OMN-4 C-0 C-1

OMN-4 C-0 C
LUJAINA AL BADI

IChO
General instructions
Cover sheet

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

‫‪G0-1‬‬‫)‪Arabic (Oman‬‬‫ﺃﻭﻟﻤﺒﻴﺎﺩ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻌﺎﻟﻤﻲ ‪ 2021‬ﺍﻟﻴﺎﺑﺎﻥ‬‫‪IChO‬‬‫‪OMN-4 C-0 G-1‬‬

‫‪IChO‬‬‫‪G0-2‬‬‫‪OMN-4 C-0 G-2‬‬‫)‪Arabic (Oman‬‬‫ﺗﻌﻠﻴﻤﺎﺕ ﻋﺎﻣﺔ‬‫• ُﻳﺴﻤﺢ ﻟﻚ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻘﻠﻢ ﻓﻘﻂ ﻟﻜﺘﺎﺑﺔ ﺍﻹﺟﺎﺑﺔ‪.‬‬‫• ﻳﺠﺐ ﺃﻥ ﺗﻜﻮﻥ ﺍﻵﻟﺔ ﺍﻟﺤﺎﺳﺒﺔ ﻏﻴﺮ ﻗﺎﺑﻠﺔ ﻟﻠﺒﺮﻣﺠﺔ‪.‬‬‫• ﻳﺘﻀﻤﻦ ﻫﺬﺍ ﺍﻻﻣﺘﺤﺎﻥ ‪ 9‬ﻣﺴﺎﺋﻞ‪.‬‬‫• ﻳﻤﻜﻨﻚ ﺣﻞ ﺍﻟﻤﺴﺎﺋﻞ ﺑﺄﻱ ﺗﺮﺗﻴﺐ‪.‬‬‫• ﺳﻴﻜﻮﻥ ﻟﺪﻳﻚ ‪ 5‬ﺳﺎﻋﺎﺕ ﻟﺤﻞ ﺟﻤﻴﻊ ﺍﻟﻤﺴﺎﺋﻞ‪.‬‬‫• ﻻ ﻳﻤﻜﻨﻚ ﺑﺪﺀ ﺍﻟﻌﻤﻞ ﺇﻻ ﺑﻌﺪ ﺇﻋﻄﺎﺀ ﺍﻷﻣﺮ ﺑﺎﻟﺒﺪﺀ‪.‬‬‫• ﻳﺠﺐ ﻛﺘﺎﺑﺔ ﺟﻤﻴﻊ ﺍﻟﻨﺘﺎﺋﺞ ﻓﻲ ﻣﺮﺑﻌﺎﺕ ﺍﻹﺟﺎﺑﺔ ﺍﻟﻤﻨﺎﺳﺒﺔ ﺑﺎﺳﺘﺨﺪﺍﻡ ﻗﻠﻢ ﻋﻠﻰ ﺃﻭﺭﺍﻕ ﺍﻹﺟﺎﺑﺔ‪ .‬ﺍﺳﺘﺨﺪﻡ ﺍﻟﺠﺰﺀ ﺍﻟﺨﻠﻔﻲ‬‫ﻣﻦ ﺃﻭﺭﺍﻕ ﺍﻷﺳﺌﻠﺔ ﺇﺫﺍ ﻛﻨﺖ ﺑﺤﺎﺟﺔ ﺇﻟﻰ ﻣﺴﻮﺩﺓ‪ .‬ﺗﺬﻛﺮ ﺃﻥ ﺍﻹﺟﺎﺑﺎﺕ ﺍﻟﻤﻜﺘﻮﺑﺔ ﺧﺎﺭﺝ ﻣﺮﺑﻌﺎﺕ ﺍﻹﺟﺎﺑﺎﺕ ﻟﻦ ﻳﺘﻢ ﺗﻘﺪﻳﺮﻫﺎ‪.‬‬‫• ﺍﻛﺘﺐ ﺍﻟﺤﺴﺎﺑﺎﺕ ﺫﺍﺕ ﺍﻟﺼﻠﺔ ﻓﻲ ﺍﻟﻤﺮﺑﻌﺎﺕ ﺍﻟﻤﻨﺎﺳﺒﺔ ﻋﻨﺪ ﺍﻟﻀﺮﻭﺭﺓ‪ .‬ﺳﻴﺘﻢ ﻣﻨﺢ ﺩﺭﺟﺎﺕ ﻛﺎﻣﻠﺔ ﻟﻺﺟﺎﺑﺎﺕ ﺍﻟﺼﺤﻴﺤﺔ ﻓﻘﻂ‬‫ﻋﻨﺪﻣﺎ ﻳﺘﻢ ﻋﺮﺽ ﻋﻤﻠﻚ‪.‬‬‫• ﺳﻴﻘﻮﻡ ﺍﻟﻤﺮﺍﻗﺐ ﺑﺎﻟﺘﺤﺬﻳﺮ ﻗﺒﻞ ‪ 30‬ﺩﻗﻴﻘﺔ ﻣﻦ ﺇﻧﺘﻬﺎﺀ ﺍﻻﻣﺘﺤﺎﻥ‪.‬‬‫• ﻳﺠﺐ ﺃﻥ ﺗﺘﻮﻗﻒ ﻋﻦ ﺍﻟﻌﻤﻞ ﻋﻨﺪ ﺇﻋﻄﺎﺀ ﺍﻷﻣﺮ ﺑﺎﻟﺘﻮﻗﻒ‪ .‬ﺳﻴﺆﺩﻱ ﻋﺪﻡ ﺍﻟﺘﻮﻗﻒ ﻋﻦ ﺍﻟﻜﺘﺎﺑﺔ ﺇﻟﻰ ﺇﺑﻄﺎﻝ ﺍﻣﺘﺤﺎﻧﻚ‪.‬‬‫• ﺍﻟﻨﺴﺨﺔ ﺍﻹﻧﺠﻠﻴﺰﻳﺔ ﺍﻟﺮﺳﻤﻴﺔ ﻣﻦ ﻫﺬﺍ ﺍﻻﺧﺘﺒﺎﺭ ﻣﺘﺎﺣﺔ ﻋﻨﺪ ﺍﻟﻄﻠﺐ ﻓﻘﻂ ﻟﻠﺘﻮﺿﻴﺢ‪.‬‬‫• ﻻ ﻳﺴﻤﺢ ﻟﻚ ﺑﻤﻐﺎﺩﺭﺓ ﻗﺎﻋﺔ ﺍﻷﻣﺘﺤﺎﻥ ﺩﻭﻥ ﺇﺫﻥ‪ .‬ﺇﺫﺍ ﻛﻨﺖ ﺑﺤﺎﺟﺔ ﺇﻟﻰ ﺃﻱ ﻣﺴﺎﻋﺪﺓ )ﺁﻟﺔ ﺣﺎﺳﺒﺔ ﻣﻌﻄﻠﺔ ‪ ،‬ﺗﺤﺘﺎﺝ‬‫ﺍﻟﺬﻫﺎﺏ ﺇﻟﻰ ﺩﻭﺭﺓ ﺍﻟﻤﻴﺎﻩ ‪ ،‬ﻭﻣﺎ ﺇﻟﻰ ﺫﻟﻚ( ‪ ،‬ﺍﺭﻓﻊ ﻳﺪﻙ ﻭﺍﻧﺘﻈﺮ ﺣﺘﻰ ﻭﺻﻮﻝ ﺍﻟﻤﺮﺍﻗﺐ‪.‬‬‫ﺣﻈﺎً ﻣﻮﻓﻘﺎً‬‫ﺟﺪﻭﻝ ﺗﻮﺯﻳﻊ ﺍﻟﺪﺭﺟﺎﺕ‬‫ﺍﻟﻌﻨﻮﺍﻥ‬‫ﻣﺠﻤﻮﻉ ﺍﻟﻨﻘﺎﻁ‬‫ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ‬‫‪35‬‬‫‪11‬‬‫‪1‬‬‫ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻋﻠﻰ ﺳﻄﺢ ﻣﻌﺪﻧﻲ‬‫‪24‬‬‫‪3‬‬‫ﻗﺎﻧﻮﻥ ﻻﻣﺒﺮﺕ ﺑﻴﺮ ؟‬‫‪22‬‬‫‪2‬‬‫ﻛﺒﺴﻮﻟﺔ ﻭﻗﺖ ﺍﻟﻨﻈﺎﺋﺮ‬

‫‪4‬‬
‫‪5‬‬
‫‪6‬‬
‫‪7‬‬
‫‪8‬‬
‫‪9‬‬

‫ﻛﻴﻤﻴﺎﺀ ﺍﻷﻛﺴﺪﺓ ﻭﺍﻻﺧﺘﺰﺍﻝ ﻟﻠﺰﻧﻚ‬‫ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﺍﻟﻐﺎﻣﺾ‬‫ﻛﻴﻤﻴﺎﺀ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺼﻠﺒﺔ ﻟﻠﻤﻌﺎﺩﻥ ﺍﻻﻧﺘﻘﺎﻟﻴﺔ‬‫‪32‬‬‫‪60‬‬‫‪45‬‬‫ﺍﻟﻠﻌﺐ ﺑﺎﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﻌﻄﺮﻳﺔ ﺍﻟﻐﻴﺮ ﺑﻨﺰﻳﻨﻴﺔ‬‫‪36‬‬‫ﻳﺤﺐ ﻭﻳﻜﺮﻩ ﺍﻟﻜﺒﺴﻮﻻﺕ‬‫‪23‬‬‫ﺍﻟﺠﺰﻳﺌﺎﺕ ﺍﻟﻌﻀﻮﻳﺔ ﺍﻟﺪﻳﻨﺎﻣﻴﻜﻴﺔ ﻭﺗﻨﺎﻗﻀﺎﺗﻬﺎ‬‫ﺍﻟﻤﺠﻤﻮﻉ‬‫‪26‬‬‫‪11‬‬‫‪8‬‬‫‪11‬‬‫‪12‬‬‫‪13‬‬‫‪13‬‬‫‪11‬‬‫‪10‬‬‫‪100‬‬

‫‪G0-3‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-0 G-3‬‬‫ﺍﻟﺜﻮﺍﺑﺖ ﻭﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﻔﻴﺰﻳﺎﺋﻴﺔ‬‫ﺍﻝﺙﻭﺍﺏﺕ‬‫ﺳﺮﻋﺔ ﺍﻟﻀﻮﺀ ﻓﻲ ﺍﻟﻔﺮﺍﻍ‬‫ﺛﺎﺑﺖ ﺑﻼﻧﻚ‬‫ﺍﻟﺸﺤﻨﺔ ﺍﻷﻭﻟﻴﺔ‬‫ﻛﺘﻠﺔ ﺍﻹﻟﻜﺘﺮﻭﻥ‬‫ﺛﺎﺑﺖ ﻛﻬﺮﺑﺎﺋﻲ‬‫ﺍﻟﺴﻤﺎﺣﻴﺔ ﻓﻲ ﺍﻟﻔﺮﺍﻍ‬‫ﺛﺎﺑﺖ ﺍﻓﻮﺟﺎﺩﺭﻭ‬‫ﺛﺎﺑﺖ ﺑﻮﻟﺘﺰﻣﺎﻥ‬‫ﺛﺎﺑﺖ ﻓﺎﺭﺍﺩﺍﻱ‬‫ﺛﺎﺑﺖ ﺍﻟﻐﺎﺯ‬‫‪𝑚e = 9.10938370 × 10−31 kg‬‬‫‪𝜀0 = 8.85418781 × 10−12 F m−1‬‬‫‪𝑁A = 6.02214076 × 1023 mol−1‬‬‫‪−1‬‬‫‪−1‬‬‫‪JK‬‬‫‪−23‬‬‫‪𝑘B = 1.380649 × 10‬‬‫‪𝐹 = 𝑁A × 𝑒 = 9.64853321233100184 × 10 C mol‬‬‫‪mol‬‬‫‪−1‬‬‫‪mol−1‬‬‫‪4‬‬‫‪−1‬‬‫‪−1‬‬‫‪𝑅 = 𝑁A × 𝑘B = 8.31446261815324 J K‬‬‫‪L atm K‬‬‫‪−2‬‬‫‪= 8.2057366081 × 10‬‬‫ﻭﺣﺪﺓ ﺍﻟﻜﺘﻞ ﺍﻟﺬﺭﻳﺔ‬‫‪𝑢 = 1 Da = 1.66053907 × 10−27 kg‬‬‫ﺍﻟﻀﻐﻂ ﺍﻟﺠﻮﻱ‬‫‪𝑝atm = 1.01325 × 105 Pa‬‬‫ﺍﻟﻀﻐﻂ ﺍﻟﻘﻴﺎﺳﻲ‬‫‪𝑝 = 1 bar = 105 Pa‬‬‫ﻧﻘﻄﺔ ﺍﻟﺼﻔﺮ ﺍﻟﺴﻠﻴﺰﻳﺔ‬‫‪0 ∘ C = 273.15 K‬‬‫ﺍﻟﺒﻴﻜﻮﻣﺘﺮ‬‫‪1 pm = 10−12 m‬‬‫ﺍﻻﻧﺠﺴﺘﺮﻭﻡ‬‫ﺷﺤﻨﺔ ﺍﻻﻟﻜﺘﺮﻭﻥ‬‫ﺍﻟﻤﻴﻜﺮﻭﻣﺘﺮ‬‫ﻧﺎﻧﻮﻣﺘﺮ‬‫ﺍﻟﺒﻴﻜﻮﻣﺘﺮ‬‫ﺑﺎﻱ‬‫ﻗﺎﻋﺪﺓ ﺍﻟﻠﻮﻏﺎﺭﺗﻢ ﺍﻟﻄﺒﻴﻌﻲ‬‫ﺭﻗﻢ ﻳﻮﻟﺮ‬‫‪1 Å = 10−10 m‬‬

‫‪G0-4‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-0 G-4‬‬‫ﺍﻝﻡﻉﺍﺩﻝﺍﺕ‬‫ﻗﺎﻧﻮﻥ ﺍﻟﻐﺎﺯ ﺍﻟﻤﺜﺎﻟﻲ‬‫ﻗﺎﻧﻮﻥ ﻛﻮﻟﻮﻣﺐ‬‫ﺍﻻﻭﻝ‬‫ﺍﻟﻘﺎﻧﻮﻥ‬‫ﻟﻠﺜﺮﻣﻮﺩﻳﻨﺎﻣﻴﻚ ﺣﻔﻆ ﺍﻟﻄﺎﻗﺔ‬‫ﺍﻻﻧﺜﺎﻟﺒﻴﺔ 𝐻‬‫ﺍﻹﻧﺘﺮﻭﺑﻴﺎ ﺍﻟﻌﺸﻮﺍﺋﻴﺔ ﻋﻠﻰ‬‫ﺃﺳﺎﺱ ﻣﺒﺪﺃ ﺑﻮﻟﺘﺰﻣﺎﻥ‬‫𝑆 ‪on based Entropy‬‬‫‪principle Boltzmann’s‬‬‫ﺍﻟﺘﻐﻴﺮ ﻓﻲ ﺍﻻﻧﺘﺮﻭﺑﻴﺔ ‪The‬‬‫‪Δ𝑆 entropy of change‬‬‫ﻃﺎﻗﺔ‬‫ﺍﻟﻤﺘﺤﺮﺭﺓ‬‫ﺟﻴﺒﺲ‬‫ﻟﻠﻄﺎﻗﺔ‬‫ﺣﺎﺻﻞ ﺍﻟﺘﻔﺎﻋﻞ 𝑄 ‪Reaction‬‬‫‪quotient‬‬‫ﺣﻴﺚ ‪ W‬ﻫﻮ ﻋﺪﺩ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺼﻐﺮﻯ‪.‬‬‫‪,‬ﺣﻴﺚ ‪ 𝑞rev‬ﺍﻟﺤﺮﺍﺭﺓ ﻟﻠﻌﻤﻠﻴﺔ ﺍﻟﻌﻜﺴﻴﺔ‬‫ﺣﻴﺚ ‪ 𝐾 ,‬ﺛﺎﺑﺖ ﺍﻟﺘﻮﺍﺯﻥ ‪ 𝑧 ،‬ﻋﺪﺩ ﺍﻻﻟﻜﺘﺮﻭﻧﺎﺕ ‪ 𝐸 ∘ ،‬ﺍﻟﺠﻬﺪ ﺍﻟﻘﻴﺎﺳﻲ ‪.‬‬‫‪reaction a For‬‬‫ﺣﻴﺚ ]‪ [A‬ﺍﻟﺘﺮﻛﻴﺰ ﺍﻟﻰ ‪A‬‬‫‪IChO‬‬‫‪G0-5‬‬‫‪OMN-4 C-0 G-5‬‬‫)‪Arabic (Oman‬‬‫ﺍﻟﺘﻐﻴﺮ ﻓﻲ ﺍﻟﺤﺮﺍﺭﺓ‬‫ﺣﻴﺚ ‪ 𝑐m‬ﺍﻟﺴﻌﺔ ﺍﻟﺤﺮﺍﺭﻳﺔ ﺍﻟﻨﻮﻋﻴﺔ‬‫ﻣﻌﺎﺩﻟﺔ ﻧﻴﺮﻧﺴﺖ‬‫ﺣﻴﺚ ‪ 𝐶ox‬ﺗﺮﻛﻴﺰ ﺍﻟﻤﺎﺩﺓ ﺍﻟﻤﺆﻛﺴﺪﺓ ‪ 𝐶red ،‬ﺗﺮﻛﻴﺰ ﺍﻟﻤﺎﺩﺓ ﺍﻟﻤﺨﺘﺰﻟﺔ‬‫ﺁﺭﻫﻴﻨﻴﻮﺱ‬‫ﻣﻌﺎﺩﻟﺔ‬‫‪equation Arrhenius‬‬‫ﺣﻴﺚ 𝑘 ﻗﻴﻤﺔ ﺛﺎﺑﺘﺔ ‪ 𝐴 ،‬ﻣﻌﺎﻣﻞ ﺃﺳﻲ ‪ 𝐸𝑎 ،‬ﻃﺎﻗﺔ ﺍﻟﺘﻨﺸﻴﻂ‬‫ﺑﻴﺮ‬‫ﻻﻣﺒﺮﺕ‬‫ﻣﻌﺎﺩﻟﺔ‬‫‪equation Lambert–Beer‬‬‫ﺣﻴﺚ ‪ A‬ﻫﻮ ﺍﻻﻣﺘﺼﺎﺹ ‪ 𝜀 ،‬ﻣﻌﺎﻣﻞ ﺍﻻﻣﺘﺼﺎﺹ ‪ 𝑙 ،‬ﻃﻮﻝ ﺍﻟﻤﺴﺎﺭ ‪ 𝑐 ،‬ﺗﺮﻛﻴﺰ ﺍﻟﻤﺤﻠﻮﻝ‬‫ﻫﻨﺪﺭﺳﻮﻥ‪-‬‬‫ﻣﻌﺎﺩﻟﺔ‬‫)‬‫ﻫﺎﺳﻠﺒﺎﻟﺘﺶ‬‫‪Henderson–Hasselbalch‬‬‫‪( equation‬‬‫ﻃﺎﻗﺔ ﺍﻟﻔﻮﺗﻮﻥ‬‫ﻣﻦ ﺃﺟﻞ ﺍﻻﺗﺰﺍﻥ‬‫ﺣﻴﺚ ‪ 𝐾a‬ﺛﺎﺑﺖ ﺍﻻﺗﺰﺍﻥ‬‫ﺣﻴﺚ 𝜈 ﺍﻟﺘﺮﺩﺩ ‪ 𝜆 ،‬ﺍﻟﻄﻮﻝ ﺍﻟﻤﻮﺟﻲ ﻟﻠﻀﻮﺀ ‪،‬‬‫ﻣﺠﻤﻮﻉ ﺍﻟﺴﻠﺴﻠﺔ ﺍﻟﻬﻨﺪﺳﻴﺔ‬‫‪When‬‬‫ﻟﺤﻞ‬‫‪when ,‬‬‫ﺍﻟﺘﻘﺮﻳﺒﻴﺔ‬‫ﺍﻟﻤﻌﺎﺩﻟﺔ‬‫ﺍﻟﻤﺴﺎﺋﻞ‬‫‪Approximation‬‬‫‪be can that equation‬‬‫‪problems solve to used‬‬IChO
OMN-4 C-0 G-6

G0-6
Arabic (Oman)‫ﺍﻟﺠﺪﻭﻝ ﺍﻟﺪﻭﺭﻱ‬‫‪G0-7‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-0 G-7‬‬‫ﺍﻻﺯﺍﺣﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﺮﻧﻴﻦ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﻟﻠﺒﺮﻭﺗﻮﻥ‬IChO
OMN-4 C-0 G-8

G0-8
Arabic (Oman)

group-substitution: alkyl one for‫ﺍﻻﺯﺍﺣﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻤﺠﻤﻮﻋﺔ ﻭﺍﺣﺪﺓ ﻣﺘﻔﺮﻋﺔ‬IChO
OMN-4 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

OMN-4 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
OMN-4 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
OMN-4 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
OMN-4 C-0 G-5

Heat change Δ𝑞
Nernst equation for redox reaction
Arrhenius equation

G0-5
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
OMN-4 C-0 G-6

Periodic Table

G0-6
English (Official)IChO
OMN-4 C-0 G-7

1

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

G0-7
English (Official)

OMN-4 C-1 C-1

OMN-4 C-1 C
LUJAINA AL BADI

IChO
Problem 1
Cover sheet

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

‫‪Q1-1‬‬‫‪IChO‬‬‫‪OMN-4 C-1 Q-1‬‬‫)‪Arabic (Oman‬‬‫ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻋﻠﻰ ﺳﻄﺢ ﻣﻌﺪﻧﻲ‬‫ﺍﻟﺴﺆﺍﻝ‬‫ﺩﺭﺟﺔ‬‫ﺣﻘﻖ‬‫‪A.1‬‬‫‪6‬‬‫‪A.2‬‬‫‪4‬‬‫‪ % 11‬ﻣﻦ ﺍﻹﺟﻤﺎﻟﻲ‬‫‪B.1‬‬‫‪5‬‬‫‪B.2‬‬‫‪3‬‬‫‪B.3‬‬‫‪3‬‬‫‪B.4‬‬‫‪3‬‬‫ﺍﻟﻤﺠﻤﻮﻉ‬‫‪24‬‬‫ﻣﻦ ﺍﻟﻤﺘﻮﻗﻊ ﺃﻥ ﻳﻜﻮﻥ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻣﺼﺪﺭ ﺍﻟﻄﺎﻗﺔ ﻓﻲ ﺍﻟﻤﺴﺘﻘﺒﻞ ﺍﻟﺬﻱ ﻻ ﻳﻌﺘﻤﺪ ﻋﻠﻰ ﺍﻟﻮﻗﻮﺩ ﺍﻷﺣﻔﻮﺭﻱ‪ .‬ﻫﻨﺎ ‪ ،‬ﺳﻮﻑ‬‫ﻧﻌﺘﺒﺮ ﻋﻤﻠﻴﺔ ﺗﺨﺰﻳﻦ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻓﻲ ﺍﻟﻤﻌﺪﻥ ‪ ،‬ﻭﺍﻟﺘﻲ ﺗﺮﺗﺒﻂ ﺑﺘﻘﻨﻴﺔ ﻧﻘﻞ ﻭﺗﺨﺰﻳﻦ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‪.‬‬‫ﺍﻟﺠﺰﺀ ‪A‬‬

‫ﺃﻭﻻ ﻧﻔﻜﺮ ﻓﻲ ﻋﻤﻠﻴﺔ ﺇﺩﻣﺼﺎﺹ‬‫ﻋﻨﺪﻣﺎ ﻳﺘﻢ ﺇﺩﻣﺼﺎﺹ ﻭﺇﻣﺘﺰﺍﺯ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻓﻲ ﺍﻟﺠﺰﺀ ﺍﻷﻛﺒﺮ ﻣﻦ ﺍﻟﻤﻌﺪﻥ ﻋﺒﺮ ﺳﻄﺤﻪ ‪ ،‬ﺩﻋﻮﻧﺎ‬‫ً‬‫ﻭﺇﻣﺘﺰﺍﺯ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻋﻠﻰ ﺳﻄﺢ ﺍﻟﻤﻌﺪﻥ ‪ ، H2 (g) → 2H(ad) ،‬ﺣﻴﺚ ﻳﺮﻣﺰ ﻟﻠﺤﺎﻟﺔ ﺍﻟﻐﺎﺯﻳﺔ )‪ (g‬ﻭﺍﻟﻤﻤﺘﺰﺓ )‪ (ad‬ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ‪.‬‬‫ﺗﻨﻔﺼﻞ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ‪ H2‬ﺍﻟﺘﻲ ﺗﺼﻞ ﺇﻟﻰ ﺍﻟﺴﻄﺢ ﺍﻟﻤﻌﺪﻧﻲ )‪ (M‬ﻋﻨﺪ ﺍﻟﺴﻄﺢ ﻭﻳﺘﻢ ﺇﻣﺘﺰﺍﺯﻫﺎ ﻋﻠﻰ ﺷﻜﻞ ﺫﺭﺍﺕ ‪H‬‬‫)ﺍﻟﺸﻜﻞ ‪ .(1‬ﻫﻨﺎ ‪ ،‬ﻳﺘﻢ ﺗﻤﺜﻴﻞ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﺎﻣﻨﺔ ﻟـ ‪ H2‬ﺑﻤﺘﻐﻴﺮﻳﻦ‪ :‬ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺍﻟﺬﺭﺍﺕ ‪ d‬ﻭﺍﻻﺭﺗﻔﺎﻉ ﺑﺎﻟﻨﺴﺒﺔ ﻟﺴﻄﺢ ﺍﻟﺬﺭﺓ‬‫ﺩﺍﺋﻤﺎ ﻋﻠﻰ ﺍﻟﺨﻂ ﺍﻟﻤﻨﻘﻂ ﺍﻟﺮﺃﺳﻲ ﻓﻲ‬‫ﺍﻟﻤﻌﺪﻥ ‪ُ . z‬ﻳﻔﺘﺮﺽ ﺃﻥ ﺍﻟﻤﺤﻮﺭ ﻋﻠﻰ ﻃﻮﻝ ﺫﺭﺗﻲ ‪ H‬ﻣﻮﺍﺯﻱ ﻟﻠﺴﻄﺢ ﻭﺃﻥ ﻣﺮﻛﺰ ﺍﻟﺜﻘﻞ‬‫ً‬‫ﺍﻟﺸﻜﻞ ‪ .1‬ﻳﻮﺿﺢ ﺍﻟﺸﻜﻞ ‪ 2‬ﻣﺨﻄﻂ ﻛﻨﺘﻮﺭ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﺎﻣﻨﺔ ﻟﻼﻧﻔﺼﺎﻝ ﻋﻨﺪ ﺍﻟﺴﻄﺢ‪ .‬ﺍﻟﻘﻴﻢ ﺍﻟﻌﺪﺩﻳﺔ ﺗﻤﺜﻞ ﺍﻟﻄﺎﻗﺔ ﺍﻟﻜﺎﻣﻨﺔ‬‫ﺑﻮﺣﺪﺍﺕ ‪ kJ‬ﻟﻜﻞ ﻣﻮﻝ ﻣﻦ ‪ .H2‬ﺗﺒﺎﻋﺪ ﺍﻷﺳﻄﺮ ﺍﻟﺼﻠﺒﺔ ﻫﻮ ‪ , 20 kJ mol−1‬ﻭﺗﺒﺎﻋﺪ ﺍﻷﺳﻄﺮ ﺍﻟﻤﺘﻘﻄﻊ ‪ ، 100 kJ mol−1‬ﻭﺍﻟﺘﺒﺎﻋﺪ‬‫ﺑﻴﻦ ﺍﻟﺨﻄﻮﻁ ﺍﻟﺼﻠﺒﺔ ﻭﺍﻟﻤﺘﻘﻄﻌﺔ ‪ .80 kJ mol−1‬ﻳﺘﻢ ﺍﻫﻤﺎﻝ ﻃﺎﻗﺔ ﺍﻫﺘﺰﺍﺯ ﺍﻟﻨﻘﻄﺔ ﺍﻟﺼﻔﺮﻳﺔ‪.‬‬‫‪Q1-2‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-1 Q-2‬‬‫ﺍﻟﺸﻜﻞ ‪ .1‬ﺗﻌﺮﻳﻒ ﺍﻟﻤﺘﻐﻴﺮﺍﺕ‪ .‬ﺍﻟﺮﺳﻢ ﻟﻴﺲ ﺣﺴﺐ ﺍﻟﻘﻴﺎﺳﺎﺕ ﺍﻟﺤﻘﻴﻘﻴﺔ‪.‬‬‫ﺍﻟﺸﻜﻞ ‪.2‬‬

