑﺰﻧﯿﺪ‪ .‬ﺍﯾﻦ ﺩﻣﺎ ﺭﺍ ﺑﺮﺍﺑﺮ ﺑﺎ ﺩﻣﺎﯼ ﻫﻮﺍ ﺩﺭ ﺩﻭﺭﺍﻥ ﺯﻧﺪﮔﯽ ﺁﻥ ﭘﻼﻧﮑﺘﻮﻥ‬‫ﺍﯾﺰﻭﺗﻮﭘﻮﻟﻮﮒ ]‪ CO2 [47‬ﺭﺍ ﺩﺭ ﻣﺤﺎﺳﺒﻪ ﺩﺭ ﻧﻈﺮ ﺑﮕﯿﺮﯾﺪ‪.‬‬‫ﺩﺭ ﻧﻈﺮ ﻣﯽﮔﯿﺮﻧﺪ‪ .‬ﻓﻘﻂ ﻣﺘﺪﺍﻭﻝﺗﺮﯾﻦ‬‫ِ‬‫‪9pt‬‬

IChO

A2-1

IRN-1 C-2 A-1

persian (Iran)‫ﮐﭙﺴﻮﻝ ﺯﻣﺎﻥ ﺍﯾﺰﻭﺗﻮﭘﯽ‬(8 pt) A.1

𝑇 → +∞ ∶ 𝐾 =,𝑇 → 0 ∶ 𝐾 =

IChO

A2-2

IRN-1 C-2 A-2

persian (Iran)(8 pt) A.2

Δ𝐻 =J mol−1

IChO

A2-3

IRN-1 C-2 A-3

persian (Iran)(10 pt) A.3

ΔD2 =

IChO
IRN-1 C-2 A-4

A2-4
persian (Iran)(9 pt) A.4

𝑇 =K

IRN-1 C-3 C
Seyed
IRN-1 C-3
C-1 Mohammad Hossein Barakati

IChO
Problem 3
Cover sheet

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

Q3-1

IChO
IRN-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
IRN-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

‫‪Q3-1‬‬‫‪IChO‬‬‫‪IRN-1 C-3 Q-1‬‬‫)‪persian (Iran‬‬‫ﻗﺎﻧﻮﻥ ﻻﻣﺒﺮﺕ‪ -‬ﺑﯿﺮ؟‬‫‪Question‬‬‫‪Points‬‬‫‪8% of the total‬‬‫‪A.1‬‬‫‪10‬‬‫‪B.1‬‬‫‪6‬‬‫‪B.2‬‬‫‪6‬‬‫‪Total‬‬‫‪22‬‬‫‪Score‬‬‫ﺩﺭ ﺍﯾﻦ ﻣﺴﺄﻟﻪ‪ ،‬ﺍﺯ ﺟﺬﺏﻫﺎﯼ ﺳﻞ ﻭ ﺣﻼﻝ ﺻﺮﻓﻨﻈﺮ ﮐﻨﯿﺪ‪ .‬ﺩﻣﺎﯼ ﺗﻤﺎﻡ ﻣﺤﻠﻮﻝﻫﺎ ﻭ ﮔﺎﺯﻫﺎ ﺩﺭ ‪ 25‬ﺩﺭﺟﻪ ﺳﺎﻧﺘﯽﮔﺮﺍﺩ ﺛﺎﺑﺖ ﻧﮕﻪ ﺩﺍﺷﺘﻪ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫ﻗﺴﻤﺖ ‪A‬‬
‫ﻣﺤﻠﻮﻝ ‪ X‬ﺑﻪﺗﺮﺗﯿﺐ ﺑﺮﺍﺑﺮﻧﺪ ﺑﺎ ‪،1.00 × 10−2 mol L−1‬‬‫ﺁﺑﯽ ‪ X‬ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ‪ HA‬ﻭ ‪ NaA‬ﺗﻬﯿﻪ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﻏﻠﻈﺖﻫﺎﯼ] ‪ [HA] ،[A−‬ﻭ ] ‪ [H+‬ﺩﺭ‬‫ﻣﺤﻠﻮﻝ ِ‬‫ِ‬‫‪ 1.00 × 10−3 mol L−1‬ﻭ ‪ .1.00 × 10−4 mol L−1‬ﺍﯾﻦ ﻏﻠﻈﺖﻫﺎ ﺍﺯ ﻃﺮﯾﻖ ﺗﻌﺎﺩﻝ ﺍﺳﯿﺪ‪-‬ﺑﺎﺯ ﺯﯾﺮ ﺑﺎ ﻫﻢ ﺍﺭﺗﺒﺎﻁ ﺩﺍﺭﻧﺪ‪:‬‬‫] ‪[A− ][H+‬‬‫]‪[HA‬‬‫)‪(1‬‬‫=𝐾‬‫‪HA ⇌ A− + H+‬‬‫ﻃﻮﻝ ﻣﺴﯿﺮ ﻧﻮﺭ ﺩﺭ ﻗﺴﻤﺖ ‪ A‬ﺍﯾﻦ ﻣﺴﺄﻟﻪ ﺑﺮﺍﺑﺮ 𝑙 ﺍﺳﺖ‪ .‬ﺍﺯ ﺗﻐﯿﯿﺮ ﺩﺍﻧﺴﯿﺘﻪ ﺑﺮ ﺍﺛﺮ ﺭﻗﯿﻖ ﺷﺪﻥ ﺻﺮﻓﻨﻈﺮ ﮐﻨﯿﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﻫﯿﭻ ﻭﺍﮐﻨﺶ ﺷﯿﻤﯿﺎﯾﯽ ﺩﯾﮕﺮﯼ‬‫ﺑﻪ ﺟﺰ ﻣﻌﺎﺩﻟﻪ )‪ (1‬ﺭﺥ ﻧﻤﯽﺩﻫﺪ‪.‬‬‫‪A.1‬‬‫ﺟﺬﺏ ﻣﺤﻠﻮﻝ ‪ X‬ﺩﺭ ﻃﻮﻝ ﻣﻮﺝ ‪ 𝜆1‬ﺑﺮﺍﺑﺮ ‪ 𝐴1‬ﺍﻧﺪﺍﺯﻩﮔﯿﺮﯼ ﻣﯽﺷﻮﺩ‪ .‬ﺳﭙﺲ ﻣﺤﻠﻮﻝ ‪ X‬ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻫﯿﺪﺭﻭﮐﻠﺮﯾﮏ‬‫ﺑﺮﺍﺑﺮ ﺣﺠﻢ ﺍﻭﻟﯿﻪﺍﺵ ﺭﻗﯿﻖ ﻣﯽﺷﻮﺩ‪ .‬ﭘﺲ ﺍﺯ ﺭﻗﯿﻖﺳﺎﺯﯼ‪ ،‬ﺟﺬﺏ ﺩﺭ ‪ 𝜆1‬ﻫﻨﻮﺯ ﺑﺮﺍﺑﺮ‬‫ﺍﺳﯿﺪ ﺑﺎ ‪ pH = 2.500‬ﺗﺎ ﺩﻭ‬‫ِ‬‫ﺟﺬﺏ ‪ HA‬ﻭ ‪ A−‬ﺭﺍ ﺩﺭ ‪ 𝜆1‬ﻧﺸﺎﻥ‬‫ﺿﺮﺍﯾﺐ‬‫ﺗﺮﺗﯿﺐ‬
‫ﺑﻪ‬
‫𝜀‬
‫ﻭ‬
‫𝜀‬
‫)‬
‫ﮐﻨﯿﺪ‬‫ﺗﻌﯿﯿﻦ‬
‫ﺭﺍ‬
‫𝜀‬
‫‪−‬‬
‫ِ‬
‫‪A‬‬
‫‪HA‬‬‫‪ 𝐴1‬ﺍﺳﺖ‪ .‬ﻧﺴﺒﺖ ‪HA /𝜀A−‬‬‫ﻣﯽﺩﻫﻨﺪ(‪.‬‬‫‪10pt‬‬‫‪Q3-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-3 Q-2‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫ﺗﻌﺎﺩﻝ ﺯﯾﺮ ﺭﺍ ﺩﺭ ﻓﺎﺯ ﮔﺎﺯﯼ ﺩﺭ ﻧﻈﺮ ﺑﮕﯿﺮﯾﺪ‪.‬‬‫‪D ⇌ 2M‬‬‫(‪)2‬‬‫ﺧﺎﻟﺺ ‪ D‬ﺑﺎ ﻓﺸﺎﺭ 𝑃 ﭘﺮ ﻣﯽﺷﻮﺩ‪ .‬ﺩﺭ ﺍﯾﻦ ﻣﺤﻔﻈﻪ ﺩﯾﻮﺍﺭ ﻣﺘﺤﺮﮎ ﺷﻔﺎﻓﯽ ﻭﺟﻮﺩ ﺩﺍﺭﺩ ﮐﻪ ﻣﺴﺎﺣﺖ ﺳﻄﺢ‬‫ﻣﺤﻔﻈﻪﺍﯼ ﺑﻪ ﺷﮑﻞ ﻣﮑﻌﺐ ﻣﺴﺘﻄﯿﻞ ﺑﺎ ﮔﺎﺯ‬‫ِ‬‫ﻣﻘﻄﻊ ﺁﻥ 𝑆 ﺍﺳﺖ )ﺷﮑﻞ ﺯﯾﺮ ﺭﺍ ﺑﺒﯿﻨﯿﺪ(‪ .‬ﺩﺭ ﺣﺎﻟﯽ ﮐﻪ ﻓﺸﺎﺭ ﮐﻞ ﺩﺭ ‪ P‬ﺛﺎﺑﺖ ﻧﮕﻪ ﺩﺍﺷﺘﻪ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﺗﻌﺎﺩﻝ ﺑﺮﻗﺮﺍﺭ ﻣﯽﺷﻮﺩ‪ .‬ﺟﺬﺏ ﮔﺎﺯ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ‬‫𝑙) 𝑉‪ 𝐴 = 𝜀(𝑛/‬ﮐﻪ 𝜀‪ 𝑉 ،𝑛 ،‬ﻭ 𝑙 ﺑﻪﺗﺮﺗﯿﺐ ﺿﺮﯾﺐ ﺟﺬﺏ‪ ،‬ﻣﻘﺪﺍﺭ ﮔﺎﺯ ﺑﺮﺣﺴﺐ ﻣﻮﻝ‪ ،‬ﺣﺠﻢ ﮔﺎﺯ ﻭ ﻃﻮﻝ ﻣﺴﯿﺮ ﻧﻮﺭ ﺭﺍ ﻧﺸﺎﻥ ﻣﯽﺩﻫﻨﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﮐﻪ‬‫ﺗﻤﺎﻡ ﺍﺟﺰﺍﯼ ﻣﺨﻠﻮﻁ ﮔﺎﺯﯼ ﻣﺎﻧﻨﺪ ﮔﺎﺯ ﺍﯾﺪﻩﺍﻝ ﺭﻓﺘﺎﺭ ﻣﯽﮐﻨﻨﺪ‪.‬‬‫ﺩﺭ ﺻﻮﺭﺕ ﻟﺰﻭﻡ ﺍﺯ ﺗﻌﺮﯾﻒﻫﺎﯼ ﺯﯾﺮ ﺍﺳﺘﻔﺎﺩﻩ ﮐﻨﯿﺪ‪.‬‬‫ﻓﺸﺎﺭ ﺟﺰﯾﯽ‬‫ﻣﻘﺪﺍﺭ ﺑﺮ ﺣﺴﺐ ﻣﻮﻝ‬
‫ﺣﺠﻢ‬

‫‪D‬‬

‫ﺣﺎﻟﺖ ﺍﻭﻟﯿﻪ‬‫ﺑﻌﺪ ﺍﺯ ﺗﻌﺎﺩﻝ‬

‫‪M‬‬

‫‪D‬‬

‫𝑃‬

‫‪0‬‬

‫‪𝑝D‬‬‫‪𝑝M‬‬‫‪𝑛0‬‬‫‪0‬‬‫‪𝑛D‬‬‫‪𝑛M‬‬‫‪𝑉0‬‬‫‪M‬‬‫𝑉‬‫‪B.1‬‬‫ﺟﺬﺏ ﮔﺎﺯ ﺩﺭ ‪ 𝜆B1‬ﮐﻪ ﺍﺯ ﺟﻬﺖ 𝑥 ) 𝑥𝑙 = 𝑙( ﺍﻧﺪﺍﺯﻩﮔﯿﺮﯼ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﻫﻢ ﺩﺭ ﺣﺎﻟﺖ ﺍﻭﻟﯿﻪ ﻭ ﻫﻢ ﭘﺲ ﺍﺯ ﺗﻌﺎﺩﻝ‪ ،‬ﺑﺮﺍﺑﺮ‬‫ﺟﺬﺏ ‪ D‬ﻭ ‪ M‬ﺭﺍ ﻧﺸﺎﻥ ﻣﯽﺩﻫﻨﺪ‪.‬‬‫‪ 𝐴B1‬ﺍﺳﺖ‪ .‬ﻧﺴﺒﺖ ‪ 𝜀D /𝜀M‬ﺭﺍ ﺩﺭ ‪ 𝜆B1‬ﺗﻌﯿﯿﻦ ﮐﻨﯿﺪ‪ 𝜀D .‬ﻭ ‪ 𝜀M‬ﺑﻪﺗﺮﺗﯿﺐ ﺿﺮﺍﯾﺐ‬‫ِ‬‫‪6pt‬‬‫‪B.2‬‬‫ﺟﺬﺏ ﮔﺎﺯ ﺩﺭ ‪ 𝜆B2‬ﮐﻪ ﺍﺯ ﺟﻬﺖ 𝑦 ﺍﻧﺪﺍﺯﻩﮔﯿﺮﯼ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﻫﻢ ﺩﺭ ﺣﺎﻟﺖ ﺍﻭﻟﯿﻪ ) ‪ (𝑙 = 𝑙𝑦0‬ﻭ ﻫﻢ ﭘﺲ ﺍﺯ ﺗﻌﺎﺩﻝ‬‫) 𝑦𝑙 = 𝑙(‪ ،‬ﺑﺮﺍﺑﺮ ‪ 𝐴B2‬ﺍﺳﺖ‪ .‬ﻧﺴﺒﺖ ‪ 𝜀D /𝜀M‬ﺭﺍ ﺩﺭ ‪ 𝜆B2‬ﺗﻌﯿﯿﻦ ﮐﻨﯿﺪ‪.‬‬‫‪6pt‬‬

‫‪A3-1‬‬‫)‪persian (Iran‬‬‫ﻗﺎﻧﻮﻥ ﻻﻣﺒﺮﺕ‪-‬ﺑﯿﺮ؟‬‫ﻗﺴﻤﺖ ‪A‬‬‫‪(10 pt) A.1‬‬‫)ﺍﺩﺍﻣﻪ ﺩﺭ ﺻﻔﺤﻪ ﺑﻌﺪ(‬‫‪IChO‬‬‫‪IRN-1 C-3 A-1‬‬

IChO
IRN-1 C-3 A-2

A3-2
persian (Iran)(cont.) A.1

𝜀HA /𝜀A− =IChO
IRN-1 C-3 A-3

A3-3
persian (Iran)B ‫ﻗﺴﻤﺖ‬(6 pt) B.1

𝜀D /𝜀M =IChO
IRN-1 C-3 A-4

A3-4
persian (Iran)(6 pt) B.2

𝜀D /𝜀M =IRN-1 C-4 C
Seyed
IRN-1 C-4
C-1 Mohammad Hossein Barakati

IChO
Problem 4
Cover sheet

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

Q4-1

IChO
IRN-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
IRN-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
IRN-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