‫‪Q1-3‬‬‫‪IChO‬‬‫‪OMN-4 C-1 Q-3‬‬‫)‪Arabic (Oman‬‬‫‪A.1‬‬‫‪A.2‬‬‫ﻟﻜﻞ ﻣﻦ ﺍﻷﺳﺌﻠﺔ ﺍﻟﺘﺎﻟﻴﺔ ﻣﻦ )‪ (i‬ﺇﻟﻰ )‪ ، (iii‬ﺇﺧﺘﺮ ﺃﻗﺮﺏ ﻗﻴﻤﺔ ﻣﻦ ‪ A‬ﺇﻟﻰ ‪G‬‬‫)‪ (i‬ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺍﻟﺬﺭﺍﺕ ﻟﺠﺰﻱﺀ ‪ 𝐻2‬ﺍﻟﻐﺎﺯﻱ‬‫)‪ (ii‬ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺍﻟﺬﺭﺍﺕ ﺑﻴﻦ ﺫﺭﺍﺕ ﺍﻟﻤﻌﺪﻥ ) ‪𝑑M‬ﻓﻲ ﺍﻟﺸﻜﻞ ‪(1‬‬‫)‪ (iii‬ﻣﺴﺎﻓﺔ ﺫﺭﺍﺕ ‪ H‬ﺍﻟﻤﻤﺘﺰﺓ ﻣﻦ ﺍﻟﺴﻄﺢ ) ‪ ℎad‬ﻓﻲ ﺍﻟﺸﻜﻞ ‪(1‬‬‫‪A. 0.03 nm‬‬‫‪B. 0.07 nm‬‬‫‪C. 0.11 nm‬‬‫‪E. 0.19 nm‬‬‫‪F. 0.23 nm‬‬‫‪G. 0.27 nm‬‬‫‪D. 0.15 nm‬‬‫ﻟﻜﻞ ﻣﻦ ﺍﻷﺳﺌﻠﺔ ﺍﻟﺘﺎﻟﻴﺔ ﻣﻦ )‪ (i‬ﺇﻟﻰ )‪ ، (ii‬ﺇﺧﺘﺮ ﺃﻗﺮﺏ ﻗﻴﻤﺔ ﻣﻦ ‪ A‬ﺇﻟﻰ ‪H.‬‬‫)‪ (i‬ﺍﻟﻄﺎﻗﺔ ﺍﻟﻤﻄﻠﻮﺑﺔ ﻟﺘﻔﻜﻚ ﻏﺎﺯ ‪ H2‬ﺇﻟﻰ ﻏﺎﺯ ‪H‬ﺣﺴﺐ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ‬‫)‪H2 (g) → 2H(g‬‬‫)‪ (ii‬ﺍﻟﻄﺎﻗﺔ ﺍﻟﻤﻨﺒﻌﺜﺔ ﺃﺛﻨﺎﺀ ﺇﻣﺘﺰﺍﺯ ﺍﻟﻐﺎﺯ ‪ H2‬ﺣﺴﺐ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ‬‫)‪H2 (g) → 2H(ad‬‬‫‪A. 20kJ mol−1‬‬‫‪E. 150kJ mol−1‬‬‫‪B. 40kJ mol−1‬‬‫‪E. 200kJ mol−1‬‬‫‪6pt‬‬‫‪C. 60kJ mol−1‬‬‫‪G. 300kJ mol−1‬‬‫‪D. 100kJ mol−1‬‬‫‪H. 400kJ mol−1‬‬‫‪4pt‬‬

‫‪Q1-4‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-1 Q-4‬‬‫ﺍﻟﺠﺰﺀ ‪B‬‬‫ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﺍﻟﺘﻲ ﺣﺪﺙ ﻟﻬﺎ ﺇﻣﺘﺰﺍﺯ ‪ adsorbed‬ﺇﻣﺎ ﺃﻥ ﺗﻤﺘﺺ ﺇﻟﻰ ﺩﺍﺧﻞ ﺍﻟﻤﻌﺪﻥ‪ ،‬ﺃﻭ ﺗﺘﺤﺪ ﻭﺗﻌﻮﺩ ﺇﻟﻰ ﺍﻟﻄﻮﺭ ﺍﻟﻐﺎﺯﻱ‬‫‪ ،‬ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻦ )‪ (1a‬ﻭ ‪ (1b).‬ﺍﻟﺮﻣﺰ ‪ (ab)H‬ﻳﺮﻣﺰ ﻟﺬﺭﺓ ﻫﻴﺪﺭﻭﺟﻴﻦ ﻣﻤﺘﺼﺔ ﻓﻲ ﺩﺍﺧﻞ ﺍﻟﻤﻌﺪﻥ‪.‬‬‫ﺳﺮﻋﺔ ﺍﻟﺘﻔﺎﻋﻼﺕ ﻟﻜﻞ ﻣﻮﻗﻊ ﺳﻄﺤﻲ ﻟﺘﺤﺮﻳﺮ ﺍﻟﻐﺎﺯ ‪ 𝑟1 ( (adsorption‬ﻭﺍﻻﻣﺘﺰﺍﺯ )‪ ،𝑟2 (desorption‬ﺍﻻﻣﺘﺼﺎﺹ)‪(absorption‬‬‫‪ 𝑟3‬ﺣﻴﺚ ] ‪𝑟1 [s−1 ], 𝑟2 [s−1 ] 𝑟3 [s−1‬‬‫ﻳﻤﻜﻦ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻨﻬﺎ ﻋﻠﻰ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﻟﻲ‪:‬‬‫ﺣﻴﺚ ] ‪ 𝑘1 [s−1 Pa−1 ], 𝑘2 [s−1 ] 𝑘3 [s−1‬ﻫﻲ ﺛﻮﺍﺑﺖ ﺳﺮﻋﺔ ﺍﻟﺘﻔﺎﻋﻞ ﻭ ‪ 𝑃H2‬ﻫﻲ ﺿﻐﻂ ‪ .H2‬ﻣﻦ ﺑﻴﻦ ﺍﻟﻤﻮﺍﻗﻊ ﺍﻟﻤﺘﺎﺣﺔ ﻋﻠﻰ‬‫ﺍﻟﺴﻄﺢ ﻳﻤﺜﻞ 𝜃 ) ‪ ( 0 ≤ 𝜃 ≤ 1‬ﺍﻟﺠﺰﺀ ﺍﻟﺬﻱ ﺗﺸﻐﻠﻪ ﺫﺭﺍﺕ ‪ H.‬ﻭﻳﻤﻜﻦ ﺇﻋﺘﺒﺎﺭ ﺃﻥ ﺍﻻﻣﺘﺰﺍﺯ ﻭﺗﺤﺮﻳﺮ ﺍﻟﻐﺎﺯ ﺳﺮﻳﻌﺎﻥ ﻣﻘﺎﺭﻧﺔ‬‫ﺑﺎﻻﻣﺘﺼﺎﺹ ‪ 𝑟1 , 𝑟2 ≫ 𝑟3‬ﻭﺃﻥ 𝜃 ﺗﻈﻞ ﺛﺎﺑﺘﺔ‪.‬‬‫‪B.1‬‬‫‪ 𝑟3‬ﻳﻤﻜﻦ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻨﻬﺎ ﻋﻠﻰ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﻟﻲ‪:‬‬‫ﻋﺒﺮ ﻋﻦ 𝐶 ﺑﺎﺳﺘﺨﺪﺍﻡ ‪ 𝑘1‬ﻭ ‪.𝑘2‬‬‫‪5pt‬‬

‫‪Q1-5‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-1 Q-5‬‬‫ﺗﻢ ﻭﺿﻊ ﻋﻴﻨﺔ ﻣﻌﺪﻧﻴﺔ ﺑﻤﺴﺎﺣﺔ ﺳﻄﺢ ‪ 𝑆 = 1.0 × 10−3 m2‬ﻓﻲ ﻭﻋﺎﺀ) ‪( =1L1.0 × 10−3 m3‬ﻣﻊ ‪(.𝑃H2 = 1.0 × 102 Pa ) H2‬‬‫ﻛﺎﻧﺖ ﻛﺜﺎﻓﺔ ﻣﻮﺍﻗﻊ ﺇﻣﺘﺰﺍﺯ ﺫﺭﺓ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻋﻠﻰ ﺍﻟﺴﻄﺢ ‪ .𝑁 = 1.3 × 1018 m−2‬ﺗﻢ ﺗﺜﺒﻴﺖ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﺍﻟﺴﻄﺢ ﻋﻨﺪ‬‫‪ .𝑇 = 400 K‬ﻣﻊ ﺍﺳﺘﻤﺮﺍﺭ ﺍﻟﺘﻔﺎﻋﻞ )‪ ، (1‬ﺍﻧﺨﻔﺾ ﺿﻐﻂ ﺍﻟﻐﺎﺯ ‪ 𝑃H2‬ﺑﻤﻌﺪﻝ ﺛﺎﺑﺖ ﻣﻘﺪﺍﺭﺓ ‪ .𝑣 = 4.0 × 10−4 Pa s−1‬ﺍﻓﺘﺮﺽ ﺃﻥ‬‫ﻏﺎﺯ ‪ H2‬ﻫﻮ ﻏﺎﺯ ﻣﺜﺎﻟﻲ ﻭﺃﻥ ﺣﺠﻢ ﻋﻴﻨﺔ ﺍﻟﻤﻌﺪﻥ ﻳﻤﻜﻦ ﺇﻫﻤﺎﻟﻬﺎ‪.‬‬‫‪B.2‬‬‫ﺍﺣﺴﺐ ﻛﻤﻴﺔ ﺫﺭﺍﺕ ‪ H‬ﻓﻲ ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ ﺍﻟﻤﻤﺘﺼﺔ ﻟﻜﻞ ﻭﺣﺪﺓ ﻣﺴﺎﺣﺔ ﻣﻦ ﺍﻟﺴﻄﺢ ﻟﻜﻞ ﻭﺣﺪﺓ‬‫ﺯﻣﻨﻴﺔ ‪𝐴 [mol s−1 m−2 ]،‬‬‫‪3pt‬‬‫‪B.3‬‬‫ﻋﻨﺪ ‪ ، 𝑇 = 400 K‬ﻓﺈﻥ 𝐶 ﺗﺴﺎﻭﻱ ‪ .1.0 × 102 Pa−1‬ﺃﺣﺴﺐ ﻗﻴﻤﺔ ‪ 𝑘3‬ﻋﻨﺪ ‪ 400K.‬ﺇﺫﺍ ﻟﻢ ﺗﺤﺼﻞ‬‫ﻋﻠﻰ ﺇﺟﺎﺑﺔ ‪ ، B.2‬ﻓﺎﺳﺘﺨﺪﻡ ‪𝐴 = 3.6 × 10−7 mol s−1 m−2‬‬‫‪3pt‬‬‫‪B.4‬‬‫ﻋﻨﺪ 𝑇 ﻣﺨﺘﻠﻔﺔ ‪ 𝐶 = 2.5 × 103 Pa−1 ،‬ﻭ ‪ 𝑘3 = 4.8 × 10−2 s−1‬ﻣﻌﻄﺎﺓ‪ .‬ﺑﺎﻟﻨﺴﺒﺔ ﻟـ ‪ r3‬ﻛﺪﺍﻟﺔ‬‫‪ 𝑃H2‬ﻓﻲ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻫﺬﻩ ‪ ،‬ﺇﺧﺘﺎﺭ ﺍﻟﻤﻨﺤﻨﻰ ﺍﻟﺼﺤﻴﺢ ﻣﻦ )‪ (a‬ﺇﻟﻰ)‪.(h‬‬‫‪3pt‬‬

Q1-1

IChO
OMN-4 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
OMN-4 C-1 Q-2

Q1-2

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

Fig.2

English (Official)

Q1-3

IChO
OMN-4 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
OMN-4 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
OMN-4 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

IChO

A1-1

OMN-4 C-1 A-1

Arabic (Oman)‫ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻓﻲ ﺍﻷﺳﻄﺢ ﺍﻟﻤﻌﺪﻧﻴﺔ‬A ‫ﺍﻟﺠﺰﺀ‬(6 pt) A.1
(iii)

(ii)

(i)

(4 pt) A.2
(ii)

(i)

IChO

A1-2

OMN-4 C-1 A-2Arabic (Oman)

B ‫ﺍﻟﺠﺰﺀ‬(5 pt) B.1𝐶=(3 pt) B.2

𝐴=

mol s−1 m−2

IChO

A1-3

OMN-4 C-1 A-3

Arabic (Oman)(3 pt) B.3

𝑘3 =

s−1
(3 pt) B.4

OMN-4 C-2 C-1

OMN-4 C-2 C
LUJAINA AL BADI

IChO
Problem 2
Cover sheet

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

‫‪Q2-1‬‬‫‪IChO‬‬‫‪OMN-4 C-2 Q-1‬‬‫)‪Arabic (Oman‬‬‫ﻛﺒﺴﻮﻟﺔ ﻭﻗﺖ ﺍﻟﻨﻈﺎﺋﺮ ‪Capsule) Time (Isotope‬‬‫ﺍﻟﺴﺆﺍﻝ‬‫ﺩﺭﺟﺔ‬‫ﺍﻟﺪﺭﺟﺔ‬‫‪A.1‬‬‫‪8‬‬‫‪ 11٪‬ﻣﻦ ﺍﻹﺟﻤﺎﻟﻲ‬‫‪A.2‬‬‫‪8‬‬‫‪A.3‬‬‫‪10‬‬‫‪A.4‬‬‫‪9‬‬‫ﺍﻟﻤﺠﻤﻮﻉ‬‫‪35‬‬‫ﺍﻟﺠﺰﺀ‬‫ﺍﻟﻜﻴﺎﻧﺎﺕ ﺍﻟﺠﺰﻳﺌﻴﺔ ﺍﻟﺘﻲ ﺗﺨﺘﻠﻒ ﻓﻘﻂ ﻓﻲ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻨﻈﺎﺋﺮﻱ ‪ ،‬ﻣﺜﻞ ‪ CH4‬ﻭ ‪ ، CH3 D‬ﺗﺴﻤﻰ ﺍﻟﻨﻈﺎﺋﺮ ‪. isotopologues‬‬‫ﺗﻌﺘﺒﺮ ﺍﻟﻨﻈﺎﺋﺮ ﺃﻥ ﻟﻬﺎ ﻧﻔﺲ ﺍﻟﺨﺼﺎﺋﺺ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‪ .‬ﻭﻣﻊ ﺫﻟﻚ ‪ ،‬ﻳﻮﺟﺪ ﺍﺧﺘﻼﻑ ﻃﻔﻴﻒ ﻓﻲ ﺍﻟﻄﺒﻴﻌﺔ‪.‬‬‫ﺍﻓﺘﺮﺽ ﺃﻥ ﺟﻤﻴﻊ ﺍﻟﻤﻮﺍﺩ ﺍﻟﻤﻮﺿﺤﺔ ﻓﻲ ﻫﺬﺍ ﺍﻟﺴﺆﺍﻝ ﻫﻲ ﻓﻲ ﺍﻟﻄﻮﺭ ﺍﻟﻐﺎﺯﻱ‪.‬‬‫ﺩﻋﻮﻧﺎ ﻧﻔﻜﺮ ﻓﻲ ﺍﻹﺗﺰﺍﻥ ﺍﻟﺘﺎﻟﻲ‪:‬‬‫ﺗﺰﺩﺍﺩ ﺍﻹﻧﺘﺮﻭﺑﻴﺎ ‪ ،𝑆 ،‬ﻣﻊ ﺯﻳﺎﺩﺓ ﻋﺪﺩ ﺍﻟﺤﺎﻻﺕ ﺍﻟﻤﺠﻬﺮﻳﺔ ﺍﻟﻤﺤﺘﻤﻠﺔ ﻟﻠﻨﻈﺎﻡ ‪:𝑊 ،‬‬‫ﺣﻴﺚ ‪ 𝑊 = 1‬ﻟـ ‪ 12 C16 O2‬ﻭ ‪ .12 C18 O2 12‬ﻓﻲ ﺍﻟﻤﻘﺎﺑﻞ ‪ 𝑊 = 2 ،‬ﻟﻠﺠﺰﻱﺀ ‪ 12 C16 O18 O‬ﻷﻥ ﺫﺭﺍﺕ ﺍﻷﻛﺴﺠﻴﻦ ﻳﻤﻜﻦ ﺗﻤﻴﻴﺰﻫﺎ‬‫ﻓﻲ ﻫﺬﺍ ﺍﻟﺠﺰﻱﺀ‪ .‬ﻻﻧﻪ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺟﺰﻳﺌﻴﻦ ‪12 C16 O18 O‬ﻓﻲ ﺍﻟﻄﺮﻑ ﺍﻷ ﻟﻤﻌﺎﺩﻟﺔ ﺍﻻﺗﺰﺍﻥ ‪ , 1‬ﻳﻜﻮﻥ ‪.𝑊 = 22 = 4‬‬

‫‪Q2-2‬‬‫)‪Arabic (Oman‬‬‫‪A.1‬‬‫‪IChO‬‬‫‪OMN-4 C-2 Q-2‬‬‫ﺗﻐﻴﻴﺮ ﺍﻟﻤﺤﺘﻮﻯ ﺍﻟﺤﺮﺍﺭﻱ ‪ ، Δ𝐻 ،‬ﻟﻠﻤﻌﺎﺩﻟﺔ ‪ 3‬ﻣﻮﺟﺐ ﺩﺍﺋﻤﺎ ﺑﻐﺾ ﺍﻟﻨﻈﺮ ﻋﻦ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‪.‬‬‫‪8pt‬‬‫ﺟﺪﺍ )ﻓﻜﺮ ﻓﻲ‪ (𝑇 → 0‬ﻭ ﻋﻨﺪ‬‫ﺍﺣﺴﺐ ﺛﻮﺍﺑﺖ ﺍﻹﺗﺰﺍﻥ 𝐾 ﻟﻠﻤﻌﺎﺩﻟﺔ ‪ 3‬ﻋﻨﺪ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﻣﻨﺨﻔﻀﺔ ً‬‫ﺟﺪﺍ )ﻓﻜﺮ ﻓﻲ∞‪ .(𝑇 → +‬ﺍﻓﺘﺮﺽ ﺃﻥ ﺍﻟﺘﻔﺎﻋﻞ ﻻ ﻳﺘﻐﻴﺮ ﻋﻨﺪ ﺩﺭﺟﺎﺕ ﺍﻟﺤﺮﺍﺭﺓ‬‫ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﻋﺎﻟﻴﺔ ً‬‫ﻫﺬﻩ ﻭﺃﻥ 𝐻‪ Δ‬ﺗﺼﺒﺢ ﻗﻴﻤﺔ ﺛﺎﺑﺘﺔ ﻋﻨﺪ ﺩﺭﺟﺎﺕ ﺍﻟﺤﺮﺍﺭﺓ ﺍﻟﻤﺮﺗﻔﻌﺔ‪.‬‬‫ﻳﻤﻜﻦ ﺗﻔﺴﻴﺮ 𝐻‪ ، Δ‬ﻟﻬﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﻣﻦ ﺧﻼﻝ ﺍﻻﻫﺘﺰﺍﺯﺍﺕ ﺍﻟﺠﺰﻳﺌﻴﺔ‬‫ﻋﻨﺪ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ 𝐾 ‪ ،𝑇 = 0‬ﻳﺘﻢ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻦ ﺍﻟﻄﺎﻗﺔ ﺍﻻﻫﺘﺰﺍﺯﻳﺔ ﻟﺠﺰﻱﺀ ﺛﻨﺎﺋﻲ ﺍﻟﺬﺭﺓ ﺗﺮﺩﺩ ﺍﻫﺘﺰﺍﺯﻩ ] ‪ 𝜈 [s−1‬ﻋﻠﻰ ﺍﻟﻨﺤﻮ‬‫ﺍﻟﺘﺎﻟﻲ‪:‬‬‫ﺣﻴﺚ ﺃﻥ 𝑘 ﻫﻮ ﺛﺎﺑﺖ ﺍﻟﻘﻮﺓ ﻭ 𝜇 ﻫﻲ ﺍﻟﻜﺘﻠﺔ ﺍﻟﻤﺨﺘﺰﻟﺔ ‪ ،‬ﻭﺍﻟﺘﻲ ﻳﺘﻢ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻨﻬﺎ ﻣﻦ ﺣﻴﺚ ﻛﺘﻠﺔ ﺍﻟﺬﺭﺗﻴﻦ ﻓﻲ ﺍﻟﺠﺰﻱﺀ ﺛﻨﺎﺋﻲ‬‫ﻭﻓﻘﺎ ﻟﻤﺎ ﻳﻠﻲ‪:‬‬‫ﺍﻟﺬﺭﺓ ‪ 𝑚1 ،‬ﻭ ‪، 𝑚2‬‬
‫ً‬‫‪A.2‬‬‫ﻳﻜﻮﻥ ﺍﻫﺘﺰﺍﺯ ‪ H2‬ﻋﻨﺪ ‪4161.0 cm−1‬ﻋﻨﺪ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻨﻪ ﺑﺎﻟﻌﺪﺩ ﺍﻟﻤﻮﺟﻲ ‪ .‬ﺍﺣﺴﺐ 𝐻‪ Δ‬ﻟﻠﻤﻌﺎﺩﻟﺔ‬‫ﺍﻟﺘﺎﻟﻴﺔ ﻋﻨﺪ 𝐾 ‪ 𝑇 = 0‬ﺑﻮﺣﺪﺍﺕ ‪. J mol−1‬‬‫ﺍﻓﺘﺮﺽ ﺃﻥ‪:‬‬‫‪.‬ﺍﻟﻄﺎﻗﺔ ﺍﻻﻫﺘﺰﺍﺯﻳﺔ ﺗﺴﺎﻫﻢ ﻓﻘﻂ ﻓﻲ ‪. ΔH‬ﻗﻴﻤﺔ 𝑘 ﻟـ ‪ H2‬ﻭ ‪ HD‬ﻭ ‪ D2‬ﻣﺘﻄﺎﺑﻘﺔ‪.‬‬‫‪.‬ﻛﺘﻠﺔ ‪ H‬ﺗﺴﺎﻭﻱ ‪ 1Da‬ﻭ ﻛﺘﻠﺔ ‪ D‬ﺗﺴﺎﻭﻱ ‪2Da‬‬‫‪8pt‬‬

‫‪Q2-3‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-2 Q-3‬‬‫ﺗﻌﺘﻤﺪ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟـ ‪ H2‬ﻭ ‪ HD‬ﻭ ‪ D2‬ﻋﻠﻰ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻓﻲ ﻧﻈﺎﻡ ﻓﻲ ﺣﺎﻟﺔ ﺍﻹﺗﺰﺍﻥ ‪ .‬ﻫﻨﺎ ‪ ،‬ﻳﺘﻢ ﺗﻌﺮﻳﻒ ‪ ΔD2‬ﻋﻠﻰ‬‫ﺃﻧﻪ ﺗﻐﻴﻴﺮ ﻓﻲ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟـ ‪.D2‬‬‫] ‪[D2‬‬‫] ‪[D‬‬‫ﻫﻨﺎ ‪ 𝑅D2 ،‬ﻳﺸﻴﺮ ﺇﻟﻰ ‪ 2‬ﻓﻲ ﺍﻟﻌﻴﻨﺔ ﻭ ‪ 𝑅D∗ 2‬ﺇﻟﻰ‬‫] ‪[H2‬‬‫] ‪[H2‬‬‫ﻋﺸﻮﺍﺋﻴﺎ ﻋﻨﺪ ∞‪. 𝑇 ⟶ +‬‬‫ً‬‫‪A.3‬‬‫ﻋﻨﺪ ∞‪ .𝑇 ⟶ +‬ﻭﺗﺠﺪﺭ ﺍﻹﺷﺎﺭﺓ ﻫﻨﺎ ﺇﻟﻰ ﺃﻥ ﺗﻮﺯﻳﻊ ﺍﻟﻨﻈﺎﺋﺮ ﻳﺼﺒﺢ‬‫ﺍﺣﺴﺐ ‪ ΔD2‬ﻣﻊ ﻧﺴﺒﺔ ﺍﻟﺘﻮﺍﺟﺪ ﺍﻟﻄﺒﻴﻌﻴﺔ )ﺍﻟﻮﻓﺮﺓ ﺍﻟﻄﺒﻴﻌﻴﺔ( ﻟـ ‪ D‬ﻋﻨﺪﻣﺎ ﺗﺘﺒﺎﺩﻝ ﺍﻟﻨﻈﺎﺋﺮ ﻓﻲ‬‫ﺣﺎﻟﺔ ﺍﻹﺗﺰﺍﻥ ﻋﻨﺪ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﺣﻴﺚ 𝐾ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ‪ 4 .‬ﻫﻲ ‪ .0.300‬ﺍﻓﺘﺮﺽ ﺃﻥ ﻧﺴﺐ ﺍﻟﻮﻓﺮﺓ‬‫ﺍﻟﻄﺒﻴﻌﻴﺔ ﻟـ ‪ D‬ﻭ ‪ H‬ﻫﻲ ‪ 1.5576 × 10–4‬ﻭ ‪ ، 1 − 1.5576 × 10–4‬ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‪.‬‬‫‪10pt‬‬‫‪Q2-4‬‬‫‪IChO‬‬‫‪OMN-4 C-2 Q-4‬‬‫)‪Arabic (Oman‬‬‫ﺑﺸﻜﻞ ﻋﺎﻡ ‪ ،‬ﺗﺰﺩﺍﺩ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﻨﻈﺎﺋﺮ ﺍﻟﻤﺴﺘﺒﺪﻟﺔ ﺍﻟﻤﺰﺩﻭﺟﺔ ‪ ،‬ﻭﺍﻟﺘﻲ ﺗﺤﺘﻮﻱ ﻋﻠﻰ ﺫﺭﺗﻴﻦ ﻣﻦ ﺍﻟﻨﻈﺎﺋﺮ ﺍﻟﺜﻘﻴﻠﺔ ﻓﻲ‬‫ﺟﺰﻱﺀ ﻭﺍﺣﺪ ‪ ،‬ﻣﻊ ﺍﻧﺨﻔﺎﺽ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‪ .‬ﺩﻋﻮﻧﺎ ﻧﻌﺘﺒﺮ ﺃﻧﻪ ﻓﻲ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﺠﺰﻳﺌﺎﺕ ‪ CO2‬ﺫﺍﺕ ﺍﻷﻭﺯﺍﻥ ﺍﻟﺠﺰﻳﺌﻴﺔ ‪ 44‬ﻭ‬‫‪ ، 47‬ﻭﺍﻟﺘﻲ ﺗﻮﺻﻒ ﺃﺩﻧﺎﻩ ﺑـ ]‪ CO2 [44‬ﻭ ]‪ .CO2 [47‬ﻳﺘﻢ ﺗﻌﺮﻳﻒ ﺍﻟﻜﻤﻴﺔ ‪ Δ47‬ﻋﻠﻰ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﻟﻲ‪:‬‬‫]]‪[CO2 [47‬‬‫]]‪[CO2 [47‬‬‫∗‬‫‪ 𝑅47‬ﺇﻟﻰ‬‫ﻓﻲ ﺍﻟﻌﻴﻨﺔ ﻭ‬‫ﻳﺸﻴﺮ ‪ 𝑅47‬ﺇﻟﻰ‬‫]]‪[CO2 [44‬‬‫]]‪[CO2 [44‬‬‫ﻣﻮﺿﺤﺔ ﺃﺩﻧﺎﻩ ؛ ﺗﺠﺎﻫﻞ ﺍﻟﻨﻈﺎﺋﺮ ﻏﻴﺮ ﺍﻟﻤﻌﺮﻭﺿﺔ ﻫﻨﺎ‪.‬‬‫ﻋﻨﺪ ∞‪ . 𝑇 → +‬ﺍﻟﻮﻓﺮﺓ ﺍﻟﻄﺒﻴﻌﻴﺔ ﻟﺬﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ﻭﺍﻷﻛﺴﺠﻴﻦ‬‫ﺍﻟﻨﻈﻴﺮ‬‫‪C‬‬‫ﺍﻟﻮﻓﺮﺓ ﻃﺒﻴﻌﻴﺔ ) ﻧﺴﺒﺔ ﺍﻟﺘﻮﺍﺟﺪ ﺍﻟﻄﺒﻴﻌﻴﺔ (‬‫ﺻﻒ‬‫ﺍﻟﻮﻓﺮﺓ ﻃﺒﻴﻌﻴﺔ ) ﻧﺴﺒﺔ ﺍﻟﺘﻮﺍﺟﺪ ﺍﻟﻄﺒﻴﻌﻴﺔ (‬‫‪O‬‬‫‪12‬‬‫‪C‬‬‫‪0.988888‬‬‫‪0.011112‬‬‫‪13‬‬‫‪16‬‬‫‪O‬‬‫‪17‬‬‫‪O‬‬‫‪0.997621‬‬‫‪0.0003790‬‬‫‪0.0020000‬‬‫‪18‬‬‫ﻳﺘﻢ ﺗﺤﺪﻳﺪ ﺍﻋﺘﻤﺎﺩ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻟـ ‪ Δ47‬ﻋﻠﻰ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﻟﻲ ‪ ،‬ﺣﻴﺚ ﻳﺘﻢ ﺇﻋﺘﺒﺎﺭ 𝑇 ﻛﺪﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﺍﻟﻤﻄﻠﻘﺔ ﺑﻮﺣﺪﺍﺕ ‪: K‬‬‫‪A.4‬‬‫ﺍﺫﺍ ﻛﺎﻥ ‪ 𝑅47‬ﻟﻠﻌﻮﺍﻟﻖ ﺍﻷﺣﻔﻮﺭﻳﺔ ﺍﻟﺘﻲ ﺗﻢ ﺍﻟﺤﺼﻮﻝ ﻋﻠﻴﻬﺎ ﻣﻦ ﻗﺎﻉ ﺍﻟﺒﺤﺮ ﻓﻲ ﺍﻟﻘﻄﺐ ﺍﻟﺠﻨﻮﺑﻲ‬‫‪ .4.50865 × 10–5‬ﻗﺪﺭ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﺑﺎﺳﺘﺨﺪﺍﻡ ‪ . 𝑅47‬ﻳﺘﻢ ﺗﻔﺴﻴﺮ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ﻫﺬﻩ ﻋﻠﻰ‬‫ﺃﻧﻬﺎ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﺍﻟﻬﻮﺍﺀ ﺧﻼﻝ ﺍﻟﻌﺼﺮ ﺍﻟﺬﻱ ﻋﺎﺷﺖ ﻓﻴﻪ ﺍﻟﻌﻮﺍﻟﻖ‪ .‬ﺿﻊ ﻓﻲ ﺍﻋﺘﺒﺎﺭﻙ ﻓﻘﻂ‬‫ﺍﻟﻨﻈﻴﺮ ﺍﻷﻛﺜﺮ ﺷﻴﻮﻋﺎً ﻝ ]‪ CO2 [47‬ﻟﻠﺤﺴﺎﺏ‪.‬‬‫‪9pt‬‬

Q2-1

IChO
OMN-4 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
OMN-4 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
OMN-4 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
OMN-4 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

IChO

A2-1

OMN-4 C-2 A-1

Arabic (Oman)‫ﻛﺒﺴﻮﻟﺔ ﻭﻗﺖ ﺍﻟﻨﻈﺎﺋﺮ‬(8 pt) A.1

𝑇 → +∞ ∶ 𝐾 =,𝑇 → 0 ∶ 𝐾 =

IChO

A2-2

OMN-4 C-2 A-2

Arabic (Oman)(8 pt) A.2

Δ𝐻 =J mol−1

IChO

A2-3

OMN-4 C-2 A-3

Arabic (Oman)(10 pt) A.3

ΔD2 =

IChO
OMN-4 C-2 A-4

A2-4
Arabic (Oman)(9 pt) A.4

𝑇 =K

OMN-4 C-3 C-1

OMN-4 C-3 C
LUJAINA AL BADI

IChO
Problem 3
Cover sheet

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

‫‪Q3-1‬‬‫‪IChO‬‬‫‪OMN-4 C-3 Q-1‬‬‫)‪Arabic (Oman‬‬‫ﻗﺎﻧﻮﻥ ﻻﻣﺒﺮﺕ ‪ -‬ﺑﻴﺮ ‪Law Lambert–Beer‬‬‫ﺍﻟﺴﺆﺍﻝ‬‫ﺍﻟﺪﺭﺟﺎﺕ‬‫ﺍﻟﺪﺭﺟﺔ‬‫‪ 8٪‬ﻣﻦ ﺍﻹﺟﻤﺎﻟﻲ‬‫‪A.1‬‬‫‪10‬‬‫‪B.1‬‬‫‪6‬‬‫‪B.2‬‬‫‪6‬‬‫ﺍﻟﻤﺠﻤﻮﻉ‬‫‪22‬‬‫ﻓﻲ ﻫﺬﻩ ﺍﻟﻤﺴﺄﻟﺔ ‪ ،‬ﺗﺠﺎﻫﻞ ﺍﻣﺘﺼﺎﺹ ﺍﻟﺨﻠﻴﺔ ﻭﺍﻟﻤﺬﻳﺐ‪ .‬ﻳﺘﻢ ﺍﻻﺣﺘﻔﺎﻅ ﺑﺪﺭﺟﺎﺕ ﺣﺮﺍﺭﺓ ﺟﻤﻴﻊ ﺍﻟﻤﺤﺎﻟﻴﻞ ﻭﺍﻟﻐﺎﺯﺍﺕ ﺛﺎﺑﺘﺔ ﻋﻨﺪ‬‫∘‬‫𝐶 ‪. 25‬‬‫ﺍﻟﺠﺰﺀ ‪A‬‬‫ﺗﻢ ﺗﺤﻀﻴﺮ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ‪ X‬ﺑﺎﺳﺘﺨﺪﺍﻡ ‪ HA‬ﻭ ‪ NaA.‬ﺍﻟﺘﺮﻛﻴﺰﺍﺕ ] ‪ ، [HA] ، [A−‬ﻭ ] ‪ [H+‬ﻓﻲ ﺍﻟﻤﺤﻠﻮﻝ ‪ X‬ﻫﻲ‬‫‪ 1.00 × 10−2 mol L−1‬ﻭ ‪ ، 1.00 × 10−3 mol L−1‬ﻭ ‪ ، 1.00 × 10−4 mol L−1‬ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ‪ ،‬ﻭﺍﻟﺘﻲ ﺗﺮﺗﺒﻂ ﻋﺒﺮ‬‫ﺍﻟﺘﻮﺍﺯﻥ ﺍﻟﺤﻤﻀﻲ ﺍﻟﻘﺎﻋﺪﻱ ‪ equilibrium acid-base‬ﺍﻟﺘﺎﻟﻲ‪:‬‬‫)‪(1‬‬‫‪𝐾 = [𝐴−][𝐻+]/[𝐻𝐴]𝐻𝐴 ↔ 𝐴 − +𝐻+‬‬‫ﻃﻮﻝ ﺍﻟﻤﺴﺎﺭ ﺍﻟﻀﻮﺋﻲ ﻫﻮ 𝑙 ﻓﻲ ﺍﻟﺠﺰﺀ ‪ . A‬ﺗﺠﺎﻫﻞ ﺗﻐﻴﺮ ﺍﻟﻜﺜﺎﻓﺔ ﻋﻨﺪ ﺍﻟﺘﺨﻔﻴﻒ‪ .‬ﺍﻓﺘﺮﺽ ﺃﻧﻪ ﻻ ﺗﻮﺟﺪ ﺗﻔﺎﻋﻼﺕ ﻛﻴﻤﻴﺎﺋﻴﺔ‬‫ﺑﺨﻼﻑ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪.1‬‬‫‪A.1‬‬‫ﻛﺎﻥ ﺍﻣﺘﺼﺎﺹ ‪ X‬ﻫﻮ ‪ 𝐴1‬ﺑﻄﻮﻝ ﻣﻮﺟﺔ ‪ .𝜆1‬ﺑﻌﺪ ﺫﻟﻚ ‪ ،‬ﺗﻢ ﺗﺨﻔﻴﻒ ﺍﻟﻤﺤﻠﻮﻝ ‪ X‬ﻣﺮﺗﻴﻦ ﺇﻟﻰ‬‫ﻟﺤﺠﻤﻪ ﺍﻷﻭﻟﻲ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺣﻤﺾ ﺍﻟﻬﻴﺪﺭﻭﻛﻠﻮﺭﻳﻚ ﺑﺎﺳﺘﺨﺪﺍﻡ ‪ . 2.500 = pH‬ﺑﻌﺪ ﺍﻟﺘﺨﻔﻴﻒ‬‫‪ ،‬ﺍﻻﻣﺘﺼﺎﺻﻴﺔ ﻻ ﺗﺰﺍﻝ ‪ 𝐴1‬ﻋﻨﺪ ‪ .𝜆1‬ﺣﺪﺩ ﺍﻟﻨﺴﺒﺔ ‪ ، 𝜀HA /𝜀A−‬ﺣﻴﺚ ﺗﻤﺜﻞ ‪ 𝜀HA‬ﻭ ‪ 𝜀A−‬ﻣﻌﺎﻣﻞ‬‫ﺍﻻﻣﺘﺼﺎﺹ ﻟـ ‪ HA‬ﻭ ‪ ، A−‬ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ‪ ،‬ﻋﻨﺪ ‪.𝜆1‬‬‫‪10pt‬‬‫‪Q3-2‬‬‫‪IChO‬‬‫‪OMN-4 C-3 Q-2‬‬‫)‪Arabic (Oman‬‬‫ﺍﻟﺠﺰﺀ ‪B‬‬‫ﺇﻋﺘﺒﺮ ﺍﻹﺗﺰﺍﻥ ﺍﻟﺘﺎﻟﻲ ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻐﺎﺯﻳﺔ‪.‬‬‫(‪)2‬‬‫𝑀‪𝐷 ↔ 2‬‬‫ُﻳﻤﻸ ﺍﻟﻐﺎﺯ ﺍﻟﻨﻘﻲ ‪ D‬ﻓﻲ ﺣﺎﻭﻳﺔ ﻣﻜﻌﺒﺔ ﻟﻬﺎ ﺟﺪﺍﺭ ﻣﺘﺤﺮﻙ ﺷﻔﺎﻑ ﻣﻊ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ﻣﻦ 𝑆 )ﺍﻧﻈﺮ ﺍﻟﺸﻜﻞ ﺃﺩﻧﺎﻩ( ﻋﻨﺪ ﺿﻐﻂ‬‫𝑃 ‪ ،‬ﻭﻳﺘﻢ ﺗﺤﻘﻴﻖ ﺍﻹﺗﺰﺍﻥ ﻋﻨﺪﺍ ﻳﺘﻢ ﺍﻻﺣﺘﻔﺎﻅ ﺑﺎﻟﻀﻐﻂ ﺍﻟﻜﻠﻲ ﻋﻨﺪ 𝑃‪ .‬ﺍﻣﺼﺘﺎﺹ ﺍﻟﻐﺎﺯ ﻫﻮ𝑙) 𝑉‪ ، 𝐴 = 𝜀(𝑛/‬ﺣﻴﺚ 𝜀 ﻭ 𝑉 ﻭ 𝑉‬
‫ﻭ 𝑙 ﻫﻲ ﻣﻌﺎﻣﻞ ﺍﻻﻣﺘﺼﺎﺹ ﻭﻛﻤﻴﺔ ﺍﻟﻐﺎﺯ ﺑﺎﻟﻤﻮﻻﺕ ﻭﺣﺠﻢ ﺍﻟﻐﺎﺯ ﻭﻃﻮﻝ ﺍﻟﻤﺴﺎﺭ ﺍﻟﻀﻮﺋﻲ ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‪ .‬ﺍﻓﺘﺮﺽ ﺃﻥ ﺟﻤﻴﻊ‬‫ﻣﻜﻮﻧﺎﺕ ﺧﻠﻴﻂ ﺍﻟﻐﺎﺯﺍﺕ ﺗﺘﺼﺮﻑ ﻛﻐﺎﺯﺍﺕ ﻣﺜﺎﻟﻴﺔ‪.‬‬‫ﺍﺳﺘﺨﺪﻡ ﺍﻟﺘﻌﺮﻳﻔﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ﻋﻨﺪ ﺍﻟﻀﺮﻭﺭﺓ‪.‬‬‫ﺍﻟﻀﻐﻂ ﺍﻟﺠﺰﺋﻲ‬‫ﺍﻟﻜﻤﻴﺔ ﺑﺎﻟﻤﻮﻻﺕ‬‫ﺍﻟﺤﺠﻢ‬‫‪D‬‬‫ﺍﻟﺤﺎﻟﺔ ﺍﻹﺑﺘﺪﺍﺋﻴﺔ‬‫𝑃 ﺍﻟﻀﻐﻂ‬‫‪ 𝑛0‬ﻋﺪﺩ ﺍﻟﻤﻮﻻﺕ ﺍﻹﺑﺘﺪﺍﺋﻴﺔ‬‫‪ 𝑉0‬ﺍﻟﺤﺠﻢ ﺍﻹﺑﺘﺪﺍﺋﻲ‬‫ﺑﻌﺪ ﺍﻹﺗﺰﺍﻥ‬

‫‪M‬‬

‫‪D‬‬

‫‪0‬‬

‫‪𝑝D‬‬‫‪𝑝M‬‬‫‪0‬‬‫‪𝑛D‬‬‫‪𝑛M‬‬‫𝑉 ﺍﻟﺤﺠﻢ‬‫‪M‬‬‫‪B.1‬‬‫ﻛﺎﻥ ﺍﻣﺘﺼﺎﺹ ﺍﻟﻐﺎﺯ ﻋﻨﺪ ‪ 𝜆B1‬ﺍﻟﻤﻘﺎﺱ ﻣﻦ ﺍﻻﺗﺠﺎﻩ ) 𝑥𝑙 = 𝑙(𝑥 ﻫﻮ ‪ 𝐴B1‬ﻓﻲ ﻛﻞ ﻣﻦ ﺍﻟﺤﺎﻟﺔ‬‫ﺍﻷﻭﻟﻴﺔ ﻭﺑﻌﺪ ﺍﻹﺗﺰﺍﻥ ‪ .‬ﺣﺪﺩ ﺍﻟﻨﺴﺒﺔ ‪ 𝜀D /𝜀M‬ﻋﻨﺪ ‪ ،𝜆B1‬ﺣﻴﺚ ﺗﻤﺜﻞ ‪ 𝜀D‬ﻭ ‪ 𝜀M‬ﻣﻌﺎﻣﻼﺕ ﺍﻣﺘﺼﺎﺹ‬‫‪ D‬ﻭ ‪ M‬ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‪.‬‬‫‪6pt‬‬‫‪B.2‬‬‫ﻛﺎﻥ ﺍﻣﺘﺼﺎﺹ ﺍﻟﻐﺎﺯ ﻋﻨﺪ ‪ 𝜆B2‬ﺍﻟﻤﻘﺎﺱ ﻣﻦ ﺍﻻﺗﺠﺎﻩ 𝑦 ﻫﻮ ‪ 𝐴B2‬ﻓﻲ ﻛﻞ ﻣﻦ ﺍﻟﺤﺎﻟﺔ ﺍﻷﺑﺘﺪﺍﺋﻴﺔ‬‫‪ 𝑙 = 𝑙𝑦0‬ﻭﺑﻌﺪ ﺍﻹﺗﺰﺍﻥ 𝑦𝑙 = 𝑙‪ .‬ﺣﺪﺩ ﺍﻟﻨﺴﺒﺔ ‪ 𝜀D /𝜀M‬ﻋﻨﺪ ‪.𝜆B2‬‬‫‪6pt‬‬

Q3-1

IChO
OMN-4 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
OMN-4 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

‫‪A3-1‬‬‫)‪Arabic (Oman‬‬‫ﻗﺎﻧﻮﻥ ﻻﻣﺒﺮﺕ‪-‬ﺑﻴﺮ‬‫ﺍﻟﺠﺰﺀ ‪A‬‬‫‪(10 pt) A.1‬‬‫)ﻭﺍﺻﻞ ﺍﻟﺤﻞ ﻓﻲ ﺍﻟﺼﻔﺤﺔ ﺍﻟﺘﺎﻟﻴﺔ(‬‫‪IChO‬‬‫‪OMN-4 C-3 A-1‬‬

IChO
OMN-4 C-3 A-2

A3-2
Arabic (Oman)(cont.) A.1

𝜀HA /𝜀A− =IChO
OMN-4 C-3 A-3

A3-3
Arabic (Oman)

B ‫ﺍﻟﺠﺰﺀ‬(6 pt) B.1

𝜀D /𝜀M =IChO
OMN-4 C-3 A-4

A3-4
Arabic (Oman)(6 pt) B.2

𝜀D /𝜀M =OMN-4 C-4 C-1

OMN-4 C-4 C
LUJAINA AL BADI

IChO
Problem 4
Cover sheet

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

‫‪Q4-1‬‬‫‪IChO‬‬‫‪OMN-4 C-4 Q-1‬‬‫)‪Arabic (Oman‬‬‫ﻛﻴﻤﻴﺎﺀ ﺍﻷﻛﺴﺪﺓ ﻭﺍﻻﺧﺘﺰﺍﻝ ﻟﻠﺰﻧﻚ ‪Zinc) of Chemistry Redox (The‬‬‫ﺍﻟﺴﺆﺍﻝ‬‫ﺍﻟﺪﺭﺟﺎﺕ‬‫ﺍﻟﺪﺭﺟﺔ‬‫‪A.1‬‬‫‪6‬‬‫‪ 11٪‬ﻣﻦ ﻣﺠﻤﻮﻉ ﺍﻟﺪﺭﺟﺎﺕ‬‫‪A.2‬‬‫‪5‬‬‫‪B.1‬‬‫‪4‬‬‫‪B.2‬‬‫‪3‬‬‫‪B.3‬‬‫‪5‬‬‫‪B.4‬‬‫‪9‬‬‫‪Total‬‬‫‪32‬‬‫ﻳﺴﺘﺨﺪﻡ ﺍﻟﺨﺎﺭﺻﻴﻦ ﻣﻨﺬ ﻓﺘﺮﺓ ﻃﻮﻳﻠﺔ ﻛﺴﺒﺎﺋﻚ ﻟﻤﻮﺍﺩ ﺍﻟﻨﺤﺎﺱ ﻭ ﺍﻟﺤﺪﻳﺪ ﺍﻟﺼﻠﺐ‪ .‬ﻳﺘﻢ ﻓﺼﻞ ﺍﻟﺰﻧﻚ ﺍﻟﻤﻮﺟﻮﺩ ﻓﻲ ﻣﻴﺎﻩ ﺍﻟﺼﺮﻑ‬‫ﺍﻟﺼﻨﺎﻋﻲ ﻋﻦ ﻃﺮﻳﻖ ﺍﻟﺘﺮﺳﻴﺐ ﻹﺯﺍﻟﺔ ﺍﻟﺴﻤﻮﻡ ﻣﻦ ﺍﻟﻤﺎﺀ ‪ ،‬ﻭﻳﺘﻢ ﺗﻘﻠﻴﻞ ﺍﻟﺮﺍﺳﺐ ﺍﻟﻨﺎﺗﺞ ﻻﺳﺘﻌﺎﺩﺗﻪ ﻭﺇﻋﺎﺩﺓ ﺍﺳﺘﺨﺪﺍﻣﻪ ﻛﺰﻧﻚ‬‫ﻣﻌﺪﻧﻲ‪.‬‬‫ﺍﻟﺠﺰﺀ ‪A‬‬

‫ﻳﺘﻢ ﺇﻋﻄﺎﺀ ﺇﺗﺰﺍﻥ ﺍﻧﺤﻼﻝ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺨﺎﺭﺻﻴﻦ)‪ Zn(OH)2 (s‬ﻋﻨﺪ ‪ 25∘ C‬ﻭﺛﻮﺍﺑﺖ ﺍﻹﺗﺰﺍﻥ ﺫﺍﺕ ﺍﻟﺼﻠﺔ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪4-1‬‬‫ﻳﺘﻢ ﺇﻋﻄﺎﺀ ﺍﻟﺬﻭﺑﺎﻧﻴﺔ ‪ ، 𝑆 ،‬ﻟﻠﺨﺎﺭﺻﻴﻦ )ﺗﺮﻛﻴﺰ ﺍﻟﺨﺎﺭﺻﻴﻦ ﻓﻲ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ﻣﺸﺒﻊ( ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪5‬‬