‫‪Q4-1‬‬‫‪IChO‬‬‫‪IRN-1 C-4 Q-1‬‬‫)‪persian (Iran‬‬‫ﺷﯿﻤﯽ ﺍﮐﺴﺎﯾﺶ‪ -‬ﮐﺎﻫﺶ ﺭﻭﯼ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪6‬‬‫‪A.2‬‬‫‪5‬‬‫‪11% of the total‬‬‫‪B.1‬‬‫‪4‬‬‫‪B.2‬‬‫‪3‬‬‫‪B.3‬‬‫‪5‬‬‫‪B.4‬‬‫‪9‬‬‫‪Total‬‬‫‪32‬‬‫‪Score‬‬‫ﺯﺩﺍﯾﯽ ﺁﺏ‪ ،‬ﺭﻭﯼ )‪ (Zn‬ﻣﻮﺟﻮﺩ ﺩﺭ ﭘﺴﺎﺏﻫﺎﯼ ﺻﻨﻌﺘﯽ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ‬‫ﺍﺯ ﻣﺪﺕﻫﺎ ﻗﺒﻞ ﺍﺯ ﺭﻭﯼ ﺩﺭ ﺁﻟﯿﺎﮊﻫﺎﯼ ﺑﺮﻧﺞ ﻭ ﺍﺳﺘﯿﻞ ﺍﺳﺘﻔﺎﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﺑﻪ ﻣﻨﻈﻮﺭ ﺳﻢ‬‫ِ‬‫ﺍﺯ ﺭﻭﺵ ﺗﻪﻧﺸﯿﻨﯽ ﺟﺪﺍ ﻣﯽﺷﻮﺩ‪ .‬ﺭﺳﻮﺑﺎﺗﯽ ﮐﻪ ﺑﻪ ﺩﺳﺖ ﻣﯽﺁﯾﺪ ﺍﺣﯿﺎ ﻣﯽﺷﻮﺩ ﺗﺎ ﺭﻭﯼ ﻓﻠﺰﯼ ﺑﺎﺯﯾﺎﺑﯽ ﻭ ﺩﻭﺑﺎﺭﻩ ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ‪.‬‬‫ﻗﺴﻤﺖ ‪A‬‬
‫ﺍﻧﺤﻼﻝ ﺭﻭﯼ ﻫﯿﺪﺭﻭﮐﺴﯿﺪ )‪ Zn(OH)2 (s‬ﺩﺭ ‪ 25 ∘ C‬ﻭ ﺛﺎﺑﺖﻫﺎﯼ ﺗﻌﺎﺩﻝ ﻣﺮﺑﻮﻃﻪ ﺩﺭ ﻣﻌﺎﺩﻻﺕ ‪ 1‬ﺗﺎ ‪ 4‬ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫ﺗﻌﺎﺩﻝ‬‫ِ‬‫ﺣﻼﻟﯿﺖ )𝑆( ﺭﻭﯼ )ﻏﻠﻈﺖ ﺭﻭﯼ ﺩﺭ ﯾﮏ ﻣﺤﻠﻮﻝ ﺁﺑﯽ ﺍﺷﺒﺎﻉﺷﺪﻩ( ﺩﺭ ﻣﻌﺎﺩﻟﻪ )‪ (5‬ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫‪Q4-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-4 Q-2‬‬‫‪A.1‬‬‫ﻫﻨﮕﺎﻣﯽ ﮐﻪ ﺗﻌﺎﺩﻝﻫﺎﯼ ﻣﻌﺎﺩﻻﺕ )‪ (1‬ﺗﺎ )‪ (4‬ﺑﺮﻗﺮﺍﺭ ﻣﯽﺷﻮﺩ‪ ،‬ﻣﺤﺪﻭﺩﻩﺍﯼ ﺍﺯ ‪ pH‬ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ ﮐﻪ ﺩﺭ ﺁﻥ‪ ،‬ﺩﺭ‬‫‪2+‬‬‫‪ ،[Zn(OH)2−‬ﺑﺰﺭﮒﺗﺮﯾﻦ ﻣﻘﺪﺍﺭ ﻣﺮﺑﻮﻁ ﺑﻪ‬‫ﺑﯿﻦ ﻏﻠﻈﺖﻫﺎﯼ‬‫ﺗﻌﺎﺩﻟﯽ ])‪ [Zn(OH)2 (aq)] ،[Zn (aq‬ﻭ ])‪4 (aq‬‬‫ِ‬‫])‪ [Zn(OH)2 (aq‬ﺑﺎﺷﺪ‪.‬‬‫‪6pt‬‬‫‪A.2‬‬‫ﻣﺤﻠﻮﻝ ﺯﯾﺮ ﺻﺎﻓﯽ‬‫ﻣﺤﻠﻮﻝ ﺁﺑﯽ ﺍﺷﺒﺎﻉﺷﺪﻩﺍﯼ ﺍﺯ )‪ Zn(OH)2 (s‬ﺑﺎ ‪ pH = 7.00‬ﺗﻬﯿﻪ ﻭ ﺻﺎﻑ ﻣﯽﺷﻮﺩ‪ .‬ﺑﻪ‬‫ِ‬‫‪ NaOH‬ﺍﺿﺎﻓﻪ ﻣﯽﺷﻮﺩ ﺗﺎ ‪ pH‬ﺁﻥ ﺗﺎ ‪ 12.00‬ﺍﻓﺰﺍﯾﺶ ﯾﺎﺑﺪ‪ .‬ﺩﺭﺻﺪ ﻣﻮﻟﯽﺍﯼ ﺍﺯ ﺭﻭﯼ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ ﮐﻪ ﺩﺭ ﺍﺛﺮ‬‫ﺍﻓﺰﺍﯾﺶ ‪ pH‬ﺍﺯ ‪ 7.00‬ﺗﺎ ‪ 12.00‬ﺭﺳﻮﺏ ﻣﯽﮐﻨﺪ‪ .‬ﺍﺯ ﺗﻐﯿﯿﺮﺍﺕ ﺣﺠﻢ ﻭ ﺩﻣﺎ ﺻﺮﻓﻨﻈﺮ ﮐﻨﯿﺪ‪.‬‬‫ِ‬‫‪5pt‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫ﺩﺭ ﻣﺮﺣﻠﻪ ﺑﻌﺪﯼ‪ ،‬ﺭﻭﯼ ﻫﯿﺪﺭﻭﮐﺴﯿﺪِ ﺑﺎﺯﯾﺎﺑﯽﺷﺪﻩ ﺣﺮﺍﺭﺕ ﺩﺍﺩﻩ ﻣﯽﺷﻮﺩ ﺗﺎ ﺑﺮ ﺍﺳﺎﺱ ﻭﺍﮐﻨﺶ ﺯﯾﺮ‪ ،‬ﺭﻭﯼ ﺍﮐﺴﯿﺪ ﺑﻪ ﺩﺳﺖ ﺁﯾﺪ‪:‬‬‫ﺭﻭﯼ ﻓﻠﺰﯼ ﺍﺣﯿﺎ ﻣﯽﺷﻮﺩ‪:‬‬‫ﺳﭙﺲ ﺭﻭﯼ ﺍﮐﺴﯿﺪ ﺩﺭ ﻭﺍﮐﻨﺶ ﺑﺎ ﻫﯿﺪﺭﻭﮊﻥ ﺑﻪ ِ‬‫‪B.1‬‬‫ﺟﺰﯾﯽ ﺑﺨﺎﺭ‬‫ﺷﺮﺍﯾﻂ ﻓﺸﺎﺭ ﺛﺎﺑﺖ ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ‪ 1‬ﺑﺎﺭ )‪ (bar‬ﭘﯿﺶ ﺑﺮﻭﺩ‪ ،‬ﻻﺯﻡ ﺍﺳﺖ ﻓﺸﺎﺭ‬‫ﺑﺮﺍﯼ ﺁﻧﮑﻪ ﻭﺍﮐﻨﺶ )‪ (7‬ﺩﺭ‬‫ِ‬‫ِ‬‫ﺁﺑﯽ ﮐﻪ ﺗﻮﻟﯿﺪ ﻣﯽﺷﻮﺩ‪ ،‬ﮐﻢ ﺷﻮﺩ‪ .‬ﺣﺪ ﺑﺎﻻﯾﯽ ﻓﺸﺎﺭ ﺟﺰﯾﯽ ﺑﺨﺎﺭ ﺁﺏ ﺭﺍ ﺑﺮﺍﯼ ﺁﻧﮑﻪ ﻭﺍﮐﻨﺶ )‪ (7‬ﺑﺘﻮﺍﻧﺪ ﺩﺭ ﺩﻣﺎﯼ ‪300‬‬‫ﺩﺭﺟﻪ ﺳﺎﻧﺘﯽﮔﺮﺍﺩ ﭘﯿﺶ ﺑﺮﻭﺩ‪ ،‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ .‬ﺩﺭ ﺍﯾﻨﺠﺎ ﺍﻧﺮﮊﯼﻫﺎﯼ ﺗﺸﮑﯿﻞ ﮔﯿﺒﺲ ﺑﺮﺍﯼ ﺭﻭﯼ ﺍﮐﺴﯿﺪ ﻭ ﺑﺨﺎﺭ ﺁﺏ ﺩﺭ‬‫ﺩﻣﺎﯼ ‪ 300‬ﺩﺭﺟﻪ ﺳﺎﻧﺘﯽﮔﺮﺍﺩ ﻭ ﻓﺸﺎﺭ ‪ 1‬ﺑﺎﺭ ﺑﺮﺍﯼ ﺗﻤﺎﻡ ﮔﻮﻧﻪﻫﺎﯼ ﮔﺎﺯﯼ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ‪:‬‬‫‪Δ𝐺ZnO (300∘ C) = −2.90 × 102 kJ mol−1‬‬‫ﻭ‬‫‪Δ𝐺H2 O (300∘ C) = −2.20 × 102 kJ mol−1‬‬‫‪4pt‬‬‫ﺭﻭﯼ ﻓﻠﺰﯼ ﺑﺮﺍﯼ ﺳﺎﺧﺘﻦ ﺍﻟﮑﺘﺮﻭﺩ ﻣﻨﻔﯽ )ﺁﻧﺪ( ﺩﺭ ﺑﺎﺗﺮﯼﻫﺎﯼ ﻓﻠﺰ‪-‬ﻫﻮﺍ ﺍﺳﺘﻔﺎﺩﻩ ﻣﯽﺷﻮﺩ‪ .‬ﺍﯾﻦ ﺍﻟﮑﺘﺮﻭﺩ ﺍﺯ ‪ Zn‬ﻭ ‪ ZnO‬ﺗﺸﮑﯿﻞ ﺷﺪﻩ ﺍﺳﺖ ﻭ ﺍﺯ ﻭﺍﮐﻨﺶ‬‫ﺍﺯ ِ‬‫ﺍﮐﺴﺎﯾﺶ‪ -‬ﮐﺎﻫﺶ ﺯﯾﺮ ﺑﺮﺍﯼ ﺗﻮﻟﯿﺪ ﺟﺮﯾﺎﻥ ﺍﻟﮑﺘﺮﯾﮑﯽ ﺍﺳﺘﻔﺎﺩﻩ ﻣﯽﮐﻨﺪ‪ .‬ﻧﯿﺮﻭﯼ ﻣﺤﺮﮐﻪ ﺍﻟﮑﺘﺮﯾﮑﯽ )‪ (e.m.f.‬ﯾﺎ ∘ 𝐸 ﺍﯾﻦ ﻭﺍﮐﻨﺶ ﺩﺭ ﺩﻣﺎﯼ ‪ 25‬ﺩﺭﺟﻪ‬‫ﺳﺎﻧﺘﯽﮔﺮﺍﺩ ﻭ ﻓﺸﺎﺭ ‪ 1‬ﺑﺎﺭ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫‪B.2‬‬‫ﯾﮏ ﺑﺎﺗﺮﯼ ﺭﻭﯼ‪-‬ﻫﻮﺍ ﺑﺎ ﺟﺮﯾﺎﻥ ‪ 20‬ﻣﯿﻠﯽﺁﻣﭙﺮ ﺩﺭ ﻣﺪﺕ ‪ 24‬ﺳﺎﻋﺖ ﺗﺨﻠﯿﻪ ﻣﯽﺷﻮﺩ‪ .‬ﺗﻐﯿﯿﺮ ﺟﺮﻡ ﺍﻟﮑﺘﺮﻭﺩ ﻣﻨﻔﯽ )ﺁﻧﺪ(‬‫ﺍﯾﻦ ﺑﺎﺗﺮﯼ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪3pt‬‬

‫‪Q4-3‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-4 Q-3‬‬‫ﮐﻮﻩ ﻓﻮﺟﯽ‬‫‪B.3‬‬‫ﺍﺛﺮ ﺗﻐﯿﯿﺮﺍﺕ ﻣﺤﯿﻄﯽ ﺩﺭ ﻧﻈﺮ ﺑﮕﯿﺮﯾﺪ‪ e.m.f. .‬ﺭﺍ ﺩﺭ ﻗﻠﻪ ﮐﻮﻩ ﻓﻮﺟﯽ‬‫ﺗﻐﯿﯿﺮ ‪ e.m.f.‬ﯾﮏ ﺑﺎﺗﺮﯼ ﺭﻭﯼ‪-‬ﻫﻮﺍ ﺭﺍ ﺩﺭ ِ‬‫ﺍﺭﺗﻔﺎﻉ‬‫ﺩﺭ‬‫ﺍﺗﻤﺴﻔﺮﯼ‬‫ﻓﺸﺎﺭ‬‫ﺑﺎﺷﺪ‪.‬‬‫ﻣﺘﺮ‬‫‪3776‬‬‫ﺍﺭﺗﻔﺎﻉ‬‫ﻭ‬‫ﻓﻮﺭﯾﻪ(‬
‫ﻣﺎﻩ‬
‫)ﺩﺭ‬
‫ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ ،‬ﺍﮔﺮ ﺩﺭ ﺍﯾﻦ ﻗﻠﻪ ﺩﻣﺎ ‪−38 ∘ C‬‬‫ِ‬‫]‪ ℎ [m‬ﻭ ﺩﻣﺎﯼ ]‪ 𝑇 [∘ C‬ﺑﺎ ﻣﻌﺎﺩﻟﻪ ﺯﯾﺮ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﻣﯽﺷﻮﺩ‪:‬‬‫‪5pt‬‬‫ﻧﺴﺒﺖ ﻣﻮﻟﯽ ﺍﮐﺴﯿﮋﻥ ﺩﺭ ﺍﺗﻤﺴﻔﺮ ‪ 21%‬ﺍﺳﺖ‪ .‬ﺗﻐﯿﯿﺮ ﺍﻧﺮﮊﯼ ﮔﯿﺒﺲ ﺑﺮﺍﯼ ﻭﺍﮐﻨﺶ )‪ (8‬ﺩﺭ ﺩﻣﺎﯼ ‪ −38 ∘ C‬ﻭ ﻓﺸﺎﺭ ‪1‬‬‫ﺑﺎﺭ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ‪Δ𝐺ZnO (−38∘ C) = −3.26 × 102 kJ mol−1‬‬‫‪B.4‬‬‫ﺗﻐﯿﯿﺮ ﺍﻧﺮﮊﯼ ﮔﯿﺒﺲ ﺭﺍ ﺑﺮﺍﯼ ﻭﺍﮐﻨﺶ )‪ (6‬ﺩﺭ ‪ 25 ∘ C‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ .‬ﺗﻮﺟﻪ ﮐﻨﯿﺪ ﮐﻪ ﭘﺘﺎﻧﺴﯿﻞﻫﺎﯼ ﮐﺎﻫﺶ ﺍﺳﺘﺎﻧﺪﺍﺭﺩ‬‫)‪ 𝐸 ∘ (Zn2+ /Zn‬ﻭ )‪ 𝐸 ∘ (O2 /H2 O‬ﺩﺭ ‪ 25 ∘ C‬ﻭ 𝑟𝑎𝑏 ‪ 1‬ﺑﻪﺗﺮﺗﯿﺐ ﺩﺭ ﻣﻌﺎﺩﻟﻪﻫﺎﯼ )‪ (10‬ﻭ )‪ (11‬ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫‪9pt‬‬IChO
IRN-1 C-4 A-1

A4-1
persian (Iran)‫ ﮐﺎﻫﺶ ﺭﻭﯼ‬-‫ﺷﯿﻤﯽ ﺍﮐﺴﺎﯾﺶ‬A ‫ﻗﺴﻤﺖ‬(6 pt) A.1< pH <IChO
IRN-1 C-4 A-2

A4-2
persian (Iran)(5 pt) A.2

%

IChO
IRN-1 C-4 A-3

A4-3
persian (Iran)B ‫ﻗﺴﻤﺖ‬(4 pt) B.1

𝑝H2 O=bar(3 pt) B.2

g

IChO
IRN-1 C-4 A-4

A4-4
persian (Iran)(5 pt) B.3

V

IChO

A4-5

IRN-1 C-4 A-5

persian (Iran)(9 pt) B.4

Δ𝐺∘ =J mol−1

IRN-1 C-5 C
Seyed
IRN-1 C-5
C-1 Mohammad Hossein Barakati

IChO
Problem 5
Cover sheet

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

Q5-1

IChO
IRN-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
IRN-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
IRN-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
IRN-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

‫‪Q5-1‬‬‫‪IChO‬‬‫‪IRN-1 C-5 Q-1‬‬‫)‪persian (Iran‬‬‫ﺳﯿﻠﯿﺴﯿﻢ ﻣﺮﻣﻮﺯ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪9‬‬‫‪A.2‬‬‫‪7‬‬‫‪12% of the total‬‬‫‪A.3‬‬‫‪6‬‬‫‪A.4‬‬‫‪10‬‬‫‪B.1‬‬‫‪5‬‬‫‪B.2‬‬‫‪15‬‬‫‪B.3‬‬‫‪8‬‬‫‪Total‬‬‫‪60‬‬‫‪Score‬‬‫ﺍﮔﺮﭼﻪ ﺳﯿﻠﯿﺴﯿﻢ ﻣﺎﻧﻨﺪ ﮐﺮﺑﻦ ﺩﺭ ﮔﺮﻭﻩ ‪ 14‬ﺍﺳﺖ‪ ،‬ﺧﻮﺍﺻﺶ ﺑﻪ ﻃﻮﺭ ﻗﺎﺑﻞ ﻣﻼﺣﻈﻪﺍﯼ ﻓﺮﻕ ﺩﺍﺭﺩ‪.‬‬‫ﻗﺴﻤﺖ ‪A‬‬

‫ﺑﺮﺧﻼﻑ ﭘﯿﻮﻧﺪ ﺳﻪ ﮔﺎﻧﻪ ﮐﺮﺑﻦ‪-‬ﮐﺮﺑﻦ‪ ،‬ﭘﯿﻮﻧﺪ ﺳﻪ ﮔﺎﻧﻪ ﺳﯿﻠﯿﺴﯿﻢ ‪-‬ﺳﯿﻠﯿﺴﯿﻢ ﺩﺭ ﺗﺮﮐﯿﺒﯽ ﺑﺎ ﻓﺮﻣﻮﻝ ‪ :R) R1 –Si ≡ Si–R1‬ﺍﺳﺘﺨﻼﻑ ﺁﻟﯽ( ﮐﺎﻣﻼ ﻓﻌﺎﻝ‬‫ﺍﺳﺖ‪ .‬ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺜﺎﻝ‪ ،‬ﺑﺎ ﺍﺗﯿﻠﻦ ﻭﺍﮐﻨﺶ ﻣﯽﺩﻫﺪ ﻭ ﯾﮏ ﻣﺤﺼﻮﻝ ﺣﻠﻘﻮﯼ ﺗﺸﮑﯿﻞ ﻣﯽﺩﻫﺪ ﮐﻪ ﯾﮏ ﺣﻠﻘﻪ ﭼﻬﺎﺭﻋﻀﻮﯼ ﺩﺍﺭﺩ‪.‬‬‫ﻭﻗﺘﯽ ‪ R1 –Si ≡ Si–R1‬ﺑﺎ ﯾﮏ ﺁﻟﮑﯿﻦ ) ‪ (R2 –C ≡ C–R2‬ﻭﺍﮐﻨﺶ ﻣﯽﺩﻫﺪ‪ ،‬ﺗﺮﮐﯿﺐ ‪ A‬ﮐﻪ ﺣﻠﻘﻪ ﭼﻬﺎﺭ ﻋﻀﻮﯼ ﺩﺍﺭﺩ ﯾﻪ ﻋﻨﻮﺍﻥ ﯾﮏ ﺣﺪﻭﺍﺳﻂ ﺍﻭﻟﯿﻪ‬‫ﺑﯿﺸﺘﺮ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﺩﯾﮕﺮ ‪ R2 –C ≡ C–R2‬ﺑﺎ ﺗﺮﮐﯿﺐ ‪ A‬ﺍﯾﺰﻭﻣﺮﻫﺎﯼ ‪ B‬ﻭ ‪ C‬ﺭﺍ ﻣﯽﺩﻫﺪ ﮐﻪ ﻫﺮ ﺩﻭﯼ ﺁﻥﻫﺎ ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ﻣﺰﺩﻭﺝ‬‫ﺗﺸﮑﯿﻞ ﻣﯽﺷﻮﺩ‪ .‬ﻭﺍﮐﻨﺶ‬‫ِ‬‫ﺣﻠﻘﻮﯼ ﺷﺒﯿﻪ ﺑﻨﺰﻥ ﺩﺍﺭﻧﺪ ﻭ‘‪ ’disilabenzenes‬ﻧﺎﻣﯿﺪﻩ ﻣﯽﺷﻮﻧﺪ‪ .‬ﻫﺮ ﺩﻭﯼ ﺁﻧﻬﺎ ﺷﺎﻣﻞ ﯾﮏ ﺣﻠﻘﻪ ﺷﺶ ﻋﻀﻮﯼ ﻫﺴﺘﻨﺪ ﻭ ﻣﯽﺗﻮﺍﻥ ﻓﺮﻣﻮﻝ ﺁﻧﻬﺎ ﺭﺍ ﺑﻪ‬‫ﺻﻮﺭﺕ ‪ (R1 –Si)2 (R2 –C)4‬ﻧﺸﺎﻥ ﺩﺍﺩ‪.‬‬‫‪Q5-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-5 Q-2‬‬‫ﻃﯿﻒ‪ 13 C NMR‬ﺑﺮﺍﯼ ﺍﺳﮑﻠﺖ ﻫﺎﯼ ﺳﺎﺧﺘﺎﺭﯼ ‪ Si2 C4‬ﺑﺎ ﺣﻠﻘﻪ ﺷﺶ ﻋﻀﻮﯼ ﺩﺭ ‪ B‬ﻭ ‪ ، C‬ﺩﻭ ﺳﯿﮕﻨﺎﻝ ﺑﺮﺍﯼ ‪ B‬ﻭ ﯾﮏ ﺳﯿﮕﻨﺎﻝ ﺑﺮﺍﯼ ‪ C‬ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ‪.‬‬‫‪A.1‬‬‫ﻓﺮﻣﻮﻝ ﻫﺎﯼ ﺳﺎﺧﺘﺎﺭﯼ ‪ B ، A‬ﻭ ‪ C‬ﺭﺍ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ‪ Si ،R2 ،R1‬ﻭ ‪ C‬ﺩﺭ ﯾﮑﯽ ﺍﺯ ﻓﺮﻡﻫﺎﯼ ﺭﺯﻭﻧﺎﻧﺴﯽ ﻣﺤﺘﻤﻞ‬‫ﺭﺳﻢ ﮐﻨﯿﺪ‪.‬‬‫‪9pt‬‬‫‪A.2‬‬‫ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺗﻐﯿﯿﺮ ﺁﻧﺘﺎﻟﭙﯽ ﺩﺭ ﺑﺮﺧﯽ ﺍﺯ ﻭﺍﮐﻨﺸﻬﺎﯼ ﻫﯿﺪﺭﻭﮊﻥﺩﺍﺭ ﺷﺪﻥ ﺳﯿﺴﺘﻢ ﻫﺎﯼ ﻏﯿﺮ ﺍﺷﺒﺎﻉ ﮐﻪ ﺩﺭ ﭘﺎﯾﯿﻦ )ﺷﮑﻞ‬‫‪ (1‬ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﺍﻧﺮﮊﯼ ﭘﺎﯾﺪﺍﺭﯼ ﺁﺭﻭﻣﺎﺗﯿﮏ )‪ (ASE‬ﺭﺍ ﺑﺮﺍﯼ ﺑﻨﺰﻥ ﻭ ﺗﺮﮐﯿﺐ ‪ (R1 = R2 = H ) C‬ﺑﻪ ﺻﻮﺭﺕ‬‫ﻣﻘﺎﺩﯾﺮ ﻣﺜﺒﺖ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪7pt‬‬‫ﺷﮑﻞ ‪1‬‬