‫‪Q4-2‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-4 Q-2‬‬‫‪A.1‬‬‫ﻋﻨﺪﻣﺎ ﺍﻹﺗﺰﺍﻥ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻣﻦ ‪ 1‬ﺇﻟﻰ ‪ 4‬ﺗﻢ ﻋﻤﻠﺔ ‪ ،‬ﺍﺣﺴﺐ ﻣﺪﻯ ‪ pH‬ﺍﻟﺬﻱ ﻳﻜﻮﻥ ﻓﻴﻪ‬‫‪2+‬‬‫‪.[Zn(OH)2−‬‬‫])‪ [Zn(OH)2 (aq‬ﻫﻮ ﺍﻷﻛﺒﺮ ﺑﻴﻦ ])‪ [Zn (aq‬ﻭ ])‪ [Zn(OH)2 (aq‬ﻭ ])‪4 (aq‬‬‫‪6pt‬‬‫‪A.2‬‬‫ﺗﻢ ﺗﺤﻀﻴﺮ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ﻣﺸﺒﻊ ﻣﻦ )‪ Zn(OH)2 (s‬ﻣﻊ ﺍﻟﺮﻗﻢ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻨﻲ ‪ 7.00 = pH‬ﻭ ﺗﻢ‬‫ﺗﺮﺷﻴﺤﻪ‪ .‬ﺗﻤﺖ ﺇﺿﺎﻓﺔ ‪ NaOH‬ﺇﻟﻰ ﻫﺬﺍ ﺍﻟﻤﺮﺷﺢ ﻟﺰﻳﺎﺩﺓ ‪ pH‬ﺇﻟﻰ ‪ .12.00‬ﺍﺣﺴﺐ ﺍﻟﻨﺴﺒﺔ‬‫ﺍﻟﻤﻮﻟﻴﺔ ﻟﻠﺨﺎﺭﺻﻴﻦ ﺍﻟﺘﻲ ﺗﺘﺮﺳﺐ ﻋﻨﺪ ﺯﻳﺎﺩﺓ ‪ pH‬ﻣﻦ ‪ 7.00‬ﺇﻟﻰ ‪ .12.00‬ﺗﺠﺎﻫﻞ ﺍﻟﺘﻐﻴﻴﺮﺍﺕ ﻓﻲ‬‫ﺍﻟﺤﺠﻢ ﻭﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‪.‬‬‫‪5pt‬‬‫ﺍﻟﺠﺰﺀ ‪B‬‬‫ﺑﻌﺪ ﺫﻟﻚ ‪ ،‬ﻳﺘﻢ ﺗﺴﺨﻴﻦ ﻫﻴﺪﺭﻭﻛﺴﻴﺪ ﺍﻟﺨﺎﺭﺻﻴﻦ ﺍﻟﻤﺴﺘﻌﺎﺩ ﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ ﺃﻛﺴﻴﺪ ﺍﻟﺨﺎﺭﺻﻴﻦ ﻭﻓﻘﺎً ﻟﻠﺘﻔﺎﻋﻞ ﺃﺩﻧﺎﻩ‪:‬‬‫ﻳﺘﻢ ﺑﻌﺪ ﺫﻟﻚ ﺍﺧﺘﺰﺍﻝ ﺃﻛﺴﻴﺪ ﺍﻟﺨﺎﺭﺻﻴﻦ ﺇﻟﻰ ﺧﺎﺭﺻﻴﻦ ﻓﻠﺰﻱ ﺑﺎﻟﺘﻔﺎﻋﻞ ﻣﻊ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ‪:‬‬‫‪B.1‬‬‫ﻟﻜﻲ ﻳﺴﺘﻤﺮ ﺍﻟﺘﻔﺎﻋﻞ )‪ (7‬ﻳﺠﺐ ﺗﺜﺒﻴﺖ ﺿﻐﻂ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻋﻨﺪ ‪ 1bar،‬ﻭﻣﻦ ﺍﻟﻀﺮﻭﺭﻱ ﺗﻘﻠﻴﻞ‬‫ﺍﻟﻀﻐﻂ ﺍﻟﺠﺰﺋﻲ ﻟﺒﺨﺎﺭ ﺍﻟﻤﺎﺀ ﺍﻟﻤﺘﻮﻟﺪ‪ .‬ﺍﺣﺴﺐ ﺍﻟﺤﺪ ﺍﻷﻋﻠﻰ ﻟﻠﻀﻐﻂ ﺍﻟﺠﺰﺋﻲ ﻟﺒﺨﺎﺭ ﺍﻟﻤﺎﺀ ﻟﻠﺴﻤﺎﺡ‬‫ﻟﻠﺘﻔﺎﻋﻞ )‪ (7‬ﻟﻼﺳﺘﻤﺮﺍﺭ ﻋﻨﺪ ‪ . 300∘ C‬ﻫﻨﺎ ‪ ،‬ﻃﺎﻗﺎﺕ ﺗﻜﻮﻳﻦ ﺟﻴﺒﺲ ﻷﻛﺴﻴﺪ ﺍﻟﺨﺎﺭﺻﻴﻦ ﻭﺑﺨﺎﺭ‬‫ﺍﻟﻤﺎﺀ ﻋﻨﺪ ‪ 300∘ C‬ﻭ ‪ 1bar‬ﻟﺠﻤﻴﻊ ﺍﻷﻧﻮﺍﻉ ﺍﻟﻐﺎﺯﻳﺔ ﻫﻲ‬‫‪ Δ𝐺ZnO (300∘ C) = −2.90 × 102 kJ mol−1‬ﻭ ‪،Δ𝐺H2 O (300∘ C) = −2.20 × 102 kJ mol−1‬‬‫ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‪.‬‬‫‪4pt‬‬‫ﻳﺴﺘﺨﺪﻡ ﺍﻟﺨﺎﺭﺻﻴﻦ ﺍﻟﻔﻠﺰﻱ ﻛﻤﺎﺩﺓ ﻗﻄﺐ ﺳﺎﻟﺐ )ﺃﻧﻮﺩ( )‪ (anode‬ﻟﺒﻄﺎﺭﻳﺎﺕ ﺍﻟﻬﻮﺍﺀ‪-‬ﺍﻟﻔﻠﺰﻱ‪ .‬ﻳﺘﻜﻮﻥ ﺍﻟﻘﻄﺐ ﻣﻦ ‪ Zn‬ﻭ ‪ZnO.‬‬‫ﻳﺴﺘﺨﺪﻡ ﺗﻔﺎﻋﻞ ﺍﻷﻛﺴﺪﺓ ﻭﺍﻻﺧﺘﺰﺍﻝ ﺍﻟﺘﺎﻟﻲ ﻟﺘﻮﻟﻴﺪ ﺍﻟﻜﻬﺮﺑﺎﺀ ﺑﺎﻟﻘﻮﺓ ﺍﻟﺪﺍﻓﻌﺔ ﺍﻟﻜﻬﺮﺑﺎﺋﻴﺔ )‪) (emf‬ﺟﻬﺪ ﺍﻟﺨﻠﻴﺔ ( ﻋﻨﺪ ‪25∘ C‬‬‫ﻭﺿﻐﻂ ‪.𝐸 ∘ ، 1bar‬‬‫‪B.2‬‬‫ﺗﻢ ﺗﻔﺮﻳﻎ ﺑﻄﺎﺭﻳﺔ ﺍﻟﺰﻧﻚ ﻭﺍﻟﻬﻮﺍﺀ ﻋﻨﺪ ‪ 20mA‬ﻟﻤﺪﺓ ‪ 24‬ﺳﺎﻋﺔ‪ .‬ﺍﺣﺴﺐ ﺍﻟﺘﻐﻴﺮ ﻓﻲ ﻛﺘﻠﺔ ﺍﻟﻘﻄﺐ‬‫ﺍﻟﺴﺎﻟﺐ )ﺍﻷﻧﻮﺩ( ﻟﻠﺒﻄﺎﺭﻳﺔ‪.‬‬‫‪3pt‬‬

‫‪Q4-3‬‬‫‪IChO‬‬‫‪OMN-4 C-4 Q-3‬‬‫)‪Arabic (Oman‬‬‫ﻗﻤﺔ ﺟﺒﻞ ﻓﻮﺟﻲ ‪Fuji Mt.‬‬‫‪B.3‬‬‫ﺍﻋﺘﺒﺮ ﺗﻐﻴﻴﺮ ﺟﻬﺪ ﺍﻟﺨﻠﻴﺔ ‪ e.m.f.‬ﻟﺒﻄﺎﺭﻳﺔ ﺍﻟﺨﺎﺭﺻﻴﻦ‪-‬ﻫﻮﺍﺀ ﺣﺴﺐ ﺍﻟﺒﻴﺌﺔ‪ .‬ﺍﺣﺴﺐ ﻗﻴﻤﺔ ﺟﻬﺪ‬‫ﺍﻟﺨﻠﻴﻪ ‪ .(𝐸 ∘ ) emf‬ﻓﻲ ﻗﻤﺔ ﺟﺒﻞ ﻓﻮﺟﻲ ‪ ،‬ﺣﻴﺚ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ )ﻓﻲ ﻓﺒﺮﺍﻳﺮ( ‪ −38 ∘ C‬ﻭﺍﻻﺭﺗﻔﺎﻉ‬‫‪ ، m 3776‬ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‪ .‬ﻳﺘﻢ ﺗﻤﺜﻴﻞ ﺍﻟﻀﻐﻂ ﺍﻟﺠﻮﻱ ﺑﻮﺍﺳﻄﺔ‬‫‪5pt‬‬‫ﻋﻠﻰ ﺍﺭﺗﻔﺎﻉ ]‪ ℎ [m‬ﻭﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ ]‪ .𝑇 [∘ C‬ﺗﺒﻠﻎ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﻮﻟﻴﺔ ﻟﻸﻛﺴﺠﻴﻦ ﻓﻲ ﺍﻟﻐﻼﻑ‬‫ﺍﻟﺠﻮﻱ‪ .21%‬ﻃﺎﻗﺔ ﺗﻐﻴﺮ ﺟﻴﺒﺲ ﻟﻠﺘﻔﺎﻋﻞ )‪ (8‬ﻫﻮ ‪Δ𝐺ZnO (−38∘ C) = −3.26 × 102 kJ mol−1‬‬‫ﻋﻨﺪ ‪ −38 ∘ C‬ﻭ ‪1bar.‬‬‫‪B.4‬‬‫ﺍﺣﺴﺐ ﺍﻟﺘﻐﻴﺮ ﻓﻲ ﻃﺎﻗﺔ ﺟﻴﺒﺲ ﻟﻠﺘﻔﺎﻋﻞ )‪ (6‬ﻋﻨﺪ ‪ . 25 ∘ C‬ﻻﺣﻆ ﺃﻥ ﺟﻬﺪ ﺍﻹﺧﺘﺰﺍﻝ ﺍﻟﻘﻴﺎﺳﻲ‬‫‪ 𝐸 (Zn/Zn ) ،‬ﻭ )‪ 𝐸 (O2 /H2 O‬ﻋﻨﺪ ‪ 25 C‬ﻭ ‪ُ 1bar‬ﺗﻌﻄﻰ ﻛـ )‪ (10‬ﻭ )‪ ، (11‬ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‪.‬‬‫‪2+‬‬

‫∘‬

‫∘‬

‫∘‬

‫‪9pt‬‬

Q4-1

IChO
OMN-4 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
OMN-4 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
OMN-4 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

‫‪A4-1‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-4 A-1‬‬‫ﻛﻴﻤﻴﺎﺀ ﺍﻷﻛﺴﺪﺓ ﻭﺍﻻﺧﺘﺰﺍﻝ ﻟﻠﺰﻧﻚ‬‫ﺍﻟﺠﺰﺀ ‪A‬‬‫‪(6 pt) A.1‬‬‫< ‪< pH‬‬IChO
OMN-4 C-4 A-2

A4-2
Arabic (Oman)(5 pt) A.2

%

IChO
OMN-4 C-4 A-3

A4-3
Arabic (Oman)

B ‫ﺍﻟﺠﺰﺀ‬(4 pt) B.1

𝑝H2 O=bar(3 pt) B.2

g

IChO
OMN-4 C-4 A-4

A4-4
Arabic (Oman)(5 pt) B.3

V

IChO

A4-5

OMN-4 C-4 A-5

Arabic (Oman)(9 pt) B.4

Δ𝐺∘ =J mol−1

OMN-4 C-5 C-1

OMN-4 C-5 C
LUJAINA AL BADI

IChO
Problem 5
Cover sheet

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

‫‪Q5-1‬‬‫‪IChO‬‬‫‪OMN-4 C-5 Q-1‬‬‫)‪Arabic (Oman‬‬‫ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﺍﻟﻐﺎﻣﺾ ‪Silicon Mysterious‬‬‫ﺍﻟﺴﺆﺍﻝ‬‫ﺍﻟﺪﺭﺟﺎﺕ‬‫ﺍﻟﺪﺭﺟﺔ‬‫‪A.1‬‬‫‪9‬‬‫‪A.2‬‬‫‪7‬‬‫‪ 12٪‬ﻣﻦ ﺇﺟﻤﺎﻟﻲ ﻣﺠﻤﻮﻉ ﺍﻟﺪﺭﺟﺎﺕ‬‫‪A.3‬‬‫‪6‬‬‫‪A.4‬‬‫‪10‬‬‫‪B.1‬‬‫‪5‬‬‫‪B.2‬‬‫‪15‬‬‫‪B.3‬‬‫‪8‬‬‫ﺍﻟﻤﺠﻤﻮﻉ‬‫‪60‬‬‫ﻋﻠﻰ ﺍﻟﺮﻏﻢ ﻣﻦ ﺃﻥ ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﻫﻮ ﺃﻳﻀﺎً ﻋﻨﺼﺮ ﻣﻦ ﺍﻟﻤﺠﻤﻮﻋﺔ ‪ 14‬ﻣﺜﻞ ﺍﻟﻜﺮﺑﻮﻥ ‪ ،‬ﺇﻻ ﺃﻥ ﺧﺼﺎﺋﺼﻪ ﺗﺨﺘﻠﻒ ﺑﺸﻜﻞ ﻛﺒﻴﺮ‪.‬‬‫ﺍﻟﺠﺰﺀ ‪A‬‬

‫ﻋﻠﻰ ﻋﻜﺲ ﺍﻟﺮﺍﺑﻄﺔ ﺍﻟﺜﻼﺛﻴﺔ ﻟﻠﻜﺮﺑﻮﻥ ﻭﺍﻟﻜﺮﺑﻮﻥ ‪ ،‬ﻓﺈﻥ ﺍﻟﺮﺍﺑﻄﺔ ﺍﻟﺜﻼﺛﻴﺔ ﻟﻠﺴﻴﻠﻴﻜﻮﻥ ﻭﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﻓﻲ ﻣﺮﻛﺐ ﺗﻤﺖ ﺻﻴﺎﻏﺘﻪ‬‫ﻋﻠﻰ ﺷﻜﻞ ‪ :R) R1 –Si ≡ Si–R1‬ﺗﻌﺘﺒﺮ ﻣﺘﻔﺮﻋﻪ ﻋﻀﻮﻱ ﺃﻟﻜﻴﻞ ( ﺷﺪﻳﺪﺓ ﺍﻟﺘﻔﺎﻋﻞ‪ .‬ﻋﻠﻰ ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ ‪ ،‬ﻳﺘﻔﺎﻋﻞ ﻣﻊ ﺍﻹﻳﺜﻴﻠﻴﻦ‬‫‪ CH2=CH2 ethylene‬ﻟﺘﻜﻮﻳﻦ ﻧﺎﺗﺞ ﺣﻠﻘﻲ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺣﻠﻘﺔ ﺭﺑﺎﻋﻴﺔ‪.‬‬‫ﻋﻨﺪﻣﺎ ﻳﺘﻢ ﺍﻟﺘﻌﺎﻣﻞ ﻣﻊ ‪ R1 –Si ≡ Si–R1‬ﺑﺎﺳﺘﺨﺪﺍﻡ ﺃﻟﻜﺎﻳﻦ‪ R2 –C ≡ C–R2 ، alkyne‬ﻳﺘﺸﻜﻞ ﺍﻟﻤﺮﻛﺐ ﺫﻭ ﺍﻟﺤﻠﻘﺔ ﺍﻟﺮﺑﺎﻋﻴﺔ ‪A‬‬
‫ﺑﺎﻋﺘﺒﺎﺭﻩ ﻭﺳﻴﻄﺎً ﺃﻭﻟﻴﺎً ‪ .‬ﻳﺘﻴﺢ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻹﺿﺎﻓﻲ ﻟﺠﺰﻱﺀ ﺁﺧﺮ ﻣﻦ ‪ R2 –C ≡ C–R2‬ﻣﻊ ‪ A‬ﻹﻧﺘﺎﺝ ﺍﻟﻤﺘﺸﺎﺑﻬﻴﻦ ‪ B‬ﻭ ‪ ، C‬ﻭﻛﻼﻫﻤﺎ‬‫ﻟﻪ ﻫﻴﺎﻛﻞ ﻣﺘﺮﺍﻓﻘﺔ ‪ conjugated‬ﺣﻠﻘﻴﺔ ﺷﺒﻴﻬﺔ ﺑﺎﻟﺒﻨﺰﻳﻦ ‪ ،‬ﺗﺴﻤﻰ ﺛﻨﺎﺋﻲ ﺳﻠﻴﻜﺎ ﺑﻨﺰﻳﻦ ”‪ ”disilabenzenes‬ﺍﻟﺘﻲ ﺗﺤﺘﻮﻱ‬‫ﻋﻠﻰ ﺣﻠﻘﺔ ﺳﺪﺍﺳﻴﺔ ﻭﻳﻤﻜﻦ ﻛﺘﺎﺑﺔ ﺻﻴﺎﻏﺘﻬﺎ ﺍﻟﻌﺎﻣﺔ ﻋﻠﻰ ﺍﻟﺸﻜﻞ ‪(R1 –Si)2 (R2 –C)4‬‬

‫‪Q5-2‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-5 Q-2‬‬‫ُﻳﻈﻬﺮ ﺗﺤﻠﻴﻞ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻨﻮﻭﻱ ﺍﻟﻤﻐﻨﺎﻃﻴﺲ ﻟﻠﻜﺮﺑﻮﻥ‪ 13 C NMR 13‬ﻟﻠﺘﺮﺍﻛﻴﺐ ﺍﻟﺒﻨﺎﺋﻴﺔ ﺫﺍﺕ ﺍﻟﺤﻠﻘﺎﺕ ﺍﻟﺴﺪﺍﺳﻴﺔ ‪ Si2 C4‬ﺇﺷﺎﺭﺗﻴﻦ‬‫)ﻗﻤﺘﻴﻦ( ﻟـﻠﻤﺮﻛﺐ ‪ B‬ﻭﺇﺷﺎﺭﺓ ﻭﺍﺣﺪﺓ )ﻗﻤﺔ ﻭﺍﺣﺪﺓ( ﻟـﻠﻤﺮﻛﺐ ‪.C‬‬‫‪A.1‬‬‫ﺍﺭﺳﻢ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﺒﻨﺎﺋﻲ ﻟﻠﻤﺮﻛﺒﺎﺕ ‪ A‬ﻭ ‪ B‬ﻭ ‪ C‬ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﺼﻴﻎ ‪ R1‬ﻭ ‪ R2‬ﻭ ‪ Si‬ﻭ ‪ ، C‬ﺑﺎﺳﺘﺨﺪﺍﻡ‬‫ﺃﺣﺪ ﺍﻟﺘﺮﺍﻛﻴﺐ ﺍﻟﺮﻧﻴﻨﻴﺔ ﺍﻟﻤﻤﻜﻨﺔ‪.‬‬‫‪9pt‬‬‫‪A.2‬‬‫ﺍﺣﺴﺐ ﻃﺎﻗﺔ ﺍﻻﺳﺘﻘﺮﺍﺭ ) ﺍﻟﺘﺜﺒﻴﺖ (ﺍﻟﻌﻄﺮﻱ )‪ (ASE‬ﻟﻠﺒﻨﺰﻳﻦ ﻭ ﺍﻟﻤﺮﻛﺐ ‪) C‬ﻓﻲ ﺣﺎﻟﺔ‬‫‪ (R1 = R2 = H‬ﻛﻘﻴﻢ ﻣﻮﺟﺒﺔ ‪ ،‬ﻣﻊ ﺍﻷﺧﺬ ﻓﻲ ﺍﻻﻋﺘﺒﺎﺭ ﺗﻐﻴﺮ ﺍﻟﻤﺤﺘﻮﻯ ﺍﻟﺤﺮﺍﺭﻱ ﻓﻲ ﺑﻌﺾ‬‫ﺗﻔﺎﻋﻼﺕ ﺍﻟﻬﺪﺭﺟﺔ ﻟﻸﻧﻈﻤﺔ ﻏﻴﺮ ﺍﻟﻤﺸﺒﻌﺔ ﺍﻟﻤﻮﺿﺤﺔ ﺃﺩﻧﺎﻩ )ﺍﻟﺸﻜﻞ ‪.(1‬‬‫‪7pt‬‬‫ﺍﻟﺸﻜﻞ ‪1‬‬

‫‪Q5-3‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-5 Q-3‬‬‫ﻋﻨﺪﻣﺎ ﻳﺘﻢ ﺗﺴﺨﻴﻦ ﻣﺤﻠﻮﻝ ﺯﻳﻠﻴﻦ ﻟﻠﻤﺮﻛﺐ ‪ ، C‬ﻓﺈﻧﻪ ﻳﻤﺮ ﺑﻌﻤﻠﻴﺔ ﻣﺸﺎﺑﻬﺔ ﻹﻋﻄﺎﺀ ﺧﻠﻴﻂ ﺇﺗﺰﺍﻥ ﻣﻦ ﺍﻟﻤﺮﻛﺒﺎﺕ ‪ D‬ﻭ ‪ .E‬ﺍﻟﻨﺴﺒﺔ‬‫ﺍﻟﻤﻮﻟﻴﺔ ﻫﻲ ‪ 𝐷 ∶ 𝐸 = 1 ∶ 20.0‬ﻋﻨﺪ ‪ 50.0‬ﺩﺭﺟﺔ ﻣﺌﻮﻳﺔ ﻭ ‪ 𝐷 ∶ 𝐸 = 1 ∶ 40.0‬ﻋﻨﺪ ‪120.0‬ﺩﺭﺟﺔ ﻣﺌﻮﻳﺔ ‪.‬‬‫‪A.3‬‬‫ﺍﺣﺴﺐ 𝐻‪ Δ‬ﻟﺘﺤﻮﻳﻞ ﺍﻟﻤﺮﻛﺐ ‪ D‬ﺇﻟﻰ ‪ .E‬ﺍﻓﺘﺮﺽ ﺃﻥ 𝐻‪ Δ‬ﻻ ﺗﻌﺘﻤﺪ ﻋﻠﻰ ﺩﺭﺟﺔ ﺍﻟﺤﺮﺍﺭﺓ‪.‬‬‫‪6pt‬‬‫ﺗﺴﺘﻤﺮ ﻋﻤﻠﻴﺔ ﺍﻟﻤﺸﺎﺑﻬﺔ ‪ isomerization‬ﻣﻦ ﺍﻟﻤﺮﻛﺐ ‪ C‬ﺇﻟﻰ ‪ D‬ﻭ ‪ E‬ﻋﺒﺮ ﺗﺤﻮﻳﻼﺕ ﺍﻟﺮﻭﺍﺑﻂ ﺍﻟﺜﻨﺎﺋﻴﺔ 𝜎 ﺇﻟﻰ ﺍﻟﺮﻭﺍﺑﻂ ﺍﻷﺣﺎﺩﻳﺔ‬‫𝜋 ﺩﻭﻥ ﻛﺴﺮ ﺃﻱ ﺭﻭﺍﺑﻂ ﺃﺣﺎﺩﻳﺔ ‪ σ.‬ﻛﺸﻒ ﺗﺤﻠﻴﻞ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻨﻮﻭﻱ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﻟﻠﻜﺮﺑﻮﻥ ‪ 13 C NMR‬ﻋﻦ ﺇﺷﺎﺭﺓ ﻭﺍﺣﺪﺓ )ﻗﻤﺔ‬‫ﻭﺍﺣﺪﺓ ( ﻟﻠﺘﺮﻛﻴﺐ ﺍﻟﺒﻨﺎﺋﻲ ﻝ ‪ Si2 C4‬ﻟﻠﻤﺮﻛﺐ ‪ D‬ﻭﺇﺷﺎﺭﺗﻴﻦ )ﻗﻤﺘﻴﻦ ( ﻟﺘﻠﻚ ﺍﻟﺨﺎﺻﺔ ﺑﺎﻟﻤﺮﻛﺐ ‪ .E‬ﻻ ﻳﺤﺘﻮﻱ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﺒﻨﺎﺋﻲ‬‫ﻟﻠﻤﺮﻛﺐ ‪ D‬ﻋﻠﻰ ﺃﻱ ﺣﻠﻘﺎﺕ ﺛﻼﺛﻴﺔ ‪ ،‬ﺑﻴﻨﻤﺎ ﻳﺤﺘﻮﻱ ﺍﻟﻤﺮﻛﺐ ‪ E‬ﻋﻠﻰ ﺣﻠﻘﺘﻴﻦ ﺛﻼﺛﻴﺔ ﺗﺸﺘﺮﻙ ﻓﻲ ﺍﻟﺤﺎﻓﺔ‪.‬‬‫‪A.4‬‬‫ﺍﺭﺳﻢ ﺍﻟﺼﻴﻎ ﺍﻟﺒﻨﺎﺋﻴﺔ ﻟﻠﻤﺮﻛﺒﻴﻦ ‪ D‬ﻭ ‪ E‬ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﺼﻴﻎ ‪C. , Si , R2 ,R1‬‬‫‪10pt‬‬‫ﺍﻟﺠﺰﺀ ‪B‬‬‫ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﻗﺎﺩﺭ ﻋﻠﻰ ﺗﻜﻮﻳﻦ ﻣﺮﻛﺒﺎﺕ ﺗﻨﺎﺳﻴﻘﺔ ﻋﺎﻟﻴﺔ ﺍﻟﺘﻨﺴﻴﻖ )< ﺃﺭﺑﻌﺔ ﺗﻔﺮﻋﺎﺕ( ﻣﻊ ﻋﻨﺎﺻﺮ ﻟﻬﺎ ﺳﺎﻟﺒﻴﺔﻛﻬﺮﺑﻴﺔ ﻣﺜﻞ ﺍﻟﻔﻠﻮﺭ‪.‬‬‫ﻧﻈﺮﺍ ﻷﻥ ﺍﻟﻔﻠﻮﺭﻳﺪﺍﺕ ﺍﻟﻔﻠﺰﻳﺔ ﻏﺎﻟﺒﺎً ﻣﺎ ﺗﺴﺘﺨﺪﻡ ﻛﻮﺍﺷﻒ ﻟﻠﻔﻠﻮﺭﺓ ‪ ،‬ﻓﺈﻥ ﻓﻠﻮﺭﻳﺪﺍﺕ ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﻋﺎﻟﻴﺔ ﺍﻟﺘﻨﺴﻴﻖ ﺗﻌﻤﻞ ﺃﻳﻀﺎً‬‫ً‬‫ﻛﻜﻮﺍﺷﻒ ﻓﻠﻮﺭﺓ‪.‬‬‫ﺗﻢ ﺇﺟﺮﺍﺀ ﺗﻔﺎﻋﻞ ﺍﻟﻔﻠﻮﺭﺓ ﻣﻦ ‪ CCl4‬ﺑﺎﺳﺘﺨﺪﺍﻡ ‪ Na2 SiF6‬ﻋﻠﻰ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﻟﻲ‪.‬‬‫ﺗﺤﻀﻴﺮ ﺍﻟﻤﺤﻠﻮﻝ ﺍﻟﻘﻴﺎﺳﻲ ‪:Na2 SiF6‬‬‫ﻃﺮﻳﻘﺔ ﺍﻟﺘﺤﻀﻴﺮ ﻟﻤﺤﻠﻮﻝ ﻗﻴﺎﺳﻲ ﻣﻦ ‪Na2 SiF6‬‬‫ﺍﻟﻤﺤﻠﻮﻝ ﺍﻟﻤﺎﺋﻲ ‪ F‬ﻳﺘﻢ ﺗﺤﻀﻴﺮﺓ ‪ 0.855𝑔 :‬ﻣﻦ ‪ ( 188.053 g mol–1 ) Na2 SiF6‬ﻣﺬﺍﺏ ﻓﻲ ﺍﻟﻤﺎﺀ )ﺍﻟﺤﺠﻢ ﺍﻟﻜﻠﻲ‪.( 200mL :‬‬‫ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ ‪ G‬ﻳﺘﻢ ﺗﺤﻀﻴﺮﺓ ‪ 6.86𝑔 :‬ﻣﻦ ‪ ( 568, 424 g mol–1 ) Ce2 (SO4 )3‬ﻣﺬﺍﺏ ﻓﻲ ﺍﻟﻤﺎﺀ )ﺍﻟﺤﺠﻢ ﺍﻟﻜﻠﻲ‪200mL). :‬‬‫· ﺇﺟﺮﺍﺀﺍﺕ‪:‬‬‫ﻣﻌﺎﻳﺮﺓ ﺍﻟﺘﺮﺳﻴﺐ ﻟﻤﺤﻠﻮﻝ ‪F‬ﺣﺠﻤﻪ )‪ mL) 50.0‬ﻋﻦ ﻃﺮﻳﻖ ﺇﺿﺎﻓﺔ ﻣﺤﻠﻮﻝ ‪ G‬ﺑﺎﻟﺘﻨﻘﻴﻂ ﻓﻲ ﻭﺟﻮﺩ ﺯﺍﻳﻠﻴﻨﻮﻝ ﺑﺮﺗﻘﺎﻟﻲ‪xylenol‬‬‫‪ ، orange‬ﻭﺍﻟﺬﻱ ﻳﺮﺗﺒﻂ ﺇﻟﻰ ‪ ، Ce3+‬ﻛﻜﺎﺷﻒ‪ .‬ﺑﻌﺪ ﺇﺿﺎﻓﺔ ‪ 18.8.mL‬ﻣﻦ ﺍﻟﻤﺤﻠﻮﻝ ‪ ، G‬ﻳﺘﻐﻴﺮ ﻟﻮﻥ ﺍﻟﻤﺤﻠﻮﻝ ﻣﻦ ﺍﻷﺻﻔﺮ ﺇﻟﻰ‬‫ﺍﻷﺭﺟﻮﺍﻧﻲ‪ .‬ﺍﻟﺮﺍﺳﺐ ﺍﻟﻤﺘﻜﻮﻥ ﻫﻮ ﻣﺮﻛﺐ ﺛﻨﺎﺋﻲ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ‪ ، Ce3+‬ﻭﻣﺮﻛﺐ ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﺍﻟﻨﺎﺗﺞ ﺍﻟﻮﺣﻴﺪ ﻫﻮ ‪.Si(OH)4‬‬‫‪B.1‬‬‫ﺍﻛﺘﺐ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻮﺯﻭﻧﺔ ﻟﺘﻔﺎﻋﻞ ‪ Na2 SiF6‬ﻣﻊ ‪.Ce2 (SO4 )3‬‬‫‪5pt‬‬‫ﺗﻔﺎﻋﻞ ‪ CCl4‬ﻣﻊ ‪:Na2 SiF6‬‬‫)ﺍﻟﻤﻮﺍﺩ ﺍﻟﻤﻔﻘﻮﺩﺓ ﻋﻦ ﻃﺮﻳﻖ ﺍﻟﺘﺒﺨﺮ ﻋﻠﻰ ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ ﻣﻬﻤﻞ ﺃﺛﻨﺎﺀ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ‪(.‬‬‫ﺗﻤﺖ ﺇﺿﺎﻓﺔ ‪ [g]) x‬ﻣﻦ ﺍﻟﻤﺎﺩﺓ ) ‪ Na2 SiF6‬ﺇﻟﻰ ‪ 500.0) CCl4‬ﺟﻢ( ﻭﺗﻢ ﺗﺴﺨﻴﻨﻪ ﺇﻟﻰ ‪˚300‬ﺩﺭﺟﺔ ﻣﺌﻮﻳﺔ ﻓﻲ ﻭﻋﺎﺀ ﺗﻔﺎﻋﻞ‬‫ﻣﺤﻜﻢ ﺍﻟﻐﻠﻖ ﻣﻘﺎﻭﻡ ﻟﻠﻀﻐﻂ‪ .‬ﺗﻤﺖ ﺇﺯﺍﻟﺔ ‪ Na2 SiF6‬ﻏﻴﺮ ﺍﻟﻤﺘﻔﺎﻋﻞ ﻭ‪ NaCl‬ﺍﻟﻨﺎﺗﺞ ﻋﻦ ﻃﺮﻳﻖ ﺍﻟﺘﺮﺷﻴﺢ‪ .‬ﺗﻢ ﺗﺨﻔﻴﻒ ﺍﻟﻤﺎﺩﺓ‬‫ﺍﻟﻤﺮﺷﺤﺔ ﺇﻟﻰ ﺣﺠﻢ ﺇﺟﻤﺎﻟﻲ ﻗﺪﺭﻩ ‪ 1.00‬ﻟﺘﺮ ﺑﺎﺳﺘﺨﺪﺍﻡ ‪) CCl4‬ﻣﺤﻠﻮﻝ ‪ .(H‬ﺃﻇﻬﺮ ﺃﻃﻴﺎﻑ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻨﻮﻭﻱ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﻝ ‪Si‬‬‫‪19‬‬‫ﻭ ‪ (NMR) 19 F‬ﻟﻠﻤﺤﻠﻮﻝ ‪ H‬ﺃﻇﻬﺮ ‪ SiF4‬ﻛﻤﺮﻛﺐ ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﺍﻟﻮﺣﻴﺪ‪ .‬ﻓﻲ ﻃﻴﻒ ‪ ، 𝐹 NMR‬ﺑﺎﻹﺿﺎﻓﺔ ﺇﻟﻰ ‪ ، SiF4‬ﻟﻮﺣﻈﺖ‬‫ﺍﻹﺷﺎﺭﺍﺕ )ﺍﻟﻘﻤﻢ( ﺍﻟﻤﻘﺎﺑﻠﺔ ﻟـ ‪ CFCl3‬ﻭ ‪ CF2 Cl2‬ﻭ ‪ CF3 Cl‬ﻭ ‪) CF4‬ﺭﺍﺟﻊ ﺍﻟﺠﺪﻭﻝ ‪ .(1‬ﺗﺘﻨﺎﺳﺐ ﻧﺴﺐ ﺍﻟﺘﻜﺎﻣﻞ ﻓﻲ ﻃﻴﻒ ‪NMR‬‬‫‪19‬‬‫𝐹 ﻣﻊ ﻋﺪﺩ ﻧﻮﻯ ﺍﻟﻔﻠﻮﺭ‪.‬‬‫‪29‬‬

‫‪Q5-4‬‬‫‪IChO‬‬‫‪OMN-4 C-5 Q-4‬‬‫)‪Arabic (Oman‬‬‫‪) data NMR 19 F‬ﺑﻴﺎﻧﺎﺕ‬‫ﺍﻟﺮﻧﻴﻦ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﻟﻠﻔﻠﻮﺭ‬‫‪(19‬‬‫ﻧﺴﺒﺔ ﺗﻜﺎﻣﻞ ﺍﻟﻤﺴﺎﺣﺔ‬‫ﺍﻟﺠﺪﻭﻝ ‪1‬‬‫‪CFCl3‬‬‫‪CF2 Cl2‬‬‫‪CF3 Cl‬‬‫‪CF4‬‬‫‪45.0‬‬‫‪65.0‬‬‫‪18.0‬‬‫‪2.0‬‬‫ﺗﺘﺤﻠﻞ ﺍﻟﻤﺎﺩﺓ ‪ SiF4‬ﺑﺎﻟﻤﺎﺀ ﻟﻴﺸﻜﻞ ‪ H2 SiF6‬ﺣﺴﺐ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪:8‬‬‫ﺗﻤﺖ ﺇﺿﺎﻓﺔ ﻣﺤﻠﻮﻝ ‪ (10 𝑚𝐿) H‬ﺇﻟﻰ ﻛﻤﻴﺔ ﺯﺍﺋﺪﺓ ﻣﻦ ﺍﻟﻤﺎﺀ ‪ ،‬ﻣﻤﺎ ﺃﺩﻯ ﺇﻟﻰ ﺍﻟﺘﺤﻠﻞ ﺍﻟﻤﺎﺋﻲ ﺍﻟﻜﺎﻣﻞ ﻟـ ‪ .SiF4‬ﺑﻌﺪ ﺍﻟﻔﺼﻞ ‪،‬‬‫ﺗﻢ ﻋﻤﻞ ﺗﻔﺎﻋﻼ ﺗﻌﺎﺩﻝ ﻟﻠﻤﺮﻛﺐ ‪ H2 SiF6‬ﺍﻟﻨﺎﺗﺞ ﻣﻦ ﺍﻟﺘﺤﻠﻞ ﺍﻟﻤﺎﺋﻲ ﻓﻲ ﺍﻟﻤﺤﻠﻮﻝ ﺍﻟﻤﺎﺋﻲ ﻭﺗﺤﻮﻳﻠﻪ ﺑﺎﻟﻜﺎﻣﻞ ﺇﻟﻰ ‪Na2 SiF6‬‬‫)ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ 𝐽(‪.‬‬‫ﺗﻢ ﺇﺫﺍﺑﺔ ﺭﺍﺳﺐ ‪ Na2 SiF6‬ﻭ ‪ NaCl‬ﻏﻴﺮ ﺍﻟﻤﺘﻔﺎﻋﻞ ‪ ،‬ﻭﺍﻟﺬﻱ ﺗﻤﺖ ﺇﺯﺍﻟﺘﻪ ﺑﺎﻟﺘﺮﺷﻴﺢ ﻓﻲ ﺍﻟﺨﻄﻮﺓ ﺍﻷﻭﻟﻴﺔ )ﺗﺤﺘﻪ ﺧﻂ( ‪ ،‬ﺗﻢ ﺇﺫﺍﺑﺘﺔ‬‫ﺗﻤﺎﻣﺎ ﻓﻲ ﺍﻟﻤﺎﺀ ﻹﻋﻄﺎﺀ ﻣﺤﻠﻮﻝ ﻣﺎﺋﻲ )ﺍﻟﻤﺤﻠﻮﻝ ‪ K‬؛ ‪ 10.0‬ﻟﺘﺮ(‪.‬‬‫ً‬‫ﺑﻌﺪ ﺫﻟﻚ ‪ ،‬ﺗﻢ ﺇﺟﺮﺍﺀ ﻣﻌﺎﻳﺮﺓ ﺇﺿﺎﻓﻴﺔ ﻟﻬﻄﻮﻝ ﺍﻷﻣﻄﺎﺭ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻤﺤﻠﻮﻝ ‪ ، G‬ﻭﻛﺎﻧﺖ ﻧﻘﺎﻁ ﺍﻟﻨﻬﺎﻳﺔ ﻟﻠﻤﻌﺎﻳﺮﺓ ﻣﻊ ‪ G‬ﻛﻤﺎ‬‫ﻳﻠﻲ‪:‬‬‫· ﻟﻠﻤﺤﻠﻮﻝ ‪) J‬ﺍﻟﻜﻤﻴﺔ ﺍﻟﻜﺎﻣﻠﺔ(‪. 61.6 𝑚𝐿 :‬‬‫· ﻟﻜﻞ ‪ 100‬ﻣﻞ ﻣﻦ ﻣﺤﻠﻮﻝ ‪ K‬ﻫﻲ ‪44.4 𝑚𝐿 :‬‬‫ﻭﺗﺠﺪﺭ ﺍﻹﺷﺎﺭﺓ ﻫﻨﺎ ﺇﻟﻰ ﺃﻥ ﺍﻟﻮﺟﻮﺩ ﺍﻟﻤﺸﺘﺮﻙ ﻟﻠﻤﺎﺩﺗﻴﻦ ‪ NaCl‬ﺃﻭ ‪ SiO2‬ﻟﻴﺲ ﻟﻪ ﺃﻱ ﺗﺄﺛﻴﺮ ﻋﻠﻰ ﻣﻌﺎﻳﺮﺓ ﺍﻟﺘﺮﺳﻴﺐ‪.‬‬‫‪B.2‬‬‫ﺍﺣﺴﺐ ﻛﺘﻠﺔ ‪ NaCl‬ﺍﻟﻤﻨﺘﺠﺔ ﻓﻲ ﻭﻋﺎﺀ ﺍﻟﺘﻔﺎﻋﻞ )ﺍﻟﻤﻌﻠﻮﻣﺎﺕ ﺍﻟﺘﻲ ﺗﺤﺘﻬﺎ ﺧﻂ( ‪ ،‬ﻭﺍﺣﺴﺐ‬‫ﺍﻟﻜﺘﻠﺔ )]𝑔[ 𝑥(ﻟـ ‪ Na2 SiF6‬ﺍﻟﻤﺴﺘﺨﺪﻣﺔ ﻛﻤﻮﺍﺩ ﺑﺪﺍﻳﺔ‪.‬‬‫‪15pt‬‬‫‪B.3‬‬‫‪ 77.8%‬ﻣﻦ ‪ CCl4‬ﺍﻟﻤﺴﺘﺨﺪﻡ ﻛﻤﺎﺩﺓ ﺑﺪﺍﻳﺔ ﻟﻢ ﻳﺘﻔﺎﻋﻞ‪ .‬ﺍﺣﺴﺐ ﻛﺘﻠﺔ‪ CF3 Cl‬ﺍﻟﻨﺎﺗﺠﺔ‪.‬‬‫‪8pt‬‬

Q5-1

IChO
OMN-4 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
OMN-4 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
OMN-4 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
OMN-4 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

IChO

A5-1

OMN-4 C-5 A-1

Arabic (Oman)

Mysterious Silicon ‫ﺍﻟﺴﻴﻠﻴﻜﻮﻥ ﺍﻟﻐﺎﻣﺾ‬A ‫ﺍﻟﺠﺰﺀ‬(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
OMN-4 C-5 A-2

A5-2
Arabic (Oman)(6 pt) A.3

kJ mol–1

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

A5-3
Arabic (Oman)

B ‫ﺍﻟﺠﺰﺀ‬(5 pt) B.1(15 pt) B.2(‫)ﺗﺎﺑﻊ ﻓﻲ ﺍﻟﺼﻔﺤﻪ ﺍﻟﺘﺎﻟﻴﻪ‬IChO
OMN-4 C-5 A-4

A5-4
Arabic (Oman)(cont.) B.2

NaCl ∶

g, Na2 SiF6 ∶

g

IChO
OMN-4 C-5 A-5

A5-5
Arabic (Oman)(8 pt) B.3

CF3 Cl ∶g

OMN-4 C-6 C-1

OMN-4 C-6 C
LUJAINA AL BADI

IChO
Problem 6
Cover sheet

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

‫‪Q6-1‬‬‫‪IChO‬‬‫‪OMN-4 C-6 Q-1‬‬‫)‪Arabic (Oman‬‬‫ﻛﻴﻤﻴﺎﺀ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺼﻠﺒﺔ ﺍﻟﻔﻠﺰﺍﺕ ﺍﻻﻧﺘﻘﺎﻟﻴﺔ ‪of Transition Metals‬‬‫‪The Solid State Chemistry‬‬‫ﺍﻟﺴﺆﺍﻝ‬‫ﺍﻟﺪﺭﺟﺎﺕ‬‫‪A.1‬‬‫‪6‬‬‫‪A.2‬‬‫‪3‬‬‫‪A.3‬‬‫‪3‬‬‫‪B.1‬‬‫‪6‬‬‫‪ 13٪‬ﻣﻦ ﺍﻹﺟﻤﺎﻟﻲ‬‫‪B.2‬‬‫‪4‬‬‫‪B.3‬‬‫‪4‬‬‫‪B.4‬‬‫‪4‬‬‫‪C.1‬‬‫‪5‬‬‫‪C.2‬‬‫‪5‬‬‫‪C.3‬‬‫‪5‬‬‫ﺍﻹﺟﻤﺎﻟﻲ‬‫‪45‬‬‫ﻧﺘﻴﺠﺔ‬‫ﺑﺮﻛﺎﻥ ﻓﻲ ﺟﺰﻳﺮﺓ ﺳﺎﻛﻮﺭﺍﺟﻴﻤﺎ‬‫ﺍﻟﺠﺰﺀ ‪A‬‬

‫ﺍﻟﻴﺎﺑﺎﻥ ﻫﻲ ﻭﺍﺣﺪﺓ ﻣﻦ ﺍﻟﺪﻭﻝ ﺍﻟﺘﻲ ﻟﺪﻳﻬﺎ ﺃﻛﺒﺮ ﻋﺪﺩ ﻣﻦ ﺍﻟﺒﺮﺍﻛﻴﻦ ﻓﻲ ﺟﻤﻴﻊ ﺃﻧﺤﺎﺀ ﺍﻟﻌﺎﻟﻢ‪ .‬ﻋﻨﺪﻣﺎ ﺗﺘﺒﻠﻮﺭ ﻣﻌﺎﺩﻥ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ‬‫ﻣﻦ ﺍﻟﺼﻬﺎﺭﺓ ﺍﻟﻤﺎﺟﻤﺎ ‪ ،‬ﻳﺤﺼﻞ ﺍﻧﺪﻣﺎﺝ ﺟﺰﺀ ﻣﻦ ﺃﻳﻮﻧﺎﺕ ﺍﻟﻔﻠﺰﺍﺕ ﺍﻻﻧﺘﻘﺎﻟﻴﺔ ) ‪ (Mn+‬ﻓﻲ ﺍﻟﺼﻬﺎﺭﺓ ﺍﻟﻤﺎﺟﻤﺎ ﻣﻊ ﻣﻌﺎﺩﻥ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ‪.‬‬‫ﺗﻜﻮﻥ ‪ Mn+‬ﺍﻟﻤﺪﺭﻭﺳﺔ ﻓﻲ ﻫﺬﻩ ﺍﻟﻤﺴﺄﻟﺔ ﻣﺮﺗﺒﻄﺔ ﺗﻨﺎﺳﻘﻴﺎ ﻣﻊ ﺃﻳﻮﻧﺎﺕ ﺍﻷﻛﺴﻴﺪ ) ‪(O2−‬ﻭﺗﺸﻜﻞ ﻋﺪﺩ ﺗﻨﺎﺳﻘﻲ ﺃﺭﺑﻌﺔ ﻓﻲ‬‫ﺷﻜﻞ ﺭﺑﺎﻋﻲ ﺍﻷﻭﺟﺔ ) ‪( 𝑇d‬ﻓﻲ ﺍﻟﺼﻬﺎﺭﺓ ﺍﻟﻤﺎﺟﻤﺎ ﻭﺗﺸﻜﻞ ﻋﺪﺩ ﺗﻨﺎﺳﻘﻲ ﺳﺘﺔ ﻓﻲ ﺷﻜﻞ ﺛﻤﺎﻧﻲ ﺍﻟﻮﺟﻮﻩ ﺍﻷﻭﺟﻪ ) ‪ (𝑂h‬ﻓﻲ‬‫ﻣﻌﺎﺩﻥ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ ﻭﻛﻼﻫﻤﺎ ﻳﻈﻬﺮ ﺗﻜﻮﻳﻦ ﺩﻭﺭﺍﻥ ﺍﻟﻜﺘﺮﻭﻧﻲ ﻣﺮﺗﻔﻊ ‪ spin. high‬ﻳﻤﻜﻦ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻦ ﻣﻌﺎﻣﻞ ﺍﻟﺘﻮﺯﻳﻊ ﻟﻼﻳﻮﻥ‬‫‪ Mn+‬ﺑﻴﻦ ﻣﻌﺎﺩﻥ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ ﻭﺍﻟﺼﻬﺎﺭﺓ ﺍﻟﻤﺎﺟﻤﺎ ‪ D‬ﺑﺎﻟﻌﻼﻗﺔ‪:‬‬‫ﺣﻴﺚ ﺃﻥ ‪ [M]s‬ﻭ ‪ [M]l‬ﺗﻤﺜﻞ ﺗﺮﺍﻛﻴﺰ ‪ Mn+‬ﻓﻲ ﻛﻞ ﻣﻦ ﻣﻌﺎﺩﻥ ﺍﻟﺴﻴﻠﻴﻜﺎﺕ ﻭﺍﻟﺼﻬﺎﺭﺓ ‪ ،‬ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‪ .‬ﻳﻮﺿﺢ ﺍﻟﺠﺪﻭﻝ ﺃﺩﻧﺎﻩ‬‫ﻗﻴﻤﺔ 𝐷 ﻟـ ‪ Mn2+ , Cr2+‬ﻛﺄﻣﺜﻠﺔ‪.‬‬‫‪Q6-2‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-6 Q-2‬‬‫ﻟﻨﻔﺘﺮﺽ ﺃﻥ ‪ ΔO‬ﻭ ‪ CFSEO‬ﻫﻤﺎ ﻃﺎﻗﺔ ﺍﻟﻔﺼﻞ ﻟﻤﺪﺍﺭﺍﺕ ﺍﻟﻔﻠﻚ ‪ d‬ﻟـ ‪ Mn+‬ﻭﻃﺎﻗﺔ ﺍﺳﺘﻘﺮﺍﺭ ﺍﻟﻤﺠﺎﻝ ﺍﻟﺒﻠﻮﺭﻱ ﻓﻲ ﺍﻟﻤﺠﺎﻝ ‪،𝑂h‬‬‫ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‪.‬‬‫ﻭﺑﻔﺮﺽ ﺃﻥ ‪ ΔT‬ﻭ ‪CFSET‬ﻳﻤﺜﻼﻧﺎﻫﻢ ﻓﻲ ﺍﻟﻤﺠﺎﻝ ‪𝑇d‬‬‫‪A.1‬‬‫ﺍﺣﺴﺐ | ‪ ΔCFSE= |CFSEO −CFSET‬ﺑﺪﻻﻟﺔ ‪ ΔO‬ﻟﻜﻦ ﻣﻦ ‪ 𝐶𝑟2+‬ﻭ ‪ 𝑀 𝑛2+‬ﻭ‬‫𝑜𝐶 ﺃﻓﺘﺮﺽ‬‫‪6pt‬‬‫‪A.2‬‬‫ﻟﻮﺣﻆ ﻭﺟﻮﺩ ﻋﻼﻗﺔ ﺧﻄﻴﺔ ﻣﻦ ﺧﻼﻝ ﺭﺳﻢ 𝐷‪ ln‬ﻣﻘﺎﺑﻞ ‪ ΔCFSE/ ΔO‬ﻓﻲ ﻧﻈﺎﻡ ﺍﻹﺣﺪﺍﺛﻴﺎﺕ‬‫ﺍﻟﺪﻳﻜﺎﺭﺗﻴﺔ ﺍﻟﻤﻮﺿﺢ ﺃﺩﻧﺎﻩ‪.‬‬‫ﻋﻴﻦ 𝐷 ﻝ ‪ Co2+‬ﺑﺎﻟﺘﻘﺪﻳﺮ‬‫‪3pt‬‬‫ﺃﻥ ‪4/9ΔO =ΔT‬‬‫‪2+‬‬‫ﺗﺘﺒﻠﻮﺭ ﺃﻛﺎﺳﻴﺪ ﺍﻟﻤﻌﺎﺩﻥ ‪ MO‬ﺣﻴﺚ ﺍﻟﻤﻌﺪﻥ ﻳﻤﺜﻞ ﺃﻱ ﻣﻦ ﺍﻟﻤﻌﺎﺩﻥ ﺍﻟﺘﺎﻟﻴﺔ ‪(𝑀 ∶ 𝐶𝑎, 𝑇 𝑖, 𝑉 .𝑀 𝑛 𝑜𝑟 𝐶𝑜)،‬ﻋﻠﻰ ﻫﻴﺌﺔ‬‫ﺑﻠﻮﺭﺓ ﺍﻟﻤﻠﺢ ﺣﻴﺚ ﻳﺘﺒﻨﻰ ‪Mn+‬ﻫﻨﺪﺳﺔ ‪𝑂h‬ﻣﻊ ﺗﺮﻛﻴﺐ ﺇﻟﻜﺘﺮﻭﻧﻲ ﻋﺎﻟﻲ ﺍﻟﺪﻭﺭﺍﻥ ‪. high spin‬ﺗﺮﺗﺒﻂ ﺍﻧﺜﺎﻟﺒﻴﺎﺕ ﻃﺎﻗﺔ ﺍﻟﺸﺒﻜﺔ‬‫ﺍﻟﺒﻠﻮﺭﻳﺔ ﻟﻬﺬﻩ ﺍﻷﻛﺎﺳﻴﺪ ﺑﺸﻜﻞ ﺃﺳﺎﺳﻲ ﺑﺘﺄﺛﻴﺮﺍﺕ ﻛﻮﻟﻮﻡ ﺑﻴﻦ ﺍﻟﺸﺤﻨﺎﺕ ﻋﻠﻰ ﺃﺳﺎﺱ ﻧﺼﻒ ﻗﻄﺮ ﻭﺷﺤﻨﺔ ﺍﻷﻳﻮﻧﺎﺕ ﻭﺑﻌﺾ‬‫ﺍﻟﻤﺴﺎﻫﻤﺎﺕ ﻣﻦ ‪ CFSE‬ﻟـ ‪ Mn+‬ﻓﻲ ﺍﻟﻤﺠﺎﻝ ‪𝑂h‬‬‫‪Q6-3‬‬‫)‪Arabic (Oman‬‬‫‪A.3‬‬‫‪IChO‬‬‫‪OMN-4 C-6 Q-3‬‬‫ﺍﺧﺘﺮ ﺍﻟﻤﺠﻤﻮﻋﺔ ﺍﻟﻤﻨﺎﺳﺒﺔ ﻣﻦ ﺍﻧﺜﺎﻟﺒﻴﺎﺕ ﻃﺎﻗﺔ ﺍﻟﺸﺒﻜﺔ ﺍﻟﺒﻠﻮﺭﻳﺔ] ‪[ kJ mol−1‬ﻣﻦ ﺃﺣﺪ ﺍﻟﺨﻴﺎﺭﺍﺕ‬‫)‪ (a‬ﺇﻟﻰ ‪(f).‬‬‫‪3pt‬‬