‫‪Q5-3‬‬‫‪IChO‬‬‫‪IRN-1 C-5 Q-3‬‬‫)‪persian (Iran‬‬‫ﻭﻗﺘﯽ ﻣﺤﻠﻮﻝ ‪ C‬ﺩﺭ ﺯﺍﯾﻠﻦ ﺣﺮﺍﺭﺕ ﺩﺍﺩﻩ ﻣﯽﺷﻮﺩ‪ ،‬ﺩﭼﺎﺭ ﺍﯾﺰﻭﻣﺮﯼ ﻣﯽﺷﻮﺩ ﻭ ﯾﮏ ﻣﺨﻠﻮﻁ ﺗﻌﺎﺩﻟﯽ ﺍﺯ ﺗﺮﮐﯿﺒﺎﺕ ‪ D‬ﻭ ‪ E‬ﺑﺪﺳﺖ ﻣﯽﺁﯾﺪ‪ .‬ﻧﺴﺒﺖ ﻣﻮﻟﯽ ‪ D‬ﺑﻪ‬‫‪ E‬ﺩﺭ ﺩﻣﺎﯼ ‪ 50.0‬ﺩﺭﺟﻪ ﺳﻠﺴﯿﻮﺱ ﻣﻌﺎﺩﻝ ‪ 1‬ﺑﻪ ‪ 40.0‬ﻭ ﻧﺴﺒﺖ ﻣﻮﻟﯽ ‪ D‬ﺑﻪ ‪ E‬ﺩﺭ ﺩﻣﺎﯼ ‪ 120.0‬ﺩﺭﺟﻪ ﺳﻠﺴﯿﻮﺱ ﻣﻌﺎﺩﻝ ‪ 1‬ﺑﻪ ‪ 20.0‬ﺍﺳﺖ‪.‬‬‫‪A.3‬‬‫‪6pt‬‬‫𝐻‪ Δ‬ﺭﺍ ﺑﺮﺍﯼ ﺗﺒﺪﯾﻞ ‪ D‬ﺑﻪ ‪ E‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ 𝐻‪ Δ‬ﺑﻪ ﺩﻣﺎ ﺑﺴﺘﮕﯽ ﻧﺪﺍﺭﺩ‪.‬‬‫ﺍﯾﺰﻭﻣﺮﯼ ﺷﺪﻥ ‪ C‬ﺑﻪ ‪ D‬ﻭ ﺑﻪ ‪ E‬ﺑﺎ ﺗﺒﺪﯾﻞ ﭘﯿﻮﻧﺪﻫﺎﯼ 𝜋 ﺑﻪ ﭘﯿﻮﻧﺪﻫﺎﯼ 𝜎 ﻭ ﺑﺪﻭﻥ ﺷﮑﺴﺘﻪ ﺷﺪﻥ ﭘﯿﻮﻧﺪﻫﺎﯼ 𝜎 ﺍﻧﺠﺎﻡ ﻣﯽﺷﻮﺩ‪ .‬ﺁﻧﺎﻟﯿﺰ ‪ 13 C NMR‬ﺑﺮﺍﯼ‬‫ﺍﺳﮑﻠﺖ ﺳﺎﺧﺘﺎﺭﯼ ‪ Si2 C4‬ﺍﺯ ‪ D‬ﯾﮏ ﺳﯿﮕﻨﺎﻝ ﻭ ﺑﺮﺍﯼ ﺍﺳﮑﻠﺖ ﺳﺎﺧﺘﺎﺭﯼ ‪ Si2 C4‬ﺍﺯ ‪ E‬ﺩﻭ ﺳﯿﮕﻨﺎﻝ ﻣﯽﺩﻫﺪ‪ .‬ﺩﺭ ﺳﺎﺧﺘﺎﺭ ‪ D‬ﺣﻠﻘﻪ ﺳﻪ ﻋﻀﻮﯼ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ‬‫ﺩﺭ ﺣﺎﻟﯿﮑﻪ ‪ E‬ﺩﻭ ﺣﻠﻘﻪ ﺳﻪ ﻋﻀﻮﯼ ﺩﺍﺭﺩ ﮐﻪ ﺩﺭ ﯾﮏ ﺿﻠﻊ ﻣﺸﺘﺮﮎ ﻫﺴﺘﻨﺪ‪.‬‬‫‪A.4‬‬‫ﻓﺮﻣﻮﻝ ﺳﺎﺧﺘﺎﺭﯼ ‪ D‬ﻭ ‪ E‬ﺭﺍ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ‪ Si ، R2 ، R1‬ﻭ ‪ C‬ﺭﺳﻢ ﮐﻨﯿﺪ‪.‬‬‫‪10pt‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫ﺳﯿﻠﯿﺴﯿﻢ ﻣﯽﺗﻮﺍﻧﺪ ﺑﺎ ﻋﻨﺎﺻﺮ ﺍﻟﮑﺘﺮﻭﻧﮕﺎﺗﯿﻮ ﻣﺎﻧﻨﺪ ﻓﻠﻮﺋﻮﺭ ﺗﺮﮐﯿﺒﺎﺗﯽ ﺑﺎ ﮐﺌﻮﺭﺩﯾﻨﺎﻧﺴﯿﻮﻥ ﺑﺎﻻ )ﺑﺎ ﺑﯿﺸﺘﺮ ﺍﺯ ﭼﻬﺎﺭ ﺍﺳﺘﺨﻼﻑ( ﺗﺸﮑﯿﻞ ﺩﻫﺪ‪ .‬ﺷﺒﯿﻪ‬‫ﻓﻠﻮﺋﻮﺭﯾﺪﻫﺎﯼ ﻓﻠﺰﯼ ﮐﻪ ﺍﻏﻠﺐ ﺑﻪ ﻋﻨﻮﺍﻥ ﻭﺍﮐﻨﺸﮕﺮﻫﺎﯼ ﻓﻠﻮﺋﻮﺭﺩﺍﺭ ﮐﺮﺩﻥ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﻣﯽ ﮔﯿﺮﻧﺪ ‪ ،‬ﻓﻠﻮﺋﻮﺭﯾﺪﻫﺎﯼ ﺳﯿﻠﯿﺴﯿﻮﻣﯽ ﺑﺎ ﮐﺌﻮﺭﺩﯾﻨﺎﺳﯿﻮﻥ‬‫ﺑﺎﻻ ﻧﯿﺰ ﺑﻪ ﻋﻨﻮﺍﻥ ﻭﺍﮐﻨﺸﮕﺮﻫﺎﯼ ﻓﻠﻮﺋﻮﺭﺩﺍﺭ ﮐﺮﺩﻥ ﻋﻤﻞ ﻣﯽﮐﻨﻨﺪ‪.‬‬‫ﻭﺍﮐﻨﺶ ﻓﻠﻮﺋﻮﺭﺩﺍﺭﮐﺮﺩﻥ ‪ CCl4‬ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ‪ Na2 SiF6‬ﺑﻪ ﺻﻮﺭﺕ ﺯﯾﺮ ﺍﻧﺠﺎﻡ ﺷﺪ‪.‬‬‫• ﺍﺳﺘﺎﻧﺪﺍﺭﺩ ﮐﺮﺩﻥ ﻣﺤﻠﻮﻝ ‪Na2 SiF6‬‬‫· ﺗﻬﯿﻪ‬‫ﻣﺤﻠﻮﻝ ﺁﺑﯽ ‪ : F‬ﻣﻘﺪﺍﺭ ‪ 0.855‬ﮔﺮﻡ ﺍﺯ ‪ 188.053 ) Na2 SiF6‬ﮔﺮﻡ ﺑﺮ ﻣﻮﻝ( ﺩﺭ ﺁﺏ ﺣﻞ ﺷﺪ )ﺣﺠﻢ ﮐﻞ ‪ 200 :‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ(‪.‬‬‫ﻣﺤﻠﻮﻝ ﺁﺑﯽ ‪ : G‬ﻣﻘﺪﺍﺭ ‪ 6.86‬ﮔﺮﻡ ﺍﺯ ‪ 568.424) Ce2 (SO4 )3‬ﮔﺮﻡ ﺑﺮ ﻣﻮﻝ( ﺩﺭ ﺁﺏ ﺣﻞ ﺷﺪ )ﺣﺠﻢ ﮐﻞ ‪ 200 :‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ(‪.‬‬‫· ﺩﺳﺘﻮﺭﺍﻟﻌﻤﻞ‬‫ﺗﯿﺘﺮﺍﺳﯿﻮﻥ ﺭﺳﻮﺑﯽ ‪ 50.0‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ ﺍﺯ ﻣﺤﻠﻮﻝ ‪ F‬ﺍﺯ ﻃﺮﯾﻖ ﻗﻄﺮﻩ ﻗﻄﺮﻩ ﺍﺿﺎﻓﻪ ﮐﺮﺩﻥ ﻣﺤﻠﻮﻝ ‪ ، G‬ﺩﺭ ﺣﻀﻮﺭ ﺷﻨﺎﺳﺎﮔﺮ ‪ orange xylenol‬ﺍﻧﺠﺎﻡ‬‫ﻣﯽﺷﻮﺩ ﮐﻪ ﺑﺎ ‪ Ce3+‬ﮐﺌﻮﺭﺩﯾﻨﻪ ﻣﯽﺷﻮﺩ‪ .‬ﺑﻌﺪ ﺍﺯ ﺍﺿﺎﻓﻪ ﮐﺮﺩﻥ ‪ 18.8‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ ﻣﺤﻠﻮﻝ ‪ ، G‬ﺭﻧﮓ ﻣﺤﻠﻮﻝ ﺍﺯ ﺯﺭﺩ ﺑﻪ ﺳﺮﺧﺎﺑﯽ ﺗﻐﯿﯿﺮ ﻣﯽﮐﻨﺪ‪ .‬ﺭﺳﻮﺏ‬‫ﺗﻮﻟﯿﺪ ﺷﺪﻩ ﯾﮏ ﺗﺮﮐﯿﺐ ﺩﻭﺗﺎﯾﯽ ﻭ ﺷﺎﻣﻞ ‪ Ce3+‬ﺍﺳﺖ ﻭ ﺗﻨﻬﺎ ﺗﺮﮐﯿﺐ ﺳﯿﻠﯿﺴﯿﻮﻣﯽ ﺣﺎﺻﻞ ‪ Si(OH)4 ،‬ﺍﺳﺖ‪.‬‬‫‪B.1‬‬‫‪5pt‬‬‫ﺑﺮﺍﯼ ﻭﺍﮐﻨﺶ ‪ Na2 SiF6‬ﺑﺎ ‪ Ce2 (SO4 )3‬ﻣﻌﺎﺩﻟﻪ ﻣﻮﺍﺯﻧﻪ ﺷﺪﻩ ﺑﻨﻮﯾﺴﯿﺪ‪.‬‬‫• ﻭﺍﮐﻨﺶ ‪ CCl4‬ﺑﺎ ‪Na2 SiF6‬‬‫)ﺍﺯ ﺩﺳﺖ ﺭﻓﺘﻦ ﻣﻮﺍﺩ ﻣﺜﻼ ﺩﺭ ﺍﺛﺮ ﺗﺒﺨﯿﺮ ﺷﺪﻥ ﺩﺭ ﺣﯿﻦ ﺍﻧﺠﺎﻡ ﺩﺳﺘﻮﺭﺍﻟﻌﻤﻞ ﺯﯾﺮ‪ ،‬ﻗﺎﺑﻞ ﺻﺮﻑ ﻧﻈﺮ ﺍﺳﺖ‪(.‬‬‫ﻣﻘﺪﺍﺭ 𝑥 ﮔﺮﻡ ﺍﺯ ‪ Na2 SiF6‬ﺑﻪ ‪ 500.0‬ﮔﺮﻡ ‪ CCl4‬ﺍﺿﺎﻓﻪ ﺷﺪﻩ ﻭ ﺩﺭ ﯾﮏ ﻇﺮﻑ ﻭﺍﮐﻨﺶ ﺩﺭ ﺑﺴﺘﻪ ﻭ ﻣﻘﺎﻭﻡ ﺩﺭ ﺑﺮﺍﺑﺮ ﻓﺸﺎﺭ ﺗﺎ ‪ 300‬ﺩﺭﺟﻪ ﺳﻠﺴﯿﻮﺱ‬‫ﺣﺮﺍﺭﺕ ﺩﺍﺩﻩ ﺷﺪ‪ .‬ﺳﭙﺲ ‪ Na2 SiF6‬ﻭﺍﮐﻨﺶ ﻧﺪﺍﺩﻩ ﻭ ‪ NaCl‬ﺗﻮﻟﯿﺪ ﺷﺪﻩ ﺑﻪ ﮐﻤﮏ ﺻﺎﻑ ﮐﺮﺩﻥ ﺟﺪﺍ ﺷﺪﻧﺪ‪ .‬ﺯﯾﺮ ﺻﺎﻓﯽ ﺑﺎ ‪ CCl4‬ﺗﺎ ﺣﺠﻢ ‪ 1.00‬ﻟﯿﺘﺮ‬‫ﺭﻗﯿﻖ ﺷﺪ )ﻣﺤﻠﻮﻝ ‪ .(H‬ﻃﯿﻒ ﻫﺎﯼ‪ 29 Si NMR‬ﻭ ‪ 19 F NMR‬ﻣﺤﻠﻮﻝ ‪ SiF4 ، H‬ﺭﺍ ﺑﻪ ﻋﻨﻮﺍﻥ ﺗﻨﻬﺎ ﺗﺮﮐﯿﺐ ﺳﯿﻠﯿﺴﯿﻮﻣﯽ ﻧﺸﺎﻥ ﺩﺍﺩ‪ .‬ﺩﺭ ﻃﯿﻒ ‪NMR‬‬‫‪19‬‬‫‪ 19 F‬ﻋﻼﻭﻩ ﺑﺮ ‪ SiF4‬ﺳﯿﮕﻨﺎﻝ ﻫﺎﯼ ﻣﺮﺑﻮﻁ ﺑﻪ ‪ CF3 Cl ، CF2 Cl2 ، CFCl3‬ﻭ ‪ CF4‬ﻣﺸﺎﻫﺪﻩ ﺷﺪﻧﺪ )ﺟﺪﻭﻝ ‪ (1‬ﻧﺴﺒﺖ ﻫﺎﯼ ﺍﻧﺘﮕﺮﺍﻟﯽ ﻃﯿﻒ ‪F NMR‬‬‫ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺗﻌﺪﺍﺩ ﻫﺴﺘﻪ ﻫﺎﯼ ﻓﻠﻮﺋﻮﺭ ﺍﺳﺖ‪.‬‬‫‪F NMR data‬‬‫‪19‬‬‫‪Integration ratio‬‬‫‪CFCl3‬‬‫ﺟﺪﻭﻝ ‪1‬‬‫‪45.0‬‬‫‪ SiF4‬ﻣﻄﺎﺑﻖ ﻣﻌﺎﺩﻟﻪ )‪ (8‬ﻫﯿﺪﺭﻭﻟﯿﺰ ﺷﺪﻩ ﻭ ‪ H2 SiF6‬ﺗﺸﮑﯿﻞ ﻣﯽﺷﻮﺩ‪.‬‬‫‪CF2 Cl2‬‬‫‪65.0‬‬‫‪CF3 Cl‬‬‫‪18.0‬‬‫‪CF4‬‬‫‪2.0‬‬