‫‪Q6-4‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-6 Q-4‬‬‫ﺍﻟﺠﺰﺀ ‪B‬‬‫ﻳﺘﺒﻠﻮﺭ ﺃﻛﺴﻴﺪ ﻣﺨﺘﻠﻂ ‪ ، A‬ﻳﺤﺘﻮﻱ ﻋﻠﻰ ‪ La3+‬ﻭ ‪ ، Cu2+‬ﻓﻲ ﺧﻠﻴﺔ ﻭﺣﺪﺓ ﺭﺑﺎﻋﻴﺔ ﺍﻟﺰﻭﺍﻳﺎ ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﻓﻲ ﺍﻟﺸﻜﻞ ‪.1‬‬‫ﻓﻲ ﺍﻟﻤﺠﺴﻢ ﺍﻟﺜﻤﺎﻧﻲ ] ‪ ، [CuO6‬ﻳﻜﻮﻥ ﻃﻮﻝ ‪ O – Cu‬ﻋﻠﻰ ﻃﻮﻝ ﺍﻟﻤﺤﻮﺭ ‪ (𝑙𝑧 ) z‬ﺃﻃﻮﻝ ﻣﻦ ﻃﻮﻝ ﺍﻟﻤﺤﻮﺭ ‪ ، (𝑙𝑥 ) x‬ﻭﻳﺘﺸﻮﻩ‬‫]] ‪ [CuO6‬ﻋﻦ ﻫﻨﺪﺳﺔ ‪ 𝑂h‬ﺍﻟﻌﺎﺩﻳﺔ‪ .‬ﻫﺬﺍ ﺍﻟﺘﺸﻮﻳﻪ ﻳﺰﻳﻞ ﺍﻻﻧﺤﻼﻝ ﺑﻴﻦ ﺍﻟﻤﺪﺍﺭﺍﺕ ‪.(d𝑥2 −𝑦2 and 𝑑𝑧2 ) eg‬‬‫ﺍﻟﺸﻜﻞ ‪1‬‬‫ﻳﻤﻜﻦ ﺗﺼﻨﻴﻊ ‪ A‬ﻋﻦ ﻃﺮﻳﻖ ﺍﻟﺘﺤﻠﻞ ﺍﻟﺤﺮﺍﺭﻱ )ﺍﻻﻧﺤﻼﻝ ﺍﻟﺤﺮﺍﺭﻱ( ﻟﻠﻤﺮﻛﺐ ‪ ، B‬ﻭﺍﻟﺬﻱ ﻳﺘﻜﻮﻥ ﻋﻦ ﻃﺮﻳﻖ ﺧﻠﻂ ﻛﻠﻮﺭﻳﺪﺍﺕ ﺍﻟﻤﻌﺎﺩﻥ‬‫ﻓﻲ ﻣﺤﻠﻮﻝ ﺃﻣﻮﻧﻴﺎ ﻣﺎﺋﻲ ﻣﺨﻔﻒ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺣﻤﺾ ﺳﻜﻮﺍﺭﻳﻚ ‪ ، C4 H2 O4‬ﺃﻱ ﺛﻨﺎﺋﻲ ﺣﺎﻣﺾ‪ُ .‬ﻳﻈﻬﺮ ﺳﻠﻮﻙ ﺍﻻﻧﺤﻼﻝ‬‫∘‬‫ﻓﻘﺪﺍ ﻓﻲ ﺍﻟﻮﺯﻥ ﺑﻨﺴﺒﺔ ‪29.1%‬ﺣﺘﻰ 𝐶 ‪200‬ﺑﺴﺒﺐ ﻓﻘﺪ ﻣﺎﺀ ﺍﻟﺘﺒﻠﻮﺭ ‪ ،‬ﻳﻠﻴﻪ ﻓﻘﺪﺍﻥ ﺁﺧﺮ ﻟﻠﻮﺯﻥ‬‫ﺍﻟﺤﺮﺍﺭﻱ ﻟـ ‪ B‬ﻓﻲ ﺍﻟﻬﻮﺍﺀ ﺍﻟﺠﺎﻑ‬‫ً‬‫ﺣﺘﻰ 𝐶 ‪ 700‬ﺑﺴﺒﺐ ﺇﻃﻼﻕ ‪ .CO2‬ﺇﺟﻤﺎﻟﻲ ﻓﻘﺪﺍﻥ ﺍﻟﻮﺯﻥ ﺃﺛﻨﺎﺀ ﺗﻜﻮﻳﻦ ‪ A‬ﻣﻦ ‪ B‬ﻫﻮ ‪ .63.6%‬ﻭﺗﺠﺪﺭ ﺍﻹﺷﺎﺭﺓ ﺇﻟﻰ ﺃﻧﻪ ﻳﺘﻢ‬‫ﺇﻃﻼﻕ ﺍﻟﻤﺎﺀ ﻭ ‪CO2‬ﻓﻘﻂ ﻓﻲ ﺗﻔﺎﻋﻞ ﺍﻻﻧﺤﻼﻝ ﺍﻟﺤﺮﺍﺭﻱ‪.‬‬‫∘‬‫‪B.1‬‬‫ﺍﻛﺘﺐ ﺍﻟﺼﻐﻴﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟـ ‪ A‬ﻭ ‪. B‬‬‫‪6pt‬‬‫‪B.2‬‬‫ﺍﺣﺴﺐ 𝑥𝑙 ﻭ 𝑧𝑙 ﺍﺳﺘﺨﺪﻡ ﺍﻟﺸﻜﻞ ‪1‬‬‫‪4pt‬‬‫‪B.3‬‬‫ﺑﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ‪Cu2+‬ﻓﻲ ﺛﻤﺎﻧﻲ ﺍﻟﺴﻄﻮﺡ ﺍﻟﻤﺸﻮﻩ ] ‪ [CuO6‬ﻓﻲ ‪ A‬ﻣﻦ ﺍﻟﺸﻜﻞ ‪، 1‬‬‫ﺍﻛﺘﺐ ﺃﺳﻤﺎﺀ ﺍﻟﻤﺪﺍﺭﺍﺕ ﺍﻟﻤﻨﻘﺴﻤﺔ ‪ (d𝑥2 −𝑦2 and d𝑧2 ) eg‬ﻓﻲ )‪ (i‬ﻭ )‪ ، (ii‬ﻭﺍﺭﺳﻢ ﺍﻟﺘﻮﺯﻳﻊ‬‫ﺍﻹﻟﻜﺘﺮﻭﻧﻲ ﻓﻲ ﺍﻟﻤﺮﺑﻊ ﺍﻟﻤﻨﻘﻂ ﻓﻲ ﻭﺭﻗﺔ ﺇﺟﺎﺑﺘﻚ‪.‬‬‫‪4pt‬‬

‫‪Q6-5‬‬‫‪IChO‬‬‫‪OMN-4 C-6 Q-5‬‬‫)‪Arabic (Oman‬‬‫ﻳﻌﺘﺒﺮ ‪ A‬ﻣﺎﺩﺓ ﻋﺎﺯﻟﺔ‪ .‬ﻭﻋﻨﺪ ﺍﺳﺘﺒﺪﺍﻝ ﺃﻳﻮﻥ ﻭﺍﺣﺪ ﻣﻦ ‪ La3+‬ﺑﺄﻳﻮﻥ ﻭﺍﺣﺪ ﻣﻦ ‪ Sr2+‬ﻳﻈﻬﺮ ﺑﺎﻟﻤﻘﺎﺑﻞ ﺛﻘﺐ ﻭﺍﺣﺪ ﻓﻲ ﺍﻟﺸﺒﻜﺔ‬‫ﺍﻟﺒﻠﻮﺭﻳﺔ ﺍﻟﺘﻲ ﻳﻤﻜﻨﻬﺎ ﻣﻦ ﺗﻮﺻﻴﻞ ﺍﻟﻜﻬﺮﺑﺎﺀ‪ .‬ﻧﺘﻴﺠﺔ ﻟﺬﻟﻚ‪ ،‬ﺃﻇﻬﺮ ‪ A‬ﺍﻟﺬﻱ ﺗﻢ ﺗﻄﻌﻴﻤﻪ ﺑﺄﻳﻮﻥ ‪ Sr2+‬ﻣﻮﺻﻠﻴﺔ ﻓﺎﺋﻘﺔ ﺗﺤﺖ‬‫‪ . 38K‬ﻋﻨﺪﻣﺎ ﺣﺪﺙ ﺗﻔﺎﻋﻞ ﺍﺳﺘﺒﺪﺍﻝ ﻟـ ‪ ، A‬ﻳﺘﻜﻮﻥ ‪2.05 × 1027‬ﻣﻦ ﺍﻟﺜﻘﻮﺏ ﻟﻜﻞ ﻣﺘﺮ ﻣﻜﻌﺐ ‪m−3‬‬‫‪B.4‬‬‫ﺍﺣﺴﺐ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﻷﻳﻮﻥ ‪ Sr2+‬ﺍﻟﻤﺴﺘﺒﺪﻝ ﻓﻲ ﻣﻮﺍﻗﻊ ﺃﻳﻮﻥ ‪La3+‬ﺑﻨﺴﺒﺔ ﻣﻮﻟﻴﺔ ﻓﻲ‬‫ﺗﻔﺎﻋﻞ ﺍﻻﺳﺘﺒﺪﺍﻝ‪ .‬ﻻﺣﻆ ﺃﻥ ﺗﻜﺎﻓﺆﺍﺕ ﺍﻷﻳﻮﻧﺎﺕ ﺍﻟﻤﻜﻮﻧﺔ ﻭﺍﻟﺒﻨﻴﺔ ﺍﻟﺒﻠﻮﺭﻳﺔ ﻟﻢ ﺗﺘﻐﻴﺮ ﺃﺛﻨﺎﺀ‬‫ﺗﻔﺎﻋﻞ ﺍﻻﺳﺘﺒﺪﺍﻝ‪.‬‬‫‪4pt‬‬‫ﺍﻟﺠﺰﺀ ‪C‬‬
‫ﻳﺘﻜﻮﻥ ‪ Cu2 (CH3 CO2 )4‬ﻣﻦ ﺍﺭﺑﻊ ‪CH3 CO2‬ﻣﺘﺮﺍﺑﻄﺔ ﺗﻨﺎﺳﻘﻴﺎ ﺇﻟﻰ ﺍﺛﻨﻴﻦ ‪ Cu‬ﻣﻦ ﺍﻟﺸﻜﻞ ‪ . 2A‬ﻳﻈﻬﺮ ‪Cu2 (CH3 CO2 )4‬‬‫‪−‬‬
‫ﻣﺴﺘﻮﻳﺎﺕ ﻋﺎﻟﻴﺔ ﻣﻦ ﺍﻟﺘﻨﺎﻇﺮ ﺍﻟﻬﻴﻜﻠﻲ‪ ،‬ﻣﻊ ﻣﺤﻮﺭﻳﻦ ﻳﻤﺮﺍﻥ ﻋﺒﺮ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ﻓﻲ ﺍﻟﻠﻴﺠﺎﻧﺪﺍﺕ )ﺍﻟﻤﺮﺗﺒﻄﺔ( ﺍﻻﺭﺑﻌﺔ ‪CH3 CO2‬‬‫‪−‬‬‫‪2+‬‬‫ﻭﻣﺤﻮﺭ ﻳﻤﺮ ﻋﺒﺮ ﺃﻳﻮﻧﻴﻦ ﻣﻦ ‪،Cu‬ﻭﺟﻤﻴﻊ ﺍﻟﻤﺤﺎﻭﺭ ﻣﻮﺟﻬﺔ ﺑﺸﻜﻞ ﻣﺘﻌﺎﻣﺪ ﺑﺎﻟﻨﺴﺒﺔ ﻟﺒﻌﻀﻬﺎ ﺍﻟﺒﻌﺾ‪ .‬ﻋﻨﺪ ﺍﺳﺘﺨﺪﺍﻡ ﻟﻴﺠﺎﻧﺪ‬‫‪−‬‬‫ﺛﻨﺎﺋﻲ ﺍﻟﻜﺮﺑﻮﻛﺴﻴﻞ ﺑﺪﻻ ‪،CH3 CO2‬ﻳﺘﻜﻮﻥ “ﻣﻌﻘﺪ ﺫﻭ ﻓﺠﻮﺍﺕ ﻗﻔﺺ“‪complex) cage).‬‬‫‪2+‬‬‫‪L1‬‬‫‪Cu4 (L1‬ﻣﻦ ﻣﺴﺘﻮﻱ ﻣﻦ ﺛﻨﺎﺋﻲ ﻛﺮﺑﻮﻛﺴﻴﻼﺕ ‪ L1‬ﺍﻟﺸﻜﻞ‪ 2B‬ﻭ ‪ Cu‬ﺍﻟﺸﻜﻞ ‪ . 2C‬ﺗﺤﺪﺩ ﺍﻟﺰﺍﻭﻳﺔ ‪θ‬‬‫ﻳﺘﻜﻮﻥ ﻣﻌﻘﺪ ﺍﻟﻘﻔﺺ ‪L1)4‬‬‫ﺑﻴﻦ ﺍﺗﺠﺎﻫﺎﺕ ﺍﻟﺘﻨﺎﺳﻖ ﻻﺛﻨﻴﻦ ﻣﻦ ﺍﻟﻜﺮﺑﻮﻛﺴﻴﻼﺕ‪ ،‬ﺍﻟﻤﺸﺎﺭ ﺇﻟﻴﻬﺎ ﺑﻮﺍﺳﻄﺔ ﺍﻷﺳﻬﻢ ﻓﻲ ﺍﻟﺸﻜﻞ ‪ ، 2B‬ﺗﺮﻛﻴﺐ ﺑﻨﻴﺔ ﻣﻌﻘﺪ‬‫∘‬‫ﺍﻟﻘﻔﺺ‪ 𝜃 = 0 .‬ﻟـ ‪. L1‬ﻻﺣﻆ ﺃﻥ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻏﻴﺮ ﻣﻮﺿﺤﺔ ﻓﻲ ﺍﻟﺸﻜﻞ ‪. 2‬‬‫‪2+‬‬‫ﺍﻟﺸﻜﻞ ‪2‬‬

‫‪Q6-6‬‬‫)‪Arabic (Oman‬‬‫‪C.1‬‬‫‪IChO‬‬‫‪OMN-4 C-6 Q-6‬‬‫ﺗﻢ ﺗﺜﺒﻴﺖ ﺍﻟﺰﺍﻭﻳﺔ ‪ θ‬ﻟﺜﻨﺎﺋﻲ ﻛﺮﺑﻮﻛﺴﻴﻼﺕ ﺍﻟﻤﺴﺘﻮﻱ ‪ L2‬ﺍﻟﻤﻮﺿﺢ ﺃﺩﻧﺎﻩ ﻋﻨﺪ ﺯﺍﻭﻳﺔ ‪ . 90‬ﺇﺫﺍ‬‫∘‬‫ﻛﺎﻥ ﺗﺮﻛﻴﺐ ﻣﻌﻘﺪ ﺍﻟﻘﻔﺺ ﺍﻟﻤﺘﻜﻮﻥ ﻣﻦ ‪ L2‬ﻭ ‪ Cu‬ﻫﻮ 𝑚)‪ 𝐶𝑢𝑛 (𝐿2‬ﺍﺫﻛﺮ ﺍﺻﻐﺮ ﻣﺠﻤﻮﻋﺔ‬‫‪−‬‬‫ﻣﻦ ﺍﻻﻋﺪﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ ﻟﻜﻞ ﻣﻦ ‪ n‬ﻭ ‪ . m‬ﺍﻓﺘﺮﺽ ﺃﻥ ﻣﺠﻤﻮﻋﺎﺕ ‪ CO2‬ﻟﻠﻴﺠﺎﻧﺪ ‪ L2‬ﺗﺮﺗﺒﻂ ﻓﻘﻂ‬‫‪2+‬‬‫ﺑﺮﺍﺑﻄﺔ ﺗﻨﺎﺳﻘﻴﺔ ﺍﻟﻰ ﺃﻳﻮﻧﺎﺕ ‪Cu‬‬‫‪2+‬‬‫‪5pt‬‬

‫‪Q6-7‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-6 Q-7‬‬‫ﻳﺤﺘﻮﻱ ﻣﻌﻘﺪ ﺍﻟﺨﺎﺭﺻﻴﻦ ‪ ، Zn4 O(CH3 CO2 )6‬ﻋﻠﻰ ﺃﺭﺑﻌﺔ ﺭﺑﺎﻋﻲ ﺍﻟﺴﻄﻮﺡ‬‫𝐴‪.(3‬‬‫‪2+‬‬‫‪ Zn‬ﻭﺳﺘﺔ‬‫‪−‬‬‫‪CH3 CO2‬ﻭ ﻭﺍﺣﺪ ‪) O2−‬ﺍﻟﺸﻜﻞ‬‫ﻓﻲ ﺍﻟﻤﻌﻘﺪ ‪ Zn4 O(CH3 CO2 )6‬ﻳﻘﻊ ‪ O2−‬ﻋﻨﺪ ﻧﻘﻄﺔ ﻣﺮﻛﺰ ﺍﻷﺻﻞ ‪ ،‬ﻭﺍﻟﻤﺤﺎﻭﺭ ﺍﻟﺜﻼﺛﺔ ﺍﻟﺘﻲ ﺗﻤﺮ ﻋﺒﺮ ﺫﺭﺍﺕ ﺍﻟﻜﺮﺑﻮﻥ ﻓﻲ‬
‫∘‬
‫‪−‬‬
‫‪ CH3 CO2‬ﻣﺘﻌﺎﻣﺪﺓ ﺑﺎﻟﻨﺴﺒﺔ ﻟﺒﻌﻀﻬﺎ ﺍﻟﺒﻌﺾ‪ .‬ﻋﻨﺪ ﺍﺳﺘﺨﺪﺍﻡ ‪) p-benzenedicarboxylate‬ﺍﻟﺸﻜﻞ 𝐵‪𝐿3 , 𝜃 = 180 3‬‬‫‪−‬‬‫‪2+‬‬‫(‬‫ﺑﺪﻻ ﻣﻦ ‪ ،CH3 CO2‬ﺗﺮﺗﺒﻂ ﻣﺠﻤﻮﻋﺎﺕ ‪ Zn‬ﺑﺒﻌﻀﻬﺎ ﺍﻟﺒﻌﺾ ﻟﺘﺸﻜﻴﻞ ﻣﺎﺩﺓ ﺻﻠﺒﺔ ﺑﻠﻮﺭﻳﺔ )‪ (X‬ﺗﺴﻤﻰ ”ﺑﻮﻟﻴﻤﺮ ﺗﻨﺴﻴﻖ‬‫ً‬‫‪L3‬‬‫‪ [Zn4 O(L3‬ﻭﻟﻬﺎ ﺑﻨﻴﺔ ﺑﻠﻮﺭﻳﺔ ﻣﻜﻌﺒﺔ ﺫﺍﺕ ﻣﺴﺎﻡ ﺑﺤﺠﻢ ﺍﻟﻨﺎﻧﻮ‪ .‬ﻳﺘﻢ ﺗﻤﺜﻴﻞ‬‫ﻣﺴﺎﻣﻲ” )ﺍﻟﺸﻜﻞ ‪ .(3𝐶 .‬ﺗﺮﻛﻴﺒﺔ ‪ X‬ﻫﻲ 𝑛] ‪L3)3‬‬‫ﺍﻟﻤﺴﺎﻡ ﺍﻟﻮﺍﺣﺪﺓ ﻋﻠﻰ ﺷﻜﻞ ﻛﺮﺓ ﻓﻲ ﺍﻟﺸﻜﻞ ‪ ، 3D‬ﻭﻳﺘﻢ ﺗﻤﺜﻴﻞ ﻛﻞ ﺭﺑﺎﻋﻲ ﺍﻟﺴﻄﻮﺡ ‪ Zn‬ﻋﻠﻰ ﺷﻜﻞ ﻣﺘﻌﺪﺩ ﺍﻟﺴﻄﻮﺡ‬‫ﺭﻣﺎﺩﻱ ﻏﺎﻣﻖ ﻓﻲ ﺍﻟﺸﻜﻞ ‪ 3C‬ﻭ ‪ 3D.‬ﻻﺣﻆ ﺃﻥ ﺫﺭﺍﺕ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻦ ﻏﻴﺮ ﻣﻮﺿﺤﺔ ﻓﻲ ﺍﻟﺸﻜﻞ ‪.3‬‬‫‪2+‬‬‫ﺍﻟﺸﻜﻞ ‪3‬‬‫‪C.2‬‬‫ﻳﺤﺘﻮﻱ ‪ X‬ﻋﻠﻰ ﺧﻠﻴﺔ ﻭﺣﺪﺓ ﺗﻜﻌﻴﺒﻴﺔ ﻃﻮﻝ ﺿﻠﻌﻬﺎ 𝑎 )ﺍﻟﺸﻜﻞ 𝐶‪ (3‬ﻭﻛﺜﺎﻓﺔ) ‪( 592.0g cm−3‬‬‫ﺍﺣﺴﺐ 𝑎 ﺑﻮﺣﺪﺓ ]‪[cm‬‬‫‪5pt‬‬‫‪C.3‬‬‫ﻳﺤﺘﻮﻱ ‪ X‬ﻋﻠﻰ ﻋﺪﺩ ﻛﺒﻴﺮ ﻣﻦ ﺍﻟﻤﺴﺎﻣﺎﺕ ‪ ،‬ﻭﻳﻤﻜﻦ ﻟـ ‪ 1g‬ﻣﻦ ‪ X‬ﺍﺳﺘﻴﻌﺎﺏ ‪ 3.0 × 102 mL‬ﻣﻦ ‪CO2‬‬‫∘‬‫ﻓﻲ ﺍﻟﻤﺴﺎﻡ ﻋﻨﺪ ‪ 1bar‬ﻭ 𝐶 ‪ . 25‬ﺍﺣﺴﺐ ﻣﺘﻮﺳﻂ ﻋﺪﺩ ﺍﻟﺠﺰﻳﺌﺎﺕ ﻣﻦ ‪ CO2‬ﻏﺎﺯ ﻟﻜﻞ ﻣﺴﺎﻡ‬‫ﻭﺍﺣﺪ‪.‬‬‫‪5pt‬‬

Q6-1

IChO
OMN-4 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
OMN-4 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
OMN-4 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
OMN-4 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
OMN-4 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
OMN-4 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

‫‪A6-1‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-6 A-1‬‬‫ﻛﻴﻤﻴﺎﺀ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺼﻠﺒﺔ ﻟﻠﻤﻌﺎﺩﻥ ﺍﻻﻧﺘﻘﺎﻟﻴﺔ‬‫ﺍﻟﺠﺰﺀ ‪A‬‬‫‪(6 pt) A.1‬‬‫∶ ‪,ΔO Cr2+‬‬‫∶ ‪, ΔO Mn2+‬‬‫∶ ‪ΔO Co2+‬‬

IChO

A6-2

OMN-4 C-6 A-2

Arabic (Oman)(3 pt) A.2

𝐷∶(3 pt) A.3

IChO

A6-3

OMN-4 C-6 A-3

Arabic (Oman)

B ‫ﺍﻟﺠﺰﺀ‬(6 pt) B.1

B∶

,A ∶
(4 pt) B.2

𝑙𝑧 =nm ,𝑙𝑥 =

nm

IChO

A6-4

OMN-4 C-6 A-4

Arabic (Oman)(4 pt) B.3(ii) ∶,(i) ∶(4 pt) B.4

%

IChO

A6-5

OMN-4 C-6 A-5

Arabic (Oman)

C ‫ﺍﻟﺠﺰﺀ‬(5 pt) C.1𝑚=,𝑛 =(5 pt) C.2

𝑎=

cm

IChO
OMN-4 C-6 A-6

A6-6
Arabic (Oman)(5 pt) C.3

OMN-4 C-7 C-1

OMN-4 C-7 C
LUJAINA AL BADI

IChO
Problem 7
Cover sheet

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

‫‪Q7-1‬‬‫‪IChO‬‬‫‪OMN-4 C-7 Q-1‬‬‫)‪Arabic (Oman‬‬‫ﺍﻟﻠﻌﺐ ﺑﻌﻄﺮﻳﺔ ﺍﻟﻤﺮﻛﺒﺎﺕ ﻏﻴﺮ ﺍﻟﺒﻨﺰﻳﻨﻴﺔ ‪Non-benzenoid with Playing‬‬‫‪Aromaticity‬‬‫ﺍﻟﺴﺆﺍﻝ‬‫ﺍﻟﺪﺭﺟﺎﺕ‬‫ﺩﺭﺟﺔ‬‫‪ 13٪‬ﻣﻦ ﺇﺟﻤﺎﻟﻲ ﺍﻟﺪﺭﺟﺎﺕ‬‫‪A.1‬‬‫‪5‬‬‫‪A.2‬‬‫‪2‬‬‫‪A.3‬‬‫‪19‬‬‫‪B.1‬‬‫‪10‬‬‫ﺍﻟﻤﺠﻤﻮﻉ‬‫‪36‬‬‫ﺍﻓﺘﺘﺢ ﺍﻟﺒﺮﻭﻓﻴﺴﻮﺭ ﻧﻮﺯﻭﻱ )‪ (1996-1902‬ﻣﺠﺎﻝ ﺍﻟﺒﺤﺚ ﻋﻦ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﻌﻄﺮﻳﺔ ﻏﻴﺮ ﺍﻟﺒﻨﺰﻳﻨﻴﺔ ‪ ،‬ﻭﺍﻟﺘﻲ ﺃﺻﺒﺤﺖ ﺍﻵﻥ ﻣﻨﺘﺸﺮﺓ‬‫ﻓﻲ ﻛﻞ ﻣﻜﺎﻥ ﻓﻲ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻌﻀﻮﻳﺔ‬‫ﺍﻟﺼﻮﺭﺓ ﺍﻟﺒﺮﻭﻓﻴﺴﻮﺭ ﻧﻮﺯﻭﻱ ‪ :‬ﺟﺎﻣﻌﺔ ﺗﻮﻫﻮﻛﻮ‪.‬‬‫ﺍﻟﺠﺰﺀ ‪A‬‬