‫‪Q5-4‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-5 Q-4‬‬‫ﻣﻘﺪﺍﺭ ‪ 10‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ ﺍﺯ ﻣﺤﻠﻮﻝ ‪ H‬ﺑﻪ ﻣﻘﺪﺍﺭ ﺍﺿﺎﻓﯽ ﺁﺏ ﺍﻓﺰﻭﺩﻩ ﺷﺪ ﮐﻪ ﻣﻨﺠﺮ ﺑﻪ ﻫﯿﺪﺭﻭﻟﯿﺰ ﮐﺎﻣﻞ ‪ SiF4‬ﮔﺮﺩﯾﺪ‪ H2 SiF6 .‬ﺗﻮﻟﯿﺪ ﺷﺪﻩ ﺩﺭ ﺍﺛﺮ ﻫﯿﺪﺭﻭﻟﯿﺰ‬‫ﺩﺭ ﻣﺤﻠﻮﻝ ﺁﺑﯽ ‪ ،‬ﭘﺲ ﺍﺯ ﺟﺪﺍﺳﺎﺯﯼ ﺧﻨﺜﯽ ﺷﺪﻩ ﻭ ﮐﺎﻣﻼ ﺑﻪ ‪ Na2 SiF6‬ﺗﺒﺪﯾﻞ ﺷﺪ )ﻣﺤﻠﻮﻝ ﺁﺑﯽ ‪(J‬‬‫ﺭﺳﻮﺏ ﻭﺍﮐﻨﺶ ﻧﺪﺍﺩﻩ ﺍﺯ ‪ Na2 SiF6‬ﻭ ﻫﻤﭽﻨﯿﻦ ‪ NaCl‬ﮐﻪ ﺍﺯ ﻃﺮﯾﻖ ﺻﺎﻑ ﮐﺮﺩﻥ ﺩﺭ ﻣﺮﺣﻠﻪ ﺍﺑﺘﺪﺍﯾﯽ ﺟﺪﺍ ﺷﺪﻩ ﺑﻮﺩﻧﺪ )ﺩﺭ ﻣﺘﻦ ﺯﯾﺮ ﺁﻥ ﺧﻂ ﮐﺸﯿﺪﻩ ﺷﺪﻩ‬‫ﺍﺳﺖ(‪ ،‬ﺑﻪ ﻃﻮﺭ ﮐﺎﻣﻞ ﺩﺭ ﺁﺏ ﺣﻞ ﺷﺪﻧﺪ ﺗﺎ ﯾﮏ ﻣﺤﻠﻮﻝ ﺁﺑﯽ ﺑﺪﺳﺖ ﺁﯾﺪ )‪ 10.0‬ﻟﯿﺘﺮ ﻣﺤﻠﻮﻝ‪( K‬‬‫ﺳﭙﺲ ﺗﯿﺘﺮﺍﺳﯿﻮﻥ ﻫﺎﯼ ﺭﺳﻮﺑﯽ ﺩﯾﮕﺮﯼ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﺤﻠﻮﻝ ‪ G‬ﺍﻧﺠﺎﻡ ﺷﺪ ﻭ ﻧﻘﺎﻁ ﭘﺎﯾﺎﻧﯽ ﺗﯿﺘﺮﺍﺳﯿﻮﻥ ﺑﺎ ‪ G‬ﺑﻪ ﺻﻮﺭﺕ ﺯﯾﺮ ﺑﻮﺩ‪:‬‬‫ﺑﺮﺍﯼ ﻣﺤﻠﻮﻝ ‪) J‬ﻣﻘﺪﺍﺭ ﮐﺎﻣﻞ ﺁﻥ( ﺑﺮﺍﺑﺮ ﺷﺪ ﺑﺎ‪ 61.6 :‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ‬‫ﺑﺮﺍﯼ ‪ 100‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ ﺍﺯ ﻣﺤﻠﻮﻝ ‪ K‬ﺑﺮﺍﺑﺮ ﺷﺪ ﺑﺎ‪ 44.4 :‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ‬‫ﻻﺯﻡ ﺑﻪ ﺫﮐﺮ ﺍﺳﺖ ‪ ،‬ﺣﻀﻮﺭ ﻫﻤﺰﻣﺎﻥ ‪ NaCl‬ﯾﺎ ‪ SiO2‬ﺗﺎﺛﯿﺮﯼ ﺑﺮ ﺗﯿﺘﺮﺍﺳﯿﻮﻥ ﺭﺳﻮﺑﯽ ﻧﺪﺍﺷﺖ‪.‬‬‫‪B.2‬‬‫ﺟﺮﻡ ‪ NaCl‬ﺗﻮﻟﯿﺪ ﺷﺪﻩ ﺩﺭ ﻇﺮﻑ ﻭﺍﮐﻨﺶ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ )ﺍﻃﻼﻋﺎﺗﯽ ﮐﻪ ﺯﯾﺮ ﺁﻥ ﺧﻂ ﮐﺸﯿﺪﻩ ﺷﺪﻩ ﺑﻮﺩ(‪ .‬ﺟﺮﻡ 𝑥‬‫ﺑﺮ ﺣﺴﺐ ﮔﺮﻡ ﻣﺮﺑﻮﻁ ﺑﻪ ‪ Na2 SiF6‬ﮐﻪ ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺎﺩﻩ ﺍﻭﻟﯿﻪ ﺍﺳﺘﻔﺎﺩﻩ ﺷﺪ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪15pt‬‬‫‪B.3‬‬‫ﺟﺮﻡ ‪ CF3 Cl‬ﺗﻮﻟﯿﺪ ﺷﺪﻩ‬‫‪8pt‬‬‫‪ 77.8‬ﺩﺭﺻﺪ ﺍﺯ ‪ CCl4‬ﮐﻪ ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺎﺩﻩ ﺍﻭﻟﯿﻪ ﺍﺳﺘﻔﺎﺩﻩ ﺷﺪ ﻭﺍﮐﻨﺶ ﻧﺪﺍﺩ‪.‬‬‫ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬

IChO

A5-1

IRN-1 C-5 A-1

persian (Iran)‫ﺳﯿﻠﯿﺴﯿﻢ ﻣﺮﻣﻮﺯ‬A ‫ﻗﺴﻤﺖ‬(9 pt) A.1
pt) 3) Cpt) 3) Bpt) 3) A(7 pt) A.2

C6 H6 ∶

kJ mol–1 , C ∶kJ mol–1

IChO
IRN-1 C-5 A-2

A5-2
persian (Iran)(6 pt) A.3

kJ mol–1

Δ𝐻 =(10 pt) A.4
pt) 5) Ept) 5) DIChO
IRN-1 C-5 A-3

A5-3
persian (Iran)B ‫ﻗﺴﻤﺖ‬(5 pt) B.1(15 pt) B.2( ‫)ﺍﺩﺍﻣﻪ ﺩﺭ ﺻﻔﺤﻪ‬IChO
IRN-1 C-5 A-4

A5-4
persian (Iran)(cont.) B.2

NaCl ∶

g, Na2 SiF6 ∶

g

IChO
IRN-1 C-5 A-5

A5-5
persian (Iran)(8 pt) B.3

CF3 Cl ∶g

IRN-1 C-6 C
Seyed
IRN-1 C-6
C-1 Mohammad Hossein Barakati

IChO
Problem 6
Cover sheet

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

Q6-1

IChO
IRN-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
IRN-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
IRN-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
IRN-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
IRN-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
IRN-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

‫‪Q6-1‬‬‫‪IChO‬‬‫‪IRN-1 C-6 Q-1‬‬‫)‪persian (Iran‬‬‫ﺷﯿﻤﯽ ﺣﺎﻟﺖ ﺟﺎﻣﺪ ﻓﻠﺰﺍﺕ ﻭﺍﺳﻄﻪ‬‫‪13% of the total‬‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪6‬‬‫‪A.2‬‬‫‪3‬‬‫‪A.3‬‬‫‪3‬‬‫‪B.1‬‬‫‪6‬‬‫‪B.2‬‬‫‪4‬‬‫‪B.3‬‬‫‪4‬‬‫‪B.4‬‬‫‪4‬‬‫‪C.1‬‬‫‪5‬‬‫‪C.2‬‬‫‪5‬‬‫‪C.3‬‬‫‪5‬‬‫‪Total‬‬‫‪45‬‬‫‪Score‬‬‫ﺁﺗﺸﻔﺸﺎﻧﯽ ﺩﺭ ﺟﺰﯾﺮﻩ ﺳﺎﮐﻮﺭﺍﺟﯿﻤﺎ‬‫ﻗﺴﻤﺖ ‪A‬‬
‫ﺳﯿﻠﯿﮑﺎﺗﯽ ﻣﻮﺟﻮﺩ ﺩﺭ ﻣﻮﺍﺩ ﻣﺬﺍﺏ ﺁﺗﺸﻔﺸﺎﻧﯽ‬‫ﮊﺍﭘﻦ ﯾﮑﯽ ﺍﺯ ﮐﺸﻮﺭﻫﺎﯾﯽ ﺍﺳﺖ ﮐﻪ ﺑﺎﻻﺗﺮﯾﻦ ﺗﻌﺪﺍﺩ ﺁﺗﺸﻔﺸﺎﻥﻫﺎ ﺭﺍ ﺩﺭ ﺟﻬﺎﻥ ﺩﺍﺭﺩ‪ .‬ﻫﻨﮕﺎﻣﯽ ﮐﻪ ﮐﺎﻧﯽﻫﺎﯼ‬‫ِ‬‫)ﻣﺎﮔﻤﺎ( ﮐﺮﯾﺴﺘﺎﻟﯽ ﻣﯽﺷﻮﻧﺪ‪ ،‬ﺑﺨﺸﯽ ﺍﺯ ﯾﻮﻥﻫﺎﯼ ﻓﻠﺰﺍﺕ ﻭﺍﺳﻄﻪ ) ‪ (Mn+‬ﺩﺭ ﻣﺎﮔﻤﺎ ﻭﺍﺭﺩِ ﮐﺎﻧﯽﻫﺎﯼ ﺳﯿﻠﯿﮑﺎﺗﯽ ﻣﯽﺷﻮﻧﺪ‪ .‬ﯾﻮﻥ ‪ Mn+‬ﻣﻮﺭﺩ ﻣﻄﺎﻟﻌﻪ ﺩﺭ ﺍﯾﻦ‬‫ﻣﺴﺎﻟﻪ ﺑﺎ ﯾﻮﻥﻫﺎﯼ ﺍﮐﺴﯿﺪ ) ‪ (O2−‬ﮐﺌﻮﺭﺩﯾﻨﻪ ﻣﯽﺷﻮﺩ‪ .‬ﺳﺎﺧﺘﺎﺭ ﻫﻨﺪﺳﯽ ﺍﻃﺮﺍﻑ ﯾﻮﻥ ﻓﻠﺰ ﺩﺭ ﻣﺎﮔﻤﺎ ﺑﻪﺻﻮﺭﺕ ﭼﻬﺎﺭ ﮐﺌﻮﺭﺩﯾﻨﺎﺳﯿﻮﻧﯽ ﻭ ﭼﻬﺎﺭﻭﺟﻬﯽ ) ‪(𝑇d‬‬‫ﻭ ﺩﺭ ﮐﺎﻧﯽﻫﺎﯼ ﺳﯿﻠﯿﮑﺎﺗﯽ ﺑﻪﺻﻮﺭﺕ ﺷﺶ ﮐﺌﻮﺭﺩﯾﻨﺎﺳﯿﻮﻧﯽ ﻭ ﻫﺸﺖﻭﺟﻬﯽ ) ‪ (𝑂h‬ﺍﺳﺖ ﮐﻪ ﻫﺮ ﺩﻭ ﺣﺎﻟﺖ ﺁﺭﺍﯾﺶ ﺍﻟﮑﺘﺮﻭﻧﯽ ﭘﺮﺍﺳﭙﯿﻦ )‪(high-spin‬‬‫ﺩﺍﺭﻧﺪ‪ .‬ﺿﺮﯾﺐ ﺗﻮﺯﯾﻊ ‪ Mn+‬ﺑﯿﻦ ﮐﺎﻧﯽﻫﺎﯼ ﺳﯿﻠﯿﮑﺎﺗﯽ ﻭ ﻣﺎﮔﻤﺎ )𝐷( ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﭼﻨﯿﻦ ﻧﻮﺷﺖ‪:‬‬‫ﺩﺭ ﺍﯾﻦ ﺭﺍﺑﻄﻪ ‪ [M]s‬ﻭ ‪ [M]l‬ﺑﻪﺗﺮﺗﯿﺐ ﻏﻠﻈﺖ ‪ Mn+‬ﺭﺍ ﺩﺭ ﮐﺎﻧﯽﻫﺎﯼ ﺳﯿﻠﯿﮑﺎﺗﯽ ﻭ ﻣﺎﮔﻤﺎ ﻧﺸﺎﻥ ﻣﯽﺩﻫﻨﺪ‪ .‬ﺩﺭ ﺟﺪﻭﻝ ﺯﯾﺮ ﻣﻘﺎﺩﯾﺮ 𝐷 ﺑﺮﺍﯼ ‪ Cr2+‬ﻭ ‪Mn2+‬‬‫ﺑﻪﻋﻨﻮﺍﻥ ﻧﻤﻮﻧﻪ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫‪Q6-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-6 Q-2‬‬‫ﻣﯿﺪﺍﻥ ‪ 𝑂h‬ﺑﺎﺷﺪ‪ .‬ﻫﻤﭽﻨﯿﻦ ‪ ΔT‬ﻭ ‪CFSET‬‬‫ﻓﺮﺽ ﮐﻨﯿﺪ ‪ ΔO‬ﻓﺎﺻﻠﻪﯼ ﺍﻧﺮﮊﯼ ﺍﻭﺭﺑﯿﺘﺎﻝﻫﺎﯼ ‪ d‬ﯾﻮﻥ ‪ Mn+‬ﻭ ‪ CFSEO‬ﺍﻧﺮﮊﯼ ﭘﺎﯾﺪﺍﺭﯼ ﻣﯿﺪﺍﻥ ﺑﻠﻮﺭ ﺩﺭ ﯾﮏ‬‫ِ‬‫ﻣﯿﺪﺍﻥ ‪ 𝑇d‬ﺩﺭ ﻧﻈﺮ ﺑﮕﯿﺮﯾﺪ‪.‬‬‫ﺭﺍ ﻫﻤﯿﻦ ﮐﻤﯿﺖﻫﺎ ﺩﺭ ﯾﮏ‬‫ِ‬‫‪A.1‬‬‫‪ |CFSEO −CFSET | =ΔCFSE‬ﺭﺍ ﺑﺮﺣﺴﺐ ‪ ΔO‬ﺑﺮﺍﯼ ﯾﻮﻥﻫﺎﯼ ‪ Mn2+ ،Cr2+‬ﻭ ‪ Co2+‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ .‬ﻓﺮﺽ‬‫ﮐﻨﯿﺪ ‪ΔT = 4/9ΔO‬‬‫‪6pt‬‬‫‪A.2‬‬‫ﺑﺎ ﺭﺳﻢ 𝐷‪ ln‬ﺩﺭ ﺑﺮﺍﺑﺮ ‪ ΔCFSE/ΔO‬ﺩﺭ ﺳﯿﺴﺘﻢ ﻣﺨﺘﺼﺎﺕ ﺩﮐﺎﺭﺗﯽ ﺯﯾﺮ‪ ،‬ﺭﺍﺑﻄﻪﺍﯼ ﺧﻄﯽ ﻣﺸﺎﻫﺪﻩ ﻣﯽﺷﻮﺩ‪.‬‬‫ﻣﻘﺪﺍﺭ 𝐷 ﺭﺍ ﺑﺮﺍﯼ ‪ Co2+‬ﺗﺨﻤﯿﻦ ﺑﺰﻧﯿﺪ‪.‬‬‫‪3pt‬‬‫ﺳﺎﺧﺘﺎﺭ ﻧﻤﮏ ﻃﻌﺎﻡ )‪ (NaCl‬ﮐﺮﯾﺴﺘﺎﻟﯽ ﻣﯽﺷﻮﻧﺪ ﮐﻪ ﺩﺭ ﺁﻥ ‪ Mn+‬ﺳﺎﺧﺘﺎﺭ‬‫ﻓﻠﺰﯼ ‪) MO‬ﮐﻪ ‪ M‬ﻣﯽﺗﻮﺍﻧﺪ ‪ Mn ، V ، Ti ، Ca‬ﯾﺎ ‪ Co‬ﺑﺎﺷﺪ( ﺩﺭ‬‫ﺍﮐﺴﯿﺪﻫﺎﯼ‬
‫ِ‬
‫ِ‬
‫ً‬
‫ﻋﻤﺪﺗﺎ ﺑﺮﻫﻢﮐﻨﺶﻫﺎﯼ ﮐﻮﻟﻨﯽ ﺗﻌﯿﯿﻦ ﻣﯽﮐﻨﺪ‪.‬‬‫ﻫﻨﺪﺳﯽ ‪ 𝑂h‬ﺑﺎ ﺁﺭﺍﯾﺶ ﺍﻟﮑﺘﺮﻭﻧﯽ ﭘﺮﺍﺳﭙﯿﻦ )‪ (high-spin‬ﺑﻪ ﺧﻮﺩ ﻣﯽﮔﯿﺮﺩ‪ .‬ﺁﻧﺘﺎﻟﭙﯽ ﺷﺒﮑﻪ ﺍﯾﻦ ﺍﮐﺴﯿﺪﻫﺎ ﺭﺍ‬‫ﺍﯾﻦ ﺑﺮﻫﻢﮐﻨﺶﻫﺎ ﺑﺎ ﺷﻌﺎﻉ ﻭ ﺑﺎﺭ ﯾﻮﻥﻫﺎ ﺗﻌﯿﯿﻦ ﻣﯽﺷﻮﺩ‪ .‬ﻫﻤﭽﻨﯿﻦ ‪ CFSE‬ﻣﺮﺑﻮﻁ ﺑﻪ ‪ Mn+‬ﺩﺭ ﻣﯿﺪﺍﻥ ‪ 𝑂h‬ﻫﻢ ﺳﻬﻢﻫﺎﯾﯽ ﺩﺭ ﺁﻧﺘﺎﻟﭙﯽ ﺷﺒﮑﻪ ﺩﺍﺭﺩ‪.‬‬‫‪A.3‬‬‫ﺍﺯ ﺑﯿﻦ ﮔﺰﯾﻨﻪﻫﺎﯼ )‪ (a‬ﺗﺎ )‪ (f‬ﺩﺭ ﺟﺪﻭﻝ ﺯﯾﺮ‪ ،‬ﻣﺠﻤﻮﻋﻪ ﻣﻨﺎﺳﺐ ﺍﺯ ﺁﻧﺘﺎﻟﭙﯽﻫﺎﯼ ﺷﺒﮑﻪ ] ‪ [kJ mol−1‬ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﮐﻨﯿﺪ‪.‬‬‫‪3pt‬‬

‫‪Q6-3‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-6 Q-3‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫‪2+‬‬‫‪3+‬‬‫ﺍﯼ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺩﺭ ﺷﮑﻞ ‪ 1‬ﮐﺮﯾﺴﺘﺎﻟﯽ ﻣﯽﺷﻮﺩ‪ .‬ﺩﺭ‬‫ﯾﮏ ﺍﮐﺴﯿﺪ ﻣﺨﻠﻮﻁ )‪ (A‬ﮐﻪ ﺷﺎﻣﻞ ﯾﻮﻥﻫﺎﯼ ‪ La‬ﻭ ‪ Cu‬ﺍﺳﺖ‪ ،‬ﺩﺭ ﺳﻠﻮﻝ ﻭﺍﺣﺪ ﭼﻬﺎﺭﮔﻮﺷﻪ ِ‬‫ﻭﺟﻬﯽ ] ‪ [CuO6‬ﻃﻮﻝ ﭘﯿﻮﻧﺪﻫﺎﯼ ‪ Cu-O‬ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻣﺤﻮﺭ ‪ (𝑙𝑧 ) z‬ﺑﻠﻨﺪﺗﺮ ﺍﺯ ﻃﻮﻝ ﭘﯿﻮﻧﺪﻫﺎﯼ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻣﺤﻮﺭ ‪ (𝑙𝑥 ) x‬ﺍﺳﺖ ﻭ ] ‪ [CuO6‬ﻧﺴﺒﺖ‬
‫ﻫﺸﺖ‬
‫ِ‬
‫ﭼﻨﺪﺣﺎﻟﺘﯽ ﺍﻭﺭﺑﯿﺘﺎﻝﻫﺎﯼ ‪ d𝑥2 −𝑦2 ) eg‬ﻭ ‪ (d𝑧2‬ﻣﯽﺷﻮﺩ‪.‬‬‫ﺭﻓﺘﻦ‬‫ﺑﯿﻦ‬‫ﺍﺯ‬‫ﺑﺎﻋﺚ‬‫ﺍﻧﺤﺮﺍﻑ‬‫ﺍﯾﻦ‬‫ﺷﻮﺩ‪.‬‬‫ﻣﯽ‬‫ﻣﻨﺤﺮﻑ‬‫ﻣﻨﻈﻢ‪،‬‬‫𝑂‬‫ﻫﻨﺪﺳﯽ‬‫ﺳﺎﺧﺘﺎﺭ‬
‫ﺑﻪ‬
‫‪h‬‬
‫ِ‬

‫ﺷﮑﻞ ‪1‬‬‫‪ A‬ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﺑﺎ ﺗﺠﺰﯾﻪ ﺣﺮﺍﺭﺗﯽ )ﭘﯿﺮﻭﻟﯿﺰ( ﮐﻤﭙﻠﮑﺲ ‪ B‬ﺳﻨﺘﺰ ﮐﺮﺩ‪ .‬ﮐﻤﭙﻠﮑﺲ ‪ B‬ﺑﺎ ﻣﺨﻠﻮﻁ ﮐﺮﺩﻥ ﮐﻠﺮﯾﺪﻫﺎﯼ ﻓﻠﺰﯼ ﺩﺭ ﻣﺤﻠﻮﻝ ﺁﺑﯽ ﺁﻣﻮﻧﯿﺎﮎ ﺭﻗﯿﻖ )ﺷﺎﻣﻞ‬‫ﭘﯿﺮﻭﻟﯿﺰ ‪ B‬ﺩﺭ ﻫﻮﺍﯼ ﺧﺸﮏ‪ ،‬ﺗﺎ ﺩﻣﺎﯼ ‪ 200‬ﺩﺭﺟﻪ ﺳﺎﻧﺘﯽﮔﺮﺍﺩ ﮐﺎﻫﺶ‬‫ﺍﺳﮑﻮﺍﺭﯾﮏ ﺍﺳﯿﺪ ﺑﺎ ﻓﺮﻣﻮﻝ ‪ C4 H2 O4‬ﮐﻪ ﯾﮏ ﺩﯼﺍﺳﯿﺪ ﺍﺳﺖ( ﺗﺸﮑﯿﻞ ﻣﯽﺷﻮﺩ‪ .‬ﺑﺎ‬‫ِ‬‫ﻭﺯﻥ ‪ 29.1‬ﺩﺭﺻﺪﯼ ﻣﺸﺎﻫﺪﻩ ﻣﯽﺷﻮﺩ ﮐﻪ ﻧﺎﺷﯽ ﺍﺯ ﺍﺯ ﺩﺳﺖ ﺭﻓﺘﻦ ﺁﺏ ﺑﻠﻮﺭﯼ ﺍﺳﺖ‪ .‬ﺑﻪ ﺩﻧﺒﺎﻝ ﺁﻥ‪ ،‬ﺗﺎ ﺩﻣﺎﯼ ‪ 700‬ﺩﺭﺟﻪ ﺳﺎﻧﺘﯽﮔﺮﺍﺩ‪ ،‬ﮐﺎﻫﺶ ﻭﺯﻥ ﺩﯾﮕﺮﯼ‬‫ِ‬‫ﻫﻢ ﺑﻪ ﺩﻟﯿﻞ ﺁﺯﺍﺩ ﺷﺪﻥ ‪ CO2‬ﺭﺥ ﻣﯽﺩﻫﺪ‪ .‬ﮐﺎﻫﺶ ﻭﺯﻥ ﮐﻠﯽ ﻃﯽ ﻓﺮﺍﯾﻨﺪ ﺗﺸﮑﯿﻞ ‪ A‬ﺍﺯ ‪ B‬ﺑﺮﺍﺑﺮ ‪ 63.6‬ﺩﺭﺻﺪ ﺍﺳﺖ‪ .‬ﺗﻮﺟﻪ ﮐﻨﯿﺪ ﮐﻪ ﻃﯽ ﻭﺍﮐﻨﺶ ﭘﯿﺮﻭﻟﯿﺰ‬‫ﻓﻘﻂ ﺁﺏ ﻭ ‪ CO2‬ﺁﺯﺍﺩ ﻣﯽﺷﻮﺩ‪.‬‬‫‪B.1‬‬‫ﻓﺮﻣﻮﻝﻫﺎﯼ ﺷﯿﻤﯿﺎﯾﯽ ‪ A‬ﻭ ‪ B‬ﺭﺍ ﺑﻨﻮﯾﺴﯿﺪ‪.‬‬‫‪6pt‬‬‫‪B.2‬‬‫ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺷﮑﻞ ‪ 𝑙𝑥 ،1‬ﻭ 𝑧𝑙 ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪4pt‬‬‫‪B.3‬‬‫ﺑﺮﺍﯼ ‪ Cu2+‬ﺩﺭ ﻫﺸﺖﻭﺟﻬﯽ ﺍﻧﺤﺮﺍﻑﯾﺎﻓﺘﻪﯼ ] ‪ [CuO6‬ﺩﺭ ﺳﺎﺧﺘﺎﺭ ‪) A‬ﺷﮑﻞ ‪ ،(1‬ﻧﺎﻡ ﺍﻭﺭﺑﯿﺘﺎﻝﻫﺎﯼ ‪ eg‬ﺷﮑﺎﻓﺘﻪﺷﺪﻩ‬‫) ‪ d𝑥2 −𝑦2‬ﻭ ‪ (d𝑧2‬ﺭﺍ ﺩﺭ ﻗﺴﻤﺖﻫﺎﯼ )‪ (i‬ﻭ )‪ (ii‬ﭘﺎﺳﺦﻧﺎﻣﻪ ﺑﻨﻮﯾﺴﯿﺪ ﻭ ﺁﺭﺍﯾﺶ ﺍﻟﮑﺘﺮﻭﻧﯽ ﺭﺍ ﺩﺭ ﮐﺎﺩﺭ ﺧﻂﭼﯿﻦ‬‫ﭘﺎﺳﺦﻧﺎﻣﻪ ﺭﺳﻢ ﮐﻨﯿﺪ‪.‬‬‫‪4pt‬‬

‫‪Q6-4‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-6 Q-4‬‬‫‪ A‬ﯾﮏ ﻋﺎﯾﻖ ﺍﺳﺖ‪ .‬ﻫﻨﮕﺎﻣﯽ ﮐﻪ ﯾﮏ ﯾﻮﻥ ‪ La3+‬ﺟﺎﯾﺶ ﺭﺍ ﺑﻪ ﯾﮏ ‪ Sr2+‬ﻣﯽﺩﻫﺪ‪ ،‬ﺩﺭ ﺷﺒﮑﻪ ﮐﺮﯾﺴﺘﺎﻟﯽ ﺣﻔﺮﻩﺍﯼ ﺍﯾﺠﺎﺩ ﻣﯽﺷﻮﺩ ﮐﻪ ﻣﯽﺗﻮﺍﻧﺪ ﺟﺮﯾﺎﻥ‬‫ﺍﻟﮑﺘﺮﯾﮑﯽ ﺭﺍ ﻫﺪﺍﯾﺖ ﮐﻨﺪ‪ .‬ﺩﺭ ﻧﺘﯿﺠﻪ‪ ،‬ﺳﺎﺧﺘﺎﺭ ‪ A‬ﺩﻭﭖﺷﺪﻩ ﺑﺎ ‪ Sr2+‬ﺩﺭ ﺩﻣﺎﯼ ﮐﻤﺘﺮ ﺍﺯ ‪ 38‬ﮐﻠﻮﯾﻦ‪ ،‬ﺧﺎﺻﯿﺖ ﺍﺑﺮﺭﺳﺎﻧﺎﯾﯽ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ‪ .‬ﻫﻨﮕﺎﻣﯽ ﮐﻪ‬‫ﻭﺍﮐﻨﺶ ﺟﺎﯾﮕﺰﯾﻨﯽ ﺑﺮﺍﯼ ‪ A‬ﺭﺥ ﻣﯽﺩﻫﺪ‪ ،‬ﺩﺭ ﻫﺮ ﻣﺘﺮ ﻣﮑﻌﺐ ﺗﻌﺪﺍﺩ ‪ 2.05 × 1027‬ﺣﻔﺮﻩ ﺍﯾﺠﺎﺩ ﻣﯽﺷﻮﺩ‪.‬‬‫‪B.4‬‬‫ﻗﺴﻤﺖ ‪C‬‬

‫ﺟﺎﯾﮕﺰﯾﻦ ‪ La3+‬ﺷﺪﻩﺍﻧﺪ ﺭﺍ ﺑﺮ ﺍﺳﺎﺱ ﻧﺴﺒﺖﻫﺎﯼ‬‫ﺩﺭﺻﺪ ﯾﻮﻥﻫﺎﯼ ‪ Sr2+‬ﮐﻪ‬‫ﻣﻮﻟﯽ ﻭﺍﮐﻨﺶ ﺟﺎﯾﮕﺰﯾﻨﯽ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫ِ‬‫ِ‬‫ﻭﺍﻻﻧﺲ ﺑﻘﯿﻪﯼ ﯾﻮﻥﻫﺎﯼ ﺗﺸﮑﯿﻞﺩﻫﻨﺪﻩﯼ ﮐﺮﯾﺴﺘﺎﻝ ﻭ ﺳﺎﺧﺘﺎﺭ ﮐﺮﯾﺴﺘﺎﻟﯽ ﺩﺭ ﺍﺛﺮ ﻭﺍﮐﻨﺶ ﺟﺎﯾﮕﺰﯾﻨﯽ‬‫ﺗﻮﺟﻪ ﮐﻨﯿﺪ ﮐﻪ‬‫ِ‬‫ﺗﻐﯿﯿﺮ ﻧﻤﯽﮐﻨﺪ‪.‬‬‫‪−‬‬‫‪4pt‬‬‫‪2+‬‬‫‪ Cu2 (CH3 CO2 )4‬ﺍﺯ ﭼﻬﺎﺭ ﯾﻮﻥ ‪ CH3 CO2‬ﺗﺸﮑﯿﻞ ﺷﺪﻩ ﺍﺳﺖ ﮐﻪ ﺑﻪ ﺩﻭ ﯾﻮﻥ ‪ Cu‬ﮐﺌﻮﺭﺩﯾﻨﻪ ﺷﺪﻩﺍﻧﺪ )ﺷﮑﻞ ‪ .( 2A‬ﺗﺮﮐﯿﺐ ‪Cu2 (CH3 CO2 )4‬‬‫‪−‬‬
‫ﮐﺮﺑﻦ ﭼﻬﺎﺭ ﮔﺮﻭﻩ ‪ CH3 CO2‬ﻋﺒﻮﺭ ﻣﯽﮐﻨﻨﺪ ﻭ ﯾﮏ ﻣﺤﻮﺭ ﺍﺯ ﺩﻭ ﯾﻮﻥ‬‫ﺗﻘﺎﺭﻥ ﺳﺎﺧﺘﺎﺭﯼ ﺑﺎﻻﯾﯽ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ‪ .‬ﺩﺭ ﺍﯾﻦ ﺳﺎﺧﺘﺎﺭ ﺩﻭ ﻣﺤﻮﺭ ﺍﺯ ﺍﺗﻢﻫﺎﯼ‬
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‫‪−‬‬

‫‪2+‬‬‫‪ Cu‬ﻣﯽﮔﺬﺭﺩ‪ .‬ﺗﻤﺎﻡ ﺍﯾﻦ ﺷﺶ ﻣﺤﻮﺭ ﺩﺭ ﻭﺿﻌﯿﺖ ﻋﻤﻮﺩ ﺑﺮ ﻫﻢ ﺟﻬﺖﮔﯿﺮﯼ ﮐﺮﺩﻩﺍﻧﺪ‪ .‬ﻫﻨﮕﺎﻣﯽ ﮐﻪ ﺑﻪ ﺟﺎﯼ ‪ CH3 CO2‬ﺍﺯ ﯾﮏ ﻟﯿﮕﺎﻧﺪ ﺩﯼﮐﺮﺑﻮﮐﺴﯿﻼﺕ‬‫‪2+‬‬‫‪L1‬‬‫ﻣﺴﻄﺢ ‪) L1‬ﺷﮑﻞ ‪ ( 2B‬ﻭ ‪ Cu‬ﺗﺸﮑﯿﻞ‬‫ﮐﺮﺑﻮﮐﺴﯿﻼﺕ‬‫‪ Cu4 (L1‬ﺍﺯ ﺩﯼ‬‫ﻗﻔﺴﯽ ‪L1)4‬‬‫ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ‪ ،‬ﯾﮏ »ﮐﻤﭙﻠﮑﺲ ﻗﻔﺴﯽ« ﺗﺸﮑﯿﻞ ﻣﯽﺷﻮﺩ‪ .‬ﮐﻤﭙﻠﮑﺲ‬
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‫ِ‬
‫ِ‬
‫ﺷﺪﻥ ﺩﻭ ﮔﺮﻭﻩ ﮐﺮﺑﻮﮐﺴﯿﻼﺕ‪ ،‬ﮐﻪ ﺩﺭ ﺷﮑﻞ ‪ 2B‬ﺑﺎ ﺩﻭ ﻋﻼﻣﺖ ﭘﯿﮑﺎﻥ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩﺍﻧﺪ‪،‬‬‫ﺷﺪﻩ ﺍﺳﺖ )ﺷﮑﻞ ‪ .( 2C‬ﺯﺍﻭﯾﻪ ‪ θ‬ﺑﯿﻦ ﺟﻬﺖﻫﺎﯼ ﮐﺌﻮﺭﯾﻨﻪ‬‫ِ‬‫ﺳﺎﺧﺘﺎﺭ ﮐﻤﭙﻠﮑﺲ ﻗﻔﺴﯽ ﺭﺍ ﺗﻌﯿﯿﻦ ﻣﯽﮐﻨﺪ‪ .‬ﺑﺮﺍﯼ ‪ L1‬ﺯﺍﻭﯾﻪ ‪ θ‬ﺑﺮﺍﺑﺮ ﺑﺎ ﺻﻔﺮ ﺩﺭﺟﻪ ﺍﺳﺖ‪ .‬ﺗﻮﺟﻪ ﮐﻨﯿﺪ ﮐﻪ ﺍﺗﻢﻫﺎﯼ ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ﺷﮑﻞ ‪ 2‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ‬‫ﻧﺸﺪﻩﺍﻧﺪ‪.‬‬‫ﺷﮑﻞ ‪2‬‬

‫‪Q6-5‬‬‫)‪persian (Iran‬‬‫‪C.1‬‬‫‪IChO‬‬‫‪IRN-1 C-6 Q-5‬‬‫ﺷﯿﻤﯿﺎﯾﯽ‬‫ﻣﺴﻄﺢ ‪) L2‬ﺷﮑﻞ ﺯﯾﺮ( ﺯﺍﻭﯾﻪ ‪ θ‬ﺩﺭ ﻣﻘﺪﺍﺭ ‪ 90‬ﺩﺭﺟﻪ ﺗﺜﺒﯿﺖ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﺍﮔﺮ ﺗﺮﮐﯿﺐ‬‫ﺑﺮﺍﯼ ﺩﯼﮐﺮﺑﻮﮐﺴﯿﻼﺕ‬‫ِ‬‫ِ‬‫‪2+‬‬‫‪L2‬‬‫‪ Cu𝑛 (L2‬ﺑﺎﺷﺪ‪ ،‬ﮐﻮﭼﮏﺗﺮﯾﻦ ﻣﻘﺎﺩﯾﺮ ﺻﺤﯿﺢ ‪ n‬ﻭ ‪m‬‬‫ﻗﻔﺴﯽ ﺗﺸﮑﯿﻞﺷﺪﻩ ﺍﺯ ‪ L2‬ﻭ ‪ Cu‬ﺑﻪﺻﻮﺭﺕ 𝑚)‪L2‬‬‫ﮐﻤﭙﻠﮑﺲ‬‫ِ‬‫‪−‬‬‫ﺭﺍ ﺑﻨﻮﯾﺴﯿﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﮐﻪ ﻓﻘﻂ ﮔﺮﻭﻩﻫﺎﯼ ‪ CO2‬ﺩﺭ ‪ L2‬ﺑﺎ ﯾﻮﻥﻫﺎﯼ‬‫ﻣﯽﺩﻫﻨﺪ‪.‬‬‫‪2+‬‬‫‪ Cu‬ﭘﯿﻮﻧﺪ ﮐﺌﻮﺭﺩﯾﻨﺎﺳﯿﻮﻧﯽ ﺗﺸﮑﯿﻞ‬‫‪5pt‬‬

‫‪Q6-6‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-6 Q-6‬‬‫‪2+‬‬‫‪−‬‬‫ﮐﻤﭙﻠﮑﺴﯽ ﺍﺯ ﺭﻭﯼ ﺑﺎ ﻓﺮﻣﻮﻝ ‪ Zn4 O(CH3 CO2 )6‬ﺷﺎﻣﻞ ﭼﻬﺎﺭ ﯾﻮﻥ ‪ Zn‬ﭼﻬﺎﺭﻭﺟﻬﯽ‪ ،‬ﺷﺶ ‪ CH3 CO2‬ﻭ ﯾﮏ ‪ O2−‬ﺍﺳﺖ )ﺷﮑﻞ ‪ .( 3A‬ﺩﺭ‬
‫‪−‬‬
‫‪−‬‬
‫ﮐﺮﺑﻦ ‪ CH3 CO2‬ﻣﯽﮔﺬﺭﻧﺪ‪ ،‬ﺑﺮ ﻫﻢ ﻋﻤﻮﺩﻧﺪ‪ .‬ﻫﻨﮕﺎﻣﯽ ﮐﻪ ﺑﻪﺟﺎﯼ ‪CH3 CO2‬‬‫ﺍﯾﻦ ﮐﻤﭙﻠﮑﺲ ‪ O2−‬ﺩﺭ ﻣﺒﺪﺃ ﻗﺮﺍﺭ ﺩﺍﺭﺩ ﻭ ﺳﻪ ﻣﺤﻮﺭﯼ ﮐﻪ ﺍﺯ ﺍﺗﻢﻫﺎﯼ‬‫ِ‬‫‪2+‬‬‫ﺍﺯ ﭘﺎﺭﺍ‪-‬ﺑﻨﺰﻥﺩﯼﮐﺮﺑﻮﮐﺴﯿﻼﺕ )ﺷﮑﻞ ‪ 3B،‬ﺳﺎﺧﺘﺎﺭ ‪ ،L3‬ﺑﺎ ‪ (180° = θ‬ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ‪ ،‬ﮐﻼﺳﺘﺮﻫﺎﯼ ‪ Zn‬ﺑﻪ ﯾﮑﺪﯾﮕﺮ ﻣﺘﺼﻞ ﻣﯽﺷﻮﻧﺪ ﻭ ﯾﮏ ﺟﺎﻣﺪ‬‫‪L3‬‬‫‪ [Zn4 O(L3‬ﺍﺳﺖ ﻭ ﯾﮏ‬‫ﮐﺮﯾﺴﺘﺎﻟﯽ )‪ (X‬ﺗﺸﮑﯿﻞ ﻣﯽﺷﻮﺩ ﮐﻪ ﻧﻮﻋﯽ »ﭘﻠﯿﻤﺮ ﮐﺌﻮﺭﺩﯾﻨﺎﺳﯿﻮﻧﯽ ﻣﺘﺨﻠﺨﻞ« ﺍﺳﺖ )ﺷﮑﻞ ‪ .( 3C‬ﺗﺮﮐﯿﺐ ‪ X‬ﺑﻪﺻﻮﺭﺕ 𝑛] ‪L3)3‬‬‫ﺳﺎﺧﺘﺎﺭ ﮐﺮﯾﺴﺘﺎﻟﯽ ﻣﮑﻌﺒﯽ ﻫﻤﺮﺍﻩ ﺑﺎ ﺣﻔﺮﻩﻫﺎﯾﯽ ﺩﺭ ﺍﻧﺪﺍﺯﻩﯼ ﻧﺎﻧﻮ ﺩﺍﺭﺩ‪ .‬ﺩﺭ ﺷﮑﻞ ‪ 3D‬ﯾﮑﯽ ﺍﺯ ﺍﯾﻦ ﺣﻔﺮﻩﻫﺎ ﺑﺎ ﯾﮏ ﮐﺮﻩ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﻫﻤﭽﻨﯿﻦ‬‫‪2+‬‬‫ﭼﻬﺎﺭﻭﺟﻬﯽ ‪ Zn‬ﺑﺎ ﺷﮑﻞﻫﺎﯼ ﭼﻨﺪﻭﺟﻬﯽ ﺑﻪ ﺭﻧﮓ ﺧﺎﮐﺴﺘﺮﯼ ﺗﯿﺮﻩ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩﺍﻧﺪ‪ .‬ﺗﻮﺟﻪ ﮐﻨﯿﺪ ﮐﻪ ﺍﺗﻢﻫﺎﯼ‬‫ﺩﺭ ﺷﮑﻞﻫﺎﯼ ‪ 3C‬ﻭ ‪ ، 3D‬ﮐﻼﺳﺘﺮﻫﺎﯼ‬‫ِ‬‫ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ﺷﮑﻞ ‪ 3‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﻧﺸﺪﻩﺍﻧﺪ‪.‬‬‫ﺷﮑﻞ ‪3‬‬‫‪C.2‬‬‫ﺗﺮﮐﯿﺐ ‪ X‬ﯾﮏ ﺳﻠﻮﻝ ﻭﺍﺣﺪ ﻣﮑﻌﺒﯽ ﺑﺎ ﻃﻮﻝ ﻟﺒﻪ ‪ a‬ﻭ ﭼﮕﺎﻟﯽ ‪ 0.592 g cm−3‬ﺩﺍﺭﺩ )ﺷﮑﻞ ‪ a .( 3C‬ﺭﺍ ﺑﺮﺣﺴﺐ‬‫‪ cm‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪5pt‬‬‫‪C.3‬‬‫ﺗﺮﮐﯿﺐ ‪ X‬ﺗﻌﺪﺍﺩ ﻗﺎﺑﻞ ﺗﻮﺟﻬﯽ ﺣﻔﺮﻩ ﺩﺍﺭﺩ ﻭ ﯾﮏ ﮔﺮﻡ ﺍﺯ ‪ X‬ﻣﯽﺗﻮﺍﻧﺪ ﺩﺭ ﻓﺸﺎﺭ ‪ 1‬ﺑﺎﺭ ﻭ ﺩﻣﺎﯼ ‪ 25‬ﺩﺭﺟﻪ ﺳﺎﻧﺘﯽﮔﺮﺍﺩ‪،‬‬‫ﻣﻘﺪﺍﺭ ‪ 3.0 × 102‬ﻣﯿﻠﯽﻟﯿﺘﺮ ﮔﺎﺯ ‪ CO2‬ﺭﺍ ﺩﺭ ﺣﻔﺮﻩﻫﺎﯾﺶ ﺟﺎﯼ ﺩﻫﺪ‪ .‬ﻣﯿﺎﻧﮕﯿﻦ ﺗﻌﺪﺍﺩ ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ‪ CO2‬ﺩﺭ ﻫﺮ‬‫ﺣﻔﺮﻩ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪5pt‬‬IChO
IRN-1 C-6 A-1