‫ﺍﻟﻤﺮﻛﺐ ‪ Lineariifolianone‬ﻫﻮ ﻣﻨﺘﺞ ﻃﺒﻴﻌﻲ ﻟﻪ ﺗﺮﻛﻴﺐ ﺑﻨﺎﺋﻲ ﻓﺮﻳﺪ ﺗﻢ ﻓﺼﻠﻪ ﻣﻦ ﺍﻟﻨﺒﺘﺔ ‪ linariifolia. Inula‬ﻣﻦ ﻓﺎﻟﻨﺴﻴﻦ‬‫‪ ، (1) valencene‬ﻳﻨﺘﺞ ﺍﻟﺘﺤﻮﻳﻞ ﻓﻲ ﺧﻄﻮﺓ ﻭﺍﺣﺪﺓ ﻳﻨﺘﺞ ﺍﻟﻤﺮﻛﺐ ‪ ، 2‬ﻗﺒﻞ ﺍﻟﺘﺤﻮﻳﻞ ﻣﻦ ﺛﻼﺙ ﺧﻄﻮﺍﺕ ﻋﺒﺮ ﺍﻟﻤﺮﻛﺐ ‪ 3‬ﻳﻨﺘﺞ‬‫ﺍﻟﻜﻴﺘﻮﻥ ‪ .4‬ﻳﺘﻢ ﺗﺤﻮﻳﻞ ‪ (5) Eremophilene‬ﺇﻟﻰ ﺍﻟﻤﺮﻛﺐ ‪ 6‬ﻋﻦ ﻃﺮﻳﻖ ﺇﺟﺮﺍﺀ ﻧﻔﺲ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﻤﻜﻮﻥ ﻣﻦ ﺃﺭﺑﻊ ﺧﻄﻮﺍﺕ‪.‬‬‫‪Q7-2‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-7 Q-2‬‬‫‪ linariifolia Inula‬ﺍﻟﻨﺒﺘﺔ ﺇﻳﻨﻮﻻ ﻟﻴﻨﺎﺭﻳﻔﻮﻟﻴﺎ‬‫‪A.1‬‬‫ﺍﺭﺳﻢ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﺒﻨﺎﺋﻲ ﻟﻠﻤﺮﻛﺒﻴﻦ ‪ 2‬ﻭ ‪ 6‬ﻭﺣﺪﺩ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺑﻮﺿﻮﺡ ﻋﻨﺪ ﺍﻟﻀﺮﻭﺭﺓ‪.‬‬‫‪5pt‬‬‫ﺑﻌﺪ ﺫﻟﻚ ‪ ،‬ﻳﺘﻢ ﺗﺤﻮﻳﻞ ﺍﻟﻜﻴﺘﻮﻥ ‪ 4‬ﺇﻟﻰ ﺍﺳﺘﺮ ‪ .15‬ﻳﺤﺘﻔﻆ ﺍﻟﻤﺮﻛﺐ ‪) 8‬ﺍﻟﻮﺯﻥ ﺍﻟﺠﺰﻳﺌﻲ‪ (188 :‬ﻳﺤﺘﻔﻆ ﺑﺠﻤﻴﻊ ﻣﺮﺍﻛﺰ ﺍﻟﺘﻨﺎﻇﺮﻳﺔ‬‫ﺍﻟﻜﺎﻳﺮ ﺍﻟﻔﺮﺍﻏﻲ ‪ (chiral) stereocenters the‬ﻓﻲ ‪ .7‬ﺍﻟﻤﺮﻛﺒﺎﺕ ‪ 9‬ﻭ ‪ 10‬ﺗﺤﺘﻮﻱ ﻋﻠﻰ ﺧﻤﺴﺔ ﻣﺮﺍﻛﺰ ﻛﺎﻳﺮﻝ ﻣﺮﻛﺰﻳﺔ ﻭﻻ ﺗﻮﺟﺪ‬‫‪Q7-3‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-7 Q-3‬‬‫ﺑﺪﻻ ﻣﻦ ‪ H2 16 O‬ﻟﺘﺼﻨﻴﻊ ﺍﻟﻤﺮﻛﺐ ‪ 13‬ﻭ ﺍﻟﻤﺮﻛﺐ‬‫ﺭﻭﺍﺑﻂ ﺛﻨﺎﺋﻴﺔ ﺑﻴﻦ ﺍﻟﻜﺮﺑﻮﻥ ﻭﺍﻟﻜﺮﺑﻮﻥ‪ .‬ﺍﻓﺘﺮﺽ ﺃﻧﻪ ﻳﺘﻢ ﺍﺳﺘﺨﺪﺍﻡ ‪H2 18 O‬‬‫ً‬‫‪18‬‬‫‪ 14‬ﻣﻦ ﺍﻟﻤﺮﻛﺐ ‪ 11‬ﻭ ‪ ، 12‬ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ‪ .‬ﺍﻟﻤﺮﻛﺒﺎﺕ ‪ 13‬ﻭ ‪ 14‬ﻋﺒﺎﺭﺓ ﻋﻦ ﻣﺸﺎﺑﻬﺎﺕ ﻧﻈﺎﺋﺮ ﻝ ‪ . O‬ﺑﺘﺠﺎﻫﻞ ﻭﺿﻊ ﺍﻟﻌﻼﻣﺎﺕ‬‫ﺍﻟﻨﻈﻴﺮﻳﺔ ‪ ،‬ﻳﻮﻓﺮ ﻛﻞ ﻣﻦ ‪ 13‬ﻭ ‪ 14‬ﻧﻔﺲ ﺍﻟﻨﺎﺗﺞ ‪ 15‬ﻣﻊ ﻛﻴﻤﻴﺎﺀ ﻓﺮﺍﻏﻴﺔ ﻣﺘﻄﺎﺑﻘﺔ‪.‬‬‫‪Q7-4‬‬‫‪IChO‬‬‫‪OMN-4 C-7 Q-4‬‬‫)‪Arabic (Oman‬‬‫‪A.2‬‬‫ﺍﺧﺘﺮ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻤﻨﺎﺳﺐ ﻟﻠﻤﺮﻛﺐ ‪.A‬‬‫‪2pt‬‬‫‪A.3‬‬‫ﺍﺭﺳﻢ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﺒﻨﺎﺋﻲ ﻟﻠﻤﺮﻛﺒﺎﺕ ﻣﻦ ‪ 8‬ﺇﻟﻰ ‪ 14‬ﻭﺣﺪﺩ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺑﻮﺿﻮﺡ ﻋﻨﺪ ﺍﻟﻀﺮﻭﺭﺓ‪.‬‬‫‪19pt‬‬‫ﺃﻳﻀﺎ ﺇﻟﻰ ﺫﺭﺍﺕ ‪O‬‬
‫ﺃﺷﺮ ً‬‫‪18‬‬‫ﺍﻟﻤﻘﺪﻣﺔ ﻟﻠﻤﺮﻛﺐ ‪ 13‬ﻭ ‪ 14‬ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﻓﻲ ﺍﻟﻤﺜﺎﻝ ﺃﺩﻧﺎﻩ‪.‬‬‫‪Q7-5‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-7 Q-5‬‬‫ﺍﻟﺠﺰﺀ ‪B‬‬‫ﻳﺘﻢ ﺗﺼﻨﻴﻊ ﺍﻟﻤﺮﻛﺐ ‪ 19‬ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺃﺩﻧﺎﻩ‪ .‬ﻓﻴﻤﺎ ﻳﺘﻌﻠﻖ ﺑﺨﺎﺻﻴﺔ ﺍﻷﺭﻭﻣﺎﺗﻴﺔ ﺍﻟﻌﻄﺮﻳﺔ ﻏﻴﺮ ﺍﻟﺒﻴﻨﺰﻳﻨﻴﺔ ‪ ،‬ﻳﻤﻜﻦ ﺍﺳﺘﺨﺪﺍﻡ‬‫ﺍﻟﻤﺮﻛﺐ ‪ 19‬ﻛﻤﻨﺸﻂ ﻟﻠﻜﺤﻮﻝ ‪ ،‬ﻭﺗﻢ ﺗﺤﻮﻳﻞ ‪ 20‬ﺇﻟﻰ ‪ 22‬ﻋﺒﺮ ﻭﺳﻴﻂ ﺯﻭﺝ ﺃﻳﻮﻧﻲ ‪ .21‬ﻋﻠﻰ ﺍﻟﺮﻏﻢ ﻣﻦ ﺃﻥ ﺗﻜﻮﻳﻦ ‪ 21‬ﻟﻮﺣﻆ‬‫ﺗﺪﺭﻳﺠﻴﺎ ﻟﻴﻌﻄﻲ ‪ 18‬ﻭ ‪.22‬‬‫ﺑﻮﺍﺳﻄﺔ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﺍﻟﻨﻮﻭﻱ ‪ 21 ،‬ﻳﺘﺤﻠﻞ‬‫ً‬‫‪B.1‬‬‫ﺍﺭﺳﻢ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﺒﻨﺎﺋﻲ ‪ 19-17‬ﻭ ‪ .21‬ﻟﻴﺲ ﻣﻦ ﺍﻟﻀﺮﻭﺭﻱ ﺗﺤﺪﻳﺪ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ‪.‬‬‫‪10pt‬‬

Q7-1

IChO
OMN-4 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
OMN-4 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
OMN-4 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
OMN-4 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
OMN-4 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

IChO
OMN-4 C-7 A-1

A7-1
Arabic (Oman)‫ﺍﻟﻠﻌﺐ ﺑﻌﻄﺮﻳﺔ ﻏﻴﺮ ﺍﻟﺒﻨﺰﻳﻨﻮﻳﺪ‬A ‫ﺍﻟﺠﺰﺀ‬(5 pt) A.1
pt) 3) 6pt) 2) 2(2 pt) A.2

IChO
OMN-4 C-7 A-2

A7-2
Arabic (Oman)(19 pt) A.3

IChO
OMN-4 C-7 A-3

A7-3
Arabic (Oman)

B ‫ﺍﻟﺠﺰﺀ‬(10 pt) B.1
pt) 2) 18

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

OMN-4 C-8 C-1

OMN-4 C-8 C
LUJAINA AL BADI

IChO
Problem 8
Cover sheet

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

‫‪Q8-1‬‬‫‪IChO‬‬‫‪OMN-4 C-8 Q-1‬‬‫)‪Arabic (Oman‬‬‫ﺍﻟﺠﺰﻳﺌﺎﺕ ﺍﻟﻌﻀﻮﻳﺔ ﺍﻟﺪﻳﻨﺎﻣﻴﻜﻴﺔ ﻭ ﺍﻟﻜﻴﺮﻳﻠﻴﺔ ‪Organic Dynamic‬‬‫‪Chirality Their and Molecules‬‬‫ﺍﻟﺴﺆﺍﻝ‬‫ﺍﻟﺪﺭﺟﺎﺕ‬‫ﺩﺭﺟﺔ‬‫‪A.1‬‬‫‪9‬‬‫‪ 11٪‬ﻣﻦ ﺇﺟﻤﺎﻟﻲ ﺍﻟﺪﺭﺟﺎﺕ‬‫‪A.2‬‬‫‪3‬‬‫‪A.3‬‬‫‪7‬‬‫ﺍﻟﺠﺰﺀ ‪A‬‬‫‪B.1‬‬‫‪3‬‬‫‪B.2‬‬‫‪4‬‬‫ﺍﻟﻤﺠﻤﻮﻉ‬‫‪26‬‬‫ﺗﺴﻤﻰ ﺍﻟﻬﻴﺪﺭﻭﻛﺮﺑﻮﻧﺎﺕ ﺍﻟﻌﻄﺮﻳﺔ ﻣﺘﻌﺪﺩﺓ ﺍﻟﺤﻠﻘﺎﺕ ﺫﺍﺕ ﺍﻟﻮﺻﻼﺕ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﻣﻦ ﻧﻮﻉ ﺍﻟﺘﻔﺮﻉ ﺃﺭﺛﻮ ﺗﺴﻤﻲ ﻥ‪-‬ﻛﺎﺭﺑﻮﻫﻴﻠﻴﺴﻴﻦ‬‫‪) n-carbohelicenes‬ﻫﻨﺎ ‪ n ،‬ﺗﻤﺜﻞ ﻋﺪﺩ ﺍﻟﺤﻠﻘﺎﺕ ﺍﻟﺴﺪﺍﺳﻴﺔ( )ﺍﻧﻈﺮ ﺃﺩﻧﺎﻩ(‪ .‬ﻳﺘﻢ ﺗﺤﻀﻴﺮ ﺑﻪ )𝐶]‪([4‬ﺑﻜﻔﺎﺀﺓ ﻋﻦ ﻃﺮﻳﻖ ﻣﺴﺎﺭ‬‫ﺑﺎﺳﺘﺨﺪﺍﻡ ﺗﻔﺎﻋﻞ ﺿﻮﺋﻲ ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺃﺩﻧﺎﻩ ‪ ،‬ﻋﺒﺮ ﻭﺳﻴﻂ )‪ (Int‬ﺍﻟﺬﻱ ﻳﺘﺄﻛﺴﺪ ﺑﺴﻬﻮﻟﺔ ﺑﻮﺍﺳﻄﺔ ﺍﻟﻴﻮﺩ‪.‬‬‫ﺗﺠﺮﻱ ﻋﻤﻠﻴﺔ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻟﻀﻮﺋﻲ ﺑﻄﺮﻳﻘﺔ ﻣﺸﺎﺑﻬﺔ ﻟﻠﻤﺜﺎﻝ ﺍﻟﺘﺎﻟﻲ‪.‬‬‫‪Q8-2‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-8 Q-2‬‬‫ﻣﻼﺣﻈﺔ‪ :‬ﻟﺠﻤﻴﻊ ﺍﻷﺳﺌﻠﺔ ‪ ، 8‬ﻳﺮﺟﻰ ﺭﺳﻢ ﺭﻭﺍﺑﻂ ﻣﻔﺮﺩﺓ ﻭﻣﺰﺩﻭﺟﺔ ﻣﺘﺒﺎﺩﻟﺔ ﻓﻲ ﺇﺟﺎﺑﺎﺗﻚ ﻋﻠﻰ ﺍﻟﻤﺸﻜﻼﺕ ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ‬‫ﻓﻲ ﺃﻣﺜﻠﺔ ‪ carbohelicene.‬ﻻ ﺗﺴﺘﺨﺪﻡ ﺍﻟﺪﺍﺋﺮﺓ ﻷﻧﻈﻤﺔ 𝜋ﺍﻟﻤﺘﻌﺎﻗﺒﺔ ﺍﻟﺘﻲ ﺗﻜﺘﺐ ﻭﺳﻂ ﺣﻠﻘﺔ ﺍﻟﺒﻨﺰﻳﻦ‪.‬‬‫‪A.1‬‬‫ﺍﺭﺳﻢ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﺒﻨﺎﺋﻲ ﻟﻠﻤﺮﻛﺒﺎﺕ ﻣﻦ ‪ A‬ﺇﻟﻰ ‪ C.‬ﻳﺠﺐ ﺍﻟﺘﻤﻴﻴﺰ ﺑﻴﻦ ﺍﻟﻤﺘﺸﺎﺑﻬﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ‬‫ﺍﻟﻤﺠﺴﻤﺔ‪.‬‬‫‪9pt‬‬‫‪A.2‬‬‫ﻣﺤﺎﻭﻻﺕ ﺗﺼﻨﻴﻊ ‪ 5carbohelicene‬ﻣﻦ ﻧﻔﺲ ﻣﻠﺢ ﺍﻟﻔﻮﺳﻔﻮﻧﻴﻮﻡ ﻭﻣﺮﻛﺐ ﺑﺪﺀ ﻣﻨﺎﺳﺐ ﺃﺩﺕ‬‫ﺑﺪﻻ ﻣﻦ ﺇﻧﺘﺎﺝ ﺍﻟﻨﺎﺗﺞ ‪ D‬ﺍﻟﺬﻱ ﻛﺎﻥ ﻭﺯﻧﻪ‬‫ﺇﻟﻰ ﺗﻜﻮﻳﻦ ﻛﻤﻴﺔ ﺿﺌﻴﻠﺔ ﻣﻦ ‪، 5carbohelicene‬‬‫ً‬‫ﺍﻟﺠﺰﻳﺌﻲ ﺃﻗﻞ ﺑﻤﻘﺪﺍﺭ ‪ 2Da‬ﻣﻦ ‪ . 5carbohelicene‬ﻳﺘﻢ ﺫﻛﺮ ﺍﻹﺯﺍﺣﺎﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ ﻟﻠﺮﻧﻴﻦ‬‫ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﻟﻠﺒﺮﻭﺗﻮﻥ ‪ NMR 1H‬ﻟﻠﻤﺮﻛﺐ ‪ D‬ﺃﺩﻧﺎﻩ‪ .‬ﺍﺭﺳﻢ ﺍﻟﺮﻛﻴﺐ ﺍﻟﺒﻨﺎﺋﻲ ﻟﻠﻤﺮﻛﺐ ‪.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])‪(2H‬‬‫‪3pt‬‬‫]‪ - [5‬ﻭﺍﻟﻜﺮﺑﻮﻫﻴﻠﻨﺎﺕ ﺍﻷﻛﺒﺮ ‪ n-arbohelicenes‬ﻟﻬﺎ ﻛﻴﺮﺍﻟﻴﺔ ﺣﻠﺰﻭﻧﻴﺔ ﻭﺍﻟﺘﺤﻮﻝ ﺍﻟﺒﻴﻨﻲ ﺑﻴﻦ ﺍﻟﻤﺘﺸﺎﺑﻬﺎﺕ ﺍﻷﻧﺎﺷﻮﻣﺮ‬‫‪ enantiomers‬ﻣﻦ ﻫﺬﻩ ﺍﻟﻤﺮﻛﺒﺎﺕ ‪ helicenes‬ﻳﻜﻮﻥ ﺑﻄﻴﺌﺎً ﺑﺸﻜﻞ ﻣﻠﺤﻮﻅ ﻓﻲ ﺩﺭﺟﺔ ﺣﺮﺍﺭﺓ ﺍﻟﻐﺮﻓﺔ‪ .‬ﻳﺘﻢ ﺗﻌﺮﻳﻒ ﺍﻟﻜﻴﺮﻳﻠﻴﺔ‬‫ﻟﻠﻤﺮﻛﺒﺎﺕ ‪ chirality‬ﻝ ‪ n-arbohelicenes‬ﻋﻠﻰ ﺃﻧﻬﺎ )‪ (M‬ﺃﻭ )‪ (P‬ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺃﺩﻧﺎﻩ‪.‬‬‫ﻳﻤﻜﻦ ﻓﺼﻞ ﺍﻟﻜﺮﺑﻮﻫﻴﻠﻴﺴﻴﻦ ‪ n-arbohelicenes‬ﺍﻟﺘﻲ ﺗﺤﺘﻮﻱ ﻋﻠﻰ ‪ n‬ﺃﻛﺒﺮ ﻣﻦ ‪ 4‬ﺑﺸﻜﻞ ﻣﺘﻤﺎﺛﻞ ﻋﻦ ﻃﺮﻳﻖ ﻛﺮﻭﻣﺎﺗﻮﻏﺮﺍﻓﻴﺎ‬‫ﺍﻟﻌﻤﻮﺩ ﺍﻟﻜﻴﺮﻳﺎﻟﻲ ‪ ،‬ﻭﺍﻟﺘﻲ ﻃﻮﺭﻫﺎ ﺍﻟﺒﺮﻭﻓﻴﺴﻮﺭ ﻳﻮﺷﻴﻮ ﺃﻭﻛﺎﻣﻮﺗﻮ‪.‬‬‫‪Q8-3‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-8 Q-3‬‬‫ﺻﻮﺭﺓ ﺍﻟﺒﺮﻭﻓﻴﺴﻮﺭ ﻳﻮﺷﻴﻮ ﺃﻭﻛﺎﻣﻮﺗﻮ‪ :.‬ﺍﻟﺤﺎﺻﻞ ﻋﻠﻰ ﺟﺎﺋﺰﺓ ﻣﺆﺳﺴﺔ ﺍﻟﻴﺎﺑﺎﻥ‬‫‪Q8-4‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-8 Q-4‬‬‫ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﻬﻴﻠﻴﺴﻨﻴﺔ ﺍﻟﻤﺘﻌﺪﺩﺓ ‪ helicenes Multiple‬ﻋﺒﺎﺭﺓ ﻋﻦ ﺟﺰﻳﺌﺎﺕ ﺗﺤﺘﻮﻱ ﻋﻠﻰ ﺍﺛﻨﻴﻦ ﺃﻭ ﺃﻛﺜﺮ ﻣﻦ ﺍﻟﺘﺮﺍﻛﻴﺐ ﺍﻟﺸﺒﻴﻬﺔ‬‫ﺑﺎﻟﻬﻴﻠﻴﻦ‪ .‬ﺇﺫﺍ ﺗﻢ ﺃﺧﺬ ﺍﻟﻜﻴﺮﺍﻟﻴﺔ ﺍﻟﺤﻠﺰﻭﻧﻴﺔ ‪ helical chirality‬ﻓﻲ ﺍﻻﻋﺘﺒﺎﺭ ‪ ،‬ﻓﺈﻥ ﺍﻟﻌﺪﻳﺪ ﻣﻦ ﺍﻟﻤﺘﺸﺎﺑﻬﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ ﻣﻮﺟﻮﺩﺓ‬‫ﻓﻲ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﻬﻴﻠﻴﺴﻨﻴﺔ ﺍﻟﻤﺘﻌﺪﺩﺓ‪ .‬ﻋﻠﻰ ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ ‪ ،‬ﻳﺤﺘﻮﻱ ﺍﻟﻤﺮﻛﺐ ‪ E‬ﻋﻠﻰ ﺛﻼﺛﺔ ﺷﻘﻮﻕ ﻣﻦ ﻧﻮﻉ ‪5carbohelicene‬‬‫ﺷﺒﻴﻬﺔ ﺑﺎﻟﻜﺮﺑﻮﻫﻴﻠﻴﺴﻴﻦ ﻓﻲ ﺟﺰﻱﺀ ﻭﺍﺣﺪ ‪ .‬ﻳﻮﺻﻒ ﺃﺣﺪ ﺍﻟﻤﺘﺸﺎﺑﻬﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺑﺄﻧﻪ ) 𝑃 ‪ (𝑃 , 𝑃 ,‬ﻛﻤﺎ ﻫﻮ ﻣﻮﺿﺢ ﺃﺩﻧﺎﻩ‪.‬‬‫‪A.3‬‬‫ﻳﻨﺘﺞ ﻋﻦ ﻋﻤﻠﻴﺔ ﺍﻟﻘﻄﻊ ﺑﻮﺳﺎﻃﺔ ﺍﻟﻨﻴﻜﻞ ﻟـ ‪-1،2‬ﺛﻨﺎﺋﻲ ﺑﺮﻭﻣﻮﺑﻴﻨﺰﻳﻦ ‪1,2-dibromobenzene‬‬‫ﻳﻨﺘﺞ ﻣﺮﻛﺐ ﺛﻼﺛﻲ ﻓﻴﻨﻠﻴﻦ ‪ . triphenylene‬ﻋﻨﺪﻣﺎ ﻳﺘﻢ ﺗﻄﺒﻴﻖ ﻧﻔﺲ ﺍﻟﺘﻔﺎﻋﻞ ﻋﻠﻰ‬‫ﻟﻠﻤﺘﺸﺎﺑﻬﺎﺕ ﻣﻦ‪ F‬ﻭ ‪ ، (P)-F‬ﻳﺘﻢ ﺍﻟﺤﺼﻮﻝ ﻋﻠﻰ ﻫﻴﻠﻴﻦ ﻣﺘﻌﺪﺩﺓ ‪ helicene multiple‬ﻭﻫﻮ‬‫ﺍﻟﻤﺮﻛﺐ ‪ . G‬ﺑﺎﻟﻨﻈﺮ ﺇﻟﻰ ﺃﻥ ﺍﻟﺘﺤﻮﻳﻞ ﺍﻟﺒﻴﻨﻲ ﺑﻴﻦ ﺍﻟﻤﺘﺸﺎﺑﻬﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ ﻻ ﻳﺤﺪﺙ ﺃﺛﻨﺎﺀ ﺍﻟﺘﻔﺎﻋﻞ‬‫‪ ،‬ﺣﺪﺩ ﺟﻤﻴﻊ ﺍﻟﻤﺘﺸﺎﺑﻬﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺍﻟﻤﻤﻜﻨﺔ ﻟﻠﻤﺮﻛﺐ ‪ G‬ﺍﻟﻤﺘﻜﻮﻧﺔ ﻓﻲ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ‪ ،‬ﺑﺪﻭﻥ‬‫ﺍﺯﺩﻭﺍﺟﻴﺔ‪ .‬ﻳﺠﺐ ﺭﺳﻢ ﻣﺘﺸﺎﺑﻪ ﻭﺍﺣﺪ ﺑﺎﻟﻜﺎﻣﻞ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻜﺎﻳﺮﻟﻴﺔ ﺍﻟﻤﺤﺪﺩ ﻛﻤﺎ ﻓﻲ ﺍﻟﻤﺜﺎﻝ‬‫ﺃﻋﻼﻩ ﺣﺘﻰ ﻳﻜﻮﻥ ﻛﻤﺮﺟﻊ ‪ ،‬ﻣﻊ ﻭﺿﻊ ﺑﻴﺎﻧﺎﺕ ﺍﻷﺭﻗﺎﻡ ﺍﻟﻌﺪﺩﻳﺔ ؛ ﻳﺠﺐ ﺃﻥ ﻳﺘﻢ ﺳﺮﺩ ﺍﻟﻤﺘﺸﺎﺑﻬﺎﺕ‬‫ﻭﻓﻘﺎ ﻟﻨﻔﺲ ﺍﻟﺘﺮﻗﻴﻢ‪ .‬ﻋﻠﻰ ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ‬‫ﺍﻟﻔﺮﺍﻏﻴﺔ ﺍﻷﺧﺮﻯ ﺑﺄﺭﻗﺎﻡ ﺍﻟﻤﻮﻗﻊ ﻭﺗﺴﻤﻴﺎﺕ ‪ M‬ﻭ ‪P‬‬‫ً‬‫‪ ،‬ﻳﺠﺐ ﺇﺩﺭﺍﺝ ﺍﻟﻤﺘﺸﺎﺑﻬﺎﺕ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺍﻷﺧﺮﻯ ﻟﻠﻤﺮﻛﺐ ‪ 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,‬‬‫‪M).‬‬‫‪7pt‬‬IChO
OMN-4 C-8 Q-5

Q8-5
Arabic (Oman)

‫‪Q8-6‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-8 Q-6‬‬‫ﺍﻟﺠﺰﺀ ‪B‬‬‫ﺳﻮﻣﺎﻧﻴﻦ ‪ Sumanene‬ﻫﻮ ﻋﺒﺎﺭﺓ ﻋﻦ ﻫﻴﺪﺭﻭﻛﺮﺑﻮﻥ ﻋﻠﻰ ﺷﻜﻞ ﻭﻋﺎﺀ ﺗﻢ ﺍﻟﻜﺸﻒ ﻋﻨﻪ ﻷﻭﻝ ﻣﺮﺓ ﻓﻲ ﺍﻟﻴﺎﺑﺎﻥ ﻓﻲ ﻋﺎﻡ ‪.2003‬‬‫ﺍﺳﻢ ”ﺳﻮﻣﺎﻧﻴﻦ” ‪ Sumanene‬ﻣﺸﺘﻖ ﻣﻦ ﺍﻟﻜﻠﻤﺔ ﺍﻟﺴﻨﺴﻜﺮﻳﺘﻴﺔ ﺍﻟﻬﻨﺪﻳﺔ ”ﺳﻮﻣﺎﻥ” ﻭﺍﻟﺘﻲ ﺗﻌﻨﻲ ﻋﺒﺎﺩ ﺍﻟﺸﻤﺲ‪.‬‬‫ﺗﻢ ﺍﻟﺘﻮﺻﻞ ﺇﻟﻰ ﺗﺼﻨﻴﻊ ﺗﻮﻟﻴﻒ ﺍﻟﺴﻮﻣﺎﻧﻴﻦ ‪ Sumanene‬ﻣﻦ ﺧﻼﻝ ﺗﺴﻠﺴﻞ ﺗﻔﺎﻋﻞ ﻳﺘﻜﻮﻥ ﻣﻦ ﻓﺘﺢ ﺍﻟﺤﻠﻘﺔ ﻭﺇﻏﻼﻕ ﺍﻟﺤﻠﻘﺔ‬‫ﻓﻲ ﺗﻔﺎﻋﻞ ﺍﻟﺘﺼﻨﻴﻊ ﺍﻟﺘﺒﺎﺩﻟﻲ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻣﺰﺩﻭﺝ ﺍﻟﺘﺒﺎﺩﻝ‪.‬‬‫ﺗﻔﺎﻋﻼﺕ ﺍﻟﺘﺼﻨﻴﻊ ﺍﻟﺘﺒﺎﺩﻟﻲ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ ﻣﺰﺩﻭﺝ ﺍﻟﺘﺒﺎﺩﻝ ﺍﻟﺘﻤﺜﻴﻠﻴﺔ ﺍﻟﻤﺤﻔﺰﺓ ﺑﻮﺍﺳﻄﺔ ﺍﻟﻌﺎﻣﻞ ﺍﻟﺤﻔﺎﺯ ﺍﻟﺮﻭﺛﻴﻨﻴﻮﻡ ) ∗𝑢𝑅( ﻣﻮﺿﺤﺔ‬‫ﺃﺩﻧﺎﻩ‪.‬‬‫‪B.1‬‬‫ﺍﺭﺳﻢ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﺒﻨﺎﺋﻲ ﻟﻠﻤﺮﻛﺐ ﺍﻟﻮﺳﻴﻂ ‪) I‬ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺍﻟﺨﺎﺻﺔ ﺑﻪ ﻏﻴﺮ ﻣﻄﻠﻮﺑﺔ(‪.‬‬‫‪3pt‬‬