A6-1
persian (Iran)‫ﺷﯿﻤﯽ ﺣﺎﻟﺖ ﺟﺎﻣﺪ ﻓﻠﺰﺍﺕ ﻭﺍﺳﻄﻪ‬A ‫ﻗﺴﻤﺖ‬(6 pt) A.1

ΔO Co2+ ∶, ΔO Mn2+ ∶,ΔO Cr2+ ∶

IChO

A6-2

IRN-1 C-6 A-2

persian (Iran)(3 pt) A.2

𝐷∶(3 pt) A.3

IChO

A6-3

IRN-1 C-6 A-3

persian (Iran)B ‫ﻗﺴﻤﺖ‬(6 pt) B.1

B∶

,A ∶
(4 pt) B.2

𝑙𝑧 =nm ,𝑙𝑥 =

nm

IChO

A6-4

IRN-1 C-6 A-4

persian (Iran)(4 pt) B.3(ii) ∶,(i) ∶(4 pt) B.4

%

IChO

A6-5

IRN-1 C-6 A-5

persian (Iran)C ‫ﻗﺴﻤﺖ‬(5 pt) C.1𝑚=,𝑛 =(5 pt) C.2

𝑎=

cm

IChO
IRN-1 C-6 A-6

A6-6
persian (Iran)(5 pt) C.3

IRN-1 C-7 C
Seyed
IRN-1 C-7
C-1 Mohammad Hossein Barakati

IChO
Problem 7
Cover sheet

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

Q7-1

IChO
IRN-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
IRN-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
IRN-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
IRN-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
IRN-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‫‪Q7-1‬‬‫‪IChO‬‬‫‪IRN-1 C-7 Q-1‬‬‫)‪persian (Iran‬‬‫ﺑﺎﺯﯼ ﺑﺎ ﺁﺭﻭﻣﺎﺗﯿﺴﯿﺘﻪ ﻏﯿﺮ ﺑﻨﺰﻧﻮﺋﯿﺪﯼ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪5‬‬‫‪13% of the total‬‬‫‪A.2‬‬‫‪2‬‬‫‪A.3‬‬‫‪19‬‬‫‪B.1‬‬‫‪10‬‬‫‪Total‬‬‫‪36‬‬‫‪Score‬‬‫ﭘﺮﻭﻓﺴﻮﺭ ﻧﻮﺯﻭ )‪ (۱۹۹۶–۱۹۰۲‬ﺯﻣﯿﻨﻪ ﺗﺤﻘﯿﻘﺎﺗﯽ ﺗﺮﮐﯿﺒﺎﺕ ﺁﺭﻭﻣﺎﺗﯿﮏ ﻏﯿﺮ ﺑﻨﺰﻧﻮﺋﯿﺪﯼ ﺭﺍ ﺍﻓﺘﺘﺎﺡ ﮐﺮﺩ ﮐﻪ ﻫﻢ ﺍﮐﻨﻮﻥ ﺩﺭ ﻫﻤﻪ ﺟﺎ ﺩﺭ ﺷﯿﻤﯽ ﺁﻟﯽ ﻭﺟﻮﺩ ﺩﺍﺭﺩ‪.‬‬‫ﻗﺴﻤﺖ ‪A‬‬‫‪ Lineariifolianone‬ﯾﮏ ﻣﺤﺼﻮﻝ ﻃﺒﯿﻌﯽ ﺑﺎ ﺳﺎﺧﺘﺎﺭ ﻣﻨﺤﺼﺮ ﺑﻪ ﻓﺮﺩ ﺍﺳﺖ‪ ،‬ﮐﻪ ﺍﺯ ﮔﯿﺎﻩ ‪ linariifolia Inula‬ﮔﺮﻓﺘﻪ ﻣﯽﺷﻮﺩ‪ .‬ﺗﺮﮐﯿﺐ‬‫‪ ، (1) valencene‬ﻃﯽ ﯾﮏ ﻭﺍﮐﻨﺶ ﺗﮏ ﻣﺮﺣﻠﻪﺍﯼ ﺑﻪ ﺗﺮﮐﯿﺐ ‪ 2‬ﺗﺒﺪﯾﻞ ﻣﯽﺷﻮﺩ‪ .‬ﺗﺮﮐﯿﺐ ‪ 2‬ﻧﯿﺰ ﻃﯽ ﺳﻪ ﻣﺮﺣﻠﻪ ﻭﺍﮐﻨﺶ ﮐﻪ ﺑﺎ ﺗﺸﮑﯿﻞ ‪ 3‬ﻫﻤﺮﺍﻩ ﺍﺳﺖ‬‫ﺑﻪ ﮐﺘﻮﻥ ‪ 4‬ﺗﺒﺪﯾﻞ ﻣﯽ ﺷﻮﺩ‪ .‬ﺗﺮﮐﯿﺐ ‪ (5) Eremophilene‬ﺑﺎ ﺍﻧﺠﺎﻡ ﻫﻤﺎﻥ ﭼﻬﺎﺭ ﻣﺮﺣﻠﻪ ﺳﻨﺘﺰ‪ ،‬ﺑﻪ ‪ 6‬ﺗﺒﺪﯾﻞ ﻣﯽ ﺷﻮﺩ‪.‬‬‫ﮔﯿﺎﻩ ‪linariifolia Inula‬‬‫‪Q7-2‬‬‫)‪persian (Iran‬‬‫‪A.1‬‬‫‪IChO‬‬‫‪IRN-1 C-7 Q-2‬‬‫ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ‪ 2‬ﻭ ‪ 6‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ ﻭ ﻫﺮﺟﺎ ﻻﺯﻡ ﺍﺳﺖ ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﺭﺍ ﺑﻪ ﻃﻮﺭ ﻭﺍﺿﺢ ﻧﻤﺎﯾﺶ ﺩﻫﯿﺪ‪.‬‬‫‪5pt‬‬‫ﺳﭙﺲ ﮐﺘﻮﻥ ‪ 4‬ﺑﻪ ﺍﺳﺘﺮ ‪ 15‬ﺗﺒﺪﯾﻞ ﻣﯽﺷﻮﺩ‪ .‬ﺗﻤﺎﻡ ﻣﺮﺍﮐﺰ ﻓﻀﺎﯾﯽ ‪ 7‬ﺩﺭ ﺗﺮﮐﯿﺐ ‪ 8‬ﮐﻪ ﻭﺯﻥ ﻣﻮﻟﮑﻮﻟﯽ ﺁﻥ ‪ 188‬ﺍﺳﺖ‪ ،‬ﺣﻔﻆ ﻣﯽﺷﻮﺩ‪ .‬ﺗﺮﮐﯿﺒﺎﺕ ‪ 9‬ﻭ ‪10‬‬‫ﺩﺍﺭﺍﯼ ﭘﻨﺞ ﻣﺮﮐﺰ ﻓﻀﺎﯾﯽ ﻫﺴﺘﻨﺪ ﻭ ﭘﯿﻮﻧﺪ ﺩﻭﮔﺎﻧﻪ ﮐﺮﺑﻦ‪-‬ﮐﺮﺑﻦ ﻧﺪﺍﺭﻧﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﺍﺯ ‪ H2 18 O‬ﺑﻪ ﺟﺎﯼ ‪ H2 16 O‬ﺑﺮﺍﯼ ﺳﻨﺘﺰ ﻟﯿﻨﺮﯾﻔﻮﻟﯿﻨﻮﻥﻫﺎﯼ ‪ 13‬ﻭ‬‫‪ 14‬ﻋﻼﻣﺖﮔﺬﺍﺭﯼ ﺷﺪﻩ ﺑﺎ ‪ ، 18 O‬ﺑﻪ ﺗﺮﺗﯿﺐ ﺍﺯ ‪ 11‬ﻭ ‪ 12‬ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ‪ .‬ﺗﺮﮐﯿﺒﺎﺕ ‪ 13‬ﻭ ‪ 14‬ﺍﯾﺰﻭﺗﻮﭘﻮﻣﺮﻫﺎﯼ ﻋﻼﻣﺖﮔﺬﺍﺭﯼ ﺷﺪﻩ ﺑﺎ ‪ 18 O‬ﻫﺴﺘﻨﺪ‪ .‬ﺍﮔﺮ‬‫ﻋﻼﻣﺖﮔﺬﺍﺭﯼ ﺍﯾﺰﻭﺗﻮﭘﯽ ﺭﺍ ﺩﺭ ﻧﻈﺮ ﻧﮕﯿﺮﯾﻢ ﻫﺮ ﺩﻭ ﺗﺮﮐﯿﺐ ‪ 13‬ﻭ ‪ 14‬ﺗﺮﮐﯿﺐ ‪ 15‬ﺭﺍ ﺑﺎ ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﯾﮑﺴﺎﻥ ﺑﺪﺳﺖ ﻣﯽﺩﻫﻨﺪ‪.‬‬IChO
IRN-1 C-7 Q-3

Q7-3
persian (Iran)‫‪Q7-4‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-7 Q-4‬‬‫‪A.2‬‬‫ﺳﺎﺧﺘﺎﺭ ﻣﻨﺎﺳﺒﯽ ﺑﺮﺍﯼ ‪ A‬ﺍﻧﺘﺨﺎﺏ ﮐﻨﯿﺪ‪.‬‬‫‪2pt‬‬‫‪A.3‬‬‫ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ﺗﺮﮐﯿﺒﺎﺕ ‪ 14–8‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ ﻭ ﻫﺮ ﺟﺎ ﻻﺯﻡ ﺍﺳﺖ ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﺭﺍ ﺑﻪ ﻃﻮﺭ ﻭﺍﺿﺢ ﻧﻤﺎﯾﺶ ﺩﻫﯿﺪ‪.‬‬‫ﺍﺗﻢﻫﺎﯼ ‪ 18 O‬ﺭﺍ ﺩﺭ ﺗﺮﮐﯿﺒﺎﺕ ‪ 13‬ﻭ ‪ 14‬ﻣﺎﻧﻨﺪ ﻣﺜﺎﻝ ﺯﯾﺮ ﻧﺸﺎﻥ ﺩﻫﯿﺪ‪.‬‬‫‪19pt‬‬‫‪Q7-5‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-7 Q-5‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫ﺗﺮﮐﯿﺐ ‪ 19‬ﺑﻪ ﺻﻮﺭﺕ ﺯﯾﺮ ﺳﻨﺘﺰ ﻣﯽﺷﻮﺩ‪ .‬ﺩﺭ ﺭﺍﺑﻄﻪ ﺑﺎ ﺁﺭﻭﻣﺎﺗﯿﺴﯿﺘﻪ ﻏﯿﺮ ﺑﻨﺰﻧﻮﺋﯿﺪﯼ‪ ،‬ﺍﺯ ﺗﺮﮐﯿﺐ ‪19‬ﻣﯽﺗﻮﺍﻥ ﺑﻪ ﻋﻨﻮﺍﻥ ﻓﻌﺎﻝﮐﻨﻨﺪﻩ ﺍﻟﮑﻞﻫﺎ ﺍﺳﺘﻔﺎﺩﻩ ﮐﺮﺩ‪،‬‬‫ﻭ ﺍﺯ ﻫﻤﯿﻦ ﻃﺮﯾﻖ ﺗﺮﮐﯿﺐ ‪ 20‬ﺍﺯ ﻃﺮﯾﻖ ﺣﺪﻭﺍﺳﻂ ﺟﻔﺖ ﯾﻮﻧﯽ ‪ 21‬ﺑﻪ ‪ 22‬ﺗﺒﺪﯾﻞ ﻣﯽﺷﻮﺩ‪ 21 .‬ﮐﻪ ﺗﺸﮑﯿﻞ ﺁﻥ ﺑﻪ ﮐﻤﮏ ‪ NMR‬ﺗﺎﯾﯿﺪ ﺷﺪﻩ ﺍﺳﺖ ﺑﺘﺪﺭﯾﺞ‬‫ﺗﺠﺰﯾﻪ ﺷﺪﻩ ﻭ ﺑﻪ ‪ 18‬ﻭ ‪ 22‬ﺗﺒﺪﯾﻞ ﻣﯽﺷﻮﺩ‪.‬‬‫‪B.1‬‬‫ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ‪ 19–17‬ﻭ ‪ 21‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ‪ .‬ﻧﻤﺎﯾﺶ ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﺿﺮﻭﺭﯼ ﻧﯿﺴﺖ‪.‬‬‫‪10pt‬‬IChO
IRN-1 C-7 A-1

A7-1
persian (Iran)‫ﺑﺎﺯﯼ ﺑﺎ ﺁﺭﻭﻣﺎﺗﯿﺴﯿﺘﻪ ﻏﯿﺮ ﺑﻨﺰﻭﺋﯿﺪﯼ‬A ‫ﻗﺴﻤﺖ‬(5 pt) A.1
pt) 3) 6pt) 2) 2(2 pt) A.2

IChO
IRN-1 C-7 A-2

A7-2
persian (Iran)(19 pt) A.3

IChO
IRN-1 C-7 A-3

A7-3
persian (Iran)B ‫ًﻗﺴﻤﺖ‬(10 pt) B.1
pt) 2) 18

pt) 2) 17pt) 3) 21pt) 3) 19

IRN-1 C-8 C
Seyed
IRN-1 C-8
C-1 Mohammad Hossein Barakati

IChO
Problem 8
Cover sheet

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

Q8-1

IChO
IRN-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
IRN-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
IRN-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
IRN-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
IRN-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