‫‪Q8-7‬‬‫)‪Arabic (Oman‬‬‫‪B.2‬‬‫‪IChO‬‬‫‪OMN-4 C-8 Q-7‬‬‫ﺑﺪﺀﺍ ﻣﻦ ﺍﻟﻤﺮﻛﺐ ‪ J‬ﺍﻟﺬﻱ ﻳﺴﺘﺨﺪﻡ ﻛﻤﺮﻛﺐ ﻣﺒﺪﺋﻲ ﻹﻧﺘﺎﺝ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻷﺧﺮﻯ ﻭﻫﺬﺍ ﻣﺮﻛﺐ ﻧﺸﻂ‬‫ً‬‫ﺑﺼﺮﻳﺎ ‪ .K‬ﻭﻻ ﺗﻌﺎﻧﻲ‬‫ﺿﻮﺋﻴﺎً ‪ ،‬ﻳﻌﻄﻲ ﺗﺴﻠﺴﻞ ﺍﻟﺘﻔﺎﻋﻞ ﻧﻔﺴﻪ ﻣﺸﺘﻖ ﺍﻟﺴﻮﻣﺎﻧﻴﻦ ﺍﻟﻨﺸﻂ‬‫ً‬‫ﺍﻟﻤﺮﺍﻛﺰ ﺍﻟﻜﻴﺎﺭﻳﻠﻴﺔ ﺍﻟﻔﺮﺍﻏﻴﺔ ﻓﻲ ‪ J‬ﻣﻦ ﺃﻱ ﺍﻧﻌﻜﺎﺱ ﺃﺛﻨﺎﺀ ﺗﻔﺎﻋﻞ ﺍﻟﺘﺼﻨﻴﻊ ﺍﻟﺘﺒﺎﺩﻟﻲ ﺍﻟﻜﻴﻤﻴﺎﺋﻲ‬‫ﻣﺰﺩﻭﺝ ﺍﻟﺘﺒﺎﺩﻝ‪ .‬ﺍﺭﺳﻢ ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﺒﻨﺎﺋﻲ ﻟﻠﻤﺮﻛﺐ ‪ K‬ﻭﺍﺳﺘﺨﺪﺍﻡ ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ ﺍﻟﻤﻨﺎﺳﺒﺔ‪.‬‬‫‪4pt‬‬

Q8-1

IChO
OMN-4 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
OMN-4 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
OMN-4 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
OMN-4 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
OMN-4 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

‫‪A8-1‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-8 A-1‬‬‫ﺍﻟﺠﺰﻳﺌﺎﺕ ﺍﻟﻌﻀﻮﻳﺔ ﺍﻟﺪﻳﻨﺎﻣﻴﻜﻴﺔ ﻭﺗﻨﺎﻗﻀﻬﺎ‬‫ﺍﻟﺠﺰﺀ ‪A‬‬‫‪(9 pt) A.1‬‬‫‪pt) 3) A‬‬‫‪(3 pt) A.2‬‬‫‪pt) 3) B‬‬‫‪pt) 3) C‬‬IChO
OMN-4 C-8 A-2

A8-2
Arabic (Oman)(7 pt) A.3

IChO
OMN-4 C-8 A-3

A8-3
Arabic (Oman)

B ‫ﺍﻟﺠﺰﺀ‬(3 pt) B.1(4 pt) B.2

OMN-4 C-9 C-1

OMN-4 C-9 C
LUJAINA AL BADI

IChO
Problem 9
Cover sheet

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

‫‪Q9-1‬‬‫‪IChO‬‬‫‪OMN-4 C-9 Q-1‬‬‫)‪Arabic (Oman‬‬‫ﻳﺤﺐ ﻭﻳﻜﺮﻩ ﺍﻟﻜﺒﺴﻮﻟﺔ ‪Capsule of Dislikes and Likes‬‬‫ﺍﻟﺴﺆﺍﻝ‬‫ﺍﻟﺪﺭﺟﺎﺕ‬‫ﺍﻟﺼﻒ‬‫‪A.1‬‬‫‪13‬‬‫‪ 10٪‬ﻣﻦ ﺇﺟﻤﺎﻟﻲ ﺍﻟﺪﺭﺟﺎﺕ‬‫‪A.2‬‬‫‪2‬‬‫‪A.3‬‬‫‪2‬‬‫‪A.4‬‬‫‪3‬‬‫‪A.5‬‬‫‪3‬‬‫ﺍﻟﻤﺠﻤﻮﻉ‬‫‪23‬‬‫ﺍﻟﺠﺰﺀ‬‫ﺍﻷﻭﻻﺩ ﺍﻟﺠﻴﺪﻭﻥ ﻻ ﻳﻔﻌﻠﻮﻥ ﺫﻟﻚ‪ ,‬ﻟﻜﻦ ﺍﺫﺍ ﻗﻤﺖ ﺑﻘﺺ ﻛﺮﺓ ﺍﻟﺘﻨﺲ ﻳﻤﻜﻨﻚ ﻓﺼﻞ ﻗﻄﻌﺘﻴﻦ ﻋﻠﻰ ﺷﻜﻞ ﺣﺮﻑ ) 𝑈( ﺑﺎﻻﻧﺠﻠﻴﺰﻳﺔ‪.‬‬‫ﻃﺒﻘﺎ ﻟﻬﺬﺓ ﺍﻟﻔﻜﺮﺓ‪ :‬ﺗﻢ ﺗﺤﻀﻴﺮ ﻣﺮﻛﺐ ‪ 1‬ﻭ ﻣﺮﻛﺐ ‪ 2‬ﻋﻠﻰ ﺷﻜﻞ ﺣﺮﻑ ) 𝑈(ﺟﺰﻳﺌﺎﺕ ﺑﺄﺣﺠﺎﻡ ﻣﺨﺘﻠﻔﺔ‪ ,‬ﺗﻢ ﺗﺤﻀﻴﺮ ﻣﺮﻛﺐ ‪3‬‬‫ﻟﻠﻤﻘﺎﺭﻧﺔ ﻣﻊ ﻣﺮﻛﺐ ‪ 1‬ﻭﺗﻤﺖ ﺩﺭﺍﺳﺔ ﻣﺪﻯ ﻗﺪﺭﺓ ﺍﻟﻤﺮﻛﺒﺎﺕ ﻋﻠﻰ ﺍﻟﻜﺒﺴﻠﺔ “ ”‪encapsulation‬‬‫‪Q9-2‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-9 Q-2‬‬‫ﻳﺘﻢ ﻋﺮﺽ ﺍﻟﻤﺴﺎﺭ ﺍﻻﺻﻄﻨﺎﻋﻲ ﺍﻟﻤﻨﺘﺠﺔ ﺇﻟﻰ ﻟﻠﻤﺮﻛﺐ ‪ 2‬ﺃﺩﻧﺎﻩ‪ .‬ﺍﻟﺘﺮﻛﻴﺐ ﺍﻟﻌﻨﺼﺮﻱ ﻟﻠﻤﺮﻛﺐ ‪ C :9‬؛ ‪ 40.49%‬ﻭ ‪ H‬؛ ‪ ، 1.70%‬ﻭ‬‫‪ O‬؛ ‪ 17.98%‬ﺑﻮﺍﺳﻄﺔ ﺍﻟﻜﺘﻠﺔ‪.‬‬‫‪Q9-3‬‬‫)‪Arabic (Oman‬‬‫‪A.1‬‬‫‪IChO‬‬‫‪OMN-4 C-9 Q-3‬‬‫ﺃﺭﺳﻢ ﺍﻟﻤﺮﻛﺒﺎﺕ ﻣﻦ ‪ 9 − 4‬ﻟﻴﺲ ﻣﻦ ﺍﻟﻀﺮﻭﺭﻱ ﺗﻮﺿﻴﺢ )ﺍﻟﻜﻴﻤﻴﺎﺀ ﺍﻟﻔﺮﺍﻏﻴﺔ(‪ ,‬ﺍﺳﺘﺨﺪﻡ ﺍﻟﺮﻣﺰ‬‫‪” “PMB‬ﻛﺈﺧﺘﺼﺎﺭ ﺑﺪﻻ ﻣﻦ ﺭﺳﻢ ﺗﺮﻛﻴﺒﺔ ﺍﻟﻤﺮﻛﺐ ﻛﺎﻣﻠﺔ ﻟﻞ ) ‪ ( p- methoxybenzyl‬ﺍﻟﻤﻮﺿﺢ‬‫ﻓﻲ ﺍﻟﻤﺨﻄﻂ ﺑﺎﻷﻋﻠﻰ‬‫‪13pt‬‬‫ﻓﻲ ﺍﻟﻄﻴﻒ ﺍﻟﻜﺘﻠﻲ ﻟﻠﻤﺮﻛﺐ ‪ ، 1‬ﻟﻮﺣﻆ ﺑﻮﺿﻮﺡ ﻗﻤﺔﺍﻷﻳﻮﻥ ﺍﻟﻤﺘﻌﻠﻘﺔ ﺑﺎﻟﺠﺰﻱﺀ ﺛﻨﺎﺋﻲ ﺍﻟﺠﺰﻱﺀ ) ‪ ، (112‬ﺑﻴﻨﻤﺎ ﻟﻢ ﻳﺘﻢ ﻣﻼﺣﻈﺔ‬‫ﻗﻤﺔﺍﻷﻳﻮﻥ ﻟـ ‪ 3 2‬ﻓﻲ ﻃﻴﻒ ﺍﻟﻤﺮﻛﺐ ‪ .3‬ﻓﻲ ﺃﻃﻴﺎﻑ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻨﻮﻭﻱ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﻟﻠﺒﺮﻭﺗﻮﻥ ‪ NMR 1H‬ﻟﻤﺤﻠﻮﻝ ‪ ، 1 2‬ﻛﻞ‬‫ﻛﻴﻤﻴﺎﺋﻴﺎ ‪ ،‬ﻭﻛﺎﻧﺖ ﺍﻹﺯﺍﺣﺔ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﻪ ﻣﺨﺘﻠﻔﺎً ﺑﺸﻜﻞ ﻛﺒﻴﺮ‬‫ﺑﺮﻭﺗﻮﻧﺎﺕ ‪ NH‬ﻟﻮﺣﻆ ﺃﻧﻬﺎ ﻣﺸﺘﻘﺔ ﻣﻦ ‪ 1‬ﻭﻟﻮﺣﻆ ﺃﻧﻬﺎ ﻣﺘﻜﺎﻓﺌﺔ‬‫ً‬‫ﻋﻦ ﺫﻟﻚ ﺍﻟﺨﺎﺹ ﺑﺒﺮﻭﺗﻮﻧﺎﺕ ‪ NH‬ﻟﻠﻤﺮﻛﺐ ‪ . 3‬ﺗﺸﻴﺮ ﻫﺬﻩ ﺍﻟﺒﻴﺎﻧﺎﺕ ﺇﻟﻰ ﺃﻥ ﺍﻟﺮﻭﺍﺑﻂ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻨﻴﺔ ﺗﺘﺸﻜﻞ ﺑﻴﻦ ﺷﻘﻮﻕ ‪NH‬‬‫ﻣﻦ ‪ 1‬ﻭﺍﻟﺬﺭﺍﺕ ‪ X‬ﻟﺠﺰﻱﺀ ﺁﺧﺮ ﻣﻦ ‪ 1‬ﺇﻟﻰ ﺗﺸﻜﻴﻞ ﻛﺒﺴﻮﻟﺔ ﺛﻨﺎﺋﻴﺔ ﺍﻟﺠﺰﻱﺀ‪.‬‬‫‪A.2‬‬‫ﺿﻊ ﺩﺍﺋﺮﺓ ﺣﻮﻝ ﻛﻞ ﺍﻟﺬﺭﺓ )ﺍﻟﺬﺭﺍﺕ( ﺍﻟﻤﻨﺎﺳﺒﺔ ‪ X‬ﻓﻲ ﺍﻟﻤﺮﻛﺐ ‪.1‬‬‫‪2pt‬‬‫‪A.3‬‬‫ﺍﻛﺘﺐ ﻋﺪﺩ ﺍﻟﺮﻭﺍﺑﻂ ﺍﻟﻬﻴﺪﺭﻭﺟﻴﻨﻴﺔ ﻓﻲ ﺍﻟﻜﺒﺴﻮﻟﺔ ﺛﻨﺎﺋﻴﺔ ﺍﻟﺠﺰﻱﺀ ) ‪.(112‬‬‫‪2pt‬‬

‫‪Q9-4‬‬‫)‪Arabic (Oman‬‬‫‪IChO‬‬‫‪OMN-4 C-9 Q-4‬‬‫ﺗﺤﺘﻮﻱ ﺍﻟﻜﺒﺴﻮﻟﺔ ﺛﻨﺎﻳﺔ ﺍﻟﺠﺰﻱﺀ ‪ (112 ) 1‬ﻋﻠﻰ ﻣﺴﺎﺣﺔ ﺩﺍﺧﻠﻴﺔ ﺣﻴﺚ ﻳﻤﻜﻦ ﺗﻐﻠﻴﻒ ﺍﻟﺠﺰﻱﺀ ﺍﻟﺼﻐﻴﺮ ﺍﻟﻤﻨﺎﺳﺐ ‪ Z.‬ﻳﺘﻢ ﺍﻟﺘﻌﺒﻴﺮ‬‫ﻋﻦ ﻫﺬﻩ ﺍﻟﻈﺎﻫﺮﺓ ﺑﺎﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬‫ﻳﺘﻢ ﺇﻋﻄﺎﺀ ﺛﺎﺑﺖ ﺍﻹﺗﺰﺍﻥ ﻟﺘﻐﻠﻴﻒ ‪ Z‬ﺑﺎﻟﻜﺒﺴﻮﻟﺔ ﺇﻟﻰ ‪ 1 2‬ﻋﻠﻰ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﻟﻲ‪:‬‬‫ﻳﻤﻜﻦ ﻣﺮﺍﻗﺒﺔ ﺗﻐﻠﻴﻒ ﺍﻟﺠﺰﻱﺀ ﻓﻲ ﻛﺒﺴﻮﻟﺔ ﺑﻮﺍﺳﻄﺔ ﺍﻟﺘﺤﻠﻴﻞ ﺍﻟﻄﻴﻔﻲ ﺑﺎﻟﺮﻧﻴﻦ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﺍﻟﻨﻮﻭﻱ‪ .‬ﻋﻠﻰ ﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ‬‫‪ ،‬ﺃﻋﻄﻰ ‪ 1 2‬ﻓﻲ ‪C6 D6‬ﺇﺷﺎﺭﺍﺕ )ﻗﻤﻢ ( ﻣﺨﺘﻠﻔﺔ ﻓﻲ ﺃﻃﻴﺎﻑ ﺍﻟﺮﻧﻴﻦ ﺍﻟﻨﻮﻭﻱ ﺍﻟﻤﻐﻨﺎﻃﻴﺴﻲ ﻟﻠﺒﺮﻭﺗﻮﻥ ‪ NMR 1H‬ﻗﺒﻞ ﻭﺑﻌﺪ‬‫ﺇﺿﺎﻓﺔ ‪.CH4‬‬‫ﺃﻳﻀﺎ ﻛﺒﺴﻮﻟﺔ ﻗﺎﺗﻤﺔ ﺻﻠﺒﺔ ﻭﺃﻛﺒﺮ ) ‪ .(222‬ﺗﻢ ﻗﻴﺎﺱ ﻃﻴﻒ ‪ NMR 1H‬ﻟﻠﻜﺒﺴﻮﻟﺔ ‪ 2 2‬ﻓﻲ ‪C6 D5 F , C6 D6‬‬
‫ﻳﺸﻜﻞ ﺍﻟﻤﺮﻛﺐ ‪ً 2‬‬‫ﻭﻣﺰﻳﺞ ﺍﻟﻤﺬﻳﺐ 𝐹 ‪ ، 𝐶6 𝐷6 /𝐶6 𝐷5‬ﻣﻊ ﺍﻟﺤﻔﺎﻅ ﻋﻠﻰ ﺟﻤﻴﻊ ﻇﺮﻭﻑ ﺍﻟﺘﻔﺎﻋﻞ ﺍﻷﺧﺮﻯ ﺛﺎﺑﺘﺔ‪ .‬ﺗﻢ ﺗﻠﺨﻴﺺ ﺍﻹﺯﺍﺣﺎﺕ ﺍﻟﻜﻴﻤﻴﺎﺋﻴﺔ‬‫ﻟﺒﺮﻭﺗﻮﻥ ‪ Ha‬ﻟﻠﻤﺮﻛﺐ ‪ 2‬ﻓﻲ ﺍﻟﻤﺬﻳﺒﺎﺕ ﺃﻋﻼﻩ ‪ ،‬ﻭﻟﻢ ﻳﺘﻢ ﻣﻼﺣﻈﺔ ﺃﻱ ﺇﺷﺎﺭﺍﺕ )ﻗﻤﻢ ( ﺃﺧﺮﻯ ﻣﻦ ‪ Ha‬ﻓﻲ ‪ ، 2‬ﺑﺎﺳﺘﺜﻨﺎﺀ ﻣﺎ ﻫﻮ‬‫ﺩﺍﺋﻤﺎ ﺑﺄﻛﺒﺮ ﻋﺪﺩ ﻣﻤﻜﻦ ﻣﻦ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻤﺬﻳﺐ ﻭﺃﻥ ﻛﻞ ﺇﺷﺎﺭﺓ )ﻗﻤﻪ‬‫ﻣﺬﻛﻮﺭ‪ .‬ﺍﻓﺘﺮﺽ ﺃﻥ ﺍﻟﺠﺰﺀ ﺍﻟﺪﺍﺧﻠﻲ ﻣﻦ ﺍﻟﻜﺒﺴﻮﻟﺔ ﻣﻤﺘﻠﺊ‬‫ً‬‫( ﺗﺘﻮﺍﻓﻖ ﻣﻊ ﻧﻮﻉ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﻜﺒﺴﻮﻟﺔ ﺍﻟﻤﻤﻠﻮﺀﺓ‪.‬‬‫ﺍﻟﻤﺬﻳﺐ‬‫‪C6 D6‬‬‫‪C6 D5 F‬‬‫‪C6 D6 /C6 D5 F‬‬‫‪A.4‬‬‫𝑎 𝐻 𝑓𝑜 )𝑚𝑝𝑝(𝛿‬‫‪4.60‬‬‫‪4.71‬‬‫‪4.60 , 4.71 , 4.82‬‬‫ﺣﺪﺩ ﻋﺪﺩ ﺟﺰﻳﺌﺎﺕ ‪ C6 D6‬ﻭ ‪ C6 D5 F‬ﺍﻟﻤﻐﻠﻔﺔ ﻭﺍﻟﻤﻜﺒﺴﻠﺔ ﻓﻲ ‪ 2 2‬ﻹﻋﻄﺎﺀ ﻛﻞ ﺇﺷﺎﺭﺓ )ﻗﻤﻪ ( ﻝ‬‫‪.Ha‬‬‫‪3pt‬‬

‫‪Q9-5‬‬‫‪IChO‬‬‫‪OMN-4 C-9 Q-5‬‬‫)‪Arabic (Oman‬‬‫ﻛﺸﻔﺖ ﻗﻴﺎﺳﺎﺕ 𝑅 𝑀 𝑁 𝐻‪ 1‬ﻓﻲ ‪ C6 D6‬ﺃﻥ ‪ 2 2‬ﻳﻤﻜﻦ ﺃﻥ ﻳﺸﺘﻤﻞ ﻋﻠﻰ ﺟﺰﻱﺀ ﻭﺍﺣﺪ ﻣﻦ ﺣﻤﺾ ‪1-adamantanecarboxylic‬‬‫)‪ ، (AdA‬ﻭﺗﻢ ﺗﺤﺪﻳﺪ ﺛﻮﺍﺑﺖ ﺍﻹﺗﺰﺍﻥ ﻟﻠﺤﻤﺾ )‪ (Ka‬ﺍﻟﺘﻲ ﻳﺘﻢ ﺍﻟﺘﻌﺒﻴﺮ ﻋﻨﻬﺎ ﺃﺩﻧﺎﻩ ﻟﺪﺭﺟﺎﺕ ﺣﺮﺍﺭﺓ ﻣﺨﺘﻠﻔﺔ‪ .‬ﻳﺸﻴﺮ ]ﺍﻟﻤﺬﻳﺐ @‬‫‪ [222‬ﺇﻟﻰ ﻧﻮﻉ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺟﺰﻱﺀ ﻭﺍﺣﺪ ﺃﻭ ﺃﻛﺜﺮ ﻣﻦ ﺟﺰﻳﺌﺎﺕ ﺍﻟﻤﺬﻳﺐ‪.‬‬‫ﺃﻳﻀﺎ ﺗﺤﺪﻳﺪ ﻗﻴﻢ ‪ 𝐾a‬ﻟـ ‪ CH4‬ﻭ ‪ 1 2‬ﺍﻟﻤﻌﻄﺎﺓ ﻛـﻤﻌﺎﺩﻟﺔ )‪ (2‬ﻋﻨﺪ ﺩﺭﺟﺎﺕ ﺣﺮﺍﺭﺓ ﻣﺨﺘﻠﻔﺔ ﻓﻲ ‪ C6 D6‬ﻭﻛﺬﻟﻚ ﺗﻢ‬‫ﻭﺑﺎﻟﻤﺜﻞ ‪ ،‬ﺗﻢ ً‬‫ﺗﺤﺪﻳﺪﻫﺎ ﺑﻮﺍﺳﻄﺔ ﻗﻴﺎﺳﺎﺕ 𝑅 𝑀 𝑁 𝐻‪ .1‬ﻳﺘﻢ ﻋﺮﺽ ﻗﻄﻊ ﺛﻮﺍﺑﺖ ﺍﻹﺗﺰﺍﻥ )ﻣﺜﻞ ‪ ln 𝐾a‬ﻣﻘﺎﺑﻞ ‪ 1‬ﻣﻘﺴﻮﻣﺎً ﻋﻠﻰ ‪ T‬ﺃﺩﻧﺎﻩ‪.‬‬‫ﻟﻢ ﻳﺘﻢ ﺗﻐﻠﻴﻒ ﺃﻱ ﺟﺰﻱﺀ ‪ C6 D6‬ﻓﻲ ‪ .112‬ﻓﻲ ﺍﻟﺴﻄﺮ ‪ ، II‬ﻳﻜﻮﻥ ﺍﻟﺘﻐﻴﺮ ﻓﻲ ﺍﻟﻌﺸﻮﺍﺋﻴﺔ )𝑆‪ (Δ‬ﻫﻮ )‪ (1‬ﻭﺗﻐﻴﺮ ﺍﻟﻤﺤﺘﻮﻯ‬‫ﺍﻟﺤﺮﺍﺭﻱ )𝐻‪ (Δ‬ﻫﻮ )‪ ، (2‬ﻣﻤﺎ ﻳﺸﻴﺮ ﺇﻟﻰ ﺃﻥ ﺍﻟﻘﻮﺓ ﺍﻟﺪﺍﻓﻌﺔ ﻟﻠﺘﻐﻠﻴﻒ ﻓﻲ ﺍﻟﺴﻄﺮ ‪ II‬ﻫﻲ )‪ .(3‬ﻟﺬﻟﻚ ‪ ،‬ﺍﻟﺴﻄﺮ ‪ I‬ﻳﻘﺎﺑﻞ )‪، (4‬‬‫ﻭﺍﻟﺴﻄﺮ ‪ II‬ﻳﻘﺎﺑﻞ )‪.(5‬‬‫‪A.5‬‬‫ﺍﺧﺘﺮ ﺍﻟﺨﻴﺎﺭﺍﺕ ﺍﻟﺼﺤﻴﺤﺔ ﻓﻲ ﺍﻟﻔﺮﺍﻏﺎﺕ )‪ (5) - (1‬ﻓﻲ ﺍﻟﻔﻘﺮﺓ ﺍﻟﺘﺎﻟﻴﺔ ﻣﻦ ‪ A‬ﻭ ‪B.‬‬‫ﺍﻟﺼﻒ‬‫‪A‬‬‫)‪(2‬‬‫‪ positive‬ﻣﻮﺟﺐ‬‫)‪(1‬‬‫)‪(3‬‬‫)‪(4‬‬‫)‪(5‬‬‫‪B‬‬‫‪ positive‬ﻣﻮﺟﺐ‬‫‪ negative‬ﺳﺎﻟﺐ‬‫𝑆‪Δ‬‬‫𝐻‪Δ‬‬‫‪CH4 and 1 2‬‬‫‪AdA and 2 2‬‬‫‪CH4 and 1 2‬‬‫‪ negative‬ﺳﺎﻟﺐ‬‫‪AdA and 2 2‬‬‫‪3pt‬‬

Q9-1

IChO
OMN-4 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
OMN-4 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
OMN-4 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
OMN-4 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
OMN-4 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

IChO
OMN-4 C-9 A-1

A9-1
Arabic (Oman)‫ﻳﺤﺐ ﻭﻳﻜﺮﻩ ﺍﻟﻜﺒﺴﻮﻟﺔ‬(13 pt) A.1
pt) 3) 5pt) 2) 4

pt) 2) 7pt) 2) 6pt) 2) 9pt) 2) 8

IChO

A9-2

OMN-4 C-9 A-2

Arabic (Oman)(2 pt) A.2(2 pt) A.3(3 pt) A.4‫ﻏﻴﺮ ﺍﻟﻤﻜﺒﺴﻠﺔ‬C6 D5 F ‫ﻋﺪﺩ ﺟﺰﻳﺌﺎﺕ‬‫ﻏﻴﺮ ﺍﻟﻤﻜﺒﺴﻠﺔ‬C6 D6 ‫ﻋﺪﺩ ﺟﺰﻳﺌﺎﺕ‬Ha of (ppm) 𝛿
ppm 60.4

ppm 71.4

ppm 82.4(3 pt) A.5
(1) ∶

(2) ∶

(3) ∶

(4) ∶

(5) ∶