‫‪Q8-1‬‬‫‪IChO‬‬‫‪IRN-1 C-8 Q-1‬‬‫)‪persian (Iran‬‬‫ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ﺁﻟﯽ ﺩﯾﻨﺎﻣﯿﮏ ﻭ ﮐﺎﯾﺮﺍﻟﯿﺘﻪ ﺁﻥﻫﺎ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪9‬‬‫‪11% of the total‬‬‫‪A.2‬‬‫‪3‬‬‫‪A.3‬‬‫‪7‬‬‫‪B.1‬‬‫‪3‬‬‫‪B.2‬‬‫‪4‬‬‫‪Total‬‬‫‪26‬‬‫‪Score‬‬‫ﻗﺴﻤﺖ ‪A‬‬
‫ﻫﯿﺪﺭﻭﮐﺮﺑﻦﻫﺎﯼ ﺁﺭﻭﻣﺎﺗﯿﮏ ﭼﻨﺪﺣﻠﻘﻪﺍﯼ ﮐﻪ ﺣﻠﻘﻪﻫﺎﯼ ﺁﻥﻫﺎ ﺑﻪ ﻃﻮﺭ ﻣﺘﻮﺍﻟﯽ ﻭ ﺑﻪ ﺷﮑﻞ ﺍﺭﺗﻮ ﺑﻪ ﻫﻢ ﻣﺘﺼﻞ ﻣﯽﺷﻮﻧﺪ‪ [n] ،‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ ﻧﺎﻣﯿﺪﻩ ﻣﯽﺷﻮﻧﺪ‪.‬‬‫‪ (n‬ﺩﺭ ﺍﯾﻨﺠﺎ ﺗﻌﺪﺍﺩ ﺣﻠﻘﻪﻫﺎﯼ ﺷﺶ ﻋﻀﻮﯼ ﺭﺍ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ( ) ﺷﮑﻞ ﭘﺎﯾﯿﻦ ﺭﺍ ﺑﺒﯿﻨﯿﺪ(‪ .‬ﺩﺭ ﯾﮏ ﺭﻭﺵ‪ [4] ،‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ )‪ ([4]C‬ﺑﻪ ﮐﻤﮏ ﯾﮏ ﻭﺍﮐﻨﺶ‬‫ﻧﻮﺭﯼ ﮐﻪ ﺩﺭ ﭘﺎﯾﯿﻦ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ﺑﺎ ﮐﺎﺭﺍﯾﯽ ﺑﺎﻻ ﺍﺯ ﻃﺮﯾﻖ ﺗﺸﮑﯿﻞ ﯾﮏ ﺣﺪﻭﺍﺳﻂ )‪ (Int‬ﺗﻬﯿﻪ ﻣﯽﺷﻮﺩ ﮐﻪ ﺍﯾﻦ ﺣﺪﻭﺍﺳﻂ ﺑﻪﺭﺍﺣﺘﯽ ﺑﻪ ﮐﻤﮏ ﯾﺪ‬‫ﺍﮐﺴﯿﺪ ﻣﯽﺷﻮﺩ‪.‬‬‫ﻭﺍﮐﻨﺶ ﻧﻮﺭﯼ ﻓﻮﻕ ﺑﻪ ﺷﯿﻮﻩﺍﯼ ﺷﺒﯿﻪ ﻭﺍﮐﻨﺶ ﺯﯾﺮ ﺍﻧﺠﺎﻡ ﻣﯽﺷﻮﺩ‪.‬‬‫‪Q8-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-8 Q-2‬‬‫ﺗﻮﺟﻪ ‪ :‬ﺩﺭ ﭘﺎﺳﺦﻫﺎﯼ ﺧﻮﺩ ﺩﺭ ﺗﻤﺎﻡ ﺑﺨﺶﻫﺎﯼ ﺳﻮﺍﻝ ‪ ، 8‬ﭘﯿﻮﻧﺪﻫﺎﯼ ﺳﺎﺩﻩ ﻭ ﺩﻭﮔﺎﻧﻪ ﻣﺘﻨﺎﻭﺏ ﺭﺍ ﺑﻪ ﺻﻮﺭﺗﯽ ﮐﻪ ﺑﺮﺍﯼ ﻣﺜﺎﻝﻫﺎﯼ ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ‬‫ﻧﻤﺎﯾﺶ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ﺭﺳﻢ ﮐﻨﯿﺪ‪ .‬ﺑﺮﺍﯼ ﺳﯿﺴﺘﻢﻫﺎﯼ ﻣﺰﺩﻭﺝ 𝜋 ﺍﺯ ﻋﻼﻣﺖ ﺩﺍﯾﺮﻩ ﺍﺳﺘﻔﺎﺩﻩ ﻧﮑﻨﯿﺪ‪.‬‬‫‪A.1‬‬‫ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ‪ A–C‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ‪ .‬ﺩﺭ ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ﺭﺳﻢ ﺷﺪﻩ ﺍﯾﺰﻣﺮﻫﺎﯼ ﻓﻀﺎﯾﯽ ﺑﺎﯾﺪ ﻗﺎﺑﻞ ﺗﺸﺨﯿﺺ ﺑﺎﺷﻨﺪ‪.‬‬‫‪9pt‬‬‫‪A.2‬‬‫ﺗﻼﺵ ﻫﺎ ﺑﺮﺍﯼ ﺳﻨﺘﺰ ]‪ [5‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ ﺍﺯ ﻫﻤﺎﻥ ﻧﻤﮏ ﻓﺴﻔﻮﻧﯿﻮﻡ ﻭ ﻣﺎﺩﻩ ﺍﻭﻟﯿﻪ ﻣﻨﺎﺳﺐ ﻓﻘﻂ ﻣﻨﺠﺮ ﺑﻪ ﺗﺸﮑﯿﻞ ﻣﻘﺪﺍﺭ‬‫ﺑﺴﯿﺎﺭ ﮐﻤﯽ ﺍﺯ ]‪ [5‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ ﺷﺪ ﻭ ﺩﺭ ﻋﻮﺽ ﻣﺤﺼﻮﻝ ‪ D‬ﮐﻪ ﻭﺯﻥ ﻣﻮﻟﮑﻮﻟﯽ ﺁﻥ ‪ 2‬ﺩﺍﻟﺘﻮﻥ )‪ (Da‬ﮐﻤﺘﺮ ﺍﺯ ]‪[5‬‬‫ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ ﺍﺳﺖ ﺑﻪﺩﺳﺖ ﺁﻣﺪ‪ .‬ﺟﺎﺑﺠﺎﯾﯽﻫﺎﯼ ﺷﯿﻤﯿﺎﯾﯽ ﭘﺮﻭﺗﻮﻥ ‪ NMR‬ﺑﺮﺍﯼ ‪ D‬ﺩﺭ ﭘﺎﯾﯿﻦ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫ﺳﺎﺧﺘﺎﺭ ‪ D‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ‪.‬‬‫‪[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]‪ [5‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ ﻭ ]‪ [n‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦﻫﺎﯼ ﺑﺰﺭﮔﺘﺮ‪ ،‬ﮐﺎﯾﺮﺍﻟﯿﺘﻪ ﻣﺎﺭﭘﯿﭽﯽ ﺩﺍﺭﻧﺪ ﻭ ﺑﻪ ﻫﻢ ﺗﺒﺪﯾﻞ ﺷﺪﻥ ﺍﻧﺎﻧﺘﯿﻮﻣﺮﻫﺎ ﺩﺭ ﺍﯾﻦ ﻣﺎﺭﭘﯿﭻﻫﺎ ﺩﺭ ﺩﻣﺎﯼ ﺍﺗﺎﻕ ﺧﯿﻠﯽ‬‫ﺁﻫﺴﺘﻪ ﺍﺳﺖ‪ .‬ﮐﺎﯾﺮﺍﻟﯿﺘﻪ ]‪ [n‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦﻫﺎ ﺑﻪ ﺻﻮﺭﺕ )‪ (M‬ﯾﺎ )‪ (P‬ﻣﺎﻧﻨﺪ ﺷﮑﻞ ﺯﯾﺮ ﻧﻤﺎﯾﺶ ﺩﺍﺩﻩ ﻣﯽﺷﻮﺩ‪.‬‬‫ﺍﻧﺎﻧﺘﯿﻮﻣﺮﻫﺎﯼ ]‪ [n‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦﻫﺎ ﺑﺎ ‪ n‬ﺑﺰﺭﮔﺘﺮ ﺍﺯ ‪ 4‬ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﺑﻪ ﮐﻤﮏ ﮐﺮﻭﻣﺎﺗﻮﮔﺮﺍﻓﯽ ﺳﺘﻮﻧﯽ ﮐﺎﯾﺮﺍﻝ ﺍﺯ ﻫﻢ ﺟﺪﺍ ﮐﺮﺩ ﮐﻪ ﺑﻮﺳﯿﻠﻪ ﭘﺮﻭﻓﺴﻮﺭ ‪Yoshio‬‬‫‪ Okamoto‬ﺗﻮﺳﻌﻪ ﺩﺍﺩﻩ ﺷﺪ‪.‬‬‫‪Q8-3‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-8 Q-3‬‬‫ﻫﻠﯿﺴﯿﻦﻫﺎﯼ ﭼﻨﺪ ﺗﺎﯾﯽ ﻣﻮﻟﮑﻮﻝﻫﺎﯾﯽ ﺷﺎﻣﻞ ﺩﻭ ﯾﺎ ﭼﻨﺪ ﺳﺎﺧﺘﺎﺭ ﺷﺒﻪ ﻣﺎﺭﭘﯿﭻ ﻫﺴﺘﻨﺪ‪ .‬ﺑﺎ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻦ ﮐﺎﯾﺮﺍﻟﯿﺘﻪ ﻣﺎﺭﭘﯿﭽﯽ ‪ ،‬ﺑﺮﺍﯼ ﯾﮏ ﻫﻠﯿﺴﯿﻦ ﭼﻨﺪ‬‫ﺗﺎﯾﯽ ﭼﻨﺪﯾﻦ ﺍﯾﺰﻭﻣﺮ ﻓﻀﺎﯾﯽ ﻭﺟﻮﺩ ﺩﺍﺭﺩ‪ .‬ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺜﺎﻝ ‪ ،‬ﺗﺮﮐﯿﺐ ‪ ، E‬ﺷﺎﻣﻞ ﺳﻪ ﺑﺨﺶ ﺷﺒﻪ ]‪ [5‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ ﺩﺭ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﺍﺳﺖ‪ .‬ﯾﮑﯽ ﺍﺯ‬‫ﺍﯾﺰﻭﻣﺮﻫﺎﯼ ﻓﻀﺎﯾﯽ ﺁﻥ ﮐﻪ ﺑﻪ ﺻﻮﺭﺕ)‪ (P,P,P‬ﻧﺎﻡﮔﺬﺍﺭﯼ ﺷﺪﻩ ﺩﺭ ﭘﺎﯾﯿﻦ ﻧﻤﺎﯾﺶ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫‪A.3‬‬‫ﺍﺯ ﺗﺮﯼﻣﺮﯾﺰﻩ ﺷﺪﻥ ‪-2،1‬ﺩﯼ ﺑﺮﻭﻣﻮ ﺑﻨﺰﻥ ﺩﺭ ﻣﺠﺎﻭﺭﺕ ﻧﯿﮑﻞ‪ ،‬ﺗﺮﯾﻔﻨﯿﻠﻦ ﺗﻮﻟﯿﺪ ﻣﯽﺷﻮﺩ‪ .‬ﻭﻗﺘﯽ ﻫﻤﯿﻦ ﻭﺍﮐﻨﺶ ﺭﻭﯼ‬‫ﯾﮏ ﺍﻧﺎﻧﺘﯿﻮﻣﺮ )‪ (P‬ﺍﺯ ﺗﺮﮐﯿﺐ‪ F‬ﺍﻧﺠﺎﻡ ﺷﻮﺩ ﻫﻠﯿﺴﯿﻦ ﭼﻨﺪ ﺗﺎﯾﯽ ‪ G‬ﺑﺎ ﻓﺮﻣﻮﻝ ﺑﺴﺘﻪ ‪ C66 H36‬ﺑﺪﺳﺖ ﻣﯽﺁﯾﺪ‪ .‬ﺑﺎ ﺩﺭ‬‫ﻧﻈﺮ ﮔﺮﻓﺘﻦ ﺍﯾﻨﮑﻪ ﺗﺒﺪﯾﻞ ﺍﯾﺰﻭﻣﺮﻫﺎﯼ ﻓﻀﺎﯾﯽ ﺑﻪ ﻫﻤﺪﯾﮕﺮ ﺩﺭ ﺣﯿﻦ ﺍﻧﺠﺎﻡ ﻭﺍﮐﻨﺶ ﺍﺗﻔﺎﻕ ﻧﻤﯽﺍﻓﺘﺪ‪ ،‬ﺗﻤﺎﻡ ﺍﯾﺰﻭﻣﺮﻫﺎﯼ‬‫ﻓﻀﺎﯾﯽ ﻣﺤﺘﻤﻞ ﺑﺮﺍﯼ ‪ G‬ﮐﻪ ﺩﺭ ﺍﯾﻦ ﻓﺮﺍﯾﻨﺪ ﺗﻮﻟﯿﺪ ﻣﯽﺷﻮﻧﺪ ﺭﺍ ﺑﺪﻭﻥ ﺗﮑﺮﺍﺭ ﺷﻨﺎﺳﺎﯾﯽ ﮐﻨﯿﺪ‪ .‬ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺮﺟﻊ ‪ ،‬ﯾﮏ‬‫ﺍﯾﺰﻭﻣﺮ ﺭﺍ ﺑﻪ ﻃﻮﺭ ﮐﺎﻣﻞ ﺭﺳﻢ ﮐﻨﯿﺪ ﻭ ﮐﺎﯾﺮﺍﻟﯿﺘﻪ ﺭﺍ ﻣﺎﻧﻨﺪ ﻣﺜﺎﻝ ﺑﺎﻻ ﻭ ﺑﺎ ﺑﻪﮐﺎﺭﮔﯿﺮﯼ ﻋﻼﻣﺖﻫﺎﯼ ﻋﺪﺩﯼ‪ ،‬ﺗﻌﯿﯿﻦ ﮐﻨﯿﺪ‪.‬‬‫ﺑﻘﯿﻪ ﺍﯾﺰﻭﻣﺮﻫﺎﯼ ﻓﻀﺎﯾﯽ ﺑﺎﯾﺪ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻫﻤﺎﻥ ﺍﻋﺪﺍﺩ ﮐﻪ ﺑﺮﺍﯼ ﻧﻤﺎﯾﺶ ﻣﻮﻗﻌﯿﺖﻫﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺷﺪ ﻭ ﻋﻼﻣﺖﻫﺎﯼ‬‫‪ M‬ﻭ ‪ P‬ﻟﯿﺴﺖ ﺷﻮﻧﺪ‪ .‬ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺜﺎﻝ‪ ،‬ﺳﺎﯾﺮ ﺍﯾﺰﻭﻣﺮﻫﺎﯼ ﻓﻀﺎﯾﯽ ‪ E‬ﺑﺎﯾﺪ ﺑﻪ ﺻﻮﺭﺕ ﺯﯾﺮ ﻟﯿﺴﺖ ﺷﻮﻧﺪ‪:‬‬‫)‪(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)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-8 Q-4‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫ﺳﻮﻣﺎﻧﻦ ﯾﮏ ﻫﯿﺪﺭﻭﮐﺮﺑﻦ ﮐﺎﺳﻪﺍﯼ ﺷﮑﻞ ﺍﺳﺖ ﮐﻪ ﺑﺮﺍﯼ ﺍﻭﻟﯿﻦ ﺑﺎﺭ ﺩﺭ ﺳﺎﻝ ‪ 2003‬ﺩﺭ ﮊﺍﭘﻦ ﮔﺰﺍﺭﺵ ﺷﺪ‪ .‬ﻧﺎﻡ ”‪ ”sumanene‬ﺑﺮﮔﺮﻓﺘﻪ ﺍﺯ ﯾﮏ ﮐﻠﻤﻪ‬‫ﺳﺎﻧﺴﮑﺮﯾﺖ‪-‬ﻫﻨﺪﯼ ”ﺳﻮﻣﺎﻥ” ﺍﺳﺖ ﮐﻪ ﺑﻪ ﻣﻌﻨﯽ ﺁﻓﺘﺎﺑﮕﺮﺩﺍﻥ ﺍﺳﺖ‪.‬‬‫ﺳﻨﺘﺰ ﺳﻮﻣﺎﻧﻦ ﺑﻪ ﮐﻤﮏ ﯾﮏ ﺗﻮﺍﻟﯽ ﺍﺯ ﻭﺍﮐﻨﺶﻫﺎ ﺍﻧﺠﺎﻡ ﺷﺪ ﮐﻪ ﺷﺎﻣﻞ ﯾﮏ ﻣﺮﺣﻠﻪ ﻭﺍﮐﻨﺶ ﻣﺘﺎﺗﺰﯾﺰ ﻫﻤﺮﺍﻩ ﺑﺎ ﺑﺎﺯ ﺷﺪﻥ ﺣﻠﻘﻪ ﻭ ﯾﮏ ﻣﺮﺣﻠﻪ ﻭﺍﮐﻨﺶ ﻣﺘﺎﺗﺰﯾﺰ‬‫ﻫﻤﺮﺍﻩ ﺑﺎ ﺑﺴﺘﻪ ﺷﺪﻥ ﺣﻠﻘﻪ ﺍﺳﺖ‪.‬‬‫ﻧﻤﻮﻧﻪﻫﺎﯾﯽ ﺍﺯ ﻭﺍﮐﻨﺶﻫﺎﯼ ﻣﺘﺎﺗﺰﯾﺰ ﮐﻪ ﺑﻪ ﮐﻤﮏ ﮐﺎﺗﺎﻟﯿﺰﻭﺭ ﺭﻭﺗﻨﯿﻮﻡ )*‪ (Ru‬ﺍﻧﺠﺎﻡ ﻣﯽﺷﻮﻧﺪ ﺩﺭ ﭘﺎﯾﯿﻦ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩﺍﻧﺪ‪.‬‬‫‪B.1‬‬‫ﺳﺎﺧﺘﺎﺭ ﺣﺪﻭﺍﺳﻂ ‪ I‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ‪) .‬ﻧﯿﺎﺯﯼ ﺑﻪ ﻧﻤﺎﯾﺶ ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﻧﯿﺴﺖ(‬‫‪3pt‬‬

‫‪Q8-5‬‬‫)‪persian (Iran‬‬‫‪B.2‬‬‫‪IChO‬‬‫‪IRN-1 C-8 Q-5‬‬‫ﺑﺎ ﺷﺮﻭﻉ ﺍﺯ ﭘﯿﺶﻣﺎﺩﻩ ﻓﻌﺎﻝ ﻧﻮﺭﯼ ‪ ، J‬ﻭ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻫﻤﺎﻥ ﺗﻮﺍﻟﯽ ﻭﺍﮐﻨﺶ ‪ ،‬ﻣﺸﺘﻖ ﻓﻌﺎﻝ ﻧﻮﺭﯼ ﺳﻮﻣﺎﻧﻦ ‪ K‬ﺑﺪﺳﺖ‬‫ﻣﯽﺁﯾﺪ‪ .‬ﻣﺮﺍﮐﺰ ﻓﻀﺎﯾﯽ ﺩﺭ ‪ J‬ﺩﺭ ﺣﯿﻦ ﻭﺍﮐﻨﺶ ﻣﺘﺎﺗﺰﯾﺰ ﺩﭼﺎﺭ ﻭﺍﺭﻭﻧﮕﯽ ﻧﻤﯽﺷﻮﻧﺪ‪ .‬ﺳﺎﺧﺘﺎﺭ ‪ K‬ﺭﺍ ﺑﺎ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻦ‬‫ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﺭﺳﻢ ﮐﻨﯿﺪ‪.‬‬‫‪4pt‬‬IChO
IRN-1 C-8 A-1

A8-1
persian (Iran)‫ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ﺁﻟﯽ ﺩﯾﻨﺎﻣﯿﮏ ﻭ ﮐﺎﯾﺮﺍﻟﯿﺘﻪ ﺁﻥﻫﺎ‬A ‫ﻗﺴﻤﺖ‬(9 pt) A.1
pt) 3) Cpt) 3) Bpt) 3) A(3 pt) A.2

IChO
IRN-1 C-8 A-2

A8-2
persian (Iran)(7 pt) A.3

IChO
IRN-1 C-8 A-3

A8-3
persian (Iran)B ‫ﻗﺴﻤﺖ‬(3 pt) B.1(4 pt) B.2

IRN-1 C-9 C
Seyed
IRN-1 C-9
C-1 Mohammad Hossein Barakati

IChO
Problem 9
Cover sheet

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

Q9-1

IChO
IRN-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
IRN-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
IRN-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
IRN-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
IRN-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

‫‪Q9-1‬‬‫‪IChO‬‬‫‪IRN-1 C-9 Q-1‬‬‫)‪persian (Iran‬‬‫ﭼﯿﺰﻫﺎﯾﯽ ﮐﻪ ﮐﭙﺴﻮﻝ ﺩﻭﺳﺖ ﺩﺍﺭﺩ ﻭ ﭼﯿﺰﻫﺎﯾﯽ ﮐﻪ ﺩﻭﺳﺖ ﻧﺪﺍﺭﺩ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪13‬‬‫‪10% of the total‬‬‫‪A.2‬‬‫‪2‬‬‫‪A.3‬‬‫‪2‬‬‫‪A.4‬‬‫‪3‬‬‫‪A.5‬‬‫‪3‬‬‫‪Total‬‬‫‪23‬‬‫‪Score‬‬‫ﺑﭽﻪﻫﺎﯼ ﺧﻮﺏ ﺍﯾﻦ ﮐﺎﺭ ﺭﺍ ﻧﻤﯽﮐﻨﻨﺪ‪ ،‬ﺍﻣﺎ ﺍﮔﺮ ﺗﻮﭖ ﺗﻨﯿﺲ ﺭﺍ ﺑﺎﺯ ﮐﻨﯿﺪ‪ ،‬ﻣﯽﺗﻮﺍﻧﯿﺪ ﺁﻥ ﺭﺍ ﺑﻪ ﺩﻭ ﻗﻄﻌﻪ ‪ U‬ﺷﮑﻞ ﺗﮑﻪ ﺗﮑﻪ ﮐﻨﯿﺪ‪.‬‬‫ﺑﺮ ﺍﺳﺎﺱ ﺍﯾﻦ ﺍﯾﺪﻩ‪ ،‬ﺗﺮﮐﯿﺒﺎﺕ ‪ 1‬ﻭ ‪ 2‬ﺑﻪﺻﻮﺭﺕ ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ‪ U‬ﺷﮑﻞ ﺑﺎ ﺍﻧﺪﺍﺯﻩﻫﺎﯼ ﻣﺨﺘﻠﻒ ﺳﻨﺘﺰ ﺷﺪﻧﺪ‪ .‬ﺗﺮﮐﯿﺐ ‪ 3‬ﺑﺮﺍﯼ ﻣﻘﺎﯾﺴﻪ ﺑﺎ ‪ 1‬ﺗﻬﯿﻪ ﺷﺪ ﻭ ﺭﻓﺘﺎﺭ‬‫ﮐﭙﺴﻮﻝﺳﺎﺯﯼ ﺍﯾﻦ ﺗﺮﮐﯿﺒﺎﺕ ﻣﻮﺭﺩ ﺑﺮﺭﺳﯽ ﻗﺮﺍﺭ ﮔﺮﻓﺖ‪.‬‬‫‪Q9-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-9 Q-2‬‬‫ﺭﻭﺵ ﺳﻨﺘﺰ ‪ 2‬ﺩﺭ ﭘﺎﯾﯿﻦ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﺩﺭﺻﺪ ﺟﺮﻣﯽ ﻋﻨﺎﺻﺮ ﺩﺭ ﺗﺮﮐﯿﺐ ‪ 9‬ﺑﻪ ﺻﻮﺭﺕ ﺯﯾﺮ ﺍﺳﺖ‪:‬‬‫‪9: C; 40.49%, H; 1.70%, and O; 17.98%‬‬

‫‪Q9-3‬‬‫)‪persian (Iran‬‬‫‪A.1‬‬‫‪IChO‬‬‫‪IRN-1 C-9 Q-3‬‬‫ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ﺗﺮﮐﯿﺒﺎﺕ ‪ 9–4‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ‪ .‬ﻣﯽﺗﻮﺍﻧﯿﺪ ﺍﺯ ﻧﻤﺎﯾﺶ ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﺻﺮﻑﻧﻈﺮ ﮐﻨﯿﺪ‪ .‬ﺍﺯ ”‪ ”PMB‬ﺑﻪ‬‫ﺟﺎﯼ ﺭﺳﻢ ﺳﺎﺧﺘﺎﺭ ﮐﺎﻣﻞ ﮔﺮﻭﻩ ‪ p-methoxybenzyl‬ﮐﻪ ﺩﺭ ﺷﻤﺎﯼ ﺑﺎﻻ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺘﻔﺎﺩﻩ ﮐﻨﯿﺪ‪.‬‬‫‪13pt‬‬‫ﺩﺭ ﻃﯿﻒ ﺟﺮﻣﯽ ‪ ،1‬ﻗﻠﻪ ﻳﻮﻧﯽ ﻣﺮﺑﻮﻁ ﺑﻪ ﺩﻳﻤﺮ ﺁﻥ ﯾﻌﻨﯽ ‪ 1 2‬ﺑﻪ ﻭﺿﻮﺡ ﻣﺸﺎﻫﺪﻩ ﺷﺪ‪ ،‬ﺩﺭ ﺣﺎﻟﯽ ﮐﻪ ﻗﻠﻪ ﻳﻮﻧﯽ ﺑﺮﺍﯼ ‪ 3 2‬ﺩﺭ ﻃﻴﻒ ‪ 3‬ﻣﺸﺎﻫﺪﻩ ﻧﺸﺪ‪ .‬ﺩﺭ ﻃﯿﻒ‬‫ﭘﺮﻭﺗﻮﻥ ‪ NMR‬ﻣﺤﻠﻮﻟﯽ ﺍﺯ ‪ ،112‬ﺗﻤﺎﻡ ﭘﺮﻭﺗﻮﻥ ﻫﺎﯼ ‪ NH‬ﮐﻪ ﺍﺯ ﺗﺮﮐﯿﺐ ‪ 1‬ﻣﯽﺁﯾﻨﺪ ﺑﻪ ﺻﻮﺭﺕ ﻣﻌﺎﺩﻝ ﺷﯿﻤﯿﺎﯾﯽ ﻣﺸﺎﻫﺪﻩ ﺷﺪﻧﺪ ﻭ ﺟﺎﺑﺠﺎﯾﯽ ﺷﯿﻤﯿﺎﯾﯽ ﺁﻧﻬﺎ‬‫ﺑﻪ ﻃﻮﺭ ﻗﺎﺑﻞ ﻣﻼﺣﻈﻪﺍﯼ ﺑﺎ ﭘﺮﻭﺗﻮﻥﻫﺎﯼ ‪ NH‬ﺩﺭ ﺗﺮﮐﯿﺐ ‪ 3‬ﻣﺘﻔﺎﻭﺕ ﺍﺳﺖ‪ .‬ﺍﯾﻦ ﺩﺍﺩﻩﻫﺎ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ ﮐﻪ ﭘﯿﻮﻧﺪﻫﺎﯼ ﻫﯿﺪﺭﻭﮊﻧﯽ ﺑﯿﻦ ‪ NH‬ﺍﺯ ﺗﺮﮐﯿﺐ ‪1‬‬‫ﻭ ﺍﺗﻢﻫﺎﯼ ‪ X‬ﺍﺯ ﻣﻮﻟﮑﻮﻝ ﺩﯾﮕﺮ ‪ 1‬ﺍﯾﺠﺎﺩ ﻣﯽﺷﻮﻧﺪ ﺗﺎ ﮐﭙﺴﻮﻝ ﺩﯾﻤﺮﯼ ﺗﺸﮑﯿﻞ ﺷﻮﺩ‪.‬‬‫‪A.2‬‬‫ﺩﻭﺭ ﻫﻤﻪ ﺍﺗﻢ)ﻫﺎﯼ( ﻣﻨﺎﺳﺐ ‪ X‬ﺩﺭ ‪ 1‬ﺩﺍﯾﺮﻩ ﺑﮑﺸﯿﺪ‪.‬‬‫‪2pt‬‬‫‪A.3‬‬‫ﺗﻌﺪﺍﺩ ﭘﯿﻮﻧﺪﻫﺎﯼ ﻫﯿﺪﺭﻭﮊﻧﯽ ﺭﺍ ﺩﺭ ﮐﭙﺴﻮﻝ ﺩﯾﻤﺮﯼ ) ‪ (112‬ﺗﻌﯿﯿﻦ ﮐﻨﯿﺪ‪.‬‬‫‪2pt‬‬

‫‪Q9-4‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-9 Q-4‬‬‫ﮐﭙﺴﻮﻝ ﺩﯾﻤﺮﯼ ‪ (112 ) 1‬ﺩﺍﺭﺍﯼ ﯾﮏ ﻓﻀﺎﯼ ﺩﺍﺧﻠﯽ ﺍﺳﺖ ﮐﻪ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﮐﻮﭼﮏ ﻣﻨﺎﺳﺐ ‪ Z‬ﻣﯽﺗﻮﺍﻧﺪ ﺩﺭ ﺁﻥ ﻣﺤﺼﻮﺭ ﺷﻮﺩ‪ .‬ﺍﯾﻦ ﭘﺪﯾﺪﻩ ﺑﺎ ﻣﻌﺎﺩﻟﻪ ﺯﯾﺮ ﺑﯿﺎﻥ‬‫ﻣﯽﺷﻮﺩ‪:‬‬‫ﺛﺎﺑﺖ ﺗﻌﺎﺩﻝ ﻣﺤﺼﻮﺭ ﺷﺪﻥ )ﮐﭙﺴﻮﻟﻪ ﺷﺪﻥ( ‪ Z‬ﺩﺭ ‪ 1 2‬ﺩﺭ ﭘﺎﯾﯿﻦ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪:‬‬‫ﻣﺤﺼﻮﺭ ﺷﺪﻥ )ﮐﭙﺴﻮﻟﻪ ﺷﺪﻥ( ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﺩﺭ ﺩﺍﺧﻞ ﮐﭙﺴﻮﻝ ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﺑﻪ ﮐﻤﮏ ﻃﯿﻒﺳﻨﺠﯽ ‪ NMR‬ﺩﻧﺒﺎﻝ ﮐﺮﺩ‪ .‬ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺜﺎﻝ‪ 1 2 ،‬ﺩﺭ ‪C6 D6‬‬
‫ﺳﯿﮕﻨﺎﻝﻫﺎﯼ ﻣﺨﺘﻠﻔﯽ ﺩﺭ ﻃﯿﻒ ﭘﺮﻭﺗﻮﻥ ‪ NMR‬ﺩﺭ ﻗﺒﻞ ﻭ ﺑﻌﺪ ﺍﺯ ﺍﺿﺎﻓﻪ ﮐﺮﺩﻥ ‪ CH4‬ﻣﯽﺩﻫﺪ‪.‬‬‫ﺗﺮﮐﯿﺐ ‪ 2‬ﻫﻢ ﯾﮏ ﮐﭙﺴﻮﻝ ﺳﺨﺖ ﻭ ﺑﺰﺭﮔﺘﺮ ﺩﯾﻤﺮﯼ ﺗﺸﮑﯿﻞ ﻣﯽﺩﻫﺪ ) ‪ .(222‬ﻃﯿﻒ ﭘﺮﻭﺗﻮﻥ ‪ NMR‬ﺑﺮﺍﯼ ‪ 2 2‬ﺩﺭ ‪ ،C6 D5 F ، C6 D6‬ﻭ ﻣﺨﻠﻮﻁ ﺣﻼﻝﻫﺎﯼ‬‫‪ C6 D6 / C6 D5 F‬ﺑﺎ ﺛﺎﺑﺖ ﻧﮕﻪ ﺩﺍﺷﺘﻦ ﺳﺎﯾﺮ ﺷﺮﺍﯾﻂ ﺍﻧﺪﺍﺯﻩﮔﯿﺮﯼ ﺷﺪ‪ .‬ﺟﺎﺑﺠﺎﯾﯽﻫﺎﯼ ﺷﯿﻤﯿﺎﯾﯽ ﺑﺮﺍﯼ ﭘﺮﻭﺗﻮﻥ ‪ Ha‬ﺍﺯ ‪ 2‬ﺩﺭ ﺣﻼﻝﻫﺎﯼ ﺑﺎﻻ ﺩﺭ ﺯﯾﺮ ﺧﻼﺻﻪ‬‫ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﻭ ﻫﯿﭻ ﺳﯿﮕﻨﺎﻝ ﺩﯾﮕﺮﯼ ﺍﺯ ‪ Ha‬ﺩﺭ ‪ ،2‬ﺑﻪ ﺟﺰ ﺳﯿﮕﻨﺎﻝﻫﺎﯼ ﻟﯿﺴﺖ ﺷﺪﻩ ﻣﺸﺎﻫﺪﻩ ﻧﻤﯽﺷﻮﺩ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﮐﻪ ﻓﻀﺎﯼ ﺩﺍﺧﻠﯽ ﮐﭙﺴﻮﻝ ﻫﻤﯿﺸﻪ ﺑﺎ‬‫ﺑﯿﺸﺘﺮﯾﻦ ﺗﻌﺪﺍﺩ ﻣﻤﮑﻦ ﺍﺯ ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ﺣﻼﻝ ﭘﺮ ﻣﯽﺷﻮﺩ ﻭ ﻫﺮ ﺳﯿﮕﻨﺎﻝ ﻣﺮﺑﻮﻁ ﺑﻪ ﯾﮏ ﮔﻮﻧﻪ ﺍﺯ ﮐﭙﺴﻮﻝ ﭘﺮ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫ﺣﻼﻝ‬‫‪C6 D6‬‬‫‪C6 D5 F‬‬‫‪C6 D5 F / C6 D6‬‬‫‪A.4‬‬‫‪𝛿(ppm) of Ha‬‬‫‪4.60‬‬‫‪4.71‬‬‫‪4.82 ,4.71 ,4.60‬‬‫ﺑﺮﺍﯼ ﻫﺮ ﮔﻮﻧﻪ ‪ 2 2‬ﮐﻪ ﯾﮏ ﺳﯿﮕﻨﺎﻝ ﺑﺮﺍﯼ ‪ Ha‬ﻣﯽﺩﻫﺪ ﺗﻌﺪﺍﺩ ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ‪C6 D6‬ﻭ ‪ C6 D5 F‬ﻣﺤﺼﻮﺭ ﺷﺪﻩ ﺩﺭ ﺁﻥ ﺭﺍ‬‫ﺗﻌﯿﯿﻦ ﮐﻨﯿﺪ‪.‬‬‫‪3pt‬‬

‫‪Q9-5‬‬‫‪IChO‬‬‫‪IRN-1 C-9 Q-5‬‬‫)‪persian (Iran‬‬‫ﺍﻧﺪﺍﺯﻩ ﮔﯿﺮﯼﻫﺎﯼ ﭘﺮﻭﺗﻮﻥ ‪ NMR‬ﺩﺭ ‪ C6 D6‬ﻧﺸﺎﻥ ﺩﺍﺩ ﮐﻪ ‪ 2 2‬ﻣﯽﺗﻮﺍﻧﺪ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﺍﺯ ‪-1‬ﺁﺩﺍﻣﺎﻧﺘﺎﻥ ﮐﺮﺑﻮﮐﺴﯿﻠﯿﮏ ﺍﺳﯿﺪ )‪ (AdA‬ﺭﺍ ﺩﺭ ﺧﻮﺩ ﺟﺎﯼ ﺩﻫﺪ ﻭ‬‫ﺛﺎﺑﺖﻫﺎﯼ ﺗﺠﻤﻊ ) ‪ (𝐾a‬ﮐﻪ ﺩﺭ ﭘﺎﯾﯿﻦ ﺑﯿﺎﻥ ﺷﺪﻩ ﺩﺭ ﺩﻣﺎﻫﺎﯼ ﻣﺨﺘﻠﻒ ﺗﻌﯿﯿﻦ ﺷﺪﻧﺪ‪ [solvent@222 ] .‬ﮔﻮﻧﻪﺍﯼ ﺷﺎﻣﻞ ﯾﮏ ﯾﺎ ﭼﻨﺪ ﻣﻮﻟﮑﻮﻝ ﺣﻼﻝ ﺭﺍ ﻧﺸﺎﻥ‬‫ﻣﯽﺩﻫﺪ‪.‬‬‫ﺑﻪ ﻃﻮﺭ ﻣﺸﺎﺑﻪ‪ ،‬ﻣﻘﺎﺩﯾﺮ ‪ 𝐾a‬ﺑﺮﺍﯼ ‪ CH4‬ﻭ ‪ 1 2‬ﺑﻪ ﺻﻮﺭﺗﯽ ﮐﻪ ﺩﺭ ﻣﻌﺎﺩﻟﻪ )‪ (2‬ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ﺩﺭ ﺩﻣﺎﻫﺎﯼ ﻣﺨﺘﻠﻒ ﺩﺭ ‪ C6 D6‬ﺑﻪ ﮐﻤﮏ ﺍﻧﺪﺍﺯﻩﮔﯿﺮﯼﻫﺎﯼ‬‫ﭘﺮﻭﺗﻮﻥ‪ NMR‬ﺗﻌﯿﯿﻦ ﺷﺪﻧﺪ‪ .‬ﻧﻤﻮﺩﺍﺭﻫﺎﯼ ﺩﻭ ﺛﺎﺑﺖ ﺗﺠﻤﻊ ) ﺑﻪ ﺻﻮﺭﺕ ‪ ln 𝐾a‬ﺩﺭ ﻣﻘﺎﺑﻞ ‪ 1‬ﺑﻪ ﺭﻭﯼ ‪ ( T‬ﺩﺭ ﺯﯾﺮ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫‪A.5‬‬‫ﮔﺰﺑﻨﻪ ﺻﺤﯿﺢ ) ‪ A‬ﯾﺎ ‪ ( B‬ﺭﺍ ﺑﺮﺍﯼ ﻫﺮ ﯾﮏ ﺍﺯ ﻣﻮﺍﺭﺩ )‪ (5)–(1‬ﺩﺭ ﭘﺎﺭﺍﮔﺮﺍﻑ ﺯﯾﺮ ﺍﻧﺘﺨﺎﺏ ﮐﻨﯿﺪ‪.‬‬‫ﻫﯿﭻ ﻣﻮﻟﮑﻮﻝ ‪ C6 D6‬ﺩﺭ ‪ 1 2‬ﻣﺤﺼﻮﺭ ﻧﻤﯽﺷﻮﺩ‪ .‬ﺩﺭ ﺧﻂ ‪ ،II‬ﺗﻐﯿﯿﺮ ﺁﻧﺘﺮﻭﭘﯽ )‪ (1) ، (ΔS‬ﺍﺳﺖ ﻭ ﺗﻐﯿﯿﺮ ﺁﻧﺘﺎﻟﭙﯽ‬‫)‪ (2) ، (ΔH‬ﺍﺳﺖ ‪ ،‬ﮐﻪ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ ﺩﺭ ﺧﻂ ‪ II‬ﻧﯿﺮﻭﯼ ﭘﯿﺶﺑﺮﻧﺪﻩ ﺑﺮﺍﯼ ﮐﭙﺴﻮﻝﺳﺎﺯﯼ )‪ (3‬ﺍﺳﺖ ‪ .‬ﺑﻨﺎﺑﺮﺍﯾﻦ ‪ ،‬ﺧﻂ‬‫‪ I‬ﻣﺮﺑﻮﻁ ﺑﻪ )‪ (4‬ﻭ ﺧﻂ ‪ II‬ﻣﺮﺑﻮﻁ ﺑﻪ ) ‪ (5‬ﺍﺳﺖ‪.‬‬
‫‪A‬‬

‫‪B‬‬

‫)‪(1‬‬‫ﻣﺜﺒﺖ‬‫ﻣﻨﻔﯽ‬‫)‪(3‬‬‫𝑆‪Δ‬‬‫𝐻‪Δ‬‬‫‪ 1 2‬ﻭ ‪CH4‬‬‫‪ 2 2‬ﻭ ‪AdA‬‬‫)‪(2‬‬‫)‪(4‬‬‫)‪(5‬‬‫ﻣﺜﺒﺖ‬‫‪ 1 2‬ﻭ ‪CH4‬‬‫ﻣﻨﻔﯽ‬‫‪ 2 2‬ﻭ ‪AdA‬‬‫‪3pt‬‬

‫‪A9-1‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-1 C-9 A-1‬‬‫ﭼﯿﺰﻫﺎﯾﯽ ﮐﻪ ﮐﭙﺴﻮﻝ ﺩﻭﺳﺖ ﺩﺍﺭﺩ ﻭ ﭼﯿﺰﻫﺎﯾﯽ ﮐﻪ ﺩﻭﺳﺖ ﻧﺪﺍﺭﺩ‬‫‪(13 pt) A.1‬‬‫‪pt) 2) 4‬‬‫‪pt) 3) 5‬‬‫‪pt) 2) 6‬‬‫‪pt) 2) 7‬‬‫‪pt) 2) 8‬‬‫‪pt) 2) 9‬‬

IChO

A9-2

IRN-1 C-9 A-2

persian (Iran)(2 pt) A.2(2 pt) A.3(3 pt) A.4
C6 D5 F ‫ﺗﻌﺪﺍﺩ‬C6 D6 ‫ﺗﻌﺪﺍﺩ‬𝛿(ppm) of Ha
4.60 ppm

4.71 ppm

4.82 ppm(3 pt) A.5
(1) ∶

(2) ∶

(3) ∶

(4) ∶

(5) ∶